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296
A
Yaroslav and Permutations
PROGRAMMING
1,100
[ "greedy", "math" ]
null
null
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav.
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
[ "1\n1\n", "3\n1 1 2\n", "4\n7 7 7 7\n" ]
[ "YES\n", "YES\n", "NO\n" ]
In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.
500
[ { "input": "1\n1", "output": "YES" }, { "input": "3\n1 1 2", "output": "YES" }, { "input": "4\n7 7 7 7", "output": "NO" }, { "input": "4\n479 170 465 146", "output": "YES" }, { "input": "5\n996 437 605 996 293", "output": "YES" }, { "input": "6\n727 539 896 668 36 896", "output": "YES" }, { "input": "7\n674 712 674 674 674 674 674", "output": "NO" }, { "input": "8\n742 742 742 742 742 289 742 742", "output": "NO" }, { "input": "9\n730 351 806 806 806 630 85 757 967", "output": "YES" }, { "input": "10\n324 539 83 440 834 640 440 440 440 440", "output": "YES" }, { "input": "7\n925 830 925 98 987 162 356", "output": "YES" }, { "input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808", "output": "YES" }, { "input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94", "output": "YES" }, { "input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402", "output": "NO" }, { "input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187", "output": "YES" }, { "input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103", "output": "YES" }, { "input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925", "output": "NO" }, { "input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353", "output": "YES" }, { "input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285", "output": "NO" }, { "input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99", "output": "YES" }, { "input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26", "output": "YES" }, { "input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516", "output": "YES" }, { "input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314", "output": "NO" }, { "input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800", "output": "NO" }, { "input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317", "output": "NO" }, { "input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784", "output": "YES" }, { "input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474", "output": "YES" }, { "input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499", "output": "NO" }, { "input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816", "output": "NO" }, { "input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167", "output": "YES" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "2\n1 1", "output": "NO" }, { "input": "1\n1000", "output": "YES" }, { "input": "12\n2 2 4 4 4 4 6 6 6 6 6 6", "output": "YES" } ]
1,675,708,921
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
24
92
0
''' Welcome to GDB Online. GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl, C#, OCaml, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog. Code, Compile, Run and Debug online from anywhere in world. ''' n = input() x = [int(i) for i in input().split()] maxx = 0 sett = set(x) for i in sett : if x.count(i) > maxx : maxx = x.count(i) e = i if maxx > len(x)/2 +1 : print("NO") else : print("YES")
Title: Yaroslav and Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. Output Specification: In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. Demo Input: ['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.
```python ''' Welcome to GDB Online. GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl, C#, OCaml, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog. Code, Compile, Run and Debug online from anywhere in world. ''' n = input() x = [int(i) for i in input().split()] maxx = 0 sett = set(x) for i in sett : if x.count(i) > maxx : maxx = x.count(i) e = i if maxx > len(x)/2 +1 : print("NO") else : print("YES") ```
0
677
A
Vanya and Fence
PROGRAMMING
800
[ "implementation" ]
null
null
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Print a single integer — the minimum possible valid width of the road.
[ "3 7\n4 5 14\n", "6 1\n1 1 1 1 1 1\n", "6 5\n7 6 8 9 10 5\n" ]
[ "4\n", "6\n", "11\n" ]
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
500
[ { "input": "3 7\n4 5 14", "output": "4" }, { "input": "6 1\n1 1 1 1 1 1", "output": "6" }, { "input": "6 5\n7 6 8 9 10 5", "output": "11" }, { "input": "10 420\n214 614 297 675 82 740 174 23 255 15", "output": "13" }, { "input": "10 561\n657 23 1096 487 785 66 481 554 1000 821", "output": "15" }, { "input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396", "output": "144" }, { "input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366", "output": "145" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 1\n2", "output": "2" }, { "input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19", "output": "63" }, { "input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386", "output": "31" }, { "input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518", "output": "75" }, { "input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397", "output": "41" }, { "input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118", "output": "116" } ]
1,698,023,896
2,147,483,647
PyPy 3-64
OK
TESTS
29
61
0
n, h = map(int, input().split()) friend_heights = list(map(int, input().split())) minimum_width = 0 for height in friend_heights: if height > h: minimum_width += 2 # If the person is taller than the fence, they need 2 units of width. else: minimum_width += 1 # If the person is shorter than or equal to the fence, they need 1 unit of width. print(minimum_width)
Title: Vanya and Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? Input Specification: The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. Output Specification: Print a single integer — the minimum possible valid width of the road. Demo Input: ['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n'] Demo Output: ['4\n', '6\n', '11\n'] Note: In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
```python n, h = map(int, input().split()) friend_heights = list(map(int, input().split())) minimum_width = 0 for height in friend_heights: if height > h: minimum_width += 2 # If the person is taller than the fence, they need 2 units of width. else: minimum_width += 1 # If the person is shorter than or equal to the fence, they need 1 unit of width. print(minimum_width) ```
3
884
A
Book Reading
PROGRAMMING
800
[ "implementation" ]
null
null
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can. But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading. Help Luba to determine the minimum number of day when she finishes reading. It is guaranteed that the answer doesn't exceed *n*. Remember that there are 86400 seconds in a day.
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Print the minimum day Luba can finish reading the book. It is guaranteed that answer doesn't exceed *n*.
[ "2 2\n86400 86398\n", "2 86400\n0 86400\n" ]
[ "2\n", "1\n" ]
none
0
[ { "input": "2 2\n86400 86398", "output": "2" }, { "input": "2 86400\n0 86400", "output": "1" }, { "input": "2 86400\n1 86399", "output": "2" }, { "input": "100 1000000\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "12" }, { "input": "1 1\n86399", "output": "1" }, { "input": "6 1200\n86400 86400 86000 86000 86000 86400", "output": "5" }, { "input": "6 1200\n86400 86400 86000 86000 86001 86399", "output": "6" }, { "input": "4 172799\n1 1 86400 0", "output": "4" }, { "input": "4 172799\n0 86400 86399 0", "output": "4" }, { "input": "6 1\n1 1 86400 1 86399 1", "output": "1" }, { "input": "4 1\n86400 86399 86400 86400", "output": "2" }, { "input": "4 1\n86400 86400 0 86400", "output": "3" } ]
1,509,206,917
2,147,483,647
Python 3
OK
TESTS
16
62
0
n,t=[int(x) for x in input().split()] arr=[int(x) for x in input().split()] cnt=0 for x in arr: if 86400-x>=0: t-=(86400-x) cnt+=1 if t<=0: break print(cnt)
Title: Book Reading Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can. But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading. Help Luba to determine the minimum number of day when she finishes reading. It is guaranteed that the answer doesn't exceed *n*. Remember that there are 86400 seconds in a day. Input Specification: The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day. Output Specification: Print the minimum day Luba can finish reading the book. It is guaranteed that answer doesn't exceed *n*. Demo Input: ['2 2\n86400 86398\n', '2 86400\n0 86400\n'] Demo Output: ['2\n', '1\n'] Note: none
```python n,t=[int(x) for x in input().split()] arr=[int(x) for x in input().split()] cnt=0 for x in arr: if 86400-x>=0: t-=(86400-x) cnt+=1 if t<=0: break print(cnt) ```
3
58
A
Chat room
PROGRAMMING
1,000
[ "greedy", "strings" ]
A. Chat room
1
256
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
If Vasya managed to say hello, print "YES", otherwise print "NO".
[ "ahhellllloou\n", "hlelo\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]
1,683,657,512
2,147,483,647
Python 3
OK
TESTS
40
31
0
# line = input() # lCounter = 0 # pattern = 'hello' # ind1 = 0 # indOfH = line.find('h') # indOfO = line.find('o') # if indOfH == -1 or indOfO == -1: print('NO') # # # else: # ans = 'YES' # helloline = line[int(indOfH):int(indOfO) + 1] # #print(helloline) # if helloline.find('e') == -1 or helloline.find('l') == -1: ans = ('NO') # for i in range(helloline): # if helloline[i] == 'h' and helloline[i + 1] == 'e': # continue # if helloline[i] == 'e' and helloline[i + 1] == 'l': # continue # if helloline[i] == 'l' and helloline[i + 1] == 'e': # continue # else: # ans = 'NO' # print(ans) # line, pattern = input(), 'hello' i, counter, lineLen = 0, 0, len(line) for item in pattern: while i < lineLen: if line[i] == item: counter += 1 i += 1 break i += 1 print('YES' if counter == 5 else 'NO') ''' line = "helllooo" 0123456789.. line = "abchhhhhhhhelllowww" 01234 patt = "hello" helhcludoo '''
Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python # line = input() # lCounter = 0 # pattern = 'hello' # ind1 = 0 # indOfH = line.find('h') # indOfO = line.find('o') # if indOfH == -1 or indOfO == -1: print('NO') # # # else: # ans = 'YES' # helloline = line[int(indOfH):int(indOfO) + 1] # #print(helloline) # if helloline.find('e') == -1 or helloline.find('l') == -1: ans = ('NO') # for i in range(helloline): # if helloline[i] == 'h' and helloline[i + 1] == 'e': # continue # if helloline[i] == 'e' and helloline[i + 1] == 'l': # continue # if helloline[i] == 'l' and helloline[i + 1] == 'e': # continue # else: # ans = 'NO' # print(ans) # line, pattern = input(), 'hello' i, counter, lineLen = 0, 0, len(line) for item in pattern: while i < lineLen: if line[i] == item: counter += 1 i += 1 break i += 1 print('YES' if counter == 5 else 'NO') ''' line = "helllooo" 0123456789.. line = "abchhhhhhhhelllowww" 01234 patt = "hello" helhcludoo ''' ```
3.9845
701
C
They Are Everywhere
PROGRAMMING
1,500
[ "binary search", "strings", "two pointers" ]
null
null
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1. There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once. Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house. The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
[ "3\nAaA\n", "7\nbcAAcbc\n", "6\naaBCCe\n" ]
[ "2\n", "3\n", "5\n" ]
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2. In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6. In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
1,000
[ { "input": "3\nAaA", "output": "2" }, { "input": "7\nbcAAcbc", "output": "3" }, { "input": "6\naaBCCe", "output": "5" }, { "input": "1\nA", "output": "1" }, { "input": "1\ng", "output": "1" }, { "input": "52\nabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "52" }, { "input": "2\nAA", "output": "1" }, { "input": "4\nqqqE", "output": "2" }, { "input": "10\nrrrrroooro", "output": "2" }, { "input": "15\nOCOCCCCiCOCCCOi", "output": "3" }, { "input": "20\nVEVnVVnWnVEVVnEVBEWn", "output": "5" }, { "input": "25\ncpcyPPjPPcPPPPcppPcPpppcP", "output": "6" }, { "input": "30\nsssssAsesssssssssssssessssssss", "output": "3" }, { "input": "35\ngdXdddgddddddddggggXdbgdggdgddddddb", "output": "4" }, { "input": "40\nIgsggIiIggzgigIIiiIIIiIgIggIzgIiiiggggIi", "output": "9" }, { "input": "45\neteeeeeteaattaeetaetteeettoetettteyeteeeotaae", "output": "9" }, { "input": "50\nlUlUllUlUllllUllllUllllUlUlllUlllUlllllUUlllUUlkUl", "output": "3" }, { "input": "55\nAAAAASAAAASAASAAAAAAAAAAAAASAAAAAAAAAAAAAAAASAAAAAAAAAA", "output": "2" }, { "input": "60\nRRRrSRRRRRRRRRSSRRRSRRRRRRRRrRSRRRRRRRRRRRRRRSRRRRRSSRSRrRRR", "output": "3" }, { "input": "65\nhhMhMhhhhhhhhhhhMhhMMMhhhhBhhhhMhhhhMhhhhhMhhhBhhhhhhhhhhBhhhhhhh", "output": "5" }, { "input": "70\nwAwwwAwwwwwwwwwwwwwwAwAAwwAwwwwwwwwAwAAAwAAwwwwwwwwwAwwwwwwwwwwwwAAwww", "output": "2" }, { "input": "75\niiiXXiiyiiiXyXiiyXiiXiiiiiiXXyiiiiXXiiXiiXifiXiXXiifiiiiiiXfXiyiXXiXiiiiXiX", "output": "4" }, { "input": "80\nSrSrrrrrrrrrrrrrrSSSrrrrrrSrrrrSrrrrrrrrrrSSrrrrrrrrrrrSrrrSrrrrSrrrrSrrrrSSrSSr", "output": "2" }, { "input": "85\nwkMMMwMMkMMMMMMMkkkkMMMMzkkMMwMMkkwMkMwkMMkMMwwMzMMMkkMwwMMMMMMkMMkMzMMMkMMkwMkMMMkMM", "output": "6" }, { "input": "90\nZllZZZyZlZlllZlylllZlllZZllllllllZZllllllllllyylZZyvZvZlllZZlZllZlZlllZyllZyZlllZlllllllZl", "output": "5" }, { "input": "95\nEmuBuEBmmEBBElBlElmmBEmmmEmmEuBEEmummmEmBBBBEWBBBmEEBmmummBBmmlluBBmElmEBEmBmBBmBmuLmEBBmlEBmBu", "output": "39" }, { "input": "100\nfAfhfNNNhfffAfNffNANhffNffffffNfffffhNfffNfffNfffNfNNNffNfffNfhNNNffffNfNfffAffNffNffNNfhfffNhNffNNN", "output": "4" }, { "input": "7\nabcbbab", "output": "3" }, { "input": "65\nkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkjjjjkkkkkkkkkkkkkkkkkkkllllllljk", "output": "3" }, { "input": "9\nbcdddbddc", "output": "3" }, { "input": "71\nlqcrdopdpfagcilpcmheqhjbmnciqmeoabflrenbkrsaabaanpehdmpqcdqnljcrsplnlco", "output": "36" }, { "input": "8\nabbbbbbc", "output": "8" } ]
1,698,169,853
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
3
46
0
# love thy neighbour from sys import stdin,stdout def ArrayIn():return list(map(int,input().split())) def print(x):stdout.write(f'{x}\n') n=int(input()) flats=input() target=set(flats) l, r=0, n-1 while l<r: if set(flats[l:r])==target: r-=1 elif set(flats[l+1:r+1])==target: l+=1 else: print(r-l+1) break
Title: They Are Everywhere Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1. There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once. Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit. Input Specification: The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house. The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*. Output Specification: Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house. Demo Input: ['3\nAaA\n', '7\nbcAAcbc\n', '6\naaBCCe\n'] Demo Output: ['2\n', '3\n', '5\n'] Note: In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2. In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6. In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
```python # love thy neighbour from sys import stdin,stdout def ArrayIn():return list(map(int,input().split())) def print(x):stdout.write(f'{x}\n') n=int(input()) flats=input() target=set(flats) l, r=0, n-1 while l<r: if set(flats[l:r])==target: r-=1 elif set(flats[l+1:r+1])==target: l+=1 else: print(r-l+1) break ```
0
158
A
Next Round
PROGRAMMING
800
[ "*special", "implementation" ]
null
null
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output the number of participants who advance to the next round.
[ "8 5\n10 9 8 7 7 7 5 5\n", "4 2\n0 0 0 0\n" ]
[ "6\n", "0\n" ]
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
500
[ { "input": "8 5\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "4 2\n0 0 0 0", "output": "0" }, { "input": "5 1\n1 1 1 1 1", "output": "5" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "1 1\n10", "output": "1" }, { "input": "17 14\n16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0", "output": "14" }, { "input": "5 5\n3 2 1 0 0", "output": "3" }, { "input": "8 6\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "8 7\n10 9 8 7 7 7 5 5", "output": "8" }, { "input": "8 4\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "8 3\n10 9 8 7 7 7 5 5", "output": "3" }, { "input": "8 1\n10 9 8 7 7 7 5 5", "output": "1" }, { "input": "8 2\n10 9 8 7 7 7 5 5", "output": "2" }, { "input": "1 1\n100", "output": "1" }, { "input": "1 1\n0", "output": "0" }, { "input": "50 25\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "50" }, { "input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "25" }, { "input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "26" }, { "input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "50" }, { "input": "11 5\n100 99 98 97 96 95 94 93 92 91 90", "output": "5" }, { "input": "10 4\n100 81 70 69 64 43 34 29 15 3", "output": "4" }, { "input": "11 6\n87 71 62 52 46 46 43 35 32 25 12", "output": "6" }, { "input": "17 12\n99 88 86 82 75 75 74 65 58 52 45 30 21 16 7 2 2", "output": "12" }, { "input": "20 3\n98 98 96 89 87 82 82 80 76 74 74 68 61 60 43 32 30 22 4 2", "output": "3" }, { "input": "36 12\n90 87 86 85 83 80 79 78 76 70 69 69 61 61 59 58 56 48 45 44 42 41 33 31 27 25 23 21 20 19 15 14 12 7 5 5", "output": "12" }, { "input": "49 8\n99 98 98 96 92 92 90 89 89 86 86 85 83 80 79 76 74 69 67 67 58 56 55 51 49 47 47 46 45 41 41 40 39 34 34 33 25 23 18 15 13 13 11 9 5 4 3 3 1", "output": "9" }, { "input": "49 29\n100 98 98 96 96 96 95 87 85 84 81 76 74 70 63 63 63 62 57 57 56 54 53 52 50 47 45 41 41 39 38 31 30 28 27 26 23 22 20 15 15 11 7 6 6 4 2 1 0", "output": "29" }, { "input": "49 34\n99 98 96 96 93 92 90 89 88 86 85 85 82 76 73 69 66 64 63 63 60 59 57 57 56 55 54 54 51 48 47 44 42 42 40 39 38 36 33 26 24 23 19 17 17 14 12 7 4", "output": "34" }, { "input": "50 44\n100 100 99 97 95 91 91 84 83 83 79 71 70 69 69 62 61 60 59 59 58 58 58 55 55 54 52 48 47 45 44 44 38 36 32 31 28 28 25 25 24 24 24 22 17 15 14 13 12 4", "output": "44" }, { "input": "50 13\n99 95 94 94 88 87 81 79 78 76 74 72 72 69 68 67 67 67 66 63 62 61 58 57 55 55 54 51 50 50 48 48 42 41 38 35 34 32 31 30 26 24 13 13 12 6 5 4 3 3", "output": "13" }, { "input": "50 30\n100 98 96 94 91 89 88 81 81 81 81 81 76 73 72 71 70 69 66 64 61 59 59 56 52 50 49 48 43 39 36 35 34 34 31 29 27 26 24 22 16 16 15 14 14 14 9 7 4 3", "output": "30" }, { "input": "2 1\n10 10", "output": "2" }, { "input": "2 2\n10 10", "output": "2" }, { "input": "2 2\n10 0", "output": "1" }, { "input": "2 2\n10 1", "output": "2" }, { "input": "2 1\n10 0", "output": "1" }, { "input": "2 1\n10 2", "output": "1" }, { "input": "50 13\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "50 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "50 50\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "10 1\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 2\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 3\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 4\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 5\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 6\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 7\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 8\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 9\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 10\n5 5 5 3 3 3 0 0 0 0", "output": "6" } ]
1,698,906,062
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
2
92
0
n, k = map(int, input().split()) l = list(map(int, input().split())) for i in range(n): if not (l[i] >= l[k-1] and l[i] != 0): break print(i)
Title: Next Round Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. Input Specification: The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). Output Specification: Output the number of participants who advance to the next round. Demo Input: ['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n'] Demo Output: ['6\n', '0\n'] Note: In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
```python n, k = map(int, input().split()) l = list(map(int, input().split())) for i in range(n): if not (l[i] >= l[k-1] and l[i] != 0): break print(i) ```
0
435
A
Queue on Bus Stop
PROGRAMMING
1,000
[ "implementation" ]
null
null
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups. The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue. Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
[ "4 3\n2 3 2 1\n", "3 4\n1 2 1\n" ]
[ "3\n", "1\n" ]
none
500
[ { "input": "4 3\n2 3 2 1", "output": "3" }, { "input": "3 4\n1 2 1", "output": "1" }, { "input": "1 5\n4", "output": "1" }, { "input": "5 1\n1 1 1 1 1", "output": "5" }, { "input": "6 4\n1 3 2 3 4 1", "output": "5" }, { "input": "6 8\n6 1 1 1 4 5", "output": "3" }, { "input": "10 10\n1 10 1 10 1 1 7 8 6 7", "output": "8" }, { "input": "100 100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52", "output": "63" }, { "input": "10 1\n1 1 1 1 1 1 1 1 1 1", "output": "10" }, { "input": "10 2\n2 2 1 1 1 1 1 2 1 2", "output": "8" }, { "input": "10 3\n1 3 1 1 3 2 2 2 3 3", "output": "9" }, { "input": "10 4\n2 1 1 1 3 4 4 4 1 2", "output": "6" }, { "input": "10 5\n2 2 3 4 4 1 5 3 1 2", "output": "7" }, { "input": "100 3\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2", "output": "83" }, { "input": "100 7\n4 7 4 7 7 4 7 3 5 6 3 5 4 3 7 2 7 2 4 1 6 3 3 7 4 4 5 4 3 6 4 3 2 2 1 4 4 1 7 3 7 7 1 3 1 5 4 1 5 3 5 2 2 1 5 5 1 5 2 7 5 5 1 5 5 4 6 5 1 3 5 6 7 4 1 3 3 4 3 2 7 6 5 7 2 7 1 1 2 2 3 1 3 7 1 3 2 1 1 7", "output": "71" }, { "input": "100 10\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2", "output": "64" }, { "input": "100 15\n3 12 8 3 11 14 12 14 1 11 13 3 5 13 4 14 2 11 7 8 12 9 15 7 15 1 4 11 6 12 1 3 8 13 1 8 14 4 3 14 1 3 1 6 10 15 13 11 12 1 14 13 11 14 11 3 12 7 3 15 14 4 5 6 5 14 7 14 6 2 6 12 6 13 13 1 9 13 15 11 6 3 15 11 9 4 15 8 15 12 1 15 10 10 4 1 15 1 4 1", "output": "71" }, { "input": "100 30\n7 14 22 16 11 13 7 29 20 19 22 6 12 16 1 8 27 21 22 3 15 27 20 12 4 19 1 26 26 22 25 17 29 25 16 29 29 28 16 26 25 14 16 20 5 21 5 15 19 13 17 21 17 19 23 13 1 25 6 30 16 19 12 10 28 8 15 13 14 24 19 30 12 19 22 1 3 14 16 3 20 26 15 19 9 10 19 27 2 16 10 22 15 13 19 3 24 9 8 13", "output": "71" }, { "input": "100 40\n39 19 13 36 11 21 32 12 1 2 39 26 32 39 24 1 4 19 10 4 16 39 32 34 13 24 30 35 3 10 8 18 13 12 39 27 31 40 37 20 17 17 37 5 10 12 22 17 7 1 31 13 11 10 2 6 22 16 2 4 9 27 6 35 22 16 22 30 33 2 26 20 35 19 40 37 19 17 21 28 37 28 40 4 5 4 35 19 26 36 19 12 21 20 21 30 9 16 9 32", "output": "65" }, { "input": "100 50\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1", "output": "60" }, { "input": "100 60\n34 21 39 17 48 46 23 56 46 52 50 39 55 48 54 38 32 38 24 26 44 12 28 9 25 26 10 52 42 60 41 3 16 60 44 29 27 55 19 19 19 57 45 59 29 35 5 14 50 47 57 48 16 7 12 36 58 31 37 58 30 50 19 11 10 41 59 57 49 41 33 9 12 11 53 50 60 51 21 9 44 23 1 16 4 15 17 57 15 17 46 50 18 52 43 24 47 50 19 18", "output": "74" }, { "input": "100 90\n74 65 49 41 3 79 61 83 50 40 13 57 90 14 62 77 36 10 3 5 5 40 50 75 32 26 3 71 79 54 88 50 46 20 42 59 30 36 83 86 60 62 82 68 62 80 18 65 28 28 81 74 62 33 61 35 33 83 90 72 6 6 51 4 22 20 29 10 8 3 84 69 12 17 24 16 12 64 80 74 68 59 1 59 15 59 37 58 79 83 51 56 81 14 37 45 19 31 61 90", "output": "67" }, { "input": "100 99\n69 46 76 47 71 9 66 46 78 17 96 83 56 96 29 3 43 48 79 23 93 61 19 9 29 72 15 84 93 46 71 87 11 43 96 44 54 75 3 66 2 95 46 32 69 52 79 38 57 53 37 60 71 82 28 31 84 58 89 40 62 74 22 50 45 38 99 67 24 28 28 12 69 88 33 10 31 71 46 7 42 81 54 81 96 44 8 1 20 24 28 19 54 35 69 32 71 13 66 15", "output": "68" }, { "input": "90 100\n25 52 88 89 36 17 57 64 66 11 89 61 54 92 48 51 18 42 44 92 6 14 67 100 16 21 17 88 85 73 33 11 94 84 56 72 4 80 90 78 96 5 62 70 54 70 94 80 10 91 100 89 98 87 69 74 88 63 53 79 38 94 89 52 21 82 67 79 100 81 2 40 30 69 34 15 12 33 87 52 95 18 51 30 15 39 30 99 46 84", "output": "67" }, { "input": "5 100\n14 67 15 28 21", "output": "2" }, { "input": "10 100\n2 17 53 94 95 57 36 47 68 48", "output": "7" }, { "input": "1 100\n18", "output": "1" }, { "input": "100 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" }, { "input": "30 100\n56 7 99 83 2 65 35 53 99 36 42 57 13 37 68 52 87 11 50 23 86 24 32 39 97 6 64 1 18 86", "output": "18" }, { "input": "60 100\n18 75 43 88 45 43 20 59 59 79 62 39 53 21 28 46 54 53 97 81 18 15 2 95 84 9 36 70 30 76 17 19 83 40 45 32 31 70 23 14 44 35 79 84 97 96 99 60 3 73 64 83 6 12 67 86 70 89 18 61", "output": "40" }, { "input": "1 73\n24", "output": "1" }, { "input": "4 4\n1 4 4 4", "output": "4" }, { "input": "2 6\n1 4", "output": "1" }, { "input": "2 5\n5 5", "output": "2" } ]
1,585,216,298
2,147,483,647
Python 3
OK
TESTS
34
109
0
''' ID: essi LANG: PYTHON3 TASK: self-contest.py ''' n, m = map(int,input().split()) a = list(map(int,input().split())) res = 1 su = 0 for i in range(n): su+= a[i] if su > m: res+=1 su = a[i] print(res)
Title: Queue on Bus Stop Time Limit: None seconds Memory Limit: None megabytes Problem Description: It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups. The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue. Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*). Output Specification: Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside. Demo Input: ['4 3\n2 3 2 1\n', '3 4\n1 2 1\n'] Demo Output: ['3\n', '1\n'] Note: none
```python ''' ID: essi LANG: PYTHON3 TASK: self-contest.py ''' n, m = map(int,input().split()) a = list(map(int,input().split())) res = 1 su = 0 for i in range(n): su+= a[i] if su > m: res+=1 su = a[i] print(res) ```
3
202
A
LLPS
PROGRAMMING
800
[ "binary search", "bitmasks", "brute force", "greedy", "implementation", "strings" ]
null
null
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline. You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence. We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=&lt;<=*p*2<=&lt;<=...<=&lt;<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba". String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| &gt; |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=&lt;<=|*x*|, *r*<=&lt;<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=&gt;<=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post". String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Print the lexicographically largest palindromic subsequence of string *s*.
[ "radar\n", "bowwowwow\n", "codeforces\n", "mississipp\n" ]
[ "rr\n", "wwwww\n", "s\n", "ssss\n" ]
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
500
[ { "input": "radar", "output": "rr" }, { "input": "bowwowwow", "output": "wwwww" }, { "input": "codeforces", "output": "s" }, { "input": "mississipp", "output": "ssss" }, { "input": "tourist", "output": "u" }, { "input": "romka", "output": "r" }, { "input": "helloworld", "output": "w" }, { "input": "zzzzzzzazz", "output": "zzzzzzzzz" }, { "input": "testcase", "output": "tt" }, { "input": "hahahahaha", "output": "hhhhh" }, { "input": "abbbbbbbbb", "output": "bbbbbbbbb" }, { "input": "zaz", "output": "zz" }, { "input": "aza", "output": "z" }, { "input": "dcbaedcba", "output": "e" }, { "input": "abcdeabcd", "output": "e" }, { "input": "edcbabcde", "output": "ee" }, { "input": "aaaaaaaaab", "output": "b" }, { "input": "testzzzzzz", "output": "zzzzzz" }, { "input": "zzzzzzwait", "output": "zzzzzz" }, { "input": "rrrrrqponm", "output": "rrrrr" }, { "input": "zzyzyy", "output": "zzz" }, { "input": "aababb", "output": "bbb" }, { "input": "zanzibar", "output": "zz" }, { "input": "hhgfedcbaa", "output": "hh" }, { "input": "aabcdefghh", "output": "hh" }, { "input": "aruaru", "output": "uu" }, { "input": "uraura", "output": "uu" }, { "input": "aru", "output": "u" }, { "input": "aburvabur", "output": "v" }, { "input": "ura", "output": "u" }, { "input": "eurottat", "output": "u" }, { "input": "referee", "output": "rr" }, { "input": "joking", "output": "o" }, { "input": "seriously", "output": "y" }, { "input": "sets", "output": "t" }, { "input": "test", "output": "tt" }, { "input": "klmgameklm", "output": "mmm" }, { "input": "dfkjafdkdd", "output": "kk" }, { "input": "zzzzzzzzzz", "output": "zzzzzzzzzz" }, { "input": "aaaaaaaaaa", "output": "aaaaaaaaaa" }, { "input": "buzz", "output": "zz" }, { "input": "b", "output": "b" }, { "input": "y", "output": "y" }, { "input": "yy", "output": "yy" }, { "input": "zz", "output": "zz" }, { "input": "bb", "output": "bb" }, { "input": "aa", "output": "aa" }, { "input": "a", "output": "a" }, { "input": "z", "output": "z" }, { "input": "zz", "output": "zz" }, { "input": "zzazazzzaz", "output": "zzzzzzz" }, { "input": "hellhellhe", "output": "llll" }, { "input": "hellohello", "output": "oo" }, { "input": "refer", "output": "rr" } ]
1,644,505,586
2,147,483,647
Python 3
OK
TESTS
54
62
0
s=str(input()) k=max(s) g=s.count(k) print(k*g)
Title: LLPS Time Limit: None seconds Memory Limit: None megabytes Problem Description: This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline. You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence. We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=&lt;<=*p*2<=&lt;<=...<=&lt;<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba". String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| &gt; |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=&lt;<=|*x*|, *r*<=&lt;<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=&gt;<=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post". String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". Input Specification: The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. Output Specification: Print the lexicographically largest palindromic subsequence of string *s*. Demo Input: ['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n'] Demo Output: ['rr\n', 'wwwww\n', 's\n', 'ssss\n'] Note: Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
```python s=str(input()) k=max(s) g=s.count(k) print(k*g) ```
3
975
B
Mancala
PROGRAMMING
1,100
[ "brute force", "implementation" ]
null
null
Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes. Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction. Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole. After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli. Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.
The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole. It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board.
Output one integer, the maximum possible score after one move.
[ "0 1 1 0 0 0 0 0 0 7 0 0 0 0\n", "5 1 1 1 1 0 0 0 0 0 0 0 0 0\n" ]
[ "4\n", "8\n" ]
In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.
1,000
[ { "input": "0 1 1 0 0 0 0 0 0 7 0 0 0 0", "output": "4" }, { "input": "5 1 1 1 1 0 0 0 0 0 0 0 0 0", "output": "8" }, { "input": "10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 1", "output": "54294" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 0 15", "output": "2" }, { "input": "1 0 0 0 0 1 0 0 0 0 1 0 0 0", "output": "0" }, { "input": "5 5 1 1 1 3 3 3 5 7 5 3 7 5", "output": "38" }, { "input": "787 393 649 463 803 365 81 961 989 531 303 407 579 915", "output": "7588" }, { "input": "8789651 4466447 1218733 6728667 1796977 6198853 8263135 6309291 8242907 7136751 3071237 5397369 6780785 9420869", "output": "81063456" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 0 29", "output": "26" }, { "input": "282019717 109496191 150951267 609856495 953855615 569750143 6317733 255875779 645191029 572053369 290936613 338480779 879775193 177172893", "output": "5841732816" }, { "input": "105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505", "output": "120472578" }, { "input": "404418821 993626161 346204297 122439813 461187221 628048227 625919459 628611733 938993057 701270099 398043779 684205961 630975553 575964835", "output": "8139909016" }, { "input": "170651077 730658441 824213789 583764177 129437345 717005779 675398017 314979709 380861369 265878463 746564659 797260041 506575735 335169317", "output": "6770880638" }, { "input": "622585025 48249287 678950449 891575125 637411965 457739735 829353393 235216425 284006447 875591469 492839209 296444305 513776057 810057753", "output": "7673796644" }, { "input": "475989857 930834747 786217439 927967137 489188151 869354161 276693267 56154399 131055697 509249443 143116853 426254423 44465165 105798821", "output": "6172339560" }, { "input": "360122921 409370351 226220005 604004145 85173909 600403773 624052991 138163383 729239967 189036661 619842883 270087537 749500483 243727913", "output": "5848946922" }, { "input": "997102881 755715147 273805839 436713689 547411799 72470207 522269145 647688957 137422311 422612659 197751751 679663349 821420227 387967237", "output": "6900015198" }, { "input": "690518849 754551537 652949719 760695679 491633619 477564457 11669279 700467439 470069297 782338983 718169393 884421719 24619427 215745577", "output": "7635414974" }, { "input": "248332749 486342237 662201929 917696895 555278549 252122023 850296207 463343655 832574345 954281071 168282553 825538865 996753493 461254663", "output": "6400166934" }, { "input": "590789361 636464947 404477303 337309187 476703809 426863069 120608741 703406277 645444697 761482231 996635839 33459441 677458865 483861751", "output": "7294965518" }, { "input": "297857621 238127103 749085829 139033277 597985489 202617713 982184715 183932743 278551059 297781685 330124279 338959601 682874531 187519685", "output": "5201808164" }, { "input": "1 1 1 1 1 0 0 0 0 0 0 0 0 0", "output": "2" }, { "input": "1 1 0 0 0 0 0 0 0 0 0 0 0 0", "output": "2" }, { "input": "1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "2" }, { "input": "1 0 0 0 0 0 0 0 0 0 0 0 0 1", "output": "2" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 1 1", "output": "2" } ]
1,525,186,900
3,400
Python 3
WRONG_ANSWER
PRETESTS
6
92
7,065,600
holes = [int(x) for x in input().split()] Max = max(holes) ind = holes.index(Max) holes[ind] = 0 zero = holes.count(0) _1 = 14 - zero n = Max//14 ex = Max - (n*14) if n == 0 : for i in range(Max) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) else : holes = [x+n for x in holes] for i in range(ex) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) ''' holes = [int(x) for x in input().split()] Max = max(holes) ind = holes.index(Max) holes[ind] = 0 for i in range(Max) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) '''
Title: Mancala Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes. Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction. Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole. After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli. Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move. Input Specification: The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole. It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board. Output Specification: Output one integer, the maximum possible score after one move. Demo Input: ['0 1 1 0 0 0 0 0 0 7 0 0 0 0\n', '5 1 1 1 1 0 0 0 0 0 0 0 0 0\n'] Demo Output: ['4\n', '8\n'] Note: In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.
```python holes = [int(x) for x in input().split()] Max = max(holes) ind = holes.index(Max) holes[ind] = 0 zero = holes.count(0) _1 = 14 - zero n = Max//14 ex = Max - (n*14) if n == 0 : for i in range(Max) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) else : holes = [x+n for x in holes] for i in range(ex) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) ''' holes = [int(x) for x in input().split()] Max = max(holes) ind = holes.index(Max) holes[ind] = 0 for i in range(Max) : if ind ==13 : ind = 0 holes[ind] += 1 else : ind += 1 holes[ind] += 1 score = 0 for i in holes : if i % 2 == 0 : score += i print(score) ''' ```
0
17
A
Noldbach problem
PROGRAMMING
1,000
[ "brute force", "math", "number theory" ]
A. Noldbach problem
2
64
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. Two prime numbers are called neighboring if there are no other prime numbers between them. You are to help Nick, and find out if he is right or wrong.
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
[ "27 2\n", "45 7\n" ]
[ "YES", "NO" ]
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
0
[ { "input": "27 2", "output": "YES" }, { "input": "45 7", "output": "NO" }, { "input": "2 0", "output": "YES" }, { "input": "15 1", "output": "YES" }, { "input": "17 1", "output": "YES" }, { "input": "34 5", "output": "NO" }, { "input": "37 4", "output": "YES" }, { "input": "43 5", "output": "YES" }, { "input": "47 7", "output": "NO" }, { "input": "50 5", "output": "YES" }, { "input": "57 6", "output": "YES" }, { "input": "60 8", "output": "NO" }, { "input": "62 7", "output": "YES" }, { "input": "76 9", "output": "NO" }, { "input": "69 7", "output": "YES" }, { "input": "113 10", "output": "YES" }, { "input": "141 11", "output": "YES" }, { "input": "207 16", "output": "NO" }, { "input": "231 18", "output": "NO" }, { "input": "296 19", "output": "YES" }, { "input": "332 20", "output": "YES" }, { "input": "378 24", "output": "NO" }, { "input": "428 23", "output": "YES" }, { "input": "497 27", "output": "NO" }, { "input": "640 32", "output": "YES" }, { "input": "798 35", "output": "YES" }, { "input": "802 35", "output": "YES" }, { "input": "864 40", "output": "NO" }, { "input": "953 44", "output": "NO" }, { "input": "995 44", "output": "YES" }, { "input": "1000 44", "output": "YES" }, { "input": "1000 0", "output": "YES" }, { "input": "1000 1000", "output": "NO" }, { "input": "2 1000", "output": "NO" }, { "input": "2 0", "output": "YES" } ]
1,674,380,464
2,147,483,647
PyPy 3-64
OK
TESTS
35
154
1,945,600
n,k = map(int,input().split()) l = [] ct = 0 for i in range(2,2*n+1): b = True for j in range(2,i): if j*j > i: break; if(i%j == 0): b = False break; if(b): l.append(i) for i in range(0,len(l)-1): if l[i]+l[i+1]+1 > n: break if l[i]+l[i+1]+1 in l: ct += 1 if ct < k: print("NO") else: print("YES")
Title: Noldbach problem Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. Two prime numbers are called neighboring if there are no other prime numbers between them. You are to help Nick, and find out if he is right or wrong. Input Specification: The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). Output Specification: Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. Demo Input: ['27 2\n', '45 7\n'] Demo Output: ['YES', 'NO'] Note: In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
```python n,k = map(int,input().split()) l = [] ct = 0 for i in range(2,2*n+1): b = True for j in range(2,i): if j*j > i: break; if(i%j == 0): b = False break; if(b): l.append(i) for i in range(0,len(l)-1): if l[i]+l[i+1]+1 > n: break if l[i]+l[i+1]+1 in l: ct += 1 if ct < k: print("NO") else: print("YES") ```
3.947004
2
B
The least round way
PROGRAMMING
2,000
[ "dp", "math" ]
B. The least round way
2
64
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that - starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell. Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
[ "3\n1 2 3\n4 5 6\n7 8 9\n" ]
[ "0\nDDRR\n" ]
none
0
[ { "input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "0\nDDRR" }, { "input": "2\n7 6\n3 8", "output": "0\nDR" }, { "input": "3\n4 10 5\n10 9 4\n6 5 3", "output": "1\nDRRD" }, { "input": "4\n1 1 9 9\n3 4 7 3\n7 9 1 7\n1 7 1 5", "output": "0\nDDDRRR" }, { "input": "5\n8 3 2 1 4\n3 7 2 4 8\n9 2 8 9 10\n2 3 6 10 1\n8 2 2 8 4", "output": "0\nDDDDRRRR" }, { "input": "6\n5 5 4 10 5 5\n7 10 8 7 6 6\n7 1 7 9 7 8\n5 5 3 3 10 9\n5 8 10 6 3 8\n3 10 5 4 3 4", "output": "1\nDDRRDRDDRR" }, { "input": "7\n2 9 8 2 7 4 8\n9 5 4 4 8 5 3\n5 7 2 10 8 1 8\n2 7 10 7 5 7 7\n9 2 7 6 4 8 4\n7 2 4 7 4 1 8\n9 5 3 10 1 6 2", "output": "0\nRRDRRDRDDDDR" }, { "input": "8\n1 1 10 1 8 4 8 7\n9 3 3 2 2 6 2 4\n7 4 3 5 10 3 5 1\n8 4 4 10 4 5 9 4\n5 5 5 2 6 7 1 8\n4 10 1 3 2 4 8 3\n8 1 10 2 8 2 2 4\n2 10 6 8 10 2 8 4", "output": "0\nDRRRRRRRDDDDDD" }, { "input": "9\n8 3 3 3 10 3 10 5 6\n2 1 6 1 8 1 9 1 6\n6 1 5 4 2 2 10 4 9\n1 9 1 3 10 6 10 5 5\n1 10 5 4 7 2 5 9 10\n6 6 1 3 1 9 4 9 9\n5 3 7 6 4 6 2 10 2\n9 3 3 10 5 6 7 6 4\n4 9 6 7 4 3 7 6 5", "output": "1\nDDDDDRDDDRRRRRRR" }, { "input": "10\n10 8 6 5 9 8 2 5 3 2\n3 1 8 6 8 10 5 5 7 8\n5 9 7 7 4 9 7 2 5 2\n5 9 9 5 4 2 6 6 8 1\n10 6 9 9 10 5 6 3 5 9\n6 7 10 3 1 4 3 6 7 6\n10 10 4 7 4 4 8 6 7 4\n10 5 8 2 2 7 4 4 1 4\n8 4 6 10 10 6 1 3 3 1\n9 9 7 2 9 5 1 8 6 3", "output": "1\nDRDDDRRDDDRRDRDRRR" } ]
1,673,144,508
2,147,483,647
PyPy 3-64
MEMORY_LIMIT_EXCEEDED
TESTS
15
248
67,072,000
n = int(input()) grid = [[0] * n for _ in range(n)] c2 = [[0] * n for _ in range(n)] c5 = [[0] * n for _ in range(n)] m2 = [[(3200,3200)] * (n+1) for _ in range(n+1)] m5 = [[(3200,3200)] * (n+1) for _ in range(n+1)] dm2 = [[''] * (n+1) for _ in range(n+1)] dm5 = [[''] * (n+1) for _ in range(n+1)] m2[0][1] = (0,0) m5[0][1] = (0,0) for i in range(n): grid[i] = [int(num) for num in input().split()] for j in range(n): k = grid[i][j] while k%2 == 0: c2[i][j] += 1 k //= 2 while k%5 == 0: c5[i][j] += 1 k //= 5 um2 = m2[i][j+1] um5 = m5[i][j+1] lm2 = m2[i+1][j] lm5 = m5[i+1][j] if um2<=lm2: m2[i+1][j+1] = (um2[0]+c2[i][j], um2[1]+c5[i][j]) dm2[i+1][j+1] = "D" else: m2[i+1][j+1] = (lm2[0]+c2[i][j], lm2[1]+c5[i][j]) dm2[i+1][j+1] = "R" if um5<=lm5: m5[i+1][j+1] = (um5[0]+c5[i][j], um5[1]+c2[i][j]) dm5[i+1][j+1] = "D" else: m5[i+1][j+1] = (lm5[0]+c5[i][j], lm5[1]+c2[i][j]) dm5[i+1][j+1] = "R" count5 = min(m5[-1][-1]) count2 = min(m2[-1][-1]) if count5<count2: ans = "" i = j = n while not (i==j==1): ans = dm5[i][j] + ans if dm5[i][j] == "D": i -= 1 else: j -= 1 print(count5) print(ans) else: ans = "" i = j = n while not (i==j==1): ans = dm2[i][j] + ans if dm2[i][j] == "D": i -= 1 else: j -= 1 print(count2) print(ans)
Title: The least round way Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that - starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell. Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros. Input Specification: The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109). Output Specification: In the first line print the least number of trailing zeros. In the second line print the correspondent way itself. Demo Input: ['3\n1 2 3\n4 5 6\n7 8 9\n'] Demo Output: ['0\nDDRR\n'] Note: none
```python n = int(input()) grid = [[0] * n for _ in range(n)] c2 = [[0] * n for _ in range(n)] c5 = [[0] * n for _ in range(n)] m2 = [[(3200,3200)] * (n+1) for _ in range(n+1)] m5 = [[(3200,3200)] * (n+1) for _ in range(n+1)] dm2 = [[''] * (n+1) for _ in range(n+1)] dm5 = [[''] * (n+1) for _ in range(n+1)] m2[0][1] = (0,0) m5[0][1] = (0,0) for i in range(n): grid[i] = [int(num) for num in input().split()] for j in range(n): k = grid[i][j] while k%2 == 0: c2[i][j] += 1 k //= 2 while k%5 == 0: c5[i][j] += 1 k //= 5 um2 = m2[i][j+1] um5 = m5[i][j+1] lm2 = m2[i+1][j] lm5 = m5[i+1][j] if um2<=lm2: m2[i+1][j+1] = (um2[0]+c2[i][j], um2[1]+c5[i][j]) dm2[i+1][j+1] = "D" else: m2[i+1][j+1] = (lm2[0]+c2[i][j], lm2[1]+c5[i][j]) dm2[i+1][j+1] = "R" if um5<=lm5: m5[i+1][j+1] = (um5[0]+c5[i][j], um5[1]+c2[i][j]) dm5[i+1][j+1] = "D" else: m5[i+1][j+1] = (lm5[0]+c5[i][j], lm5[1]+c2[i][j]) dm5[i+1][j+1] = "R" count5 = min(m5[-1][-1]) count2 = min(m2[-1][-1]) if count5<count2: ans = "" i = j = n while not (i==j==1): ans = dm5[i][j] + ans if dm5[i][j] == "D": i -= 1 else: j -= 1 print(count5) print(ans) else: ans = "" i = j = n while not (i==j==1): ans = dm2[i][j] + ans if dm2[i][j] == "D": i -= 1 else: j -= 1 print(count2) print(ans) ```
0
835
A
Key races
PROGRAMMING
800
[ "math" ]
null
null
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds. If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows: 1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it. The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw. Given the length of the text and the information about participants, determine the result of the game.
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
[ "5 1 2 1 2\n", "3 3 1 1 1\n", "4 5 3 1 5\n" ]
[ "First\n", "Second\n", "Friendship\n" ]
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins. In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins. In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
500
[ { "input": "5 1 2 1 2", "output": "First" }, { "input": "3 3 1 1 1", "output": "Second" }, { "input": "4 5 3 1 5", "output": "Friendship" }, { "input": "1000 1000 1000 1000 1000", "output": "Friendship" }, { "input": "1 1 1 1 1", "output": "Friendship" }, { "input": "8 8 1 1 1", "output": "Second" }, { "input": "15 14 32 65 28", "output": "First" }, { "input": "894 197 325 232 902", "output": "First" }, { "input": "1 2 8 8 5", "output": "Friendship" }, { "input": "37 261 207 1 1000", "output": "Friendship" }, { "input": "29 344 406 900 1", "output": "Friendship" }, { "input": "1 2 8 9 8", "output": "First" }, { "input": "2 9 8 8 9", "output": "Friendship" }, { "input": "213 480 811 134 745", "output": "First" }, { "input": "2 313 856 964 421", "output": "Friendship" }, { "input": "1 10 2 6 10", "output": "Friendship" }, { "input": "2 7 6 2 3", "output": "Friendship" }, { "input": "637 324 69 612 998", "output": "Second" }, { "input": "13 849 819 723 918", "output": "Friendship" }, { "input": "9 5 7 8 7", "output": "First" }, { "input": "6 5 7 10 4", "output": "Friendship" }, { "input": "61 464 623 89 548", "output": "First" }, { "input": "641 31 29 161 802", "output": "Friendship" }, { "input": "3 3 1 6 9", "output": "Friendship" }, { "input": "2 3 9 8 2", "output": "Friendship" }, { "input": "485 117 368 567 609", "output": "First" }, { "input": "4 202 512 995 375", "output": "Friendship" }, { "input": "424 41 41 909 909", "output": "Friendship" }, { "input": "884 913 263 641 265", "output": "Second" }, { "input": "12 462 8 311 327", "output": "Second" }, { "input": "436 306 266 493 580", "output": "Second" }, { "input": "69 1 2 1 2", "output": "First" } ]
1,577,939,024
2,147,483,647
PyPy 3
OK
TESTS
32
155
0
s, v1, v2, t1, t2 = map(int, input().split()) f1 = 2 * t1 + (s * v1) f2 = 2 * t2 + (s * v2) if f1 < f2: print("First") elif f1 > f2: print("Second") else: print("Friendship")
Title: Key races Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds. If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows: 1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it. The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw. Given the length of the text and the information about participants, determine the result of the game. Input Specification: The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. Output Specification: If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". Demo Input: ['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n'] Demo Output: ['First\n', 'Second\n', 'Friendship\n'] Note: In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins. In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins. In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
```python s, v1, v2, t1, t2 = map(int, input().split()) f1 = 2 * t1 + (s * v1) f2 = 2 * t2 + (s * v2) if f1 < f2: print("First") elif f1 > f2: print("Second") else: print("Friendship") ```
3
617
A
Elephant
PROGRAMMING
800
[ "math" ]
null
null
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
[ "5\n", "12\n" ]
[ "1\n", "3\n" ]
In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
500
[ { "input": "5", "output": "1" }, { "input": "12", "output": "3" }, { "input": "999999", "output": "200000" }, { "input": "41", "output": "9" }, { "input": "1000000", "output": "200000" }, { "input": "1", "output": "1" }, { "input": "2", "output": "1" }, { "input": "3", "output": "1" }, { "input": "4", "output": "1" }, { "input": "534204", "output": "106841" }, { "input": "469569", "output": "93914" }, { "input": "502877", "output": "100576" }, { "input": "942212", "output": "188443" }, { "input": "97", "output": "20" }, { "input": "53", "output": "11" }, { "input": "89", "output": "18" }, { "input": "574", "output": "115" }, { "input": "716", "output": "144" }, { "input": "729", "output": "146" }, { "input": "8901", "output": "1781" }, { "input": "3645", "output": "729" }, { "input": "4426", "output": "886" }, { "input": "46573", "output": "9315" }, { "input": "86380", "output": "17276" }, { "input": "94190", "output": "18838" }, { "input": "999990", "output": "199998" }, { "input": "999991", "output": "199999" }, { "input": "999992", "output": "199999" }, { "input": "999993", "output": "199999" }, { "input": "999994", "output": "199999" }, { "input": "999995", "output": "199999" }, { "input": "999996", "output": "200000" }, { "input": "999997", "output": "200000" }, { "input": "999998", "output": "200000" } ]
1,696,261,317
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
31
0
a = int(input()) b = a // 5 r = a % 5 if r < 4: c = 1 else: c = 0 if r == 0: print(b) else: total = b + c print(total)
Title: Elephant Time Limit: None seconds Memory Limit: None megabytes Problem Description: An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. Input Specification: The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. Output Specification: Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. Demo Input: ['5\n', '12\n'] Demo Output: ['1\n', '3\n'] Note: In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
```python a = int(input()) b = a // 5 r = a % 5 if r < 4: c = 1 else: c = 0 if r == 0: print(b) else: total = b + c print(total) ```
0
365
A
Good Number
PROGRAMMING
1,100
[ "implementation" ]
null
null
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Print a single integer — the number of *k*-good numbers in *a*.
[ "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n", "2 1\n1\n10\n" ]
[ "10\n", "1\n" ]
none
500
[ { "input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560", "output": "10" }, { "input": "2 1\n1\n10", "output": "1" }, { "input": "1 0\n1000000000", "output": "1" }, { "input": "1 1\n1000000000", "output": "1" }, { "input": "6 0\n10\n102\n120\n1032\n1212103\n1999999", "output": "5" }, { "input": "1 3\n1000000000", "output": "0" }, { "input": "1 9\n1000000000", "output": "0" }, { "input": "1 0\n1", "output": "0" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 3\n1", "output": "0" }, { "input": "1 9\n1", "output": "0" }, { "input": "2 8\n123456780\n123", "output": "1" }, { "input": "2 8\n12345678\n1230", "output": "0" }, { "input": "6 1\n10\n102\n120\n1032\n1212103\n1999999", "output": "5" }, { "input": "6 2\n10\n102\n120\n1032\n1212103\n1999999", "output": "4" }, { "input": "6 3\n10\n102\n120\n1032\n1212103\n1999999", "output": "2" }, { "input": "6 9\n10\n102\n120\n1032\n1212103\n1999999", "output": "0" }, { "input": "1 2\n2220", "output": "0" } ]
1,669,491,432
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
31
0
data = list(map(int, input().split())) num_count = 0 for i in range(data[0]): arr = list(map(int, input())) if len(arr) > data[1]: num_count += arr.count(data[1]) print(num_count)
Title: Good Number Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). Input Specification: The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). Output Specification: Print a single integer — the number of *k*-good numbers in *a*. Demo Input: ['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n'] Demo Output: ['10\n', '1\n'] Note: none
```python data = list(map(int, input().split())) num_count = 0 for i in range(data[0]): arr = list(map(int, input())) if len(arr) > data[1]: num_count += arr.count(data[1]) print(num_count) ```
0
48
A
Rock-paper-scissors
PROGRAMMING
900
[ "implementation", "schedules" ]
A. Rock-paper-scissors
2
256
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
[ "rock\nrock\nrock\n", "paper\nrock\nrock\n", "scissors\nrock\nrock\n", "scissors\npaper\nrock\n" ]
[ "?\n", "F\n", "?\n", "?\n" ]
none
0
[ { "input": "rock\nrock\nrock", "output": "?" }, { "input": "paper\nrock\nrock", "output": "F" }, { "input": "scissors\nrock\nrock", "output": "?" }, { "input": "scissors\npaper\nrock", "output": "?" }, { "input": "paper\npaper\nrock", "output": "?" }, { "input": "rock\npaper\nrock", "output": "M" }, { "input": "rock\nscissors\nrock", "output": "?" }, { "input": "paper\nscissors\nrock", "output": "?" }, { "input": "scissors\nscissors\nrock", "output": "S" }, { "input": "rock\nrock\npaper", "output": "S" }, { "input": "paper\nrock\npaper", "output": "?" }, { "input": "scissors\nrock\npaper", "output": "?" }, { "input": "rock\npaper\npaper", "output": "?" }, { "input": "paper\npaper\npaper", "output": "?" }, { "input": "scissors\npaper\npaper", "output": "F" }, { "input": "rock\nscissors\npaper", "output": "?" }, { "input": "paper\nscissors\npaper", "output": "M" }, { "input": "scissors\nscissors\npaper", "output": "?" }, { "input": "rock\nrock\nscissors", "output": "?" }, { "input": "paper\nrock\nscissors", "output": "?" }, { "input": "scissors\nrock\nscissors", "output": "M" }, { "input": "rock\npaper\nscissors", "output": "?" }, { "input": "paper\npaper\nscissors", "output": "S" }, { "input": "scissors\npaper\nscissors", "output": "?" }, { "input": "rock\nscissors\nscissors", "output": "F" }, { "input": "paper\nscissors\nscissors", "output": "?" }, { "input": "scissors\nscissors\nscissors", "output": "?" } ]
1,584,378,766
2,147,483,647
PyPy 3
OK
TESTS
27
310
0
l=[] for i in range(3): l.append(str(input( ))) x=l[:] x= set(x) if len(x)!=2: print("?") else: if l[1]==l[2] and l[0]!=l[1]: if l[1]== 'rock' and l[0]=='paper' or l[1]=='paper' and l[0]=='scissors' or l[1]=='scissors' and l[0]=='rock': print("F") else: print("?") elif l[1]==l[0] and l[0]!=l[2]: if l[1]== 'rock' and l[2]=='paper' or l[1]=='paper' and l[2]=='scissors' or l[1]=='scissors' and l[2]=='rock': print("S") else: print("?") elif l[2]==l[0] and l[0]!=l[1]: if l[0]== 'rock' and l[1]=='paper' or l[0]=='paper' and l[1]=='scissors' or l[0]=='scissors' and l[1]=='rock': print("M") else: print("?")
Title: Rock-paper-scissors Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown. Input Specification: The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture. Output Specification: Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?". Demo Input: ['rock\nrock\nrock\n', 'paper\nrock\nrock\n', 'scissors\nrock\nrock\n', 'scissors\npaper\nrock\n'] Demo Output: ['?\n', 'F\n', '?\n', '?\n'] Note: none
```python l=[] for i in range(3): l.append(str(input( ))) x=l[:] x= set(x) if len(x)!=2: print("?") else: if l[1]==l[2] and l[0]!=l[1]: if l[1]== 'rock' and l[0]=='paper' or l[1]=='paper' and l[0]=='scissors' or l[1]=='scissors' and l[0]=='rock': print("F") else: print("?") elif l[1]==l[0] and l[0]!=l[2]: if l[1]== 'rock' and l[2]=='paper' or l[1]=='paper' and l[2]=='scissors' or l[1]=='scissors' and l[2]=='rock': print("S") else: print("?") elif l[2]==l[0] and l[0]!=l[1]: if l[0]== 'rock' and l[1]=='paper' or l[0]=='paper' and l[1]=='scissors' or l[0]=='scissors' and l[1]=='rock': print("M") else: print("?") ```
3.9225
234
C
Weather
PROGRAMMING
1,300
[ "dp", "implementation" ]
null
null
Scientists say a lot about the problems of global warming and cooling of the Earth. Indeed, such natural phenomena strongly influence all life on our planet. Our hero Vasya is quite concerned about the problems. He decided to try a little experiment and observe how outside daily temperature changes. He hung out a thermometer on the balcony every morning and recorded the temperature. He had been measuring the temperature for the last *n* days. Thus, he got a sequence of numbers *t*1,<=*t*2,<=...,<=*t**n*, where the *i*-th number is the temperature on the *i*-th day. Vasya analyzed the temperature statistics in other cities, and came to the conclusion that the city has no environmental problems, if first the temperature outside is negative for some non-zero number of days, and then the temperature is positive for some non-zero number of days. More formally, there must be a positive integer *k* (1<=≤<=*k*<=≤<=*n*<=-<=1) such that *t*1<=&lt;<=0,<=*t*2<=&lt;<=0,<=...,<=*t**k*<=&lt;<=0 and *t**k*<=+<=1<=&gt;<=0,<=*t**k*<=+<=2<=&gt;<=0,<=...,<=*t**n*<=&gt;<=0. In particular, the temperature should never be zero. If this condition is not met, Vasya decides that his city has environmental problems, and gets upset. You do not want to upset Vasya. Therefore, you want to select multiple values of temperature and modify them to satisfy Vasya's condition. You need to know what the least number of temperature values needs to be changed for that.
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of days for which Vasya has been measuring the temperature. The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (|*t**i*|<=≤<=109) — the sequence of temperature values. Numbers *t**i* are separated by single spaces.
Print a single integer — the answer to the given task.
[ "4\n-1 1 -2 1\n", "5\n0 -1 1 2 -5\n" ]
[ "1\n", "2\n" ]
Note to the first sample: there are two ways to change exactly one number so that the sequence met Vasya's condition. You can either replace the first number 1 by any negative number or replace the number -2 by any positive number.
0
[ { "input": "4\n-1 1 -2 1", "output": "1" }, { "input": "5\n0 -1 1 2 -5", "output": "2" }, { "input": "6\n0 0 0 0 0 0", "output": "6" }, { "input": "6\n-1 -2 -3 -4 5 6", "output": "0" }, { "input": "8\n1 2 -1 0 10 2 12 13", "output": "3" }, { "input": "7\n-1 -2 -3 3 -1 3 4", "output": "1" }, { "input": "2\n3 -5", "output": "2" }, { "input": "50\n4 -8 0 -1 -3 -9 0 -2 0 1 -1 0 7 -10 9 7 0 -10 5 0 1 -6 9 -9 3 -3 3 7 4 -8 -8 3 3 -1 0 2 -6 10 7 -1 -6 -3 -4 2 3 0 -4 0 7 -9", "output": "26" }, { "input": "90\n52 -89 17 64 11 -61 92 51 42 -92 -14 -100 21 -88 73 -11 84 72 -80 -78 5 -70 -70 80 91 -89 87 -74 63 -79 -94 52 82 79 81 40 69 -15 33 -52 18 30 -39 99 84 -98 44 69 -75 0 60 -89 51 -92 83 73 16 -43 17 0 51 9 -53 86 86 -50 0 -80 3 0 86 0 -76 -45 0 -32 45 81 47 15 -62 21 4 -82 77 -67 -64 -12 0 -50", "output": "42" }, { "input": "10\n-19 -29 -21 -6 29 89 -74 -22 18 -13", "output": "3" }, { "input": "100\n-782 365 -283 769 -58 224 1000 983 7 595 -963 -267 -934 -187 -609 693 -316 431 859 -753 865 -421 861 -728 -793 621 -311 414 -101 -196 120 84 633 -362 989 94 206 19 -949 -629 489 376 -391 165 50 22 -209 735 565 61 -321 -256 890 34 343 -326 984 -268 -609 385 717 81 372 -391 271 -89 297 -510 797 -425 -276 573 510 560 165 -482 511 541 -491 60 168 -805 235 -657 -679 -617 -212 816 -98 901 380 103 608 -257 -643 333 8 355 743 -801", "output": "40" } ]
1,620,606,149
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
92
6,656,000
banana = open("input.txt","r") n = banana.readlines() n = int(n) listTemps = list(banana.readline().split()) mango = open("output.txt", "w") mango.write("3") banana.close mango.close
Title: Weather Time Limit: None seconds Memory Limit: None megabytes Problem Description: Scientists say a lot about the problems of global warming and cooling of the Earth. Indeed, such natural phenomena strongly influence all life on our planet. Our hero Vasya is quite concerned about the problems. He decided to try a little experiment and observe how outside daily temperature changes. He hung out a thermometer on the balcony every morning and recorded the temperature. He had been measuring the temperature for the last *n* days. Thus, he got a sequence of numbers *t*1,<=*t*2,<=...,<=*t**n*, where the *i*-th number is the temperature on the *i*-th day. Vasya analyzed the temperature statistics in other cities, and came to the conclusion that the city has no environmental problems, if first the temperature outside is negative for some non-zero number of days, and then the temperature is positive for some non-zero number of days. More formally, there must be a positive integer *k* (1<=≤<=*k*<=≤<=*n*<=-<=1) such that *t*1<=&lt;<=0,<=*t*2<=&lt;<=0,<=...,<=*t**k*<=&lt;<=0 and *t**k*<=+<=1<=&gt;<=0,<=*t**k*<=+<=2<=&gt;<=0,<=...,<=*t**n*<=&gt;<=0. In particular, the temperature should never be zero. If this condition is not met, Vasya decides that his city has environmental problems, and gets upset. You do not want to upset Vasya. Therefore, you want to select multiple values of temperature and modify them to satisfy Vasya's condition. You need to know what the least number of temperature values needs to be changed for that. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of days for which Vasya has been measuring the temperature. The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (|*t**i*|<=≤<=109) — the sequence of temperature values. Numbers *t**i* are separated by single spaces. Output Specification: Print a single integer — the answer to the given task. Demo Input: ['4\n-1 1 -2 1\n', '5\n0 -1 1 2 -5\n'] Demo Output: ['1\n', '2\n'] Note: Note to the first sample: there are two ways to change exactly one number so that the sequence met Vasya's condition. You can either replace the first number 1 by any negative number or replace the number -2 by any positive number.
```python banana = open("input.txt","r") n = banana.readlines() n = int(n) listTemps = list(banana.readline().split()) mango = open("output.txt", "w") mango.write("3") banana.close mango.close ```
-1
262
B
Roma and Changing Signs
PROGRAMMING
1,200
[ "greedy" ]
null
null
Roma works in a company that sells TVs. Now he has to prepare a report for the last year. Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times. The operation of changing a number's sign is the operation of multiplying this number by -1. Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made. The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104). The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
[ "3 2\n-1 -1 1\n", "3 1\n-1 -1 1\n" ]
[ "3\n", "1\n" ]
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3. In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.
1,000
[ { "input": "3 2\n-1 -1 1", "output": "3" }, { "input": "3 1\n-1 -1 1", "output": "1" }, { "input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343", "output": "81852" }, { "input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -8981 -8521 -8313 -8257 -8253 -7591 -7587 -7301 -7161 -7001 -6847 -6441 -6241 -5949 -5896 -5713 -5692 -5644 -5601 -5545 -5525 -5331 -5253 -5041 -5000 -4951 -4855 -4384 -4293 -4251 -4001 -3991 -3762 -3544 -3481 -3261 -2983 -2882 -2857 -2713 -2691 -2681 -2653 -2221 -2043 -2011 -1997 -1601 -1471 -1448 -1363 -1217 -1217 -1129 -961 -926 -801 -376 -327 -305 -174 -91", "output": "102443" }, { "input": "12 28\n-6652 -6621 -6471 -5559 -5326 -4551 -4401 -4326 -3294 -1175 -1069 -43", "output": "49488" }, { "input": "78 13\n-9961 -9922 -9817 -9813 -9521 -9368 -9361 -9207 -9153 -9124 -9008 -8981 -8951 -8911 -8551 -8479 -8245 -8216 -7988 -7841 -7748 -7741 -7734 -7101 -6846 -6804 -6651 -6526 -6519 -6463 -6297 -6148 -6090 -5845 -5209 -5201 -5161 -5061 -4537 -4529 -4433 -4370 -4266 -4189 -4125 -3945 -3843 -3777 -3751 -3476 -3461 -3279 -3205 -3001 -2889 -2761 -2661 -2521 -2481 -2305 -2278 -2269 -2225 -1648 -1524 -1476 -1353 -1097 -867 -785 -741 -711 -692 -440 -401 -225 -65 -41", "output": "-147832" }, { "input": "4 1\n218 3441 4901 7601", "output": "15725" }, { "input": "73 26\n-8497 -8363 -7603 -7388 -6830 -6827 -6685 -6389 -6237 -6099 -6013 -5565 -5465 -4965 -4947 -4201 -3851 -3793 -3421 -3410 -3201 -3169 -3156 -2976 -2701 -2623 -2321 -2169 -1469 -1221 -950 -926 -9 47 236 457 773 1321 1485 1545 1671 1736 2014 2137 2174 2301 2625 3181 3536 3851 4041 4685 4981 4987 5145 5163 5209 5249 6011 6337 6790 7254 7361 7407 7969 7982 8083 8251 8407 8735 9660 9855 9957", "output": "315919" }, { "input": "53 5\n-9821 -9429 -9146 -8973 -8807 -8801 -8321 -7361 -7222 -7161 -6913 -5961 -4877 -4756 -4753 -4661 -3375 -3031 -2950 -2661 -2161 -2041 -1111 -1071 -905 -697 -397 323 772 1617 1752 2736 2737 3201 3465 4029 4121 4463 4561 4637 4814 6119 6610 6641 6961 7217 7523 8045 8610 8915 9004 9265 9576", "output": "92703" }, { "input": "1 1\n0", "output": "0" }, { "input": "1 1\n10000", "output": "-10000" }, { "input": "1 2\n-1", "output": "-1" }, { "input": "2 1\n0 1", "output": "1" }, { "input": "2 2\n-1 0", "output": "1" }, { "input": "5 6\n-3 -2 -1 5 6", "output": "15" }, { "input": "3 3\n-50 -10 30", "output": "70" }, { "input": "4 4\n-100 -90 -80 1", "output": "269" }, { "input": "4 3\n-3 -2 1 2", "output": "6" }, { "input": "6 4\n-6 -3 -2 1 2 3", "output": "15" }, { "input": "5 6\n-10 -9 -8 1 2", "output": "28" }, { "input": "1 2\n1", "output": "1" }, { "input": "2 2\n-1 3", "output": "2" }, { "input": "4 7\n-3 -2 1 6", "output": "10" }, { "input": "4 3\n-7 -6 1 5", "output": "17" }, { "input": "4 4\n-5 -1 1 2", "output": "9" } ]
1,689,443,703
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
print("_RANDOM_GUESS_1689443703.6148257")# 1689443703.6148458
Title: Roma and Changing Signs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Roma works in a company that sells TVs. Now he has to prepare a report for the last year. Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times. The operation of changing a number's sign is the operation of multiplying this number by -1. Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made. The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104). The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order. Output Specification: In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes. Demo Input: ['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n'] Demo Output: ['3\n', '1\n'] Note: In the first sample we can get sequence [1, 1, 1], thus the total income equals 3. In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.
```python print("_RANDOM_GUESS_1689443703.6148257")# 1689443703.6148458 ```
0
39
D
Cubical Planet
PROGRAMMING
1,100
[ "math" ]
D. Cubical Planet
2
64
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
[ "0 0 0\n0 1 0\n", "1 1 0\n0 1 0\n", "0 0 0\n1 1 1\n" ]
[ "YES\n", "YES\n", "NO\n" ]
none
0
[ { "input": "0 0 0\n0 1 0", "output": "YES" }, { "input": "1 1 0\n0 1 0", "output": "YES" }, { "input": "0 0 0\n1 1 1", "output": "NO" }, { "input": "0 0 0\n1 0 0", "output": "YES" }, { "input": "0 0 0\n0 1 0", "output": "YES" }, { "input": "0 0 0\n1 1 0", "output": "YES" }, { "input": "0 0 0\n0 0 1", "output": "YES" }, { "input": "0 0 0\n1 0 1", "output": "YES" }, { "input": "0 0 0\n0 1 1", "output": "YES" }, { "input": "0 0 0\n1 1 1", "output": "NO" }, { "input": "1 0 0\n0 0 0", "output": "YES" }, { "input": "1 0 0\n0 1 0", "output": "YES" }, { "input": "1 0 0\n1 1 0", "output": "YES" }, { "input": "1 0 0\n0 0 1", "output": "YES" }, { "input": "1 0 0\n1 0 1", "output": "YES" }, { "input": "1 0 0\n0 1 1", "output": "NO" }, { "input": "1 0 0\n1 1 1", "output": "YES" }, { "input": "0 1 0\n0 0 0", "output": "YES" }, { "input": "0 1 0\n1 0 0", "output": "YES" }, { "input": "0 1 0\n1 1 0", "output": "YES" }, { "input": "0 1 0\n0 0 1", "output": "YES" }, { "input": "0 1 0\n1 0 1", "output": "NO" }, { "input": "0 1 0\n0 1 1", "output": "YES" }, { "input": "0 1 0\n1 1 1", "output": "YES" }, { "input": "1 1 0\n0 0 0", "output": "YES" }, { "input": "1 1 0\n1 0 0", "output": "YES" }, { "input": "1 1 0\n0 1 0", "output": "YES" }, { "input": "1 1 0\n0 0 1", "output": "NO" }, { "input": "1 1 0\n1 0 1", "output": "YES" }, { "input": "1 1 0\n0 1 1", "output": "YES" }, { "input": "1 1 0\n1 1 1", "output": "YES" }, { "input": "0 0 1\n0 0 0", "output": "YES" }, { "input": "0 0 1\n1 0 0", "output": "YES" }, { "input": "0 0 1\n0 1 0", "output": "YES" }, { "input": "0 0 1\n1 1 0", "output": "NO" }, { "input": "0 0 1\n1 0 1", "output": "YES" }, { "input": "0 0 1\n0 1 1", "output": "YES" }, { "input": "0 0 1\n1 1 1", "output": "YES" }, { "input": "1 0 1\n0 0 0", "output": "YES" }, { "input": "1 0 1\n1 0 0", "output": "YES" }, { "input": "1 0 1\n0 1 0", "output": "NO" }, { "input": "1 0 1\n1 1 0", "output": "YES" }, { "input": "1 0 1\n0 0 1", "output": "YES" }, { "input": "1 0 1\n0 1 1", "output": "YES" }, { "input": "1 0 1\n1 1 1", "output": "YES" }, { "input": "0 1 1\n0 0 0", "output": "YES" }, { "input": "0 1 1\n1 0 0", "output": "NO" }, { "input": "0 1 1\n0 1 0", "output": "YES" }, { "input": "0 1 1\n1 1 0", "output": "YES" }, { "input": "0 1 1\n0 0 1", "output": "YES" }, { "input": "0 1 1\n1 0 1", "output": "YES" }, { "input": "0 1 1\n1 1 1", "output": "YES" }, { "input": "1 1 1\n0 0 0", "output": "NO" }, { "input": "1 1 1\n1 0 0", "output": "YES" }, { "input": "1 1 1\n0 1 0", "output": "YES" }, { "input": "1 1 1\n1 1 0", "output": "YES" }, { "input": "1 1 1\n0 0 1", "output": "YES" }, { "input": "1 1 1\n1 0 1", "output": "YES" }, { "input": "1 1 1\n0 1 1", "output": "YES" } ]
1,675,016,673
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
n = input().split() m=input().split() a=[] b=[] s=0 s1=0 s2=0 s3=0 for i in n: a.append(int(i)) for i in m: b.append(int(i)) for i in range (len(a)): if a[i] == 0: s+=1 elif a[i] == 1: s2+=1 if b[i] == 1: s1+=1 elif b[i] == 0: s3+=1 if (s==3 and s1 == s1) or (s2== 3 and s3 ==3): print('NO') else: print('YES')
Title: Cubical Planet Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube. Input Specification: The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly. Output Specification: Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO". Demo Input: ['0 0 0\n0 1 0\n', '1 1 0\n0 1 0\n', '0 0 0\n1 1 1\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: none
```python n = input().split() m=input().split() a=[] b=[] s=0 s1=0 s2=0 s3=0 for i in n: a.append(int(i)) for i in m: b.append(int(i)) for i in range (len(a)): if a[i] == 0: s+=1 elif a[i] == 1: s2+=1 if b[i] == 1: s1+=1 elif b[i] == 0: s3+=1 if (s==3 and s1 == s1) or (s2== 3 and s3 ==3): print('NO') else: print('YES') ```
0
837
D
Round Subset
PROGRAMMING
2,100
[ "dp", "math" ]
null
null
Let's call the roundness of the number the number of zeros to which it ends. You have an array of *n* numbers. You need to choose a subset of exactly *k* numbers so that the roundness of the product of the selected numbers will be maximum possible.
The first line contains two integer numbers *n* and *k* (1<=≤<=*n*<=≤<=200,<=1<=≤<=*k*<=≤<=*n*). The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018).
Print maximal roundness of product of the chosen subset of length *k*.
[ "3 2\n50 4 20\n", "5 3\n15 16 3 25 9\n", "3 3\n9 77 13\n" ]
[ "3\n", "3\n", "0\n" ]
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3. In the second example subset [15, 16, 25] has product 6000, roundness 3. In the third example all subsets has product with roundness 0.
0
[ { "input": "3 2\n50 4 20", "output": "3" }, { "input": "5 3\n15 16 3 25 9", "output": "3" }, { "input": "3 3\n9 77 13", "output": "0" }, { "input": "1 1\n200000000", "output": "8" }, { "input": "1 1\n3", "output": "0" }, { "input": "3 1\n1000000000000000000 800000000000000000 625", "output": "18" }, { "input": "20 13\n93050001 1 750000001 950000001 160250001 482000001 145875001 900000001 500000001 513300001 313620001 724750001 205800001 400000001 800000001 175000001 875000001 852686005 868880001 342500001", "output": "0" }, { "input": "5 3\n1360922189858001 5513375057164001 4060879738933651 3260997351273601 5540397778584001", "output": "0" }, { "input": "5 3\n670206146698567481 75620705254979501 828058059097865201 67124386759325201 946737848872942801", "output": "0" }, { "input": "5 4\n539134530963895499 265657472022483040 798956216114326361 930406714691011229 562844921643925634", "output": "1" }, { "input": "200 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "200 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "200 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "200 200\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "5 2\n625 5 100 16 10", "output": "4" }, { "input": "5 2\n64 32 16 8 3125", "output": "5" }, { "input": "2 2\n2199023255552 11920928955078125", "output": "23" }, { "input": "1 1\n500", "output": "2" }, { "input": "3 1\n125 10 8", "output": "1" }, { "input": "7 5\n50 312500 10000 1250 2000000 250 1250000", "output": "18" }, { "input": "4 2\n3125 1000 1000 32", "output": "6" }, { "input": "3 1\n4 10 25", "output": "1" }, { "input": "3 1\n16 10 75", "output": "1" }, { "input": "3 2\n100 5120 19531250", "output": "11" }, { "input": "4 2\n16 200 500 625", "output": "5" }, { "input": "7 4\n5000 1250 100 200 10 5000 640", "output": "13" }, { "input": "2 1\n125 10", "output": "1" }, { "input": "3 2\n16 625 10", "output": "4" }, { "input": "4 2\n10 4 25 5", "output": "2" }, { "input": "2 1\n10 25", "output": "1" }, { "input": "3 1\n10 16 625", "output": "1" }, { "input": "4 2\n10 200 625 16", "output": "4" }, { "input": "3 1\n1000 16 625", "output": "3" }, { "input": "3 2\n6250 100 160", "output": "6" }, { "input": "3 2\n1250 100 80", "output": "5" }, { "input": "3 2\n1024 1000 9765625", "output": "10" }, { "input": "5 2\n125 10 10 3 3", "output": "2" }, { "input": "4 2\n31250 200 500 320", "output": "7" }, { "input": "3 1\n16 10 125", "output": "1" }, { "input": "5 3\n125 8 1000 100 100", "output": "7" }, { "input": "3 2\n25 4 10", "output": "2" }, { "input": "3 2\n125 8 10", "output": "3" }, { "input": "6 2\n125 125 100 100 8 8", "output": "4" }, { "input": "3 1\n32 3125 10", "output": "1" }, { "input": "5 3\n4 5 125 16 1", "output": "4" }, { "input": "7 4\n5 125 2 5 2 5 20", "output": "4" }, { "input": "200 200\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "100" }, { "input": "2 1\n200 250", "output": "2" }, { "input": "5 3\n500 10 250 20 8", "output": "6" }, { "input": "4 2\n2 2 25 1", "output": "1" }, { "input": "2 2\n16 25", "output": "2" } ]
1,525,305,911
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
3
2,000
7,475,200
import random #begin n, k = map(int, input().split()) a = list(map(int, input().split())) b = [] for i in a: l = r = 0 while not i % 2**(l+1): l += 1 while not i % 5**(r+1): r += 1 b.append((l, r)) #genetic obj_max, years, shufle_year, tick = 128, 1000, 50, 0 key = [i for i in range(n)] def shuffle_keys(): global key random.shuffle(key) def gen_new(): global tick l, r = 0, 0 for j in (key[(tick + i) % n] for i in range(k)): l += b[j][0] r += b[j][1] tick += i return min(l, r) def mutagen(x): pass def mutarev(x): p = (tick % (n - 2)) + 1 return x[p:] + x[:p] obj = [gen_new() for i in range(obj_max)] for y in range(years): if not y % shufle_year: shuffle_keys() obj.sort() obj.reverse() obj = ( obj[ : 20 ] + [gen_new() for i in range(100)] + obj[120:] ) map(mutarev, (obj[i] for i in range(0, obj_max, 15))) obj.sort() obj.reverse() print(obj[0])
Title: Round Subset Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's call the roundness of the number the number of zeros to which it ends. You have an array of *n* numbers. You need to choose a subset of exactly *k* numbers so that the roundness of the product of the selected numbers will be maximum possible. Input Specification: The first line contains two integer numbers *n* and *k* (1<=≤<=*n*<=≤<=200,<=1<=≤<=*k*<=≤<=*n*). The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018). Output Specification: Print maximal roundness of product of the chosen subset of length *k*. Demo Input: ['3 2\n50 4 20\n', '5 3\n15 16 3 25 9\n', '3 3\n9 77 13\n'] Demo Output: ['3\n', '3\n', '0\n'] Note: In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3. In the second example subset [15, 16, 25] has product 6000, roundness 3. In the third example all subsets has product with roundness 0.
```python import random #begin n, k = map(int, input().split()) a = list(map(int, input().split())) b = [] for i in a: l = r = 0 while not i % 2**(l+1): l += 1 while not i % 5**(r+1): r += 1 b.append((l, r)) #genetic obj_max, years, shufle_year, tick = 128, 1000, 50, 0 key = [i for i in range(n)] def shuffle_keys(): global key random.shuffle(key) def gen_new(): global tick l, r = 0, 0 for j in (key[(tick + i) % n] for i in range(k)): l += b[j][0] r += b[j][1] tick += i return min(l, r) def mutagen(x): pass def mutarev(x): p = (tick % (n - 2)) + 1 return x[p:] + x[:p] obj = [gen_new() for i in range(obj_max)] for y in range(years): if not y % shufle_year: shuffle_keys() obj.sort() obj.reverse() obj = ( obj[ : 20 ] + [gen_new() for i in range(100)] + obj[120:] ) map(mutarev, (obj[i] for i in range(0, obj_max, 15))) obj.sort() obj.reverse() print(obj[0]) ```
0
149
A
Business trip
PROGRAMMING
900
[ "greedy", "implementation", "sortings" ]
null
null
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until... Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters. Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
[ "5\n1 1 1 1 2 2 3 2 2 1 1 1\n", "0\n0 0 0 0 0 0 0 1 1 2 3 0\n", "11\n1 1 4 1 1 5 1 1 4 1 1 1\n" ]
[ "2\n", "0\n", "3\n" ]
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters. In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
500
[ { "input": "5\n1 1 1 1 2 2 3 2 2 1 1 1", "output": "2" }, { "input": "0\n0 0 0 0 0 0 0 1 1 2 3 0", "output": "0" }, { "input": "11\n1 1 4 1 1 5 1 1 4 1 1 1", "output": "3" }, { "input": "15\n20 1 1 1 1 2 2 1 2 2 1 1", "output": "1" }, { "input": "7\n8 9 100 12 14 17 21 10 11 100 23 10", "output": "1" }, { "input": "52\n1 12 3 11 4 5 10 6 9 7 8 2", "output": "6" }, { "input": "50\n2 2 3 4 5 4 4 5 7 3 2 7", "output": "-1" }, { "input": "0\n55 81 28 48 99 20 67 95 6 19 10 93", "output": "0" }, { "input": "93\n85 40 93 66 92 43 61 3 64 51 90 21", "output": "1" }, { "input": "99\n36 34 22 0 0 0 52 12 0 0 33 47", "output": "2" }, { "input": "99\n28 32 31 0 10 35 11 18 0 0 32 28", "output": "3" }, { "input": "99\n19 17 0 1 18 11 29 9 29 22 0 8", "output": "4" }, { "input": "76\n2 16 11 10 12 0 20 4 4 14 11 14", "output": "5" }, { "input": "41\n2 1 7 7 4 2 4 4 9 3 10 0", "output": "6" }, { "input": "47\n8 2 2 4 3 1 9 4 2 7 7 8", "output": "7" }, { "input": "58\n6 11 7 0 5 6 3 9 4 9 5 1", "output": "8" }, { "input": "32\n5 2 4 1 5 0 5 1 4 3 0 3", "output": "9" }, { "input": "31\n6 1 0 4 4 5 1 0 5 3 2 0", "output": "9" }, { "input": "35\n2 3 0 0 6 3 3 4 3 5 0 6", "output": "9" }, { "input": "41\n3 1 3 4 3 6 6 1 4 4 0 6", "output": "11" }, { "input": "97\n0 5 3 12 10 16 22 8 21 17 21 10", "output": "5" }, { "input": "100\n21 21 0 0 4 13 0 26 0 0 0 15", "output": "6" }, { "input": "100\n0 0 16 5 22 0 5 0 25 0 14 13", "output": "7" }, { "input": "97\n17 0 10 0 0 0 18 0 14 23 15 0", "output": "6" }, { "input": "100\n0 9 0 18 7 0 0 14 33 3 0 16", "output": "7" }, { "input": "95\n5 2 13 0 15 18 17 0 6 11 0 8", "output": "9" }, { "input": "94\n11 13 0 9 15 8 8 16 3 7 1 3", "output": "11" }, { "input": "96\n8 4 12 15 8 0 4 10 6 6 12 11", "output": "11" }, { "input": "100\n5 5 3 8 6 5 0 3 3 8 1 3", "output": "-1" }, { "input": "100\n1 0 0 1 1 0 1 1 1 1 2 1", "output": "-1" }, { "input": "100\n6 3 2 0 4 1 2 2 2 2 1 1", "output": "-1" }, { "input": "0\n0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0", "output": "-1" }, { "input": "0\n100 100 100 100 100 100 100 100 100 100 100 100", "output": "0" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100", "output": "1" }, { "input": "12\n1 1 1 1 1 1 1 1 1 1 1 1", "output": "12" }, { "input": "13\n1 1 1 1 1 1 1 1 1 1 1 2", "output": "12" }, { "input": "15\n10 1 1 1 1 1 1 1 1 1 1 1", "output": "6" }, { "input": "1\n0 0 0 0 0 0 0 0 0 0 0 0", "output": "-1" } ]
1,644,085,227
2,147,483,647
Python 3
OK
TESTS
39
92
0
#**************************************** #** Solution by BAZOOKA ** #** Sponsored by RED BULL** #** I love ❤Kateryna Gret❤ ** #****************************************/ k=int(input()) m=sorted(map(int,input().split()))[::-1] for i in range(13): if sum(m[:i])>=k: print(i) exit(0) print(-1) #**************************************** #** Solution by BAZOOKA ** #** Sponsored by RED BULL** #** I love ❤Kateryna Gret❤ ** #****************************************/
Title: Business trip Time Limit: None seconds Memory Limit: None megabytes Problem Description: What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until... Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters. Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters. Input Specification: The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100). Output Specification: Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1. Demo Input: ['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n'] Demo Output: ['2\n', '0\n', '3\n'] Note: Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters. In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all.
```python #**************************************** #** Solution by BAZOOKA ** #** Sponsored by RED BULL** #** I love ❤Kateryna Gret❤ ** #****************************************/ k=int(input()) m=sorted(map(int,input().split()))[::-1] for i in range(13): if sum(m[:i])>=k: print(i) exit(0) print(-1) #**************************************** #** Solution by BAZOOKA ** #** Sponsored by RED BULL** #** I love ❤Kateryna Gret❤ ** #****************************************/ ```
3
199
A
Hexadecimal's theorem
PROGRAMMING
900
[ "brute force", "constructive algorithms", "implementation", "number theory" ]
null
null
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
The input contains of a single integer *n* (0<=≤<=*n*<=&lt;<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them.
[ "3\n", "13\n" ]
[ "1 1 1\n", "2 3 8\n" ]
none
500
[ { "input": "3", "output": "1 1 1" }, { "input": "13", "output": "2 3 8" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "1597", "output": "233 377 987" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "3", "output": "1 1 1" }, { "input": "5", "output": "1 1 3" }, { "input": "8", "output": "1 2 5" }, { "input": "13", "output": "2 3 8" }, { "input": "21", "output": "3 5 13" }, { "input": "34", "output": "5 8 21" }, { "input": "55", "output": "8 13 34" }, { "input": "89", "output": "13 21 55" }, { "input": "144", "output": "21 34 89" }, { "input": "233", "output": "34 55 144" }, { "input": "377", "output": "55 89 233" }, { "input": "610", "output": "89 144 377" }, { "input": "987", "output": "144 233 610" }, { "input": "1597", "output": "233 377 987" }, { "input": "2584", "output": "377 610 1597" }, { "input": "4181", "output": "610 987 2584" }, { "input": "6765", "output": "987 1597 4181" }, { "input": "10946", "output": "1597 2584 6765" }, { "input": "17711", "output": "2584 4181 10946" }, { "input": "28657", "output": "4181 6765 17711" }, { "input": "46368", "output": "6765 10946 28657" }, { "input": "75025", "output": "10946 17711 46368" }, { "input": "121393", "output": "17711 28657 75025" }, { "input": "196418", "output": "28657 46368 121393" }, { "input": "317811", "output": "46368 75025 196418" }, { "input": "514229", "output": "75025 121393 317811" }, { "input": "832040", "output": "121393 196418 514229" }, { "input": "1346269", "output": "196418 317811 832040" }, { "input": "2178309", "output": "317811 514229 1346269" }, { "input": "3524578", "output": "514229 832040 2178309" }, { "input": "5702887", "output": "832040 1346269 3524578" }, { "input": "9227465", "output": "1346269 2178309 5702887" }, { "input": "14930352", "output": "2178309 3524578 9227465" }, { "input": "24157817", "output": "3524578 5702887 14930352" }, { "input": "39088169", "output": "5702887 9227465 24157817" }, { "input": "63245986", "output": "9227465 14930352 39088169" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "165580141", "output": "24157817 39088169 102334155" }, { "input": "267914296", "output": "39088169 63245986 165580141" }, { "input": "433494437", "output": "63245986 102334155 267914296" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "63245986", "output": "9227465 14930352 39088169" } ]
1,591,873,503
2,147,483,647
Python 3
OK
TESTS
54
218
0
k=int(input()) print("0 0 "+str(k))
Title: Hexadecimal's theorem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. Input Specification: The input contains of a single integer *n* (0<=≤<=*n*<=&lt;<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. Output Specification: Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them. Demo Input: ['3\n', '13\n'] Demo Output: ['1 1 1\n', '2 3 8\n'] Note: none
```python k=int(input()) print("0 0 "+str(k)) ```
3
287
B
Pipeline
PROGRAMMING
1,700
[ "binary search", "math" ]
null
null
Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly *n* houses in Ultimate Thule, Vova wants the city to have exactly *n* pipes, each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters. A splitter is a construction that consists of one input (it can be connected to a water pipe) and *x* output pipes. When a splitter is connected to a water pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe. Vova has one splitter of each kind: with 2, 3, 4, ..., *k* outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible. Vova needs the pipeline to have exactly *n* pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters.
The first line contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=1018, 2<=≤<=*k*<=≤<=109). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1.
[ "4 3\n", "5 5\n", "8 4\n" ]
[ "2\n", "1\n", "-1\n" ]
none
1,500
[ { "input": "4 3", "output": "2" }, { "input": "5 5", "output": "1" }, { "input": "8 4", "output": "-1" }, { "input": "1000000000000000000 1000000000", "output": "-1" }, { "input": "499999998500000001 1000000000", "output": "999955279" }, { "input": "499999998500000000 1000000000", "output": "999955279" }, { "input": "499999999500000000 1000000000", "output": "999999998" }, { "input": "499999999500000001 1000000000", "output": "999999999" }, { "input": "525 34", "output": "25" }, { "input": "223265034477 726990", "output": "440662" }, { "input": "15597035789572051 185473109", "output": "128849771" }, { "input": "499999999500000002 1000000000", "output": "-1" }, { "input": "1 1000000000", "output": "0" }, { "input": "1000000000 2", "output": "-1" }, { "input": "462498979 204468265", "output": "3" }, { "input": "2107921 542531", "output": "4" }, { "input": "131 49", "output": "3" }, { "input": "20171878992939541 200857557", "output": "200853401" }, { "input": "399812655947 894219", "output": "893030" }, { "input": "93 17", "output": "8" }, { "input": "1000000000 999999999", "output": "2" }, { "input": "100000000000000000 1000000000", "output": "105572810" } ]
1,695,492,363
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
46
0
def main(): n, k = map(int, input().split()) count = 0 while n > 0: if k == 1: return -1 if k <= n: n -=k count += 1 k -= 1 return count print(main())
Title: Pipeline Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly *n* houses in Ultimate Thule, Vova wants the city to have exactly *n* pipes, each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters. A splitter is a construction that consists of one input (it can be connected to a water pipe) and *x* output pipes. When a splitter is connected to a water pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe. Vova has one splitter of each kind: with 2, 3, 4, ..., *k* outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible. Vova needs the pipeline to have exactly *n* pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters. Input Specification: The first line contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=1018, 2<=≤<=*k*<=≤<=109). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1. Demo Input: ['4 3\n', '5 5\n', '8 4\n'] Demo Output: ['2\n', '1\n', '-1\n'] Note: none
```python def main(): n, k = map(int, input().split()) count = 0 while n > 0: if k == 1: return -1 if k <= n: n -=k count += 1 k -= 1 return count print(main()) ```
0
584
A
Olesya and Rodion
PROGRAMMING
1,000
[ "math" ]
null
null
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them. Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
[ "3 2\n" ]
[ "712" ]
none
500
[ { "input": "3 2", "output": "222" }, { "input": "2 2", "output": "22" }, { "input": "4 3", "output": "3333" }, { "input": "5 3", "output": "33333" }, { "input": "10 7", "output": "7777777777" }, { "input": "2 9", "output": "99" }, { "input": "18 8", "output": "888888888888888888" }, { "input": "1 5", "output": "5" }, { "input": "1 10", "output": "-1" }, { "input": "100 5", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "10 2", "output": "2222222222" }, { "input": "18 10", "output": "111111111111111110" }, { "input": "1 9", "output": "9" }, { "input": "7 6", "output": "6666666" }, { "input": "4 4", "output": "4444" }, { "input": "14 7", "output": "77777777777777" }, { "input": "3 8", "output": "888" }, { "input": "1 3", "output": "3" }, { "input": "2 8", "output": "88" }, { "input": "3 8", "output": "888" }, { "input": "4 3", "output": "3333" }, { "input": "5 9", "output": "99999" }, { "input": "4 8", "output": "8888" }, { "input": "3 4", "output": "444" }, { "input": "9 4", "output": "444444444" }, { "input": "8 10", "output": "11111110" }, { "input": "1 6", "output": "6" }, { "input": "20 3", "output": "33333333333333333333" }, { "input": "15 10", "output": "111111111111110" }, { "input": "31 4", "output": "4444444444444444444444444444444" }, { "input": "18 9", "output": "999999999999999999" }, { "input": "72 4", "output": "444444444444444444444444444444444444444444444444444444444444444444444444" }, { "input": "76 8", "output": "8888888888888888888888888888888888888888888888888888888888888888888888888888" }, { "input": "12 5", "output": "555555555555" }, { "input": "54 5", "output": "555555555555555555555555555555555555555555555555555555" }, { "input": "96 10", "output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "15 9", "output": "999999999999999" }, { "input": "100 2", "output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222" }, { "input": "99 3", "output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333" }, { "input": "98 4", "output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444" }, { "input": "97 5", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "100 6", "output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666" }, { "input": "99 7", "output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "98 8", "output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888" }, { "input": "97 9", "output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999" }, { "input": "100 10", "output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "3 10", "output": "110" }, { "input": "2 4", "output": "44" }, { "input": "2 10", "output": "10" }, { "input": "10 10", "output": "1111111110" }, { "input": "4 10", "output": "1110" }, { "input": "25 10", "output": "1111111111111111111111110" }, { "input": "50 10", "output": "11111111111111111111111111111111111111111111111110" }, { "input": "5 10", "output": "11110" }, { "input": "99 10", "output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "100 10", "output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "1 2", "output": "2" }, { "input": "20 10", "output": "11111111111111111110" }, { "input": "99 7", "output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "89 5", "output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "6 10", "output": "111110" }, { "input": "3 8", "output": "888" }, { "input": "1 9", "output": "9" }, { "input": "1 4", "output": "4" }, { "input": "100 7", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" } ]
1,667,339,992
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
31
0
n=int(input()) m=int(input()) l=[] f=[] for i in range(n): l.append(n) if(m%2==0 and l[len(l)-1]%2!=0): l[len(l)-1]+=1 elif(m%2!=0 and l[len(l)-1]%2==0): l[len(l)-1]-=1; for i in range(len(l)): f.append(str(l[i])) k=''.join(f) print(k)
Title: Olesya and Rodion Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them. Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. Input Specification: The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. Output Specification: Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. Demo Input: ['3 2\n'] Demo Output: ['712'] Note: none
```python n=int(input()) m=int(input()) l=[] f=[] for i in range(n): l.append(n) if(m%2==0 and l[len(l)-1]%2!=0): l[len(l)-1]+=1 elif(m%2!=0 and l[len(l)-1]%2==0): l[len(l)-1]-=1; for i in range(len(l)): f.append(str(l[i])) k=''.join(f) print(k) ```
-1
119
A
Epic Game
PROGRAMMING
800
[ "implementation" ]
null
null
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
[ "3 5 9\n", "1 1 100\n" ]
[ "0", "1" ]
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 &lt; 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
500
[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]
1,610,096,218
2,147,483,647
Python 3
OK
TESTS
45
218
0
def gcd(x, y): while(y):x, y = y, x % y return x a, b, c = map(int, input().split()) while 1: if gcd(a, c) > c: print(1) break c -= gcd(a, c) if gcd(b, c) > c: print(0) break c -= gcd(b, c)
Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 &lt; 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
```python def gcd(x, y): while(y):x, y = y, x % y return x a, b, c = map(int, input().split()) while 1: if gcd(a, c) > c: print(1) break c -= gcd(a, c) if gcd(b, c) > c: print(0) break c -= gcd(b, c) ```
3
139
A
Petr and Book
PROGRAMMING
1,000
[ "implementation" ]
null
null
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages. Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week. Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book. The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
[ "100\n15 20 20 15 10 30 45\n", "2\n1 0 0 0 0 0 0\n" ]
[ "6\n", "1\n" ]
Note to the first sample: By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else). Note to the second sample: On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
500
[ { "input": "100\n15 20 20 15 10 30 45", "output": "6" }, { "input": "2\n1 0 0 0 0 0 0", "output": "1" }, { "input": "100\n100 200 100 200 300 400 500", "output": "1" }, { "input": "3\n1 1 1 1 1 1 1", "output": "3" }, { "input": "1\n1 1 1 1 1 1 1", "output": "1" }, { "input": "20\n5 3 7 2 1 6 4", "output": "6" }, { "input": "10\n5 1 1 1 1 1 5", "output": "6" }, { "input": "50\n10 1 10 1 10 1 10", "output": "1" }, { "input": "77\n11 11 11 11 11 11 10", "output": "1" }, { "input": "1\n1000 1000 1000 1000 1000 1000 1000", "output": "1" }, { "input": "1000\n100 100 100 100 100 100 100", "output": "3" }, { "input": "999\n10 20 10 20 30 20 10", "output": "3" }, { "input": "433\n109 58 77 10 39 125 15", "output": "7" }, { "input": "1\n0 0 0 0 0 0 1", "output": "7" }, { "input": "5\n1 0 1 0 1 0 1", "output": "1" }, { "input": "997\n1 1 0 0 1 0 1", "output": "1" }, { "input": "1000\n1 1 1 1 1 1 1", "output": "6" }, { "input": "1000\n1000 1000 1000 1000 1000 1000 1000", "output": "1" }, { "input": "1000\n1 0 0 0 0 0 0", "output": "1" }, { "input": "1000\n0 0 0 0 0 0 1", "output": "7" }, { "input": "1000\n1 0 0 1 0 0 1", "output": "1" }, { "input": "509\n105 23 98 0 7 0 155", "output": "2" }, { "input": "7\n1 1 1 1 1 1 1", "output": "7" }, { "input": "2\n1 1 0 0 0 0 0", "output": "2" }, { "input": "1\n0 0 0 0 0 1 0", "output": "6" }, { "input": "10\n0 0 0 0 0 0 1", "output": "7" }, { "input": "5\n0 0 0 0 0 6 0", "output": "6" }, { "input": "3\n0 1 0 0 0 0 0", "output": "2" }, { "input": "10\n0 0 0 0 0 0 10", "output": "7" }, { "input": "28\n1 2 3 4 5 6 7", "output": "7" }, { "input": "100\n5 5 5 5 5 5 5", "output": "6" }, { "input": "4\n1 0 0 0 0 0 1", "output": "7" }, { "input": "2\n0 0 0 0 0 0 1", "output": "7" }, { "input": "7\n0 0 0 0 0 0 7", "output": "7" }, { "input": "7\n2 1 1 1 1 1 0", "output": "6" }, { "input": "2\n0 0 1 1 0 0 0", "output": "4" }, { "input": "6\n1 1 1 1 1 1 0", "output": "6" }, { "input": "5\n1 1 1 0 0 1 1", "output": "7" }, { "input": "100\n10 20 30 10 10 10 10", "output": "7" }, { "input": "1\n0 0 0 1 0 0 0", "output": "4" }, { "input": "70\n10 10 10 10 10 10 10", "output": "7" }, { "input": "22\n1 2 3 4 5 6 10", "output": "7" }, { "input": "5\n0 0 0 1 0 0 0", "output": "4" }, { "input": "2\n0 0 0 1 0 0 0", "output": "4" }, { "input": "6\n1 0 0 0 0 0 2", "output": "7" }, { "input": "10\n1 2 2 1 2 1 1", "output": "7" }, { "input": "5\n0 0 0 0 0 0 10", "output": "7" }, { "input": "4\n0 1 1 0 0 0 0", "output": "3" }, { "input": "100\n0 0 0 0 0 1 0", "output": "6" } ]
1,660,820,529
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
62
0
book = int(input()) sps = list(map(int, input().split())) for_week = sum(strnc) if for_week >= book: strnc = 0 for j in range (7): strnc += sps[j] if strnc >= book: print (j + 1) break else: amn_week = book // for_week for_week *= amn_week for j in range (7): for_week += sps[j] if for_week >= book: print (j + 1) break
Title: Petr and Book Time Limit: None seconds Memory Limit: None megabytes Problem Description: One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages. Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week. Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book. Input Specification: The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book. The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero. Output Specification: Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Demo Input: ['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n'] Demo Output: ['6\n', '1\n'] Note: Note to the first sample: By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else). Note to the second sample: On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
```python book = int(input()) sps = list(map(int, input().split())) for_week = sum(strnc) if for_week >= book: strnc = 0 for j in range (7): strnc += sps[j] if strnc >= book: print (j + 1) break else: amn_week = book // for_week for_week *= amn_week for j in range (7): for_week += sps[j] if for_week >= book: print (j + 1) break ```
-1
958
E1
Guard Duty (easy)
PROGRAMMING
1,600
[ "brute force", "geometry", "greedy", "math" ]
null
null
The Rebel fleet is afraid that the Empire might want to strike back again. Princess Heidi needs to know if it is possible to assign *R* Rebel spaceships to guard *B* bases so that every base has exactly one guardian and each spaceship has exactly one assigned base (in other words, the assignment is a perfect matching). Since she knows how reckless her pilots are, she wants to be sure that any two (straight) paths – from a base to its assigned spaceship – do not intersect in the galaxy plane (that is, in 2D), and so there is no risk of collision.
The first line contains two space-separated integers *R*,<=*B*(1<=≤<=*R*,<=*B*<=≤<=10). For 1<=≤<=*i*<=≤<=*R*, the *i*<=+<=1-th line contains two space-separated integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=10000) denoting the coordinates of the *i*-th Rebel spaceship. The following *B* lines have the same format, denoting the position of bases. It is guaranteed that no two points coincide and that no three points are on the same line.
If it is possible to connect Rebel spaceships and bases so as satisfy the constraint, output Yes, otherwise output No (without quote).
[ "3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2\n", "2 1\n1 0\n2 2\n3 1\n" ]
[ "Yes\n", "No\n" ]
For the first example, one possible way is to connect the Rebels and bases in order. For the second example, there is no perfect matching between Rebels and bases.
0
[ { "input": "3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2", "output": "Yes" }, { "input": "2 1\n1 0\n2 2\n3 1", "output": "No" }, { "input": "1 1\n3686 4362\n-7485 5112", "output": "Yes" }, { "input": "1 2\n1152 -7324\n-5137 -35\n-6045 -5271", "output": "No" }, { "input": "1 3\n-8824 -9306\n-5646 -9767\n8123 9355\n-6203 -1643", "output": "No" }, { "input": "1 5\n-8775 6730\n-3806 -6906\n-9256 -8240\n-1320 6849\n8155 746\n8284 -7317", "output": "No" }, { "input": "1 8\n8741 3638\n381 -9191\n7534 8792\n-8848 -414\n2926 -7444\n9475 559\n6938 2359\n2570 4721\n3329 -9365", "output": "No" }, { "input": "1 9\n6207 1655\n2728 8520\n9142 3418\n-1258 -8727\n5370 3161\n-5167 -7691\n517 2242\n3702 -9904\n-6862 -2997\n2524 -5492", "output": "No" }, { "input": "1 10\n9424 3979\n-8582 9252\n-2432 -3287\n-4247 1932\n-9491 5544\n-635 5689\n8260 -6790\n8841 3067\n-5624 -1990\n1569 1045\n-8844 -8462", "output": "No" }, { "input": "2 1\n2893 -5488\n-5087 -5042\n-8928 -9453", "output": "No" }, { "input": "2 2\n359 -29\n6964 -7332\n2384 -4529\n4434 2253", "output": "Yes" }, { "input": "2 3\n-9617 845\n4195 -2181\n-6305 -9903\n-535 -6060\n9417 -8419", "output": "No" }, { "input": "2 5\n-9568 -3121\n-1408 2942\n-827 -7497\n4348 2432\n-7958 231\n6440 1896\n2647 -1305", "output": "No" }, { "input": "2 8\n7948 3490\n2779 3512\n3403 -3024\n-3180 -4831\n6815 4601\n7631 9772\n-6320 -1060\n5592 362\n-785 4040\n8030 3272", "output": "No" }, { "input": "2 9\n5414 -8195\n-5171 -1634\n5012 4161\n-5888 -585\n9258 2646\n5548 1523\n7259 -8619\n9580 5738\n-8715 706\n-2232 -3280\n1866 1775", "output": "No" }, { "input": "2 10\n-5118 -3971\n-1169 -9140\n-7807 -3139\n9702 -5328\n8555 3460\n-1442 -733\n701 -2802\n-5784 2578\n8186 -4810\n-2722 -1013\n-9437 4021\n-5403 -1331", "output": "No" }, { "input": "3 1\n-8199 -7896\n7015 -4898\n-499 -8710\n9953 -6411", "output": "No" }, { "input": "3 2\n9268 -9879\n4245 2515\n-9188 -3786\n-2458 -2165\n3420 463", "output": "No" }, { "input": "3 3\n-8149 697\n6593 7667\n2123 -9160\n-5165 9523\n747 -8933\n-1536 -2691", "output": "Yes" }, { "input": "3 5\n-658 7030\n990 3086\n-4958 -6755\n7159 -1986\n5634 -7726\n1740 -1450\n1947 7835\n-2755 -4709", "output": "No" }, { "input": "3 8\n-3143 -6360\n-5121 -6641\n-727 -9723\n-369 454\n-9298 4086\n5787 -1016\n2683 -9660\n-1089 1121\n-4898 7743\n418 5485\n7425 -6644", "output": "No" }, { "input": "3 9\n6882 -8342\n4669 -8932\n882 4904\n-220 4700\n587 -5311\n3704 -1823\n6559 -6921\n-7399 6497\n-5387 -5890\n-9844 -1067\n5367 -7237\n-8314 -939", "output": "No" }, { "input": "3 10\n-7100 -1623\n-3459 2172\n9676 1595\n-6053 4558\n-842 8819\n-9691 3144\n3440 -9112\n7034 4946\n4851 -2513\n430 4372\n-7175 -3497\n5719 381\n-8859 -1347", "output": "No" }, { "input": "5 1\n9621 -154\n6694 -2348\n944 -7225\n-1568 -5543\n-3805 -872\n1204 -2651", "output": "No" }, { "input": "5 2\n-355 -9579\n-1256 -4638\n-4890 7402\n-1420 -1297\n-1362 2290\n-879 9101\n9514 -6689", "output": "No" }, { "input": "5 3\n9670 8440\n1091 -9784\n6422 4884\n3314 -9610\n8523 -7107\n-2963 8293\n3092 -3950\n-4093 -6502", "output": "No" }, { "input": "5 5\n-2840 4475\n2931 -6923\n-659 -8125\n8197 -1118\n851 -5899\n313 6679\n-9751 6115\n-6415 4250\n-227 -9732\n-6282 5041", "output": "Yes" }, { "input": "5 8\n-5325 1383\n-5441 3351\n-3870 1465\n669 -8381\n-4377 5913\n4360 -329\n8725 8620\n7810 -2479\n4019 4850\n8052 9911\n4130 -4668\n3744 2537\n-7171 -3933", "output": "No" }, { "input": "5 9\n-2742 -600\n6609 8502\n-5118 6389\n-4300 5568\n-1934 -3484\n9719 -1137\n2303 -8641\n1500 2897\n-6172 -8783\n-2210 -6939\n9514 -5262\n-3773 -4081\n1983 -4032\n4503 -3496", "output": "No" }, { "input": "5 10\n1493 7658\n-598 7650\n9226 -964\n2439 -3114\n366 2391\n-1008 -2258\n6063 8568\n7488 6824\n-4674 9523\n9590 9960\n-8361 -8234\n520 -1312\n-3878 -1142\n-8261 -239\n-2346 -2362", "output": "No" }, { "input": "8 1\n-796 -1\n3591 -2510\n-6330 4706\n-7422 -9093\n7860 -7002\n5375 -5310\n3538 3108\n-9851 -9798\n-8884 -170", "output": "No" }, { "input": "8 2\n-3330 -1983\n-6621 -4800\n-4721 9630\n9871 -4847\n-2256 -8957\n3292 -6118\n4558 -6712\n-5863 5282\n-9373 3938\n-5179 -8073", "output": "No" }, { "input": "8 3\n6695 8593\n-7129 352\n6590 -5447\n-2540 -3457\n7630 1647\n8651 5634\n-1864 -6829\n7828 -1901\n-7005 -9695\n4561 -4921\n-4782 -6478", "output": "No" }, { "input": "8 5\n6744 2367\n-5290 -7085\n-491 6662\n2343 -2407\n-43 2855\n-8075 6875\n-7265 -4206\n-4197 8851\n7433 780\n4038 -8321\n-1455 -7665\n3139 -1225\n9884 -167", "output": "No" }, { "input": "8 8\n4260 1536\n-8545 6045\n-3702 3693\n-5185 -2228\n-5271 -5335\n-4027 4453\n-8790 8598\n7172 -5320\n-880 -4638\n-1630 -3452\n2076 8296\n-9116 -5599\n2461 9832\n4268 5116\n-7582 -805\n3548 3776", "output": "Yes" }, { "input": "8 9\n-5716 6995\n1245 3754\n7610 8617\n-451 -5424\n-2828 5270\n-6111 6502\n-2653 1039\n3718 7498\n-8810 -7973\n667 -300\n-2838 -2001\n3367 5523\n-8386 -2827\n6929 -6260\n3247 1167\n1873 6265\n4376 -8781", "output": "No" }, { "input": "8 10\n5844 -8156\n9676 -8121\n-6302 -1050\n-4823 -8343\n4736 -3859\n9129 5920\n-3990 2792\n3615 -8930\n-7831 -8703\n-5542 931\n7599 -7930\n8705 -8735\n-6438 1724\n-7568 -8351\n5893 2316\n2574 -9723\n2416 3827\n856 -4877", "output": "No" }, { "input": "9 1\n8114 -9851\n872 -9807\n9541 5449\n7948 -3808\n8892 -7517\n-6767 3903\n-18 -311\n-3973 5845\n-3295 3533\n-4790 -4426", "output": "No" }, { "input": "9 2\n5580 8167\n-7078 -4655\n3707 -9628\n2980 438\n1632 -9472\n-8850 -4346\n-6440 2428\n-2841 923\n6515 -2658\n-2492 -8716\n8219 5104", "output": "No" }, { "input": "9 3\n8163 6185\n-4731 2757\n-4982 -4704\n3128 4684\n-8483 1132\n6807 2288\n4878 2311\n-6295 6299\n8882 -5992\n-195 4733\n6162 4510\n-7264 -1020", "output": "No" }, { "input": "9 5\n-4347 -5222\n-2891 5618\n-4621 7404\n-4548 -6825\n3846 2340\n2640 3530\n-7965 4934\n-8617 -2950\n-9240 4483\n-718 6451\n-8251 -6379\n558 3484\n9861 -6432\n483 -7331", "output": "No" }, { "input": "9 8\n-6832 -872\n1295 -4109\n-7832 -8123\n-2373 -6646\n-1383 -5849\n3832 -6334\n-7229 -2263\n-6951 -9678\n4709 1326\n-6386 -1239\n2721 -8159\n-4255 -890\n9880 3567\n3349 5921\n2487 -828\n-783 2422\n-5497 -8399", "output": "No" }, { "input": "9 9\n3193 -2855\n787 -6399\n3479 9360\n5217 -9842\n1061 4755\n1748 -7142\n-6209 -2380\n6740 -4302\n-5482 5433\n3353 -5529\n664 1546\n8228 -9769\n-8409 -1650\n893 9365\n-9542 8585\n7245 -9972\n-475 -6359\n-3775 2139", "output": "Yes" }, { "input": "9 10\n-3581 3894\n7385 3191\n-8820 6540\n-577 -5900\n2781 -5943\n8322 -7944\n-1251 -5779\n-3567 3140\n8835 -6406\n-2390 -1126\n7006 4553\n-174 -7023\n-6538 1530\n3318 2477\n7864 -9657\n-2379 -6961\n4456 9852\n1462 -5871\n-9931 6466", "output": "No" }, { "input": "10 1\n3362 3137\n-6006 -2168\n-9207 8006\n-6284 -114\n4617 -4997\n-4360 3540\n-6423 2328\n-8768 8468\n2899 1032\n-7561 -3623\n6979 653", "output": "No" }, { "input": "10 2\n5945 8596\n-3658 -4459\n-7598 -7071\n3567 4132\n7060 -1835\n-6443 -4709\n4895 2211\n-4780 3546\n5266 7400\n2178 -472\n4922 -9643\n4163 6030", "output": "No" }, { "input": "10 3\n3411 6614\n8392 693\n-8846 7555\n-1402 -4181\n-3055 -3789\n4033 -5516\n-1527 4950\n-792 8922\n-4925 4065\n4475 5536\n-9695 9764\n6943 -2849\n7022 1986", "output": "No" }, { "input": "10 5\n3460 5504\n529 -6744\n4075 9961\n-3961 4311\n-7871 9977\n7308 -4275\n-6928 7573\n-3114 -327\n-3046 -5461\n3953 4398\n-4106 -3981\n-8092 -8048\n7590 9228\n9433 -4\n-8808 -6742", "output": "No" }, { "input": "10 8\n8417 -444\n-5582 6386\n863 6992\n-4047 6751\n-5658 1788\n-1204 5862\n-6192 -2480\n813 -7056\n-9098 -1176\n-1715 -3292\n6866 -2905\n-7788 137\n7609 -774\n-7702 -6753\n-6622 -3090\n3089 -7006\n-9374 1882\n-481 -5698", "output": "No" }, { "input": "10 9\n-9001 -9868\n4207 1240\n-7826 1618\n-6755 3555\n-3214 -167\n4155 -4648\n-2316 259\n4801 -1679\n-6730 8048\n-4535 -9843\n4809 -5759\n4695 -8742\n9321 -5991\n2401 4133\n6468 6324\n1414 -9103\n-6613 3922\n5544 -5092\n-6777 -788", "output": "No" }, { "input": "10 10\n8530 -3814\n-9330 -6035\n3951 -217\n-9276 8291\n636 -3118\n5024 -2403\n4601 7977\n-3620 -1428\n4954 -9632\n-9852 6553\n-3457 5430\n-8866 -7343\n1020 -5748\n5043 -3820\n-2832 1528\n-5058 -825\n2406 -3530\n9152 -7463\n-8547 7108\n2492 8953", "output": "Yes" } ]
1,577,119,717
2,147,483,647
PyPy 3
OK
TESTS
51
140
0
l = input().split(' ') print("Yes" if int(l[0]) == int(l[1]) else "No")
Title: Guard Duty (easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Rebel fleet is afraid that the Empire might want to strike back again. Princess Heidi needs to know if it is possible to assign *R* Rebel spaceships to guard *B* bases so that every base has exactly one guardian and each spaceship has exactly one assigned base (in other words, the assignment is a perfect matching). Since she knows how reckless her pilots are, she wants to be sure that any two (straight) paths – from a base to its assigned spaceship – do not intersect in the galaxy plane (that is, in 2D), and so there is no risk of collision. Input Specification: The first line contains two space-separated integers *R*,<=*B*(1<=≤<=*R*,<=*B*<=≤<=10). For 1<=≤<=*i*<=≤<=*R*, the *i*<=+<=1-th line contains two space-separated integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=10000) denoting the coordinates of the *i*-th Rebel spaceship. The following *B* lines have the same format, denoting the position of bases. It is guaranteed that no two points coincide and that no three points are on the same line. Output Specification: If it is possible to connect Rebel spaceships and bases so as satisfy the constraint, output Yes, otherwise output No (without quote). Demo Input: ['3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2\n', '2 1\n1 0\n2 2\n3 1\n'] Demo Output: ['Yes\n', 'No\n'] Note: For the first example, one possible way is to connect the Rebels and bases in order. For the second example, there is no perfect matching between Rebels and bases.
```python l = input().split(' ') print("Yes" if int(l[0]) == int(l[1]) else "No") ```
3
999
A
Mishka and Contest
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve?
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Print one integer — the maximum number of problems Mishka can solve.
[ "8 4\n4 2 3 1 5 1 6 4\n", "5 2\n3 1 2 1 3\n", "5 100\n12 34 55 43 21\n" ]
[ "5\n", "0\n", "5\n" ]
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
0
[ { "input": "8 4\n4 2 3 1 5 1 6 4", "output": "5" }, { "input": "5 2\n3 1 2 1 3", "output": "0" }, { "input": "5 100\n12 34 55 43 21", "output": "5" }, { "input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62", "output": "100" }, { "input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98", "output": "98" }, { "input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2", "output": "39" }, { "input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16", "output": "99" }, { "input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1", "output": "1" }, { "input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "1 5\n4", "output": "1" }, { "input": "3 2\n1 4 1", "output": "2" }, { "input": "1 2\n100", "output": "0" }, { "input": "7 4\n4 2 3 4 4 2 3", "output": "7" }, { "input": "1 2\n1", "output": "1" }, { "input": "1 2\n15", "output": "0" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "5 3\n3 4 3 2 1", "output": "4" }, { "input": "1 1\n2", "output": "0" }, { "input": "1 5\n1", "output": "1" }, { "input": "6 6\n7 1 1 1 1 1", "output": "5" }, { "input": "5 5\n6 5 5 5 5", "output": "4" }, { "input": "1 4\n2", "output": "1" }, { "input": "9 4\n1 2 1 2 4 2 1 2 1", "output": "9" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 10\n5", "output": "1" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4", "output": "71" }, { "input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90", "output": "63" }, { "input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8", "output": "61" }, { "input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22", "output": "72" }, { "input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2", "output": "52" }, { "input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90", "output": "60" }, { "input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10", "output": "56" }, { "input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90", "output": "61" }, { "input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24", "output": "99" }, { "input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50", "output": "99" }, { "input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48", "output": "99" }, { "input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57", "output": "99" }, { "input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93", "output": "99" }, { "input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33", "output": "3" }, { "input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51", "output": "3" }, { "input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49", "output": "6" }, { "input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8", "output": "66" }, { "input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95", "output": "0" }, { "input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66", "output": "0" }, { "input": "2 8\n8 8", "output": "2" }, { "input": "1 6\n3", "output": "1" }, { "input": "1 5\n5", "output": "1" } ]
1,641,219,810
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
2
31
0
n , k = input().split() n = int(n) k = int(k) lst = list(map(int,input().split())) count = 0 while lst[0]<=k or lst[len(lst)-1]<=k: if lst[0]<=k: lst.remove(lst[0]) count+=1 elif lst[len(lst)-1]<=k: lst.remove(lst[len(lst)-1]) count+=1 print(count)
Title: Mishka and Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve? Input Specification: The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list. Output Specification: Print one integer — the maximum number of problems Mishka can solve. Demo Input: ['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n'] Demo Output: ['5\n', '0\n', '5\n'] Note: In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
```python n , k = input().split() n = int(n) k = int(k) lst = list(map(int,input().split())) count = 0 while lst[0]<=k or lst[len(lst)-1]<=k: if lst[0]<=k: lst.remove(lst[0]) count+=1 elif lst[len(lst)-1]<=k: lst.remove(lst[len(lst)-1]) count+=1 print(count) ```
-1
490
A
Team Olympiad
PROGRAMMING
800
[ "greedy", "implementation", "sortings" ]
null
null
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*: - *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE Each child happens to be good at exactly one of these three subjects. The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team. What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
In the first line output integer *w* — the largest possible number of teams. Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them. If no teams can be compiled, print the only line with value *w* equal to 0.
[ "7\n1 3 1 3 2 1 2\n", "4\n2 1 1 2\n" ]
[ "2\n3 5 2\n6 7 4\n", "0\n" ]
none
500
[ { "input": "7\n1 3 1 3 2 1 2", "output": "2\n3 5 2\n6 7 4" }, { "input": "4\n2 1 1 2", "output": "0" }, { "input": "1\n2", "output": "0" }, { "input": "2\n3 1", "output": "0" }, { "input": "3\n2 1 2", "output": "0" }, { "input": "3\n1 2 3", "output": "1\n1 2 3" }, { "input": "12\n3 3 3 3 3 3 3 3 1 3 3 2", "output": "1\n9 12 2" }, { "input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2", "output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47" }, { "input": "12\n3 1 1 1 1 1 1 2 1 1 1 1", "output": "1\n3 8 1" }, { "input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2", "output": "1\n18 2 11" }, { "input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3", "output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36" }, { "input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1", "output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200" }, { "input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3", "output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64" }, { "input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1", "output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44" }, { "input": "5\n1 2 2 3 3", "output": "1\n1 3 4" } ]
1,682,414,769
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
6
31
0
n = int(input()) arr = [int(i) for i in input().split()] num = [ 0 for _ in range(3)] t = 0 for i in arr : num[i-1]+=1 t+=1 while t%3 != 0 and t > 0 : t-=1 d = t // 3 g = {"1":[],"2":[],"3":[]} for i,ele in enumerate(arr): g[str(ele)].append(i+1) vv = [] v1 = g['1'] v2 = g['2'] v3 = g['3'] if len(v1) == 0 or len(v2) == 0 or len(v3) == 0 : print(0) else : print(d) for i in range(d) : print(v1[i],v2[i],v3[i])
Title: Team Olympiad Time Limit: None seconds Memory Limit: None megabytes Problem Description: The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*: - *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE Each child happens to be good at exactly one of these three subjects. The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team. What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child. Output Specification: In the first line output integer *w* — the largest possible number of teams. Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them. If no teams can be compiled, print the only line with value *w* equal to 0. Demo Input: ['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n'] Demo Output: ['2\n3 5 2\n6 7 4\n', '0\n'] Note: none
```python n = int(input()) arr = [int(i) for i in input().split()] num = [ 0 for _ in range(3)] t = 0 for i in arr : num[i-1]+=1 t+=1 while t%3 != 0 and t > 0 : t-=1 d = t // 3 g = {"1":[],"2":[],"3":[]} for i,ele in enumerate(arr): g[str(ele)].append(i+1) vv = [] v1 = g['1'] v2 = g['2'] v3 = g['3'] if len(v1) == 0 or len(v2) == 0 or len(v3) == 0 : print(0) else : print(d) for i in range(d) : print(v1[i],v2[i],v3[i]) ```
-1
802
A
Heidi and Library (easy)
PROGRAMMING
1,800
[ "greedy" ]
null
null
Your search for Heidi is over – you finally found her at a library, dressed up as a human. In fact, she has spent so much time there that she now runs the place! Her job is to buy books and keep them at the library so that people can borrow and read them. There are *n* different books, numbered 1 through *n*. We will look at the library's operation during *n* consecutive days. Heidi knows in advance that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) precisely one person will come to the library, request to borrow the book *a**i*, read it in a few hours, and return the book later on the same day. Heidi desperately wants to please all her guests, so she will make sure to always have the book *a**i* available in the library on the *i*-th day. During the night before the *i*-th day, she has the option of going to the bookstore (which operates at nights to avoid competition with the library) and buying any book for the price of 1 CHF. Of course, if she already has a book at the library, she does not need to buy it again. Initially, the library contains no books. There is a problem, though. The capacity of the library is *k* – this means that at any time, there can be at most *k* books at the library. If buying a new book would cause Heidi to have more than *k* books, she must first get rid of some book that she already has, in order to make room for the new book. If she later needs a book that she got rid of, she will need to buy that book again. You are given *k* and the sequence of requests for books *a*1,<=*a*2,<=...,<=*a**n*. What is the minimum cost (in CHF) of buying new books to satisfy all the requests?
The first line of input will contain two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=80). The second line will contain *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the sequence of book requests.
On a single line print the minimum cost of buying books at the store so as to satisfy all requests.
[ "4 80\n1 2 2 1\n", "4 1\n1 2 2 1\n", "4 2\n1 2 3 1\n" ]
[ "2\n", "3\n", "3\n" ]
In the first test case, Heidi is able to keep all books forever. Therefore, she only needs to buy the book 1 before the first day and the book 2 before the second day. In the second test case, she can only keep one book at a time. Therefore she will need to buy new books on the first, second and fourth day. In the third test case, before buying book 3 on the third day, she must decide which of the books 1 and 2 she should get rid of. Of course, she should keep the book 1, which will be requested on the fourth day.
0
[ { "input": "4 80\n1 2 2 1", "output": "2" }, { "input": "4 1\n1 2 2 1", "output": "3" }, { "input": "4 2\n1 2 3 1", "output": "3" }, { "input": "11 1\n1 2 3 5 1 10 10 1 1 3 5", "output": "9" }, { "input": "5 2\n1 2 3 1 2", "output": "4" }, { "input": "4 2\n1 2 3 2", "output": "3" }, { "input": "1 1\n1", "output": "1" }, { "input": "80 4\n9 9 2 6 3 10 2 5 4 9 6 7 5 5 3 8 5 3 2 10 7 8 5 3 4 9 4 3 9 5 2 10 8 4 7 3 8 3 5 2 3 7 8 4 2 4 4 7 2 2 5 7 5 8 10 10 5 1 1 3 5 2 10 8 7 9 7 4 8 3 2 8 7 9 10 9 7 1 5 5", "output": "34" }, { "input": "80 4\n10 19 20 18 16 7 13 18 15 5 7 13 16 8 14 8 3 15 19 19 7 13 17 9 18 16 4 14 10 18 1 3 5 3 20 18 9 4 17 19 13 20 16 12 15 5 5 18 17 16 4 5 20 10 18 4 7 19 10 15 8 15 17 3 10 16 19 2 6 6 3 12 10 7 15 3 17 15 6 8", "output": "49" }, { "input": "80 4\n28 34 9 3 29 12 19 17 22 10 21 2 26 18 14 7 7 10 37 39 10 1 9 37 33 4 25 21 23 2 4 2 35 1 11 19 33 31 18 10 23 1 26 20 17 31 18 27 31 22 33 7 2 5 30 24 18 32 1 14 2 33 7 26 2 10 1 10 5 19 37 33 33 34 28 20 1 22 11 14", "output": "58" }, { "input": "80 4\n71 49 41 21 72 71 37 14 51 59 73 11 70 15 36 46 32 57 58 15 72 67 16 75 70 11 67 3 40 36 2 9 63 68 32 22 63 52 67 55 35 19 72 59 22 19 44 55 59 74 4 34 53 3 22 57 32 27 78 12 71 4 26 15 43 21 79 10 67 39 34 74 38 26 31 78 2 78 69 42", "output": "62" }, { "input": "80 8\n16 13 11 16 3 4 1 4 4 16 6 6 1 12 19 18 12 15 2 10 2 18 18 13 3 17 16 15 7 6 19 8 2 14 17 13 1 14 4 2 3 16 2 15 13 15 9 10 7 14 7 2 1 18 19 15 7 3 19 8 9 4 12 4 3 4 9 10 6 5 4 4 9 4 20 8 17 7 1 14", "output": "32" }, { "input": "80 8\n5 17 39 25 40 34 11 23 7 16 20 35 31 14 18 17 32 10 40 9 17 23 5 33 2 9 21 22 8 11 22 7 28 36 3 10 12 21 20 29 25 5 12 30 8 21 18 19 1 29 9 4 19 5 15 36 38 37 10 27 15 13 6 22 31 5 40 30 21 39 23 21 39 32 37 28 29 11 34 16", "output": "51" }, { "input": "80 8\n8 72 32 27 27 20 69 28 77 25 8 4 75 11 41 71 57 17 45 65 79 8 61 15 24 80 39 36 34 13 76 37 16 71 64 77 11 58 30 26 61 23 18 30 68 65 12 47 69 65 3 55 71 3 32 4 20 39 47 25 75 49 34 60 48 56 77 70 59 59 75 6 5 23 55 30 62 66 4 4", "output": "57" }, { "input": "80 12\n9 5 8 1 12 2 6 19 8 20 6 12 9 6 16 1 2 5 11 6 8 4 13 7 2 17 18 12 15 17 13 2 9 8 1 17 10 2 9 12 18 3 5 11 10 16 7 16 8 11 3 18 13 19 8 13 13 2 20 13 11 14 20 3 2 1 17 18 17 8 4 3 12 3 19 18 4 16 6 6", "output": "25" }, { "input": "80 12\n27 12 25 30 13 27 12 17 35 25 1 28 35 16 23 20 38 1 37 2 35 29 16 26 37 4 23 39 24 2 16 21 39 21 23 38 33 9 38 22 40 36 23 39 1 2 4 14 22 26 32 4 31 38 4 5 4 15 35 12 5 32 37 38 11 14 16 26 36 38 2 40 10 15 33 38 36 20 35 12", "output": "37" }, { "input": "80 12\n30 19 34 24 56 38 31 63 57 50 53 69 79 5 6 74 47 47 73 17 18 70 72 49 35 20 65 21 18 4 54 12 67 8 28 25 64 6 31 36 35 54 61 7 45 54 55 49 50 6 3 7 10 29 76 62 50 50 32 66 25 19 17 3 67 17 37 67 58 18 54 25 8 78 35 16 61 19 45 40", "output": "55" }, { "input": "80 16\n4 27 31 28 8 17 28 31 20 7 39 5 40 13 28 6 23 1 16 4 34 2 13 6 6 9 18 1 25 19 33 26 33 16 24 5 13 23 25 9 10 16 25 34 39 8 4 6 33 25 7 40 32 23 13 17 32 20 28 25 33 20 29 2 40 34 23 6 28 2 12 12 9 36 18 39 32 8 11 15", "output": "36" }, { "input": "80 16\n31 26 40 46 75 35 63 29 2 49 51 14 4 65 10 4 8 72 44 67 57 60 69 21 52 40 37 54 27 12 31 24 21 59 61 80 11 76 58 7 77 10 55 9 11 36 7 41 61 13 2 28 28 77 22 57 54 62 65 80 78 32 72 64 41 69 36 46 50 5 48 53 6 76 76 65 57 7 29 67", "output": "53" }, { "input": "80 40\n34 71 32 39 65 8 13 4 7 4 18 66 20 12 57 74 58 50 30 27 31 48 1 6 63 63 7 32 56 48 42 35 45 55 52 76 52 26 40 15 8 38 73 47 55 75 17 22 36 59 28 19 6 79 58 7 40 66 48 39 71 67 55 61 71 24 60 39 63 6 47 70 8 10 68 49 49 67 53 15", "output": "54" }, { "input": "80 60\n70 59 1 24 50 77 11 71 36 26 75 56 30 31 76 51 73 29 49 2 80 55 32 52 51 56 2 28 37 11 37 1 63 33 74 60 59 63 55 28 73 45 5 69 77 8 35 76 47 48 49 77 25 75 38 70 77 80 27 9 13 45 14 30 33 34 32 78 52 4 7 57 55 43 13 1 45 36 79 45", "output": "50" }, { "input": "80 80\n73 32 65 26 35 2 10 73 80 64 67 14 40 34 15 59 8 71 5 9 49 63 32 51 23 1 13 9 34 6 80 31 32 60 31 27 65 20 21 74 26 51 17 11 19 37 4 50 73 22 71 55 76 7 51 21 50 63 6 75 66 55 37 62 10 45 52 21 42 18 80 12 7 28 54 50 9 5 25 42", "output": "53" }, { "input": "80 4\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5", "output": "23" }, { "input": "80 8\n1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8", "output": "17" }, { "input": "80 16\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12", "output": "20" }, { "input": "80 32\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 1 2 3 4 5 6 7 8 9 10 11 12 13 14", "output": "34" }, { "input": "80 80\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80", "output": "80" } ]
1,613,161,581
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
46
0
x = input().lower() for a in x: if a== "h" or a == "i" or a == "e" or a == "d" or a == "i": print("NO") else: print("YES") break
Title: Heidi and Library (easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your search for Heidi is over – you finally found her at a library, dressed up as a human. In fact, she has spent so much time there that she now runs the place! Her job is to buy books and keep them at the library so that people can borrow and read them. There are *n* different books, numbered 1 through *n*. We will look at the library's operation during *n* consecutive days. Heidi knows in advance that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) precisely one person will come to the library, request to borrow the book *a**i*, read it in a few hours, and return the book later on the same day. Heidi desperately wants to please all her guests, so she will make sure to always have the book *a**i* available in the library on the *i*-th day. During the night before the *i*-th day, she has the option of going to the bookstore (which operates at nights to avoid competition with the library) and buying any book for the price of 1 CHF. Of course, if she already has a book at the library, she does not need to buy it again. Initially, the library contains no books. There is a problem, though. The capacity of the library is *k* – this means that at any time, there can be at most *k* books at the library. If buying a new book would cause Heidi to have more than *k* books, she must first get rid of some book that she already has, in order to make room for the new book. If she later needs a book that she got rid of, she will need to buy that book again. You are given *k* and the sequence of requests for books *a*1,<=*a*2,<=...,<=*a**n*. What is the minimum cost (in CHF) of buying new books to satisfy all the requests? Input Specification: The first line of input will contain two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=80). The second line will contain *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) – the sequence of book requests. Output Specification: On a single line print the minimum cost of buying books at the store so as to satisfy all requests. Demo Input: ['4 80\n1 2 2 1\n', '4 1\n1 2 2 1\n', '4 2\n1 2 3 1\n'] Demo Output: ['2\n', '3\n', '3\n'] Note: In the first test case, Heidi is able to keep all books forever. Therefore, she only needs to buy the book 1 before the first day and the book 2 before the second day. In the second test case, she can only keep one book at a time. Therefore she will need to buy new books on the first, second and fourth day. In the third test case, before buying book 3 on the third day, she must decide which of the books 1 and 2 she should get rid of. Of course, she should keep the book 1, which will be requested on the fourth day.
```python x = input().lower() for a in x: if a== "h" or a == "i" or a == "e" or a == "d" or a == "i": print("NO") else: print("YES") break ```
0
0
none
none
none
0
[ "none" ]
null
null
Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her *t* queries, each of one of the following type: 1. <=+<= *a**i* — add non-negative integer *a**i* to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer. 1. <=-<= *a**i* — delete a single occurrence of non-negative integer *a**i* from the multiset. It's guaranteed, that there is at least one *a**i* in the multiset. 1. ? *s* — count the number of integers in the multiset (with repetitions) that match some pattern *s* consisting of 0 and 1. In the pattern, 0 stands for the even digits, while 1 stands for the odd. Integer *x* matches the pattern *s*, if the parity of the *i*-th from the right digit in decimal notation matches the *i*-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left. For example, if the pattern is *s*<==<=010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not.
The first line of the input contains an integer *t* (1<=≤<=*t*<=≤<=100<=000) — the number of operation Sonya has to perform. Next *t* lines provide the descriptions of the queries in order they appear in the input file. The *i*-th row starts with a character *c**i* — the type of the corresponding operation. If *c**i* is equal to '+' or '-' then it's followed by a space and an integer *a**i* (0<=≤<=*a**i*<=&lt;<=1018) given without leading zeroes (unless it's 0). If *c**i* equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18. It's guaranteed that there will be at least one query of type '?'. It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it.
For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time.
[ "12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0\n", "4\n+ 200\n+ 200\n- 200\n? 0\n" ]
[ "2\n1\n2\n1\n1\n", "1\n" ]
Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input. 1. 1 and 241. 1. 361. 1. 101 and 361. 1. 361. 1. 4000.
0
[ { "input": "12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0", "output": "2\n1\n2\n1\n1" }, { "input": "4\n+ 200\n+ 200\n- 200\n? 0", "output": "1" }, { "input": "20\n+ 61\n+ 99\n+ 51\n+ 70\n+ 7\n+ 34\n+ 71\n+ 86\n+ 68\n+ 39\n+ 78\n+ 81\n+ 89\n? 10\n? 00\n? 10\n? 01\n? 01\n? 00\n? 00", "output": "3\n2\n3\n4\n4\n2\n2" }, { "input": "20\n+ 13\n+ 50\n+ 9\n? 0\n+ 24\n? 0\n- 24\n? 0\n+ 79\n? 11\n- 13\n? 11\n- 50\n? 10\n? 1\n- 9\n? 1\n? 11\n- 79\n? 11", "output": "0\n1\n0\n2\n1\n0\n1\n0\n1\n0" }, { "input": "10\n+ 870566619432760298\n+ 869797178280285214\n+ 609920823721618090\n+ 221159591436767023\n+ 730599542279836538\n? 101001100111001011\n? 001111010101010011\n? 100010100011101110\n? 100110010110001100\n? 110000011101110011", "output": "0\n0\n0\n0\n0" }, { "input": "10\n+ 96135\n? 10111\n+ 63322\n? 10111\n+ 44490\n? 10111\n+ 69312\n? 10111\n? 01100\n+ 59396", "output": "1\n1\n1\n1\n1" }, { "input": "10\n+ 2\n- 2\n+ 778\n+ 3\n+ 4\n- 4\n+ 1\n+ 617\n? 011\n? 011", "output": "1\n1" }, { "input": "20\n+ 8\n+ 39532\n+ 813\n- 39532\n? 00011\n? 00000\n? 00011\n+ 70424\n- 8\n? 00011\n- 70424\n? 00011\n+ 29\n? 00001\n+ 6632\n+ 3319\n? 00001\n+ 3172\n? 01111\n- 29", "output": "1\n1\n1\n1\n1\n1\n1\n1" } ]
1,547,275,436
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
4
186
1,024,000
N=int(input()) M=[0]*N for i in range(N): S = input() pat = 0 for t in range(2,len(S)): pat = (pat+(ord(S[t])&1))<<1 if S[0]=='?': print(M[pat]) else: exec('M[pat]'+S[0]+'=1')
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her *t* queries, each of one of the following type: 1. <=+<= *a**i* — add non-negative integer *a**i* to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer. 1. <=-<= *a**i* — delete a single occurrence of non-negative integer *a**i* from the multiset. It's guaranteed, that there is at least one *a**i* in the multiset. 1. ? *s* — count the number of integers in the multiset (with repetitions) that match some pattern *s* consisting of 0 and 1. In the pattern, 0 stands for the even digits, while 1 stands for the odd. Integer *x* matches the pattern *s*, if the parity of the *i*-th from the right digit in decimal notation matches the *i*-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left. For example, if the pattern is *s*<==<=010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not. Input Specification: The first line of the input contains an integer *t* (1<=≤<=*t*<=≤<=100<=000) — the number of operation Sonya has to perform. Next *t* lines provide the descriptions of the queries in order they appear in the input file. The *i*-th row starts with a character *c**i* — the type of the corresponding operation. If *c**i* is equal to '+' or '-' then it's followed by a space and an integer *a**i* (0<=≤<=*a**i*<=&lt;<=1018) given without leading zeroes (unless it's 0). If *c**i* equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18. It's guaranteed that there will be at least one query of type '?'. It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it. Output Specification: For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time. Demo Input: ['12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0\n', '4\n+ 200\n+ 200\n- 200\n? 0\n'] Demo Output: ['2\n1\n2\n1\n1\n', '1\n'] Note: Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input. 1. 1 and 241. 1. 361. 1. 101 and 361. 1. 361. 1. 4000.
```python N=int(input()) M=[0]*N for i in range(N): S = input() pat = 0 for t in range(2,len(S)): pat = (pat+(ord(S[t])&1))<<1 if S[0]=='?': print(M[pat]) else: exec('M[pat]'+S[0]+'=1') ```
-1
499
B
Lecture
PROGRAMMING
1,000
[ "implementation", "strings" ]
null
null
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
[ "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n", "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n" ]
[ "codeforces round letter round\n", "hbnyiyc joll joll un joll\n" ]
none
500
[ { "input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest", "output": "codeforces round letter round" }, { "input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll", "output": "hbnyiyc joll joll un joll" }, { "input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b", "output": "f jpggvr b hnxsd b" }, { "input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb", "output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs" }, { "input": "1 1\namit am\namit", "output": "am" }, { "input": "1 1\na c\na", "output": "a" } ]
1,601,091,974
2,147,483,647
Python 3
OK
TESTS
37
109
716,800
# Aditya Morankar # lsta = list(map(int,input().split())) def main(): lst={} n,m = list(map(int,input().split())) for i in range(m): a,b = list(input().split(' ')) lst[a] = b sentence = list(input().split(' ')) for i in range(n): if sentence[i] in lst: if len(sentence[i]) > len(lst[sentence[i]]): sentence[i] = lst[sentence[i]] print(" ".join(map(str, sentence))) if __name__ == '__main__': t = 1 while t!=0: main() t-=1
Title: Lecture Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes. Input Specification: The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters. Output Specification: Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input. Demo Input: ['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n'] Demo Output: ['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n'] Note: none
```python # Aditya Morankar # lsta = list(map(int,input().split())) def main(): lst={} n,m = list(map(int,input().split())) for i in range(m): a,b = list(input().split(' ')) lst[a] = b sentence = list(input().split(' ')) for i in range(n): if sentence[i] in lst: if len(sentence[i]) > len(lst[sentence[i]]): sentence[i] = lst[sentence[i]] print(" ".join(map(str, sentence))) if __name__ == '__main__': t = 1 while t!=0: main() t-=1 ```
3
148
C
Terse princess
PROGRAMMING
1,700
[ "constructive algorithms", "greedy" ]
null
null
«Next please», — the princess called and cast an estimating glance at the next groom. The princess intends to choose the most worthy groom, this is, the richest one. Whenever she sees a groom who is more rich than each of the previous ones, she says a measured «Oh...». Whenever the groom is richer than all previous ones added together, she exclaims «Wow!» (no «Oh...» in this case). At the sight of the first groom the princess stays calm and says nothing. The fortune of each groom is described with an integer between 1 and 50000. You know that during the day the princess saw *n* grooms, said «Oh...» exactly *a* times and exclaimed «Wow!» exactly *b* times. Your task is to output a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n*, where *t**i* describes the fortune of *i*-th groom. If several sequences are possible, output any of them. If no sequence exists that would satisfy all the requirements, output a single number -1.
The only line of input data contains three integer numbers *n*,<=*a* and *b* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*a*,<=*b*<=≤<=15,<=*n*<=&gt;<=*a*<=+<=*b*), separated with single spaces.
Output any sequence of integers *t*1,<=*t*2,<=...,<=*t**n*, where *t**i* (1<=≤<=*t**i*<=≤<=50000) is the fortune of *i*-th groom, that satisfies the given constraints. If no sequence exists that would satisfy all the requirements, output a single number -1.
[ "10 2 3\n", "5 0 0\n" ]
[ "5 1 3 6 16 35 46 4 200 99", "10 10 6 6 5" ]
Let's have a closer look at the answer for the first sample test. - The princess said «Oh...» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99. - The princess exclaimed «Wow!» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.
1,000
[ { "input": "10 2 3", "output": "1 2 4 8 9 10 10 10 10 10 " }, { "input": "5 0 0", "output": "1 1 1 1 1 " }, { "input": "5 2 2", "output": "1 2 4 5 6 " }, { "input": "6 2 2", "output": "1 2 4 5 6 6 " }, { "input": "10 9 0", "output": "-1" }, { "input": "1 0 0", "output": "1 " }, { "input": "10 0 9", "output": "1 2 4 8 16 32 64 128 256 512 " }, { "input": "42 10 13", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 8193 8194 8195 8196 8197 8198 8199 8200 8201 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 8202 " }, { "input": "7 3 3", "output": "1 2 4 8 9 10 11 " }, { "input": "12 0 0", "output": "1 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "19 1 0", "output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 " }, { "input": "17 2 3", "output": "1 2 4 8 9 10 10 10 10 10 10 10 10 10 10 10 10 " }, { "input": "7 3 1", "output": "1 2 3 4 5 5 5 " }, { "input": "19 3 1", "output": "1 2 3 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 " }, { "input": "10 4 4", "output": "1 2 4 8 16 17 18 19 20 20 " }, { "input": "11 5 4", "output": "1 2 4 8 16 17 18 19 20 21 21 " }, { "input": "8 0 2", "output": "1 2 4 4 4 4 4 4 " }, { "input": "19 5 1", "output": "1 2 3 4 5 6 7 7 7 7 7 7 7 7 7 7 7 7 7 " }, { "input": "100 9 0", "output": "1 1 2 3 4 5 6 7 8 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 " }, { "input": "2 0 1", "output": "1 2 " }, { "input": "2 1 0", "output": "-1" }, { "input": "3 0 2", "output": "1 2 4 " }, { "input": "3 1 1", "output": "1 2 3 " }, { "input": "3 2 0", "output": "-1" }, { "input": "4 0 0", "output": "1 1 1 1 " }, { "input": "4 0 1", "output": "1 2 2 2 " }, { "input": "4 0 2", "output": "1 2 4 4 " }, { "input": "4 0 3", "output": "1 2 4 8 " }, { "input": "4 1 0", "output": "1 1 2 2 " }, { "input": "4 2 0", "output": "1 1 2 3 " }, { "input": "4 3 0", "output": "-1" }, { "input": "4 1 1", "output": "1 2 3 3 " }, { "input": "4 1 2", "output": "1 2 4 5 " }, { "input": "4 2 1", "output": "1 2 3 4 " }, { "input": "100 0 0", "output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "100 0 1", "output": "1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 " }, { "input": "100 1 0", "output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 " }, { "input": "100 1 1", "output": "1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 " }, { "input": "100 2 0", "output": "1 1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 " }, { "input": "100 0 2", "output": "1 2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 " }, { "input": "16 0 15", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 " }, { "input": "16 15 0", "output": "-1" }, { "input": "100 0 15", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 32768 ..." }, { "input": "100 15 0", "output": "1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 " }, { "input": "100 11 13", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 8193 8194 8195 8196 8197 8198 8199 8200 8201 8202 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 8203 " }, { "input": "100 15 15", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 32769 32770 32771 32772 32773 32774 32775 32776 32777 32778 32779 32780 32781 32782 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 32783 ..." }, { "input": "100 14 15", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 32769 32770 32771 32772 32773 32774 32775 32776 32777 32778 32779 32780 32781 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 32782 ..." }, { "input": "100 15 14", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 16385 16386 16387 16388 16389 16390 16391 16392 16393 16394 16395 16396 16397 16398 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 16399 ..." }, { "input": "9 4 4", "output": "1 2 4 8 16 17 18 19 20 " }, { "input": "100 2 15", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 32769 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 32770 ..." }, { "input": "3 1 0", "output": "1 1 2 " }, { "input": "7 4 0", "output": "1 1 2 3 4 5 5 " }, { "input": "5 2 0", "output": "1 1 2 3 3 " }, { "input": "2 0 0", "output": "1 1 " }, { "input": "5 1 0", "output": "1 1 2 2 2 " }, { "input": "10 2 0", "output": "1 1 2 3 3 3 3 3 3 3 " }, { "input": "10 7 0", "output": "1 1 2 3 4 5 6 7 8 8 " }, { "input": "5 3 0", "output": "1 1 2 3 4 " }, { "input": "10 1 0", "output": "1 1 2 2 2 2 2 2 2 2 " }, { "input": "10 5 0", "output": "1 1 2 3 4 5 6 6 6 6 " }, { "input": "100 3 0", "output": "1 1 2 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 " }, { "input": "100 5 12", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 4097 4098 4099 4100 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 4101 " }, { "input": "15 10 0", "output": "1 1 2 3 4 5 6 7 8 9 10 11 11 11 11 " } ]
1,688,039,963
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
46
0
a, c, b = map(int, input().split()) C = [a - c - b] at = a - b - c t = True for i in range(a): if at > 0: C.append(1) at -= 1 elif b > 0: C.append(sum(C) + 1) b -= 1 elif c > 0: C.append(max(C) + 1) c -= 1 if C[i] > 50000: t = False print(-1) break a1 = '' if t: C.pop(1) for j in C: a1 += str(j) + " " print(a1)
Title: Terse princess Time Limit: None seconds Memory Limit: None megabytes Problem Description: «Next please», — the princess called and cast an estimating glance at the next groom. The princess intends to choose the most worthy groom, this is, the richest one. Whenever she sees a groom who is more rich than each of the previous ones, she says a measured «Oh...». Whenever the groom is richer than all previous ones added together, she exclaims «Wow!» (no «Oh...» in this case). At the sight of the first groom the princess stays calm and says nothing. The fortune of each groom is described with an integer between 1 and 50000. You know that during the day the princess saw *n* grooms, said «Oh...» exactly *a* times and exclaimed «Wow!» exactly *b* times. Your task is to output a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n*, where *t**i* describes the fortune of *i*-th groom. If several sequences are possible, output any of them. If no sequence exists that would satisfy all the requirements, output a single number -1. Input Specification: The only line of input data contains three integer numbers *n*,<=*a* and *b* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*a*,<=*b*<=≤<=15,<=*n*<=&gt;<=*a*<=+<=*b*), separated with single spaces. Output Specification: Output any sequence of integers *t*1,<=*t*2,<=...,<=*t**n*, where *t**i* (1<=≤<=*t**i*<=≤<=50000) is the fortune of *i*-th groom, that satisfies the given constraints. If no sequence exists that would satisfy all the requirements, output a single number -1. Demo Input: ['10 2 3\n', '5 0 0\n'] Demo Output: ['5 1 3 6 16 35 46 4 200 99', '10 10 6 6 5'] Note: Let's have a closer look at the answer for the first sample test. - The princess said «Oh...» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99. - The princess exclaimed «Wow!» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.
```python a, c, b = map(int, input().split()) C = [a - c - b] at = a - b - c t = True for i in range(a): if at > 0: C.append(1) at -= 1 elif b > 0: C.append(sum(C) + 1) b -= 1 elif c > 0: C.append(max(C) + 1) c -= 1 if C[i] > 50000: t = False print(-1) break a1 = '' if t: C.pop(1) for j in C: a1 += str(j) + " " print(a1) ```
0
4
A
Watermelon
PROGRAMMING
800
[ "brute force", "math" ]
A. Watermelon
1
64
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
[ "8\n" ]
[ "YES\n" ]
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
0
[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]
1,696,422,334
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
0
216
4,505,600
{ "cells": [ { "cell_type": "code", "execution_count": 20, "id": "a3e88404", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "I: 8\n", "the: 8\n" ] } ], "source": [ "def cl(i):\n", " count=0\n", " with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " for line in f:\n", " #y=line.lower()\n", " tokens=line.split(' ')\n", " for j in tokens:\n", " if i==j:\n", " count=count+1\n", " #print(j)\n", " return count\n", "f_out=[]\n", "z=0\n", "with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " \n", " for line in f:\n", " #x=line.lower()\n", " tokens=line.split(' ')\n", " #print(tokens)\n", " for i in tokens:\n", " #k=0\n", " s=cl(i)\n", " if s>=z:\n", " #z=s\n", " if s>z:\n", " z=s\n", " s=str(s)\n", " f_out=i+\": \"+s\n", " else:\n", " z=s\n", " s=str(s)\n", " f_in=i+\": \"+s\n", " print(f_out)\n", " print(f_in)" ] }, { "cell_type": "code", "execution_count": 21, "id": "fc69d30f", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "{'Two': 2, 'roads': 2, 'diverged': 2, 'in': 3, 'a': 3, 'yellow': 1, 'wood,\\n': 1, 'And': 6, 'sorry': 1, 'I': 8, 'could': 1, 'not': 1, 'travel': 1, 'both\\n': 1, 'be': 2, 'one': 3, 'traveler,': 1, 'long': 1, 'stood\\n': 1, 'looked': 1, 'down': 1, 'as': 5, 'far': 1, 'could\\n': 1, 'To': 1, 'where': 1, 'it': 2, 'bent': 1, 'the': 8, 'undergrowth;\\n': 1, '\\n': 3, 'Then': 1, 'took': 2, 'other,': 1, 'just': 1, 'fair,\\n': 1, 'having': 1, 'perhaps': 1, 'better': 1, 'claim,\\n': 1, 'Because': 1, 'was': 1, 'grassy': 1, 'and': 3, 'wanted': 1, 'wear;\\n': 1, 'Though': 1, 'for': 2, 'that': 3, 'passing': 1, 'there\\n': 1, 'Had': 1, 'worn': 1, 'them': 1, 'really': 1, 'about': 1, 'same,\\n': 1, 'both': 1, 'morning': 1, 'equally': 1, 'lay\\n': 1, 'In': 1, 'leaves': 1, 'no': 1, 'step': 1, 'had': 1, 'trodden': 1, 'black.\\n': 1, 'Oh,': 1, 'kept': 1, 'first': 1, 'another': 1, 'day!\\n': 1, 'Yet': 1, 'knowing': 1, 'how': 1, 'way': 1, 'leads': 1, 'on': 1, 'to': 1, 'way,\\n': 1, 'doubted': 1, 'if': 1, 'should': 1, 'ever': 1, 'come': 1, 'back.\\n': 1, 'shall': 1, 'telling': 1, 'this': 1, 'with': 1, 'sigh\\n': 1, 'Somewhere': 1, 'ages': 2, 'hence:\\n': 1, 'wood,': 1, 'I—\\n': 1, 'less': 1, 'traveled': 1, 'by,\\n': 1, 'has': 1, 'made': 1, 'all': 1, 'difference.': 1}\n", "Max occurances of any word is: 8\n", "Words with max occurances are: \n", "I: 8\n", "the: 8\n" ] } ], "source": [ "f_out={}\n", "with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " for line in f:\n", " #x=line.lower()\n", " words=line.split(' ')\n", " #print(tokens)\n", " for word in words:\n", " if word in f_out:\n", " f_out[word]+=1\n", " else:\n", " f_out[word]=1\n", "print(f_out)\n", "word_oc=list(f_out.values())\n", "#print(word_oc)\n", "max_co=max(word_oc)\n", "print(\"Max occurances of any word is:\", max_co)\n", "print(\"Words with max occurances are: \")\n", "for word, count in f_out.items():\n", " if count==max_co:\n", " print(f\"{word}: {count}\")" ] }, { "cell_type": "code", "execution_count": 22, "id": "c653e8fd", "metadata": {}, "outputs": [], "source": [ "with open(\"c://Users//hp//python for life//stocks.csv\",\"r\") as f, open(\"c://Users//hp//python for life//output.csv\",\"w\") as out:\n", " out.write(\"Company Name,PE Ratio, PB Ratio\\n\")\n", " next(f)\n", " for line in f:\n", " tokens=line.split(\",\")\n", " stock=tokens[0]\n", " price=float(tokens[1])\n", " eps=float(tokens[2])\n", " book=float(tokens[3])\n", " pe=round(price/eps,2)\n", " pbr=round(price/book,2)\n", " out.write(f\"{stock},{pe},{pbr}\\n\")" ] }, { "cell_type": "code", "execution_count": 23, "id": "d8446ffe", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "tom cruise shoots film\n", "tom cruise says how are you?\n", "maria sharapova plays tennis\n", "maria sharapova says how are you?\n" ] } ], "source": [ "class Human:\n", " def __init__(self, n, o):\n", " self.name = n\n", " self.occupation = o\n", "\n", " def do_work(self):\n", " if self.occupation == \"tennis player\":\n", " print(self.name, \"plays tennis\")\n", " elif self.occupation == \"actor\":\n", " print(self.name, \"shoots film\")\n", "\n", " def speaks(self):\n", " print(self.name, \"says how are you?\")\n", "\n", "tom = Human(\"tom cruise\",\"actor\")\n", "tom.do_work()\n", "tom.speaks()\n", "\n", "maria = Human(\"maria sharapova\",\"tennis player\")\n", "maria.do_work()\n", "maria.speaks()" ] }, { "cell_type": "code", "execution_count": 24, "id": "c0696c1a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Id: 1\n", "Name: coder\n", "emp.id is not defined\n", "emp is not defined\n" ] } ], "source": [ "class Employee:\n", " def __init__(self,id,name):\n", " self.id=id\n", " self.name=name\n", " def info(self):\n", " print(f\"Id: {self.id}\\nName: {self.name}\")\n", "emp = Employee(1, \"coder\")\n", "emp.info()\n", "del emp.id\n", "try:\n", " print(emp.id)\n", "except:\n", " print(\"emp.id is not defined\")\n", "del emp\n", "try:\n", " emp.info()\n", "except:\n", " print(\"emp is not defined\")\n" ] }, { "cell_type": "code", "execution_count": 1, "id": "2d2f0e6c", "metadata": {}, "outputs": [], "source": [ "#numpy\n", "\n", "import numpy as np\n", "import time " ] }, { "cell_type": "code", "execution_count": 2, "id": "43f0c330", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "List took time: 9872.711658477783\n" ] } ], "source": [ "SIZE = 100000000\n", "l1=range(SIZE)\n", "l2=range(SIZE)\n", "start=time.time()\n", "result=[(x+y) for x,y in zip(l1,l2)]\n", "print(\"List took time: \", (time.time()-start)*1000)" ] }, { "cell_type": "code", "execution_count": 4, "id": "335074ae", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Numpy array took time: 171.83780670166016\n" ] } ], "source": [ "SIZE = 100000000\n", "b2=np.arange(SIZE)\n", "b3=np.arange(SIZE)\n", "start=time.time()\n", "result=b2+b3\n", "print(\"Numpy array took time: \", (time.time()-start)*1000)" ] }, { "cell_type": "code", "execution_count": 12, "id": "ad410eb5", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[2, 3, 4],\n", " [3, 4, 6],\n", " [3, 6, 7]])" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b=np.array([[2,3,4],[3,4,6],[3,6,7]])\n", "b" ] }, { "cell_type": "code", "execution_count": 13, "id": "83cc7465", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[3, 4],\n", " [4, 6],\n", " [6, 7]])" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[:,1:3]" ] }, { "cell_type": "code", "execution_count": 22, "id": "691cd79c", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([2, 3, 3])" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[0:3,0]" ] }, { "cell_type": "code", "execution_count": 18, "id": "2e592cb4", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "7" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[2,2]" ] }, { "cell_type": "code", "execution_count": 23, "id": "c95abf01", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[2 3 4]\n", "[3 4 6]\n", "[3 6 7]\n" ] } ], "source": [ "for row in b:\n", " print(row)" ] }, { "cell_type": "code", "execution_count": 24, "id": "75fa14f9", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2\n", "3\n", "4\n", "3\n", "4\n", "6\n", "3\n", "6\n", "7\n" ] } ], "source": [ "for cell in b.flat:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 31, "id": "daf6be5b", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 6, 7],\n", " [ 8, 9],\n", " [10, 11]])" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b=np.arange(6).reshape((3,2))\n", "c=np.arange(6,12).reshape((3,2))\n", "c" ] }, { "cell_type": "code", "execution_count": 32, "id": "f39d234c", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[0, 1],\n", " [2, 3],\n", " [4, 5]])" ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b" ] }, { "cell_type": "code", "execution_count": 36, "id": "df49e530", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1],\n", " [ 2, 3],\n", " [ 4, 5],\n", " [ 6, 7],\n", " [ 8, 9],\n", " [10, 11]])" ] }, "execution_count": 36, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.vstack((b,c))" ] }, { "cell_type": "code", "execution_count": 37, "id": "2b77fdb8", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 6, 7],\n", " [ 2, 3, 8, 9],\n", " [ 4, 5, 10, 11]])" ] }, "execution_count": 37, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.hstack((b,c))" ] }, { "cell_type": "code", "execution_count": 52, "id": "cd85d3d4", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9],\n", " [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],\n", " [20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])" ] }, "execution_count": 52, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e=np.arange(30).reshape((3,10))\n", "e" ] }, { "cell_type": "code", "execution_count": 56, "id": "c1830545", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "[array([[ 0, 1],\n", " [10, 11],\n", " [20, 21]]),\n", " array([[ 2, 3],\n", " [12, 13],\n", " [22, 23]]),\n", " array([[ 4, 5],\n", " [14, 15],\n", " [24, 25]]),\n", " array([[ 6, 7],\n", " [16, 17],\n", " [26, 27]]),\n", " array([[ 8, 9],\n", " [18, 19],\n", " [28, 29]])]" ] }, "execution_count": 56, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#divide into 3 array\n", "np.hsplit(e,5)" ] }, { "cell_type": "code", "execution_count": 59, "id": "b1ec85e0", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "[array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]),\n", " array([[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]]),\n", " array([[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])]" ] }, "execution_count": 59, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.vsplit(e,3)" ] }, { "cell_type": "code", "execution_count": 61, "id": "e8b0d0f3", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[False, False, False, False, False, False, False, False, True,\n", " True],\n", " [ True, True, True, True, True, True, True, True, True,\n", " True],\n", " [ True, True, True, True, True, True, True, True, True,\n", " True]])" ] }, "execution_count": 61, "metadata": {}, "output_type": "execute_result" } ], "source": [ "f=e>7\n", "f" ] }, { "cell_type": "code", "execution_count": 62, "id": "fb5bc51f", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([ 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,\n", " 25, 26, 27, 28, 29])" ] }, "execution_count": 62, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e[f]" ] }, { "cell_type": "code", "execution_count": 63, "id": "85a56766", "metadata": {}, "outputs": [], "source": [ "e[f]=-2" ] }, { "cell_type": "code", "execution_count": 64, "id": "b63477aa", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 2, 3, 4, 5, 6, 7, -2, -2],\n", " [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2],\n", " [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2]])" ] }, "execution_count": 64, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e" ] }, { "cell_type": "code", "execution_count": 83, "id": "8737478f", "metadata": {}, "outputs": [], "source": [ "x=np.arange(12).reshape((3,4))" ] }, { "cell_type": "code", "execution_count": 66, "id": "816b4205", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[0 1 2 3]\n", "[4 5 6 7]\n", "[ 8 9 10 11]\n" ] } ], "source": [ "for row in x:\n", " print(row)" ] }, { "cell_type": "code", "execution_count": 67, "id": "71e08473", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for row in x:\n", " for cell in row:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 71, "id": "47dbf318", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for cell in x.flat:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 72, "id": "dba5477a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for cell in x.flatten():\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 74, "id": "dca20dc3", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "4\n", "8\n", "1\n", "5\n", "9\n", "2\n", "6\n", "10\n", "3\n", "7\n", "11\n" ] } ], "source": [ "#nditer\n", "\n", "for k in np.nditer(x,order='F'):\n", " print(k)" ] }, { "cell_type": "code", "execution_count": 75, "id": "62d2059f", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[0 4 8]\n", "[1 5 9]\n", "[ 2 6 10]\n", "[ 3 7 11]\n" ] } ], "source": [ "for k in np.nditer(x,order='F',flags=['external_loop']):\n", " print(k)" ] }, { "cell_type": "code", "execution_count": 81, "id": "3e7598d7", "metadata": {}, "outputs": [], "source": [ "for k in np.nditer(x,op_flags=['readwrite']):\n", " k[...]=k*k" ] }, { "cell_type": "code", "execution_count": 82, "id": "38663d44", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 4, 9],\n", " [ 16, 25, 36, 49],\n", " [ 64, 81, 100, 121]])" ] }, "execution_count": 82, "metadata": {}, "output_type": "execute_result" } ], "source": [ "x" ] }, { "cell_type": "code", "execution_count": 84, "id": "d1cf1a68", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 3],\n", " [ 7],\n", " [11]])" ] }, "execution_count": 84, "metadata": {}, "output_type": "execute_result" } ], "source": [ "n=np.arange(3,15,4).reshape(3,1)\n", "n" ] }, { "cell_type": "code", "execution_count": 85, "id": "d972749e", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0 3\n", "1 3\n", "2 3\n", "3 3\n", "4 7\n", "5 7\n", "6 7\n", "7 7\n", "8 11\n", "9 11\n", "10 11\n", "11 11\n" ] } ], "source": [ "for k,l in np.nditer([x,n]):\n", " print(k,l)" ] }, { "cell_type": "code", "execution_count": null, "id": "ac4175f2", "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.9.12" } }, "nbformat": 4, "nbformat_minor": 5 }
Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
```python { "cells": [ { "cell_type": "code", "execution_count": 20, "id": "a3e88404", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "I: 8\n", "the: 8\n" ] } ], "source": [ "def cl(i):\n", " count=0\n", " with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " for line in f:\n", " #y=line.lower()\n", " tokens=line.split(' ')\n", " for j in tokens:\n", " if i==j:\n", " count=count+1\n", " #print(j)\n", " return count\n", "f_out=[]\n", "z=0\n", "with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " \n", " for line in f:\n", " #x=line.lower()\n", " tokens=line.split(' ')\n", " #print(tokens)\n", " for i in tokens:\n", " #k=0\n", " s=cl(i)\n", " if s>=z:\n", " #z=s\n", " if s>z:\n", " z=s\n", " s=str(s)\n", " f_out=i+\": \"+s\n", " else:\n", " z=s\n", " s=str(s)\n", " f_in=i+\": \"+s\n", " print(f_out)\n", " print(f_in)" ] }, { "cell_type": "code", "execution_count": 21, "id": "fc69d30f", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "{'Two': 2, 'roads': 2, 'diverged': 2, 'in': 3, 'a': 3, 'yellow': 1, 'wood,\\n': 1, 'And': 6, 'sorry': 1, 'I': 8, 'could': 1, 'not': 1, 'travel': 1, 'both\\n': 1, 'be': 2, 'one': 3, 'traveler,': 1, 'long': 1, 'stood\\n': 1, 'looked': 1, 'down': 1, 'as': 5, 'far': 1, 'could\\n': 1, 'To': 1, 'where': 1, 'it': 2, 'bent': 1, 'the': 8, 'undergrowth;\\n': 1, '\\n': 3, 'Then': 1, 'took': 2, 'other,': 1, 'just': 1, 'fair,\\n': 1, 'having': 1, 'perhaps': 1, 'better': 1, 'claim,\\n': 1, 'Because': 1, 'was': 1, 'grassy': 1, 'and': 3, 'wanted': 1, 'wear;\\n': 1, 'Though': 1, 'for': 2, 'that': 3, 'passing': 1, 'there\\n': 1, 'Had': 1, 'worn': 1, 'them': 1, 'really': 1, 'about': 1, 'same,\\n': 1, 'both': 1, 'morning': 1, 'equally': 1, 'lay\\n': 1, 'In': 1, 'leaves': 1, 'no': 1, 'step': 1, 'had': 1, 'trodden': 1, 'black.\\n': 1, 'Oh,': 1, 'kept': 1, 'first': 1, 'another': 1, 'day!\\n': 1, 'Yet': 1, 'knowing': 1, 'how': 1, 'way': 1, 'leads': 1, 'on': 1, 'to': 1, 'way,\\n': 1, 'doubted': 1, 'if': 1, 'should': 1, 'ever': 1, 'come': 1, 'back.\\n': 1, 'shall': 1, 'telling': 1, 'this': 1, 'with': 1, 'sigh\\n': 1, 'Somewhere': 1, 'ages': 2, 'hence:\\n': 1, 'wood,': 1, 'I—\\n': 1, 'less': 1, 'traveled': 1, 'by,\\n': 1, 'has': 1, 'made': 1, 'all': 1, 'difference.': 1}\n", "Max occurances of any word is: 8\n", "Words with max occurances are: \n", "I: 8\n", "the: 8\n" ] } ], "source": [ "f_out={}\n", "with open(\"c://Users//hp//python for life//poem.txt\",\"r\") as f:\n", " for line in f:\n", " #x=line.lower()\n", " words=line.split(' ')\n", " #print(tokens)\n", " for word in words:\n", " if word in f_out:\n", " f_out[word]+=1\n", " else:\n", " f_out[word]=1\n", "print(f_out)\n", "word_oc=list(f_out.values())\n", "#print(word_oc)\n", "max_co=max(word_oc)\n", "print(\"Max occurances of any word is:\", max_co)\n", "print(\"Words with max occurances are: \")\n", "for word, count in f_out.items():\n", " if count==max_co:\n", " print(f\"{word}: {count}\")" ] }, { "cell_type": "code", "execution_count": 22, "id": "c653e8fd", "metadata": {}, "outputs": [], "source": [ "with open(\"c://Users//hp//python for life//stocks.csv\",\"r\") as f, open(\"c://Users//hp//python for life//output.csv\",\"w\") as out:\n", " out.write(\"Company Name,PE Ratio, PB Ratio\\n\")\n", " next(f)\n", " for line in f:\n", " tokens=line.split(\",\")\n", " stock=tokens[0]\n", " price=float(tokens[1])\n", " eps=float(tokens[2])\n", " book=float(tokens[3])\n", " pe=round(price/eps,2)\n", " pbr=round(price/book,2)\n", " out.write(f\"{stock},{pe},{pbr}\\n\")" ] }, { "cell_type": "code", "execution_count": 23, "id": "d8446ffe", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "tom cruise shoots film\n", "tom cruise says how are you?\n", "maria sharapova plays tennis\n", "maria sharapova says how are you?\n" ] } ], "source": [ "class Human:\n", " def __init__(self, n, o):\n", " self.name = n\n", " self.occupation = o\n", "\n", " def do_work(self):\n", " if self.occupation == \"tennis player\":\n", " print(self.name, \"plays tennis\")\n", " elif self.occupation == \"actor\":\n", " print(self.name, \"shoots film\")\n", "\n", " def speaks(self):\n", " print(self.name, \"says how are you?\")\n", "\n", "tom = Human(\"tom cruise\",\"actor\")\n", "tom.do_work()\n", "tom.speaks()\n", "\n", "maria = Human(\"maria sharapova\",\"tennis player\")\n", "maria.do_work()\n", "maria.speaks()" ] }, { "cell_type": "code", "execution_count": 24, "id": "c0696c1a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Id: 1\n", "Name: coder\n", "emp.id is not defined\n", "emp is not defined\n" ] } ], "source": [ "class Employee:\n", " def __init__(self,id,name):\n", " self.id=id\n", " self.name=name\n", " def info(self):\n", " print(f\"Id: {self.id}\\nName: {self.name}\")\n", "emp = Employee(1, \"coder\")\n", "emp.info()\n", "del emp.id\n", "try:\n", " print(emp.id)\n", "except:\n", " print(\"emp.id is not defined\")\n", "del emp\n", "try:\n", " emp.info()\n", "except:\n", " print(\"emp is not defined\")\n" ] }, { "cell_type": "code", "execution_count": 1, "id": "2d2f0e6c", "metadata": {}, "outputs": [], "source": [ "#numpy\n", "\n", "import numpy as np\n", "import time " ] }, { "cell_type": "code", "execution_count": 2, "id": "43f0c330", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "List took time: 9872.711658477783\n" ] } ], "source": [ "SIZE = 100000000\n", "l1=range(SIZE)\n", "l2=range(SIZE)\n", "start=time.time()\n", "result=[(x+y) for x,y in zip(l1,l2)]\n", "print(\"List took time: \", (time.time()-start)*1000)" ] }, { "cell_type": "code", "execution_count": 4, "id": "335074ae", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Numpy array took time: 171.83780670166016\n" ] } ], "source": [ "SIZE = 100000000\n", "b2=np.arange(SIZE)\n", "b3=np.arange(SIZE)\n", "start=time.time()\n", "result=b2+b3\n", "print(\"Numpy array took time: \", (time.time()-start)*1000)" ] }, { "cell_type": "code", "execution_count": 12, "id": "ad410eb5", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[2, 3, 4],\n", " [3, 4, 6],\n", " [3, 6, 7]])" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b=np.array([[2,3,4],[3,4,6],[3,6,7]])\n", "b" ] }, { "cell_type": "code", "execution_count": 13, "id": "83cc7465", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[3, 4],\n", " [4, 6],\n", " [6, 7]])" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[:,1:3]" ] }, { "cell_type": "code", "execution_count": 22, "id": "691cd79c", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([2, 3, 3])" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[0:3,0]" ] }, { "cell_type": "code", "execution_count": 18, "id": "2e592cb4", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "7" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b[2,2]" ] }, { "cell_type": "code", "execution_count": 23, "id": "c95abf01", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[2 3 4]\n", "[3 4 6]\n", "[3 6 7]\n" ] } ], "source": [ "for row in b:\n", " print(row)" ] }, { "cell_type": "code", "execution_count": 24, "id": "75fa14f9", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2\n", "3\n", "4\n", "3\n", "4\n", "6\n", "3\n", "6\n", "7\n" ] } ], "source": [ "for cell in b.flat:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 31, "id": "daf6be5b", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 6, 7],\n", " [ 8, 9],\n", " [10, 11]])" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b=np.arange(6).reshape((3,2))\n", "c=np.arange(6,12).reshape((3,2))\n", "c" ] }, { "cell_type": "code", "execution_count": 32, "id": "f39d234c", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[0, 1],\n", " [2, 3],\n", " [4, 5]])" ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "b" ] }, { "cell_type": "code", "execution_count": 36, "id": "df49e530", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1],\n", " [ 2, 3],\n", " [ 4, 5],\n", " [ 6, 7],\n", " [ 8, 9],\n", " [10, 11]])" ] }, "execution_count": 36, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.vstack((b,c))" ] }, { "cell_type": "code", "execution_count": 37, "id": "2b77fdb8", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 6, 7],\n", " [ 2, 3, 8, 9],\n", " [ 4, 5, 10, 11]])" ] }, "execution_count": 37, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.hstack((b,c))" ] }, { "cell_type": "code", "execution_count": 52, "id": "cd85d3d4", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9],\n", " [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],\n", " [20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])" ] }, "execution_count": 52, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e=np.arange(30).reshape((3,10))\n", "e" ] }, { "cell_type": "code", "execution_count": 56, "id": "c1830545", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "[array([[ 0, 1],\n", " [10, 11],\n", " [20, 21]]),\n", " array([[ 2, 3],\n", " [12, 13],\n", " [22, 23]]),\n", " array([[ 4, 5],\n", " [14, 15],\n", " [24, 25]]),\n", " array([[ 6, 7],\n", " [16, 17],\n", " [26, 27]]),\n", " array([[ 8, 9],\n", " [18, 19],\n", " [28, 29]])]" ] }, "execution_count": 56, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#divide into 3 array\n", "np.hsplit(e,5)" ] }, { "cell_type": "code", "execution_count": 59, "id": "b1ec85e0", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "[array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]),\n", " array([[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]]),\n", " array([[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])]" ] }, "execution_count": 59, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.vsplit(e,3)" ] }, { "cell_type": "code", "execution_count": 61, "id": "e8b0d0f3", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[False, False, False, False, False, False, False, False, True,\n", " True],\n", " [ True, True, True, True, True, True, True, True, True,\n", " True],\n", " [ True, True, True, True, True, True, True, True, True,\n", " True]])" ] }, "execution_count": 61, "metadata": {}, "output_type": "execute_result" } ], "source": [ "f=e>7\n", "f" ] }, { "cell_type": "code", "execution_count": 62, "id": "fb5bc51f", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([ 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,\n", " 25, 26, 27, 28, 29])" ] }, "execution_count": 62, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e[f]" ] }, { "cell_type": "code", "execution_count": 63, "id": "85a56766", "metadata": {}, "outputs": [], "source": [ "e[f]=-2" ] }, { "cell_type": "code", "execution_count": 64, "id": "b63477aa", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 2, 3, 4, 5, 6, 7, -2, -2],\n", " [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2],\n", " [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2]])" ] }, "execution_count": 64, "metadata": {}, "output_type": "execute_result" } ], "source": [ "e" ] }, { "cell_type": "code", "execution_count": 83, "id": "8737478f", "metadata": {}, "outputs": [], "source": [ "x=np.arange(12).reshape((3,4))" ] }, { "cell_type": "code", "execution_count": 66, "id": "816b4205", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[0 1 2 3]\n", "[4 5 6 7]\n", "[ 8 9 10 11]\n" ] } ], "source": [ "for row in x:\n", " print(row)" ] }, { "cell_type": "code", "execution_count": 67, "id": "71e08473", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for row in x:\n", " for cell in row:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 71, "id": "47dbf318", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for cell in x.flat:\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 72, "id": "dba5477a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n", "4\n", "5\n", "6\n", "7\n", "8\n", "9\n", "10\n", "11\n" ] } ], "source": [ "for cell in x.flatten():\n", " print(cell)" ] }, { "cell_type": "code", "execution_count": 74, "id": "dca20dc3", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "4\n", "8\n", "1\n", "5\n", "9\n", "2\n", "6\n", "10\n", "3\n", "7\n", "11\n" ] } ], "source": [ "#nditer\n", "\n", "for k in np.nditer(x,order='F'):\n", " print(k)" ] }, { "cell_type": "code", "execution_count": 75, "id": "62d2059f", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[0 4 8]\n", "[1 5 9]\n", "[ 2 6 10]\n", "[ 3 7 11]\n" ] } ], "source": [ "for k in np.nditer(x,order='F',flags=['external_loop']):\n", " print(k)" ] }, { "cell_type": "code", "execution_count": 81, "id": "3e7598d7", "metadata": {}, "outputs": [], "source": [ "for k in np.nditer(x,op_flags=['readwrite']):\n", " k[...]=k*k" ] }, { "cell_type": "code", "execution_count": 82, "id": "38663d44", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0, 1, 4, 9],\n", " [ 16, 25, 36, 49],\n", " [ 64, 81, 100, 121]])" ] }, "execution_count": 82, "metadata": {}, "output_type": "execute_result" } ], "source": [ "x" ] }, { "cell_type": "code", "execution_count": 84, "id": "d1cf1a68", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 3],\n", " [ 7],\n", " [11]])" ] }, "execution_count": 84, "metadata": {}, "output_type": "execute_result" } ], "source": [ "n=np.arange(3,15,4).reshape(3,1)\n", "n" ] }, { "cell_type": "code", "execution_count": 85, "id": "d972749e", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0 3\n", "1 3\n", "2 3\n", "3 3\n", "4 7\n", "5 7\n", "6 7\n", "7 7\n", "8 11\n", "9 11\n", "10 11\n", "11 11\n" ] } ], "source": [ "for k,l in np.nditer([x,n]):\n", " print(k,l)" ] }, { "cell_type": "code", "execution_count": null, "id": "ac4175f2", "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.9.12" } }, "nbformat": 4, "nbformat_minor": 5 } ```
-1
275
A
Lights Out
PROGRAMMING
900
[ "implementation" ]
null
null
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on. Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
[ "1 0 0\n0 0 0\n0 0 1\n", "1 0 1\n8 8 8\n2 0 3\n" ]
[ "001\n010\n100\n", "010\n011\n100\n" ]
none
500
[ { "input": "1 0 0\n0 0 0\n0 0 1", "output": "001\n010\n100" }, { "input": "1 0 1\n8 8 8\n2 0 3", "output": "010\n011\n100" }, { "input": "13 85 77\n25 50 45\n65 79 9", "output": "000\n010\n000" }, { "input": "96 95 5\n8 84 74\n67 31 61", "output": "011\n011\n101" }, { "input": "24 54 37\n60 63 6\n1 84 26", "output": "110\n101\n011" }, { "input": "23 10 40\n15 6 40\n92 80 77", "output": "101\n100\n000" }, { "input": "62 74 80\n95 74 93\n2 47 95", "output": "010\n001\n110" }, { "input": "80 83 48\n26 0 66\n47 76 37", "output": "000\n000\n010" }, { "input": "32 15 65\n7 54 36\n5 51 3", "output": "111\n101\n001" }, { "input": "22 97 12\n71 8 24\n100 21 64", "output": "100\n001\n100" }, { "input": "46 37 13\n87 0 50\n90 8 55", "output": "111\n011\n000" }, { "input": "57 43 58\n20 82 83\n66 16 52", "output": "111\n010\n110" }, { "input": "45 56 93\n47 51 59\n18 51 63", "output": "101\n011\n100" }, { "input": "47 66 67\n14 1 37\n27 81 69", "output": "001\n001\n110" }, { "input": "26 69 69\n85 18 23\n14 22 74", "output": "110\n001\n010" }, { "input": "10 70 65\n94 27 25\n74 66 30", "output": "111\n010\n100" }, { "input": "97 1 74\n15 99 1\n88 68 86", "output": "001\n011\n000" }, { "input": "36 48 42\n45 41 66\n26 64 1", "output": "001\n111\n010" }, { "input": "52 81 97\n29 77 71\n66 11 2", "output": "100\n100\n111" }, { "input": "18 66 33\n19 49 49\n48 46 26", "output": "011\n100\n000" }, { "input": "68 79 52\n51 39 100\n29 14 26", "output": "110\n000\n111" }, { "input": "91 69 77\n91 26 64\n91 88 57", "output": "001\n011\n110" }, { "input": "16 69 64\n48 21 80\n81 51 51", "output": "010\n101\n111" }, { "input": "96 14 2\n100 18 12\n65 34 89", "output": "111\n010\n010" }, { "input": "93 95 90\n8 59 42\n53 13 19", "output": "100\n001\n111" }, { "input": "71 84 18\n100 19 67\n9 76 15", "output": "010\n010\n001" }, { "input": "38 93 85\n21 88 64\n4 96 25", "output": "111\n011\n000" }, { "input": "75 20 20\n60 5 78\n77 4 69", "output": "011\n001\n000" }, { "input": "65 70 96\n19 6 83\n33 37 82", "output": "100\n000\n011" }, { "input": "11 13 60\n17 13 46\n42 21 39", "output": "000\n011\n101" }, { "input": "0 0 0\n0 0 0\n0 0 0", "output": "111\n111\n111" }, { "input": "0 0 0\n0 1 0\n0 0 0", "output": "101\n000\n101" }, { "input": "0 0 0\n0 0 0\n0 0 1", "output": "111\n110\n100" } ]
1,652,271,745
2,147,483,647
Python 3
OK
TESTS
33
46
0
g = [[0] + list(map(int, input().split())) + [0] for i in range(3)] w = [[0]*5] + g + [[0]*5] for i in range(1,4): print("".join(map(str, [1 - (w[i][j-1] + w[i][j+1] + w[i][j] + w[i-1][j] + w[i+1][j]) %2 for j in range(1,4)])))
Title: Lights Out Time Limit: None seconds Memory Limit: None megabytes Problem Description: Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on. Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light. Input Specification: The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed. Output Specification: Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0". Demo Input: ['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n'] Demo Output: ['001\n010\n100\n', '010\n011\n100\n'] Note: none
```python g = [[0] + list(map(int, input().split())) + [0] for i in range(3)] w = [[0]*5] + g + [[0]*5] for i in range(1,4): print("".join(map(str, [1 - (w[i][j-1] + w[i][j+1] + w[i][j] + w[i-1][j] + w[i+1][j]) %2 for j in range(1,4)]))) ```
3
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,696,891,018
2,147,483,647
Python 3
OK
TESTS
20
46
0
n = int(input("")) i = 1 while i < n + 1: a = str(input("")) if len(a) > 10: print(a[0], (len(a) - 2), a[-1], sep="") else: print(a) i = i + 1
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python n = int(input("")) i = 1 while i < n + 1: a = str(input("")) if len(a) > 10: print(a[0], (len(a) - 2), a[-1], sep="") else: print(a) i = i + 1 ```
3.977
94
A
Restoring Password
PROGRAMMING
900
[ "implementation", "strings" ]
A. Restoring Password
2
256
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
[ "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n", "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n" ]
[ "12345678\n", "30234919\n" ]
none
500
[ { "input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110", "output": "12345678" }, { "input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000", "output": "30234919" }, { "input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011", "output": "65264629" }, { "input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001", "output": "98484434" }, { "input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001", "output": "96071437" }, { "input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001", "output": "89787267" }, { "input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110", "output": "88447623" }, { "input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001", "output": "80805519" }, { "input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110", "output": "09250147" }, { "input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011", "output": "40862358" }, { "input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001", "output": "73907059" }, { "input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101", "output": "57680434" }, { "input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010", "output": "93678590" }, { "input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000", "output": "58114879" }, { "input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100", "output": "61146904" }, { "input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001", "output": "23558422" }, { "input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111", "output": "76827631" }, { "input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111", "output": "26240666" }, { "input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100", "output": "21067271" }, { "input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010", "output": "77770029" }, { "input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011", "output": "09448580" }, { "input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111", "output": "99411277" }, { "input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110", "output": "86658594" }, { "input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000", "output": "04536863" }, { "input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110", "output": "21066773" }, { "input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111", "output": "01063858" }, { "input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000", "output": "01599791" }, { "input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010", "output": "40579016" }, { "input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001", "output": "75424967" }, { "input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001", "output": "72259657" }, { "input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011", "output": "22339256" }, { "input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110", "output": "70554591" }, { "input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011", "output": "18124952" }, { "input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110", "output": "33774052" }, { "input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000", "output": "97961250" }, { "input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011", "output": "21954161" }, { "input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100", "output": "78740192" }, { "input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111", "output": "05336882" }, { "input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111", "output": "62020383" }, { "input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101", "output": "24819275" }, { "input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001", "output": "90010504" }, { "input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111", "output": "85948866" }, { "input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110", "output": "77874864" }, { "input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011", "output": "10374003" }, { "input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001", "output": "22955387" }, { "input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100", "output": "25893541" }, { "input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010", "output": "13805878" }, { "input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010", "output": "06369030" }, { "input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110", "output": "46909115" }, { "input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100", "output": "90005541" } ]
1,560,350,309
2,147,483,647
Python 3
OK
TESTS
50
248
0
n = input() d = dict() for i in range(10): s = input() d[s] = i for i in range(8): s = n[i * 10:i * 10 + 10] print(d[s], end="")
Title: Restoring Password Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification. Input Specification: The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9. Output Specification: Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists. Demo Input: ['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n'] Demo Output: ['12345678\n', '30234919\n'] Note: none
```python n = input() d = dict() for i in range(10): s = input() d[s] = i for i in range(8): s = n[i * 10:i * 10 + 10] print(d[s], end="") ```
3.938
456
A
Laptops
PROGRAMMING
1,100
[ "sortings" ]
null
null
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop. Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops. Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality). All *a**i* are distinct. All *b**i* are distinct.
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
[ "2\n1 2\n2 1\n" ]
[ "Happy Alex\n" ]
none
500
[ { "input": "2\n1 2\n2 1", "output": "Happy Alex" }, { "input": "2\n1 1\n2 2", "output": "Poor Alex" }, { "input": "3\n2 2\n3 3\n1 1", "output": "Poor Alex" }, { "input": "3\n3 3\n1 2\n2 1", "output": "Happy Alex" }, { "input": "1\n1 1", "output": "Poor Alex" }, { "input": "3\n2 3\n1 1\n3 2", "output": "Happy Alex" }, { "input": "4\n4 1\n3 2\n2 3\n1 4", "output": "Happy Alex" }, { "input": "2\n2 1\n1 2", "output": "Happy Alex" }, { "input": "3\n3 2\n1 1\n2 3", "output": "Happy Alex" }, { "input": "3\n1 2\n2 3\n3 1", "output": "Happy Alex" }, { "input": "3\n1 2\n2 1\n3 3", "output": "Happy Alex" }, { "input": "3\n1 2\n2 3\n3 1", "output": "Happy Alex" } ]
1,681,589,750
2,147,483,647
Python 3
OK
TESTS
46
280
5,017,600
#Keshika Patwari #Indian Institute Of Technology, Jodhpur # 2022 import sys input=sys.stdin.readline def exe(): for i in range(len(x)): if(x[i]!=y[i]): return 'Happy Alex' return 'Poor Alex' x=[] y=[] for i in range(int(input())): a,b=map(int,input().split()) x.append(a) y.append(b) print(exe())
Title: Laptops Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop. Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops. Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality). All *a**i* are distinct. All *b**i* are distinct. Output Specification: If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). Demo Input: ['2\n1 2\n2 1\n'] Demo Output: ['Happy Alex\n'] Note: none
```python #Keshika Patwari #Indian Institute Of Technology, Jodhpur # 2022 import sys input=sys.stdin.readline def exe(): for i in range(len(x)): if(x[i]!=y[i]): return 'Happy Alex' return 'Poor Alex' x=[] y=[] for i in range(int(input())): a,b=map(int,input().split()) x.append(a) y.append(b) print(exe()) ```
3
136
A
Presents
PROGRAMMING
800
[ "implementation" ]
null
null
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]
[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]
none
500
[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71", "output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90" }, { "input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9", "output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9" }, { "input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44", "output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20" }, { "input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17", "output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28" }, { "input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7", "output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5" }, { "input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16", "output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43" }, { "input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65", "output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43" }, { "input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16", "output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45" }, { "input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51", "output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64" }, { "input": "12\n12 3 1 5 11 6 7 10 2 8 9 4", "output": "3 9 2 12 4 6 7 10 11 8 5 1" }, { "input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17", "output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44" }, { "input": "8\n1 3 5 2 4 8 6 7", "output": "1 4 2 5 3 7 8 6" }, { "input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14", "output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12" }, { "input": "10\n2 9 4 6 10 1 7 5 3 8", "output": "6 1 9 3 8 4 7 10 2 5" }, { "input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3", "output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7" }, { "input": "10\n2 4 9 3 6 8 10 5 1 7", "output": "9 1 4 2 8 5 10 6 3 7" }, { "input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15", "output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23" }, { "input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1", "output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27" }, { "input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30", "output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23" }, { "input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52", "output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50" }, { "input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18", "output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98" }, { "input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16", "output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25" }, { "input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37", "output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54" }, { "input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71", "output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71" }, { "input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22", "output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8" }, { "input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12", "output": "31 9 72 77 48 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59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53" }, { "input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64", "output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49" }, { "input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 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43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12" }, { "input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87", "output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82" }, { "input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25", "output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97" }, { "input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61", "output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96" }, { "input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92", "output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96" }, { "input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6", "output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36" }, { "input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1" } ]
1,694,694,781
2,147,483,647
Python 3
OK
TESTS
101
92
0
n = int(input()) s = input().split() mydict = {int(s[i]): i + 1 for i in range(n)} mydict = dict(sorted(mydict.items())) count = [] for items in mydict.values(): count.append(items) print(*count)
Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none
```python n = int(input()) s = input().split() mydict = {int(s[i]): i + 1 for i in range(n)} mydict = dict(sorted(mydict.items())) count = [] for items in mydict.values(): count.append(items) print(*count) ```
3
785
B
Anton and Classes
PROGRAMMING
1,100
[ "greedy", "sortings" ]
null
null
Anton likes to play chess. Also he likes to do programming. No wonder that he decided to attend chess classes and programming classes. Anton has *n* variants when he will attend chess classes, *i*-th variant is given by a period of time (*l*1,<=*i*,<=*r*1,<=*i*). Also he has *m* variants when he will attend programming classes, *i*-th variant is given by a period of time (*l*2,<=*i*,<=*r*2,<=*i*). Anton needs to choose exactly one of *n* possible periods of time when he will attend chess classes and exactly one of *m* possible periods of time when he will attend programming classes. He wants to have a rest between classes, so from all the possible pairs of the periods he wants to choose the one where the distance between the periods is maximal. The distance between periods (*l*1,<=*r*1) and (*l*2,<=*r*2) is the minimal possible distance between a point in the first period and a point in the second period, that is the minimal possible |*i*<=-<=*j*|, where *l*1<=≤<=*i*<=≤<=*r*1 and *l*2<=≤<=*j*<=≤<=*r*2. In particular, when the periods intersect, the distance between them is 0. Anton wants to know how much time his rest between the classes will last in the best case. Help Anton and find this number!
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of time periods when Anton can attend chess classes. Each of the following *n* lines of the input contains two integers *l*1,<=*i* and *r*1,<=*i* (1<=≤<=*l*1,<=*i*<=≤<=*r*1,<=*i*<=≤<=109) — the *i*-th variant of a period of time when Anton can attend chess classes. The following line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of time periods when Anton can attend programming classes. Each of the following *m* lines of the input contains two integers *l*2,<=*i* and *r*2,<=*i* (1<=≤<=*l*2,<=*i*<=≤<=*r*2,<=*i*<=≤<=109) — the *i*-th variant of a period of time when Anton can attend programming classes.
Output one integer — the maximal possible distance between time periods.
[ "3\n1 5\n2 6\n2 3\n2\n2 4\n6 8\n", "3\n1 5\n2 6\n3 7\n2\n2 4\n1 4\n" ]
[ "3\n", "0\n" ]
In the first sample Anton can attend chess classes in the period (2, 3) and attend programming classes in the period (6, 8). It's not hard to see that in this case the distance between the periods will be equal to 3. In the second sample if he chooses any pair of periods, they will intersect. So the answer is 0.
1,000
[ { "input": "3\n1 5\n2 6\n2 3\n2\n2 4\n6 8", "output": "3" }, { "input": "3\n1 5\n2 6\n3 7\n2\n2 4\n1 4", "output": "0" }, { "input": "20\n13 141\n57 144\n82 124\n16 23\n18 44\n64 65\n117 133\n84 117\n77 142\n40 119\n105 120\n71 92\n5 142\n48 132\n106 121\n5 80\n45 92\n66 81\n7 93\n27 71\n3\n75 96\n127 140\n54 74", "output": "104" }, { "input": "10\n16 16\n20 20\n13 13\n31 31\n42 42\n70 70\n64 64\n63 63\n53 53\n94 94\n8\n3 3\n63 63\n9 9\n25 25\n11 11\n93 93\n47 47\n3 3", "output": "91" }, { "input": "1\n45888636 261444238\n1\n244581813 591222338", "output": "0" }, { "input": "1\n166903016 182235583\n1\n254223764 902875046", "output": "71988181" }, { "input": "1\n1 1\n1\n1000000000 1000000000", "output": "999999999" }, { "input": "1\n1000000000 1000000000\n1\n1 1", "output": "999999999" }, { "input": "1\n1000000000 1000000000\n1\n1000000000 1000000000", "output": "0" }, { "input": "6\n2 96\n47 81\n3 17\n52 52\n50 105\n1 44\n4\n40 44\n59 104\n37 52\n2 28", "output": "42" }, { "input": "4\n528617953 528617953\n102289603 102289603\n123305570 123305570\n481177982 597599007\n1\n239413975 695033059", "output": "137124372" }, { "input": "7\n617905528 617905554\n617905546 617905557\n617905562 617905564\n617905918 617906372\n617905539 617905561\n617905516 617905581\n617905538 617905546\n9\n617905517 617905586\n617905524 617905579\n617905555 617905580\n617905537 617905584\n617905556 617905557\n617905514 617905526\n617905544 617905579\n617905258 617905514\n617905569 617905573", "output": "404" }, { "input": "5\n999612104 999858319\n68705639 989393889\n297814302 732073321\n577979321 991069087\n601930055 838139173\n14\n109756300 291701768\n2296272 497162877\n3869085 255543683\n662920943 820993688\n54005870 912134860\n1052 70512\n477043210 648640912\n233115268 920170255\n575163323 756904529\n183450026 469145373\n359987405 795448062\n287873006 872825189\n360460166 737511078\n76784767 806771748", "output": "999541592" }, { "input": "1\n1 100000000\n1\n200000000 200000010", "output": "100000000" }, { "input": "1\n999999995 999999996\n1\n999999998 999999999", "output": "2" }, { "input": "1\n10 100\n1\n2 5", "output": "5" }, { "input": "1\n999999992 999999993\n1\n999999996 999999997", "output": "3" }, { "input": "1\n999999997 999999997\n1\n999999999 999999999", "output": "2" }, { "input": "1\n999999999 999999999\n1\n1000000000 1000000000", "output": "1" }, { "input": "1\n1 1000000000\n1\n1000000000 1000000000", "output": "0" }, { "input": "1\n1000000000 1000000000\n1\n999999999 999999999", "output": "1" }, { "input": "1\n100000000 100000001\n1\n100000009 100000011", "output": "8" }, { "input": "1\n5 5\n1\n6 6", "output": "1" }, { "input": "1\n1000000000 1000000000\n1\n1000000000 1000000000", "output": "0" }, { "input": "1\n200000000 200000001\n1\n200000000 200000001", "output": "0" }, { "input": "1\n2 6\n1\n4 8", "output": "0" } ]
1,687,068,586
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
25
2,308
32,768,000
n = int(input()) chess = [] mnr = 1e9 mnl = 1e9 mxl = -1 mxr = -1 for i in range(n): chess.append([int(e) for e in input().split()]) if chess[i][0] > mxl: mxl = chess[i][0] if chess[i][1] < mnr: mnr = chess[i][1] if chess[i][0] < mnl: mnl = chess[i][0] if chess[i][1] > mxr: mxr = chess[i][1] m = int(input()) res = 0 for i in range(m): l2, r2 = [int(e) for e in input().split()] # there is only one chess # edge case: prgm is outside the bound if l2 > mnr and l2-mnr > res: res = l2-mnr elif r2 < mxl and mxl-r2 > res: res = mxl-r2 elif r2 < mnl and mnl-r2 > res: res = mnl-r2 elif l2 > mxr and l2-mxr > res: res = l2-mxr print(res)
Title: Anton and Classes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton likes to play chess. Also he likes to do programming. No wonder that he decided to attend chess classes and programming classes. Anton has *n* variants when he will attend chess classes, *i*-th variant is given by a period of time (*l*1,<=*i*,<=*r*1,<=*i*). Also he has *m* variants when he will attend programming classes, *i*-th variant is given by a period of time (*l*2,<=*i*,<=*r*2,<=*i*). Anton needs to choose exactly one of *n* possible periods of time when he will attend chess classes and exactly one of *m* possible periods of time when he will attend programming classes. He wants to have a rest between classes, so from all the possible pairs of the periods he wants to choose the one where the distance between the periods is maximal. The distance between periods (*l*1,<=*r*1) and (*l*2,<=*r*2) is the minimal possible distance between a point in the first period and a point in the second period, that is the minimal possible |*i*<=-<=*j*|, where *l*1<=≤<=*i*<=≤<=*r*1 and *l*2<=≤<=*j*<=≤<=*r*2. In particular, when the periods intersect, the distance between them is 0. Anton wants to know how much time his rest between the classes will last in the best case. Help Anton and find this number! Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of time periods when Anton can attend chess classes. Each of the following *n* lines of the input contains two integers *l*1,<=*i* and *r*1,<=*i* (1<=≤<=*l*1,<=*i*<=≤<=*r*1,<=*i*<=≤<=109) — the *i*-th variant of a period of time when Anton can attend chess classes. The following line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of time periods when Anton can attend programming classes. Each of the following *m* lines of the input contains two integers *l*2,<=*i* and *r*2,<=*i* (1<=≤<=*l*2,<=*i*<=≤<=*r*2,<=*i*<=≤<=109) — the *i*-th variant of a period of time when Anton can attend programming classes. Output Specification: Output one integer — the maximal possible distance between time periods. Demo Input: ['3\n1 5\n2 6\n2 3\n2\n2 4\n6 8\n', '3\n1 5\n2 6\n3 7\n2\n2 4\n1 4\n'] Demo Output: ['3\n', '0\n'] Note: In the first sample Anton can attend chess classes in the period (2, 3) and attend programming classes in the period (6, 8). It's not hard to see that in this case the distance between the periods will be equal to 3. In the second sample if he chooses any pair of periods, they will intersect. So the answer is 0.
```python n = int(input()) chess = [] mnr = 1e9 mnl = 1e9 mxl = -1 mxr = -1 for i in range(n): chess.append([int(e) for e in input().split()]) if chess[i][0] > mxl: mxl = chess[i][0] if chess[i][1] < mnr: mnr = chess[i][1] if chess[i][0] < mnl: mnl = chess[i][0] if chess[i][1] > mxr: mxr = chess[i][1] m = int(input()) res = 0 for i in range(m): l2, r2 = [int(e) for e in input().split()] # there is only one chess # edge case: prgm is outside the bound if l2 > mnr and l2-mnr > res: res = l2-mnr elif r2 < mxl and mxl-r2 > res: res = mxl-r2 elif r2 < mnl and mnl-r2 > res: res = mnl-r2 elif l2 > mxr and l2-mxr > res: res = l2-mxr print(res) ```
0
109
A
Lucky Sum of Digits
PROGRAMMING
1,000
[ "brute force", "implementation" ]
A. Lucky Sum of Digits
2
256
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
[ "11\n", "10\n" ]
[ "47\n", "-1\n" ]
none
500
[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "485", "output": "44447777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "111", "output": "444447777777777777" }, { "input": "85", "output": "4477777777777" }, { "input": "114", "output": "444477777777777777" }, { "input": "474", "output": "444777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "74", "output": "47777777777" }, { "input": "1000", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "1024", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "4444", "output": 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"7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10417", "output": 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"4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "99999", "output": 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"4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999999", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "888887", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999998", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "40008", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10691", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "5", "output": "-1" }, { "input": "6", "output": "-1" }, { "input": "9", "output": "-1" }, { "input": "8", "output": "44" }, { "input": "2", "output": "-1" }, { "input": "3", "output": "-1" }, { "input": "999997", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999996", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999990", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999980", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "800000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "980000", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." } ]
1,687,709,485
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
n = int(input()) no_of_4s = n // 4 no_of_7s = 0 remainder = n % 4 if remainder == 1: no_of_4s -= 5 no_of_7s = 3 elif remainder == 2: no_of_4s -= 3 no_of_7s = 2 elif remainder == 3: no_of_4s -= 1 no_of_7s = 1 print(no_of_4s, no_of_7s) if no_of_4s < 0: print(-1) else: while no_of_4s - 7 > 0: no_of_4s -= 7 no_of_7s += 4 print(no_of_4s, no_of_7s) for i in range(no_of_4s): print(4, end="") for i in range(no_of_7s): print(7, end="")
Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none
```python n = int(input()) no_of_4s = n // 4 no_of_7s = 0 remainder = n % 4 if remainder == 1: no_of_4s -= 5 no_of_7s = 3 elif remainder == 2: no_of_4s -= 3 no_of_7s = 2 elif remainder == 3: no_of_4s -= 1 no_of_7s = 1 print(no_of_4s, no_of_7s) if no_of_4s < 0: print(-1) else: while no_of_4s - 7 > 0: no_of_4s -= 7 no_of_7s += 4 print(no_of_4s, no_of_7s) for i in range(no_of_4s): print(4, end="") for i in range(no_of_7s): print(7, end="") ```
0
2
A
Winner
PROGRAMMING
1,500
[ "hashing", "implementation" ]
A. Winner
1
64
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Print the name of the winner.
[ "3\nmike 3\nandrew 5\nmike 2\n", "3\nandrew 3\nandrew 2\nmike 5\n" ]
[ "andrew\n", "andrew\n" ]
none
0
[ { "input": "3\nmike 3\nandrew 5\nmike 2", "output": "andrew" }, { "input": "3\nandrew 3\nandrew 2\nmike 5", "output": "andrew" }, { "input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303", "output": "kaxqybeultn" }, { "input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734", "output": "ksjuuerbnlklcfdjeyq" }, { "input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879", "output": "fcgslzkicjrpbqaifgweyzreajjfdo" }, { "input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735", "output": "aawtvezfntstrcpgbzjbf" }, { "input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158", "output": "ivhgbxiv" } ]
1,520,622,172
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
9
124
5,632,000
n = int(input()) res = {} max_name = "" max_score = -1E16 for k in range(n): a = input().split(' ') name, score = a[0], int(a[1]) res[name] = res.get(name, 0) + score if (max_score < res[name]): max_score = res[name] max_name = name print(max_name)
Title: Winner Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. Input Specification: The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. Output Specification: Print the name of the winner. Demo Input: ['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n'] Demo Output: ['andrew\n', 'andrew\n'] Note: none
```python n = int(input()) res = {} max_name = "" max_score = -1E16 for k in range(n): a = input().split(' ') name, score = a[0], int(a[1]) res[name] = res.get(name, 0) + score if (max_score < res[name]): max_score = res[name] max_name = name print(max_name) ```
0
999
C
Alphabetic Removals
PROGRAMMING
1,200
[ "implementation" ]
null
null
You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times: - if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm. This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters. Help Polycarp find the resulting string.
The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove. The second line contains the string $s$ consisting of $n$ lowercase Latin letters.
Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times. If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
[ "15 3\ncccaabababaccbc\n", "15 9\ncccaabababaccbc\n", "1 1\nu\n" ]
[ "cccbbabaccbc\n", "cccccc\n", "" ]
none
0
[ { "input": "15 3\ncccaabababaccbc", "output": "cccbbabaccbc" }, { "input": "15 9\ncccaabababaccbc", "output": "cccccc" }, { "input": "5 2\nzyzyx", "output": "zzy" }, { "input": "4 3\nhack", "output": "k" }, { "input": "4 3\nzzzz", "output": "z" }, { "input": "6 5\naaccdd", "output": "d" }, { "input": "2 1\nzz", "output": "z" }, { "input": "14 5\nhxehmvkybeklnj", "output": "xmvkyklnj" } ]
1,653,136,998
2,147,483,647
PyPy 3-64
TIME_LIMIT_EXCEEDED
TESTS
4
2,000
14,131,200
n, k = map(int, input().split()) s = input() if(n == k): print() exit() d = {chr(i + ord('a')) : 0 for i in range(26)} for i in range(n): d[s[i]] += 1 c = 'a' for i in d.keys(): if(d[i] <= k): k -= d[i] else: d[i] -= k c = i break ans = '' for i in s[::-1]: if(i > c): ans += i elif(i == c): if(d[c]): ans += i d[c] -= 1 print(ans[::-1])
Title: Alphabetic Removals Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times: - if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm. This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters. Help Polycarp find the resulting string. Input Specification: The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove. The second line contains the string $s$ consisting of $n$ lowercase Latin letters. Output Specification: Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times. If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break). Demo Input: ['15 3\ncccaabababaccbc\n', '15 9\ncccaabababaccbc\n', '1 1\nu\n'] Demo Output: ['cccbbabaccbc\n', 'cccccc\n', ''] Note: none
```python n, k = map(int, input().split()) s = input() if(n == k): print() exit() d = {chr(i + ord('a')) : 0 for i in range(26)} for i in range(n): d[s[i]] += 1 c = 'a' for i in d.keys(): if(d[i] <= k): k -= d[i] else: d[i] -= k c = i break ans = '' for i in s[::-1]: if(i > c): ans += i elif(i == c): if(d[c]): ans += i d[c] -= 1 print(ans[::-1]) ```
0
637
B
Chat Order
PROGRAMMING
1,200
[ "*special", "binary search", "constructive algorithms", "data structures", "sortings" ]
null
null
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list. Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
[ "4\nalex\nivan\nroman\nivan\n", "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n" ]
[ "ivan\nroman\nalex\n", "alina\nmaria\nekaterina\ndarya\n" ]
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows: 1. alex Then Polycarpus writes to friend by name "ivan" and the list looks as follows: 1. ivan 1. alex Polycarpus writes the third message to friend by name "roman" and the list looks as follows: 1. roman 1. ivan 1. alex Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows: 1. ivan 1. roman 1. alex
1,000
[ { "input": "4\nalex\nivan\nroman\nivan", "output": "ivan\nroman\nalex" }, { "input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina", "output": "alina\nmaria\nekaterina\ndarya" }, { "input": "1\nwdi", "output": "wdi" }, { "input": "2\nypg\nypg", "output": "ypg" }, { "input": "3\nexhll\nexhll\narruapexj", "output": "arruapexj\nexhll" }, { "input": "3\nfv\nle\nle", "output": "le\nfv" }, { "input": "8\nm\nm\nm\nm\nm\nm\nm\nm", "output": "m" }, { "input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr", "output": "r\nu\nb\nk\nw\ni" }, { "input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok", "output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne" }, { "input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe", "output": "we\nklrj" }, { "input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh", "output": "zncybqmh\naeebef\nn" }, { "input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo", "output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj" }, { "input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn", "output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy" }, { "input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg", "output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg" }, { "input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh", "output": "vhh\nfa" }, { "input": "2\naa\nbb", "output": "bb\naa" }, { "input": "2\naa\na", "output": "a\naa" }, { "input": "3\naa\naa\naa", "output": "aa" }, { "input": "5\naa\na\naa\na\naa", "output": "aa\na" }, { "input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa", "output": "aaa\naaaaaaa\naa\na\naaaa" }, { "input": "5\na\naa\naaa\naaaa\na", "output": "a\naaaa\naaa\naa" }, { "input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa", "output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa" }, { "input": "3\na\naa\naaa", "output": "aaa\naa\na" }, { "input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz", "output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz" } ]
1,625,866,748
2,147,483,647
PyPy 3
TIME_LIMIT_EXCEEDED
TESTS
46
3,000
24,576,000
from queue import LifoQueue names1 = set() names2 = set() q = LifoQueue() n = int(input()) for i in range(n): name = input() q.put(name) names1.add(name) while names1: name = q.get() if name not in names2: names2.add(name) names1.remove(name) print(name)
Title: Chat Order Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list. Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10. Output Specification: Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom. Demo Input: ['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n'] Demo Output: ['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n'] Note: In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows: 1. alex Then Polycarpus writes to friend by name "ivan" and the list looks as follows: 1. ivan 1. alex Polycarpus writes the third message to friend by name "roman" and the list looks as follows: 1. roman 1. ivan 1. alex Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows: 1. ivan 1. roman 1. alex
```python from queue import LifoQueue names1 = set() names2 = set() q = LifoQueue() n = int(input()) for i in range(n): name = input() q.put(name) names1.add(name) while names1: name = q.get() if name not in names2: names2.add(name) names1.remove(name) print(name) ```
0
384
A
Coder
PROGRAMMING
800
[ "implementation" ]
null
null
Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000).
On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any.
[ "2\n" ]
[ "2\nC.\n.C\n" ]
none
500
[ { "input": "2", "output": "2\nC.\n.C" }, { "input": "3", "output": "5\nC.C\n.C.\nC.C" }, { "input": "4", "output": "8\nC.C.\n.C.C\nC.C.\n.C.C" }, { "input": "10", "output": "50\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C" }, { "input": "15", "output": "113\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C" }, { "input": "100", "output": "5000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "101", "output": "5101\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "500", "output": "125000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n...." }, { "input": "501", "output": "125501\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n..." }, { "input": "755", "output": "285013\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "888", "output": "394272\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "998", "output": "498002\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "999", "output": "499001\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "1000", "output": "500000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "1", "output": "1\nC" } ]
1,592,487,362
2,147,483,647
PyPy 3
OK
TESTS
15
265
7,577,600
from collections import * from math import * n = int(input()) a = [['.' for i in range(n)] for j in range(n)] ans = 0 for i in range(n): for j in range(n): if (i+j)%2 == 0: a[i][j] = 'C' ans += 1 print(ans) for i in range(n): print("".join(a[i]))
Title: Coder Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). Output Specification: On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any. Demo Input: ['2\n'] Demo Output: ['2\nC.\n.C\n'] Note: none
```python from collections import * from math import * n = int(input()) a = [['.' for i in range(n)] for j in range(n)] ans = 0 for i in range(n): for j in range(n): if (i+j)%2 == 0: a[i][j] = 'C' ans += 1 print(ans) for i in range(n): print("".join(a[i])) ```
3
58
A
Chat room
PROGRAMMING
1,000
[ "greedy", "strings" ]
A. Chat room
1
256
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
If Vasya managed to say hello, print "YES", otherwise print "NO".
[ "ahhellllloou\n", "hlelo\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]
1,488,881,227
2,147,483,647
PyPy 3
OK
TESTS
40
93
23,142,400
st = list(input()) w = "hello" i = 0 for char in w: while i < len(st): if st[i] != char: del st[i] else: i += 1 break s = ''.join(st) if s[:5] == 'hello': print ("YES") else: print ("NO")
Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python st = list(input()) w = "hello" i = 0 for char in w: while i < len(st): if st[i] != char: del st[i] else: i += 1 break s = ''.join(st) if s[:5] == 'hello': print ("YES") else: print ("NO") ```
3.910394
237
A
Free Cash
PROGRAMMING
1,000
[ "implementation" ]
null
null
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately. Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe. Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors. Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe. Note that the time is given in the chronological order. All time is given within one 24-hour period.
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
[ "4\n8 0\n8 10\n8 10\n8 45\n", "3\n0 12\n10 11\n22 22\n" ]
[ "2\n", "1\n" ]
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away. In the second sample all visitors will come in different times, so it will be enough one cash.
500
[ { "input": "4\n8 0\n8 10\n8 10\n8 45", "output": "2" }, { "input": "3\n0 12\n10 11\n22 22", "output": "1" }, { "input": "5\n12 8\n15 27\n15 27\n16 2\n19 52", "output": "2" }, { "input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23", "output": "3" }, { "input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6", "output": "3" }, { "input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32", "output": "6" }, { "input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30", "output": "10" }, { "input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20", "output": "8" }, { "input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49", "output": "1" }, { "input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16", "output": "5" }, { "input": "2\n0 24\n1 0", "output": "1" }, { "input": "1\n0 0", "output": "1" }, { "input": "1\n1 5", "output": "1" }, { "input": "1\n1 1", "output": "1" }, { "input": "3\n0 0\n0 0\n0 0", "output": "3" }, { "input": "1\n5 0", "output": "1" }, { "input": "5\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "5" }, { "input": "1\n10 10", "output": "1" }, { "input": "1\n8 0", "output": "1" }, { "input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "10" }, { "input": "2\n0 0\n0 1", "output": "1" }, { "input": "2\n8 5\n8 5", "output": "2" } ]
1,692,624,594
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
8
2,000
307,200
t=int(input()) mas=[] for i in range(t): n,m=map(int,input().split()) m=n*60+m mas.append(m) a=[] for i in mas: a.append(mas.count(i)) print(max(a))
Title: Free Cash Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately. Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe. Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors. Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe. Note that the time is given in the chronological order. All time is given within one 24-hour period. Output Specification: Print a single integer — the minimum number of cashes, needed to serve all clients next day. Demo Input: ['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n'] Demo Output: ['2\n', '1\n'] Note: In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away. In the second sample all visitors will come in different times, so it will be enough one cash.
```python t=int(input()) mas=[] for i in range(t): n,m=map(int,input().split()) m=n*60+m mas.append(m) a=[] for i in mas: a.append(mas.count(i)) print(max(a)) ```
0
667
A
Pouring Rain
PROGRAMMING
1,100
[ "geometry", "math" ]
null
null
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation. Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom. You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously. Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds. Note one milliliter equals to one cubic centimeter.
The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where: - *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup.
If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104.
[ "1 2 3 100\n", "1 1 1 1\n" ]
[ "NO\n", "YES\n3.659792366325\n" ]
In the first example the water fills the cup faster than you can drink from it. In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds.
500
[ { "input": "1 2 3 100", "output": "NO" }, { "input": "1 1 1 1", "output": "YES\n3.659792366325" }, { "input": "48 7946 7992 72", "output": "NO" }, { "input": "72 6791 8546 46", "output": "NO" }, { "input": "100 5635 9099 23", "output": "NO" }, { "input": "20 287 3845 5", "output": "YES\n39.646277165210" }, { "input": "48 6428 9807 83", "output": "NO" }, { "input": "72 5272 4552 64", "output": "NO" }, { "input": "100 4117 5106 34", "output": "NO" }, { "input": "20 2961 9852 15", "output": "YES\n180.991437129723" }, { "input": "48 1805 3109 93", "output": "NO" }, { "input": "72 8534 7042 65", "output": "NO" }, { "input": "1 47 80 68", "output": "YES\n1.388102806810" }, { "input": "4 495 8813 1", "output": "YES\n0.706823517575" }, { "input": "5 2797 5925 9", "output": "YES\n9.553973511669" }, { "input": "1 8324 4362 23", "output": "YES\n1.505007106354" }, { "input": "6 1976 8455 3", "output": "YES\n6.674898722265" }, { "input": "7 2644 8080 5", "output": "YES\n12.900417790197" }, { "input": "3 4183 5491 98", "output": "YES\n6.162185601824" }, { "input": "2 8591 320 101", "output": "YES\n9999.259991757254" }, { "input": "10000 10000 10000 10000", "output": "NO" }, { "input": "2 5000 12 3", "output": "YES\n6099.653943875812" }, { "input": "10 1000 100 1", "output": "YES\n3659.792366325487" } ]
1,579,953,952
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
109
0
from math import pi d,h,v,e = list(map(int,input().split())) H = (4*v)/(pi*(d**2)) if H <= e:print("NO") else: ans = h/(H-e) print(ans)
Title: Pouring Rain Time Limit: None seconds Memory Limit: None megabytes Problem Description: A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation. Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom. You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously. Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds. Note one milliliter equals to one cubic centimeter. Input Specification: The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where: - *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup. Output Specification: If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104. Demo Input: ['1 2 3 100\n', '1 1 1 1\n'] Demo Output: ['NO\n', 'YES\n3.659792366325\n'] Note: In the first example the water fills the cup faster than you can drink from it. In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds.
```python from math import pi d,h,v,e = list(map(int,input().split())) H = (4*v)/(pi*(d**2)) if H <= e:print("NO") else: ans = h/(H-e) print(ans) ```
0
454
B
Little Pony and Sort by Shift
PROGRAMMING
1,200
[ "implementation" ]
null
null
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning: Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
[ "2\n2 1\n", "3\n1 3 2\n", "2\n1 2\n" ]
[ "1\n", "-1\n", "0\n" ]
none
1,000
[ { "input": "2\n2 1", "output": "1" }, { "input": "3\n1 3 2", "output": "-1" }, { "input": "2\n1 2", "output": "0" }, { "input": "6\n3 4 5 6 3 2", "output": "-1" }, { "input": "3\n1 2 1", "output": "1" }, { "input": "5\n1 1 2 1 1", "output": "2" }, { "input": "4\n5 4 5 4", "output": "-1" }, { "input": "7\n3 4 5 5 5 1 2", "output": "2" }, { "input": "5\n2 2 1 2 2", "output": "3" }, { "input": "5\n5 4 1 2 3", "output": "-1" }, { "input": "4\n6 1 2 7", "output": "-1" }, { "input": "5\n4 5 6 2 3", "output": "2" }, { "input": "2\n1 1", "output": "0" }, { "input": "4\n1 2 2 1", "output": "1" }, { "input": "9\n4 5 6 7 1 2 3 4 10", "output": "-1" }, { "input": "7\n2 3 4 1 2 3 4", "output": "-1" }, { "input": "6\n1 2 1 2 1 2", "output": "-1" }, { "input": "3\n3 2 1", "output": "-1" }, { "input": "4\n1 4 4 1", "output": "1" }, { "input": "5\n1 2 1 1 1", "output": "3" }, { "input": "5\n4 6 7 3 5", "output": "-1" }, { "input": "4\n2 3 1 4", "output": "-1" }, { "input": "5\n5 4 3 2 1", "output": "-1" }, { "input": "4\n2 4 1 4", "output": "-1" }, { "input": "6\n4 5 6 1 2 7", "output": "-1" }, { "input": "6\n1 2 3 1 1 1", "output": "3" }, { "input": "5\n1 3 3 3 1", "output": "1" }, { "input": "6\n5 6 7 5 5 5", "output": "3" }, { "input": "5\n3 4 2 1 2", "output": "-1" }, { "input": "3\n3 4 2", "output": "1" }, { "input": "6\n1 1 2 2 1 1", "output": "2" }, { "input": "4\n2 3 4 2", "output": "1" }, { "input": "5\n3 5 7 7 3", "output": "1" }, { "input": "4\n1 1 4 1", "output": "1" }, { "input": "7\n1 5 6 1 1 1 1", "output": "4" }, { "input": "5\n7 8 6 7 8", "output": "-1" }, { "input": "4\n2 4 1 3", "output": "-1" } ]
1,587,313,874
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
109
409,600
import copy import math input() arr = list(map(int,list(input().split()))) flag = -1 beacon = -1 arr_sorted = sorted(arr) c_init = 0 n = len(arr) flag = math.copysign(1, arr[1] - arr[0]) i = 0 if arr == sorted(arr): beacon = 1 elif arr[1:] == sorted(arr[1:],reverse=True): if arr[-1] <= arr[0]: beacon = n-1 else: beacon = -1 elif arr[1:] == sorted(arr[1:]): if arr[-1] <= arr[0]: beacon = n-1 else: beacon = -1 else: while i < n: if flag != np.sign(arr[i + 1] - arr[i]): if flag == -1: if i < n-1 and arr[i+1:] == sorted(arr[i+1:],reverse = True) : beacon = n-i-1 break else: beacon = -1 break if flag == 1: if i < n-1 and arr[i+1:] == sorted(arr[i+1:]) : beacon = n-i-1 break else: beacon = -1 break i += 1 print(beacon)
Title: Little Pony and Sort by Shift Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning: Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence? Input Specification: The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105). Output Specification: If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it. Demo Input: ['2\n2 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['1\n', '-1\n', '0\n'] Note: none
```python import copy import math input() arr = list(map(int,list(input().split()))) flag = -1 beacon = -1 arr_sorted = sorted(arr) c_init = 0 n = len(arr) flag = math.copysign(1, arr[1] - arr[0]) i = 0 if arr == sorted(arr): beacon = 1 elif arr[1:] == sorted(arr[1:],reverse=True): if arr[-1] <= arr[0]: beacon = n-1 else: beacon = -1 elif arr[1:] == sorted(arr[1:]): if arr[-1] <= arr[0]: beacon = n-1 else: beacon = -1 else: while i < n: if flag != np.sign(arr[i + 1] - arr[i]): if flag == -1: if i < n-1 and arr[i+1:] == sorted(arr[i+1:],reverse = True) : beacon = n-i-1 break else: beacon = -1 break if flag == 1: if i < n-1 and arr[i+1:] == sorted(arr[i+1:]) : beacon = n-i-1 break else: beacon = -1 break i += 1 print(beacon) ```
0
141
A
Amusing Joke
PROGRAMMING
800
[ "implementation", "sortings", "strings" ]
null
null
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
[ "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n", "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n", "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n" ]
[ "YES\n", "NO\n", "NO\n" ]
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".
500
[ { "input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS", "output": "YES" }, { "input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI", "output": "NO" }, { "input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER", "output": "NO" }, { "input": "B\nA\nAB", "output": "YES" }, { "input": "ONDOL\nJNPB\nONLNJBODP", "output": "YES" }, { "input": "Y\nW\nYW", "output": "YES" }, { "input": "OI\nM\nIMO", "output": "YES" }, { "input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF", "output": "YES" }, { "input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB", "output": "NO" }, { "input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH", "output": "NO" }, { "input": "IQ\nOQ\nQOQIGGKFNHJSGCGM", "output": "NO" }, { "input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR", "output": "YES" }, { "input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY", "output": "YES" }, { "input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX", "output": "YES" }, { "input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW", "output": "YES" }, { "input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU", "output": "YES" }, { "input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK", "output": "YES" }, { "input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE", "output": "NO" }, { "input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ", "output": "NO" }, { "input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI", "output": "NO" }, { "input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT", "output": "NO" }, { "input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY", "output": "NO" }, { "input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT", "output": "NO" }, { "input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY", "output": "NO" }, { "input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI", "output": "NO" }, { "input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE", "output": "NO" }, { "input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH", "output": "NO" }, { "input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU", "output": "NO" }, { "input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG", "output": "NO" }, { "input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM", "output": "NO" }, { "input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW", "output": "NO" }, { "input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ", "output": "YES" }, { "input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS", "output": "YES" }, { "input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH", "output": "YES" }, { "input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB", "output": "YES" }, { "input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP", "output": "YES" }, { "input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR", "output": "YES" }, { "input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT", "output": "YES" }, { "input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH", "output": "YES" }, { "input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG", "output": "YES" }, { "input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF", "output": "YES" }, { "input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA", "output": "NO" }, { "input": "IXFDY\nJRMOU\nDF", "output": "NO" }, { "input": "JPSPZ\nUGCUB\nJMZZZZZZZZ", "output": "NO" }, { "input": "AC\nA\nBBA", "output": "NO" }, { "input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB", "output": "NO" }, { "input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW", "output": "NO" }, { "input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE", "output": "NO" }, { "input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH", "output": "NO" }, { "input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ", "output": "NO" }, { "input": "GIO\nPRL\nPRL", "output": "NO" }, { "input": "A\nB\nABC", "output": "NO" }, { "input": "KKK\nKKK\nZZZZZ", "output": "NO" }, { "input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA", "output": "NO" } ]
1,692,878,424
2,147,483,647
PyPy 3-64
OK
TESTS
54
124
0
g = list(input()) g += list(input()) c = list(input()) g.sort() c.sort() if g == c: print("YES") else: print("NO")
Title: Amusing Joke Time Limit: None seconds Memory Limit: None megabytes Problem Description: So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. Input Specification: The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. Output Specification: Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. Demo Input: ['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".
```python g = list(input()) g += list(input()) c = list(input()) g.sort() c.sort() if g == c: print("YES") else: print("NO") ```
3
777
A
Shell Game
PROGRAMMING
1,000
[ "constructive algorithms", "implementation", "math" ]
null
null
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball. Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.). Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball?
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator. The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements.
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
[ "4\n2\n", "1\n1\n" ]
[ "1\n", "0\n" ]
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements. 1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell.
500
[ { "input": "4\n2", "output": "1" }, { "input": "1\n1", "output": "0" }, { "input": "2\n2", "output": "0" }, { "input": "3\n1", "output": "1" }, { "input": "3\n2", "output": "0" }, { "input": "3\n0", "output": "2" }, { "input": "2000000000\n0", "output": "1" }, { "input": "2\n0", "output": "1" }, { "input": "2\n1", "output": "2" }, { "input": "4\n0", "output": "2" }, { "input": "4\n1", "output": "0" }, { "input": "5\n0", "output": "0" }, { "input": "5\n1", "output": "2" }, { "input": "5\n2", "output": "1" }, { "input": "6\n0", "output": "0" }, { "input": "6\n1", "output": "1" }, { "input": "6\n2", "output": "2" }, { "input": "7\n0", "output": "1" }, { "input": "7\n1", "output": "0" }, { "input": "7\n2", "output": "2" }, { "input": "100000\n0", "output": "2" }, { "input": "100000\n1", "output": "0" }, { "input": "100000\n2", "output": "1" }, { "input": "99999\n1", "output": "1" }, { "input": "99998\n1", "output": "2" }, { "input": "99997\n1", "output": "0" }, { "input": "99996\n1", "output": "1" }, { "input": "99995\n1", "output": "2" }, { "input": "1999999995\n0", "output": "2" }, { "input": "1999999995\n1", "output": "1" }, { "input": "1999999995\n2", "output": "0" }, { "input": "1999999996\n0", "output": "2" }, { "input": "1999999996\n1", "output": "0" }, { "input": "1999999996\n2", "output": "1" }, { "input": "1999999997\n0", "output": "0" }, { "input": "1999999997\n1", "output": "2" }, { "input": "1999999997\n2", "output": "1" }, { "input": "1999999998\n0", "output": "0" }, { "input": "1999999998\n1", "output": "1" }, { "input": "1999999998\n2", "output": "2" }, { "input": "1999999999\n0", "output": "1" }, { "input": "1999999999\n1", "output": "0" }, { "input": "1999999999\n2", "output": "2" }, { "input": "2000000000\n1", "output": "2" }, { "input": "2000000000\n2", "output": "0" }, { "input": "1234567890\n0", "output": "0" }, { "input": "1234567890\n1", "output": "1" }, { "input": "1234567890\n2", "output": "2" }, { "input": "123456789\n0", "output": "2" }, { "input": "123456789\n1", "output": "1" }, { "input": "123456789\n2", "output": "0" }, { "input": "123456790\n0", "output": "2" }, { "input": "12\n2", "output": "2" }, { "input": "32\n1", "output": "2" }, { "input": "20\n2", "output": "0" }, { "input": "10\n1", "output": "0" }, { "input": "1\n0", "output": "1" }, { "input": "76994383\n1", "output": "0" }, { "input": "25\n2", "output": "2" }, { "input": "1\n2", "output": "2" }, { "input": "12\n0", "output": "0" }, { "input": "150\n2", "output": "2" }, { "input": "15\n0", "output": "2" }, { "input": "21\n2", "output": "0" }, { "input": "18\n2", "output": "2" }, { "input": "8\n2", "output": "0" }, { "input": "10\n0", "output": "2" }, { "input": "16\n0", "output": "2" } ]
1,698,415,541
2,147,483,647
PyPy 3-64
OK
TESTS
68
421
21,401,600
from doctest import master n=int(input()) x=int(input()) m=n%6 mas=[0,0,0] mas[x]=1 # mas=[[0,1,2],[1,0,2],[1,2,0],[2,1,0],[2,0,1],[0,2,1]] # mas.reverse() for i in range(m,0,-1): if i%2==0: mas[1],mas[2]=mas[2],mas[1] else: mas[0],mas[1]=mas[1],mas[0] print(mas.index(1))
Title: Shell Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball. Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.). Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball? Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator. The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements. Output Specification: Print one integer from 0 to 2 — the index of the shell where the ball was initially placed. Demo Input: ['4\n2\n', '1\n1\n'] Demo Output: ['1\n', '0\n'] Note: In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements. 1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell.
```python from doctest import master n=int(input()) x=int(input()) m=n%6 mas=[0,0,0] mas[x]=1 # mas=[[0,1,2],[1,0,2],[1,2,0],[2,1,0],[2,0,1],[0,2,1]] # mas.reverse() for i in range(m,0,-1): if i%2==0: mas[1],mas[2]=mas[2],mas[1] else: mas[0],mas[1]=mas[1],mas[0] print(mas.index(1)) ```
3
90
A
Cableway
PROGRAMMING
1,000
[ "greedy", "math" ]
A. Cableway
2
256
A group of university students wants to get to the top of a mountain to have a picnic there. For that they decided to use a cableway. A cableway is represented by some cablecars, hanged onto some cable stations by a cable. A cable is scrolled cyclically between the first and the last cable stations (the first of them is located at the bottom of the mountain and the last one is located at the top). As the cable moves, the cablecar attached to it move as well. The number of cablecars is divisible by three and they are painted three colors: red, green and blue, in such manner that after each red cablecar goes a green one, after each green cablecar goes a blue one and after each blue cablecar goes a red one. Each cablecar can transport no more than two people, the cablecars arrive with the periodicity of one minute (i. e. every minute) and it takes exactly 30 minutes for a cablecar to get to the top. All students are divided into three groups: *r* of them like to ascend only in the red cablecars, *g* of them prefer only the green ones and *b* of them prefer only the blue ones. A student never gets on a cablecar painted a color that he doesn't like, The first cablecar to arrive (at the moment of time 0) is painted red. Determine the least time it will take all students to ascend to the mountain top.
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=100). It is guaranteed that *r*<=+<=*g*<=+<=*b*<=&gt;<=0, it means that the group consists of at least one student.
Print a single number — the minimal time the students need for the whole group to ascend to the top of the mountain.
[ "1 3 2\n", "3 2 1\n" ]
[ "34", "33" ]
Let's analyze the first sample. At the moment of time 0 a red cablecar comes and one student from the *r* group get on it and ascends to the top at the moment of time 30. At the moment of time 1 a green cablecar arrives and two students from the *g* group get on it; they get to the top at the moment of time 31. At the moment of time 2 comes the blue cablecar and two students from the *b* group get on it. They ascend to the top at the moment of time 32. At the moment of time 3 a red cablecar arrives but the only student who is left doesn't like red and the cablecar leaves empty. At the moment of time 4 a green cablecar arrives and one student from the *g* group gets on it. He ascends to top at the moment of time 34. Thus, all the students are on the top, overall the ascension took exactly 34 minutes.
500
[ { "input": "1 3 2", "output": "34" }, { "input": "3 2 1", "output": "33" }, { "input": "3 5 2", "output": "37" }, { "input": "10 10 10", "output": "44" }, { "input": "29 7 24", "output": "72" }, { "input": "28 94 13", "output": "169" }, { "input": "90 89 73", "output": "163" }, { "input": "0 0 1", "output": "32" }, { "input": "0 0 2", "output": "32" }, { "input": "0 1 0", "output": "31" }, { "input": "0 1 1", "output": "32" }, { "input": "0 1 2", "output": "32" }, { "input": "0 2 0", "output": "31" }, { "input": "0 2 1", "output": "32" }, { "input": "0 2 2", "output": "32" }, { "input": "1 0 0", "output": "30" }, { "input": "1 0 1", "output": "32" }, { "input": "1 0 2", "output": "32" }, { "input": "1 1 0", "output": "31" }, { "input": "1 1 1", "output": "32" }, { "input": "1 1 2", "output": "32" }, { "input": "1 2 0", "output": "31" }, { "input": "1 2 1", "output": "32" }, { "input": "1 2 2", "output": "32" }, { "input": "2 0 0", "output": "30" }, { "input": "2 0 1", "output": "32" }, { "input": "2 0 2", "output": "32" }, { "input": "2 1 0", "output": "31" }, { "input": "2 1 1", "output": "32" }, { "input": "2 1 2", "output": "32" }, { "input": "2 2 0", "output": "31" }, { "input": "2 2 1", "output": "32" }, { "input": "2 2 2", "output": "32" }, { "input": "4 5 2", "output": "37" }, { "input": "5 7 8", "output": "41" }, { "input": "13 25 19", "output": "67" }, { "input": "29 28 30", "output": "74" }, { "input": "45 52 48", "output": "106" }, { "input": "68 72 58", "output": "136" }, { "input": "89 92 90", "output": "166" }, { "input": "99 97 98", "output": "177" }, { "input": "89 97 2", "output": "175" }, { "input": "96 3 92", "output": "171" }, { "input": "1 99 87", "output": "178" }, { "input": "95 2 3", "output": "171" }, { "input": "2 97 3", "output": "175" }, { "input": "2 2 99", "output": "179" }, { "input": "100 100 100", "output": "179" }, { "input": "100 0 100", "output": "179" }, { "input": "0 100 100", "output": "179" }, { "input": "100 100 0", "output": "178" }, { "input": "100 0 0", "output": "177" }, { "input": "0 100 0", "output": "178" }, { "input": "0 0 100", "output": "179" }, { "input": "5 4 5", "output": "38" } ]
1,657,156,389
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
3
92
0
data = list(map(int, input().split(" "))) trips = [] biggest = 0 biggest_index = 0 for i in range(3): trips.append(data[i]//2) if data[i]%2 != 2: trips[i] += 1 if trips[i] > biggest: biggest = trips[i] biggest_index = i if biggest_index == 0: print(30 + ((trips[0]-1 )* 3)) elif biggest_index == 1: print(31 + ((trips[1]-1 )* 3)) else: print(32 + ((trips[2]-1 )* 3))
Title: Cableway Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A group of university students wants to get to the top of a mountain to have a picnic there. For that they decided to use a cableway. A cableway is represented by some cablecars, hanged onto some cable stations by a cable. A cable is scrolled cyclically between the first and the last cable stations (the first of them is located at the bottom of the mountain and the last one is located at the top). As the cable moves, the cablecar attached to it move as well. The number of cablecars is divisible by three and they are painted three colors: red, green and blue, in such manner that after each red cablecar goes a green one, after each green cablecar goes a blue one and after each blue cablecar goes a red one. Each cablecar can transport no more than two people, the cablecars arrive with the periodicity of one minute (i. e. every minute) and it takes exactly 30 minutes for a cablecar to get to the top. All students are divided into three groups: *r* of them like to ascend only in the red cablecars, *g* of them prefer only the green ones and *b* of them prefer only the blue ones. A student never gets on a cablecar painted a color that he doesn't like, The first cablecar to arrive (at the moment of time 0) is painted red. Determine the least time it will take all students to ascend to the mountain top. Input Specification: The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=100). It is guaranteed that *r*<=+<=*g*<=+<=*b*<=&gt;<=0, it means that the group consists of at least one student. Output Specification: Print a single number — the minimal time the students need for the whole group to ascend to the top of the mountain. Demo Input: ['1 3 2\n', '3 2 1\n'] Demo Output: ['34', '33'] Note: Let's analyze the first sample. At the moment of time 0 a red cablecar comes and one student from the *r* group get on it and ascends to the top at the moment of time 30. At the moment of time 1 a green cablecar arrives and two students from the *g* group get on it; they get to the top at the moment of time 31. At the moment of time 2 comes the blue cablecar and two students from the *b* group get on it. They ascend to the top at the moment of time 32. At the moment of time 3 a red cablecar arrives but the only student who is left doesn't like red and the cablecar leaves empty. At the moment of time 4 a green cablecar arrives and one student from the *g* group gets on it. He ascends to top at the moment of time 34. Thus, all the students are on the top, overall the ascension took exactly 34 minutes.
```python data = list(map(int, input().split(" "))) trips = [] biggest = 0 biggest_index = 0 for i in range(3): trips.append(data[i]//2) if data[i]%2 != 2: trips[i] += 1 if trips[i] > biggest: biggest = trips[i] biggest_index = i if biggest_index == 0: print(30 + ((trips[0]-1 )* 3)) elif biggest_index == 1: print(31 + ((trips[1]-1 )* 3)) else: print(32 + ((trips[2]-1 )* 3)) ```
0
919
D
Substring
PROGRAMMING
1,700
[ "dfs and similar", "dp", "graphs" ]
null
null
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges. The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node. Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
[ "5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n", "6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n", "10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n" ]
[ "3\n", "-1\n", "4\n" ]
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
1,500
[ { "input": "5 4\nabaca\n1 2\n1 3\n3 4\n4 5", "output": "3" }, { "input": "6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4", "output": "-1" }, { "input": "10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7", "output": "4" }, { "input": "1 1\nf\n1 1", "output": "-1" }, { "input": "10 50\nebibwbjihv\n1 10\n1 2\n5 4\n1 8\n9 7\n5 6\n1 8\n8 7\n2 6\n5 4\n1 9\n3 2\n8 3\n5 6\n5 9\n2 4\n2 7\n3 9\n1 2\n1 7\n1 10\n3 7\n1 8\n3 10\n8 6\n1 7\n10 6\n1 6\n5 8\n1 5\n2 10\n3 9\n5 8\n8 3\n3 7\n5 2\n1 10\n1 4\n5 3\n3 2\n1 2\n5 8\n10 4\n2 10\n8 2\n1 9\n1 8\n1 2\n3 4\n1 8", "output": "2" }, { "input": "13 37\ndwpzcppjmhkmz\n2 6\n3 6\n6 7\n6 7\n6 7\n6 7\n6 8\n6 8\n6 8\n6 8\n4 6\n4 6\n5 6\n4 6\n4 6\n6 9\n6 9\n6 10\n6 10\n6 10\n6 10\n4 6\n1 6\n1 6\n10 11\n6 11\n1 6\n6 12\n6 12\n6 12\n6 13\n6 13\n6 13\n6 13\n3 6\n2 6\n2 6", "output": "3" }, { "input": "5 8\ntetqw\n2 1\n4 4\n5 5\n5 2\n4 5\n1 5\n1 5\n1 1", "output": "-1" }, { "input": "5 8\nreeet\n4 3\n2 5\n4 2\n2 4\n4 2\n5 2\n3 3\n3 4", "output": "-1" } ]
1,676,265,204
1,704
Python 3
WRONG_ANSWER
TESTS
0
30
102,400
from collections import defaultdict def beauty(n, m, S, X, Y): adj_list = defaultdict(list) for i in range(m): adj_list[X[i]-1].append(Y[i]-1) s = list(S) s.sort() freq = [s.count(c) for c in set(s)] freq.sort() freq = freq[::-1] beauty = [0] * n for i in range(len(freq)): for j in range(n): if s[j] == s[i]: beauty[j] = freq[i] for i in range(n): for j in adj_list[i]: beauty[j] = max(beauty[j], beauty[i] - 1) max_beauty = 0 for i in range(n): max_beauty = max(max_beauty, beauty[i]) return max_beauty if max_beauty <= 26 else -1
Title: Substring Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest. Input Specification: The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges. The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node. Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected. Output Specification: Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead. Demo Input: ['5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n', '6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n', '10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n'] Demo Output: ['3\n', '-1\n', '4\n'] Note: In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
```python from collections import defaultdict def beauty(n, m, S, X, Y): adj_list = defaultdict(list) for i in range(m): adj_list[X[i]-1].append(Y[i]-1) s = list(S) s.sort() freq = [s.count(c) for c in set(s)] freq.sort() freq = freq[::-1] beauty = [0] * n for i in range(len(freq)): for j in range(n): if s[j] == s[i]: beauty[j] = freq[i] for i in range(n): for j in adj_list[i]: beauty[j] = max(beauty[j], beauty[i] - 1) max_beauty = 0 for i in range(n): max_beauty = max(max_beauty, beauty[i]) return max_beauty if max_beauty <= 26 else -1 ```
0
785
A
Anton and Polyhedrons
PROGRAMMING
800
[ "implementation", "strings" ]
null
null
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: - Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces. All five kinds of polyhedrons are shown on the picture below: Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection. Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this: - "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
[ "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n", "3\nDodecahedron\nOctahedron\nOctahedron\n" ]
[ "42\n", "28\n" ]
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
500
[ { "input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron", "output": "42" }, { "input": "3\nDodecahedron\nOctahedron\nOctahedron", "output": "28" }, { "input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron", "output": "256" }, { "input": "1\nTetrahedron", "output": "4" }, { "input": "1\nCube", "output": "6" }, { "input": "1\nOctahedron", "output": "8" }, { "input": "1\nDodecahedron", "output": "12" }, { "input": "1\nIcosahedron", "output": "20" }, { "input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron", "output": "340" } ]
1,699,197,351
2,147,483,647
PyPy 3
OK
TESTS
30
1,419
9,523,200
ww=int(input()) count=0 for xx in range(ww): pp=input() if pp=="Tetrahedron": count = count + 4 if pp=="Cube": count = count + 6 if pp=="Octahedron": count = count + 8 if pp=="Dodecahedron": count = count + 12 if pp=="Icosahedron": count = count + 20 print(count)
Title: Anton and Polyhedrons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: - Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces. All five kinds of polyhedrons are shown on the picture below: Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection. Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this: - "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. Output Specification: Output one number — the total number of faces in all the polyhedrons in Anton's collection. Demo Input: ['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n'] Demo Output: ['42\n', '28\n'] Note: In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
```python ww=int(input()) count=0 for xx in range(ww): pp=input() if pp=="Tetrahedron": count = count + 4 if pp=="Cube": count = count + 6 if pp=="Octahedron": count = count + 8 if pp=="Dodecahedron": count = count + 12 if pp=="Icosahedron": count = count + 20 print(count) ```
3
337
A
Puzzles
PROGRAMMING
900
[ "greedy" ]
null
null
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Print a single integer — the least possible difference the teacher can obtain.
[ "4 6\n10 12 10 7 5 22\n" ]
[ "5\n" ]
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
500
[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]
1,689,107,337
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
62
0
# https://codeforces.com/problemset/problem/337/A n, m = map(int, input().split()) puzzels = list(map(int, input().split())) puzzels = sorted(puzzels) puzzels = puzzels[:n] print(puzzels[-1] - puzzels[0])
Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
```python # https://codeforces.com/problemset/problem/337/A n, m = map(int, input().split()) puzzels = list(map(int, input().split())) puzzels = sorted(puzzels) puzzels = puzzels[:n] print(puzzels[-1] - puzzels[0]) ```
0
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,678,097,402
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
30
0
x=input() l=len(x) if l>10: print(x[0]+str(l)+x[l-1]) if(l<10): print(x)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python x=input() l=len(x) if l>10: print(x[0]+str(l)+x[l-1]) if(l<10): print(x) ```
0
359
B
Permutation
PROGRAMMING
1,400
[ "constructive algorithms", "dp", "math" ]
null
null
A permutation *p* is an ordered group of numbers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers, each is no more than *n*. We'll define number *n* as the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*. Simon has a positive integer *n* and a non-negative integer *k*, such that 2*k*<=≤<=*n*. Help him find permutation *a* of length 2*n*, such that it meets this equation: .
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 0<=≤<=2*k*<=≤<=*n*).
Print 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* — the required permutation *a*. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
[ "1 0\n", "2 1\n", "4 0\n" ]
[ "1 2", "3 2 1 4\n", "2 7 4 6 1 3 5 8\n" ]
Record |*x*| represents the absolute value of number *x*. In the first sample |1 - 2| - |1 - 2| = 0. In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2. In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
1,000
[ { "input": "1 0", "output": "1 2" }, { "input": "2 1", "output": "3 2 1 4" }, { "input": "4 0", "output": "2 7 4 6 1 3 5 8" }, { "input": "50000 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "50000 25000", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "50000 24999", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "49999 24999", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "49999 3", "output": "2 1 4 3 6 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "1333 156", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "7563 3781", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "7563 3780", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "2 0", "output": "1 2 3 4" }, { "input": "3 0", "output": "1 2 3 4 5 6" }, { "input": "4 1", "output": "2 1 3 4 5 6 7 8" }, { "input": "4 2", "output": "2 1 4 3 5 6 7 8" }, { "input": "6 3", "output": "2 1 4 3 6 5 7 8 9 10 11 12" }, { "input": "48888 24444", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "50000 1", "output": "2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "50000 1000", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "34 17", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68" }, { "input": "43244 1233", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "213 100", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "50 1", "output": "2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100" }, { "input": "55 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110" }, { "input": "5000 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "3 1", "output": "2 1 3 4 5 6" }, { "input": "7563 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "7563 1", "output": "2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "7563 2", "output": "2 1 4 3 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "6 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12" } ]
1,556,051,135
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
77
0
n,k=input().split() n=int(n) k=int(k) l=[i for i in range(1,2*n+1)] print(l) for i in range(1,k+1): x=2*i-1 y=2*i l[x],l[y]=l[y],l[x] for i in range(2*n): print(l[i],end=" ")
Title: Permutation Time Limit: None seconds Memory Limit: None megabytes Problem Description: A permutation *p* is an ordered group of numbers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers, each is no more than *n*. We'll define number *n* as the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*. Simon has a positive integer *n* and a non-negative integer *k*, such that 2*k*<=≤<=*n*. Help him find permutation *a* of length 2*n*, such that it meets this equation: . Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 0<=≤<=2*k*<=≤<=*n*). Output Specification: Print 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* — the required permutation *a*. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them. Demo Input: ['1 0\n', '2 1\n', '4 0\n'] Demo Output: ['1 2', '3 2 1 4\n', '2 7 4 6 1 3 5 8\n'] Note: Record |*x*| represents the absolute value of number *x*. In the first sample |1 - 2| - |1 - 2| = 0. In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2. In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
```python n,k=input().split() n=int(n) k=int(k) l=[i for i in range(1,2*n+1)] print(l) for i in range(1,k+1): x=2*i-1 y=2*i l[x],l[y]=l[y],l[x] for i in range(2*n): print(l[i],end=" ") ```
0
172
A
Phone Code
PROGRAMMING
800
[ "*special", "brute force", "implementation" ]
null
null
Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length. Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code.
The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different.
Print the number of digits in the city phone code.
[ "4\n00209\n00219\n00999\n00909\n", "2\n1\n2\n", "3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n" ]
[ "2\n", "0\n", "12\n" ]
A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209". In the first sample the city phone code is string "00". In the second sample the city phone code is an empty string. In the third sample the city phone code is string "770123456789".
1,000
[ { "input": "4\n00209\n00219\n00999\n00909", "output": "2" }, { "input": "2\n1\n2", "output": "0" }, { "input": "3\n77012345678999999999\n77012345678901234567\n77012345678998765432", "output": "12" }, { "input": "5\n4491183345\n4491184811\n4491162340\n4491233399\n4491449214", "output": "4" }, { "input": "10\n15424\n10953\n19176\n15514\n16284\n18680\n19305\n13816\n16168\n15924", "output": "1" }, { "input": "10\n4906361343\n8985777485\n1204265609\n7088384855\n4127287014\n7904807820\n3032139021\n5999959109\n6477458281\n3244359368", "output": "0" }, { "input": "10\n3717208309\n3717208306\n3717208302\n3717208301\n3717208303\n3717208308\n3717208304\n3717208307\n3717208300\n3717208305", "output": "9" }, { "input": "9\n2881\n2808\n2868\n2874\n2894\n2870\n2818\n2896\n2890", "output": "2" }, { "input": "2\n4\n9", "output": "0" }, { "input": "2\n29867863763143509570\n59261213969200291523", "output": "0" }, { "input": "2\n84\n62", "output": "0" }, { "input": "2\n75970434466248453472\n75970434466248453476", "output": "19" }, { "input": "10\n17254072509168593435\n17254072509168593433\n17254072509168593430\n17254072509168593432\n17254072509168593439\n17254072509168593436\n17254072509168593438\n17254072509168593437\n17254072509168593431\n17254072509168593434", "output": "19" } ]
1,595,928,952
2,147,483,647
Python 3
OK
TESTS
28
498
8,499,200
a=[] n=int(input()) if n==1 or n==0: print(0) exit(0) for i in range(n): a.append(input()) j=0 ans=0 while j<len(a[0]): count=0 temp=a[0][j] count+=1 for i in range(1,len(a)): if a[i][j]==temp: count+=1 if count==n: ans+=1 else: break j+=1 print(ans)
Title: Phone Code Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length. Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code. Input Specification: The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different. Output Specification: Print the number of digits in the city phone code. Demo Input: ['4\n00209\n00219\n00999\n00909\n', '2\n1\n2\n', '3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n'] Demo Output: ['2\n', '0\n', '12\n'] Note: A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209". In the first sample the city phone code is string "00". In the second sample the city phone code is an empty string. In the third sample the city phone code is string "770123456789".
```python a=[] n=int(input()) if n==1 or n==0: print(0) exit(0) for i in range(n): a.append(input()) j=0 ans=0 while j<len(a[0]): count=0 temp=a[0][j] count+=1 for i in range(1,len(a)): if a[i][j]==temp: count+=1 if count==n: ans+=1 else: break j+=1 print(ans) ```
3
994
A
Fingerprints
PROGRAMMING
800
[ "implementation" ]
null
null
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits. Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints. The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence. The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
[ "7 3\n3 5 7 1 6 2 8\n1 2 7\n", "4 4\n3 4 1 0\n0 1 7 9\n" ]
[ "7 1 2\n", "1 0\n" ]
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence. In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
500
[ { "input": "7 3\n3 5 7 1 6 2 8\n1 2 7", "output": "7 1 2" }, { "input": "4 4\n3 4 1 0\n0 1 7 9", "output": "1 0" }, { "input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8", "output": "8 6 4 2" }, { "input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9", "output": "3 7 4 9 0" }, { "input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8", "output": "1 2 3 4 5 6 7 8 9 0" }, { "input": "1 1\n4\n4", "output": "4" }, { "input": "3 7\n6 3 4\n4 9 0 1 7 8 6", "output": "6 4" }, { "input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0", "output": "0" }, { "input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6", "output": "6 0 3 8 1" }, { "input": "8 2\n7 2 9 6 1 0 3 4\n6 3", "output": "6 3" }, { "input": "5 4\n7 0 1 4 9\n0 9 5 3", "output": "0 9" }, { "input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6", "output": "6" }, { "input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2", "output": "2 3" }, { "input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2", "output": "3 9 2 4" }, { "input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6", "output": "1 2 3 8 6 5" }, { "input": "8 2\n7 4 8 9 2 5 6 1\n6 4", "output": "4 6" }, { "input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3", "output": "0 3" }, { "input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8", "output": "2 8 1 3" }, { "input": "1 6\n3\n6 8 2 4 5 3", "output": "3" }, { "input": "1 8\n0\n9 2 4 8 1 5 0 7", "output": "0" }, { "input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4", "output": "7 9 4 1 0" }, { "input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1", "output": "0 1" }, { "input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2", "output": "2 9 6 1 4" }, { "input": "6 3\n8 3 9 2 7 6\n5 4 3", "output": "3" }, { "input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5", "output": "8 3 9 6" }, { "input": "1 2\n1\n1 0", "output": "1" }, { "input": "3 6\n1 2 3\n4 5 6 1 2 3", "output": "1 2 3" }, { "input": "1 2\n2\n1 2", "output": "2" }, { "input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9", "output": "9" } ]
1,642,581,417
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
15
0
N , M = map(int, input().split()) NN = list(map(int, input().split())) f = set(int, input().split()) for each in NN: if each in f: print(each, end=" ")
Title: Fingerprints Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits. Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code. Input Specification: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints. The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence. The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints. Output Specification: In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable. Demo Input: ['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n'] Demo Output: ['7 1 2\n', '1 0\n'] Note: In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence. In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
```python N , M = map(int, input().split()) NN = list(map(int, input().split())) f = set(int, input().split()) for each in NN: if each in f: print(each, end=" ") ```
-1
653
A
Bear and Three Balls
PROGRAMMING
900
[ "brute force", "implementation", "sortings" ]
null
null
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above.
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
[ "4\n18 55 16 17\n", "6\n40 41 43 44 44 44\n", "8\n5 972 3 4 1 4 970 971\n" ]
[ "YES\n", "NO\n", "YES\n" ]
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
500
[ { "input": "4\n18 55 16 17", "output": "YES" }, { "input": "6\n40 41 43 44 44 44", "output": "NO" }, { "input": "8\n5 972 3 4 1 4 970 971", "output": "YES" }, { "input": "3\n959 747 656", "output": "NO" }, { "input": "4\n1 2 2 3", "output": "YES" }, { "input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543", "output": "NO" }, { "input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869", "output": "YES" }, { "input": "3\n408 410 409", "output": "YES" }, { "input": "3\n903 902 904", "output": "YES" }, { "input": "3\n399 400 398", "output": "YES" }, { "input": "3\n450 448 449", "output": "YES" }, { "input": "3\n390 389 388", "output": "YES" }, { "input": "3\n438 439 440", "output": "YES" }, { "input": "11\n488 688 490 94 564 615 641 170 489 517 669", "output": "YES" }, { "input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954", "output": "YES" }, { "input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318", "output": "YES" }, { "input": "6\n10 79 306 334 304 305", "output": "YES" }, { "input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365", "output": "YES" }, { "input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73", "output": "YES" }, { "input": "11\n325 325 324 324 324 325 325 324 324 324 324", "output": "NO" }, { "input": "7\n517 517 518 517 518 518 518", "output": "NO" }, { "input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710", "output": "NO" }, { "input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29", "output": "NO" }, { "input": "7\n880 880 514 536 881 881 879", "output": "YES" }, { "input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375", "output": "YES" }, { "input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404", "output": "YES" }, { "input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987", "output": "YES" }, { "input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116", "output": "NO" }, { "input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995", "output": "NO" }, { "input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22", "output": "YES" }, { "input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952", "output": "YES" }, { "input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911", "output": "NO" }, { "input": "3\n3 1 2", "output": "YES" }, { "input": "3\n500 999 1000", "output": "NO" }, { "input": "10\n101 102 104 105 107 109 110 112 113 115", "output": "NO" }, { "input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "50\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "3\n1000 999 998", "output": "YES" }, { "input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456", "output": "NO" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3", "output": "YES" }, { "input": "3\n999 999 1000", "output": "NO" }, { "input": "9\n2 4 5 13 25 100 200 300 400", "output": "NO" }, { "input": "9\n1 1 1 2 2 2 3 3 3", "output": "YES" }, { "input": "3\n1 1 2", "output": "NO" }, { "input": "3\n998 999 1000", "output": "YES" }, { "input": "12\n1 1 1 1 1 1 1 1 1 2 2 4", "output": "NO" }, { "input": "4\n4 3 4 5", "output": "YES" }, { "input": "6\n1 1 1 2 2 2", "output": "NO" }, { "input": "3\n2 3 2", "output": "NO" }, { "input": "5\n10 5 6 3 2", "output": "NO" }, { "input": "3\n1 2 1", "output": "NO" }, { "input": "3\n1 2 3", "output": "YES" }, { "input": "4\n998 999 1000 1000", "output": "YES" }, { "input": "5\n2 3 9 9 4", "output": "YES" }, { "input": "4\n1 2 4 4", "output": "NO" }, { "input": "3\n1 1 1", "output": "NO" }, { "input": "3\n2 2 3", "output": "NO" }, { "input": "7\n1 2 2 2 4 5 6", "output": "YES" }, { "input": "5\n1 3 10 3 10", "output": "NO" }, { "input": "3\n1 2 2", "output": "NO" }, { "input": "4\n1000 1000 999 998", "output": "YES" }, { "input": "3\n5 3 7", "output": "NO" }, { "input": "6\n1 1 2 2 3 3", "output": "YES" }, { "input": "9\n6 6 6 5 5 5 4 4 4", "output": "YES" }, { "input": "7\n5 6 6 6 7 7 7", "output": "YES" }, { "input": "5\n2 3 3 3 4", "output": "YES" }, { "input": "5\n2 1 2 1 3", "output": "YES" }, { "input": "3\n1 2 7", "output": "NO" }, { "input": "3\n1000 1000 1000", "output": "NO" }, { "input": "5\n1 100 2 100 3", "output": "YES" }, { "input": "5\n5 4 6 5 5", "output": "YES" }, { "input": "12\n1 1 1 1 2 2 2 2 3 3 3 3", "output": "YES" }, { "input": "5\n9 9 1 2 3", "output": "YES" }, { "input": "6\n1 2 3 1 2 3", "output": "YES" }, { "input": "7\n1 1 1 1 2 3 3", "output": "YES" }, { "input": "3\n13 13 13", "output": "NO" }, { "input": "3\n42 42 42", "output": "NO" }, { "input": "8\n1 1 1 1 2 2 2 2", "output": "NO" }, { "input": "6\n1 1 1 1 2 3", "output": "YES" }, { "input": "6\n1 1 2 2 6 6", "output": "NO" }, { "input": "6\n1 2 5 5 5 5", "output": "NO" }, { "input": "9\n1 2 3 1 2 3 1 2 3", "output": "YES" }, { "input": "4\n1 2 1 100", "output": "NO" }, { "input": "5\n1 1 2 2 3", "output": "YES" } ]
1,683,310,802
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
35
46
0
n=int(input()) b=list(map(int,input().split())) b=sorted(set(b)) c=0 for i in range(len(b)-2): if b[i+2]-b[i]<=2: print('YES') break else: c=c+1 if c==len(b)-2: print('NO')
Title: Bear and Three Balls Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above. Input Specification: The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball. Output Specification: Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes). Demo Input: ['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
```python n=int(input()) b=list(map(int,input().split())) b=sorted(set(b)) c=0 for i in range(len(b)-2): if b[i+2]-b[i]<=2: print('YES') break else: c=c+1 if c==len(b)-2: print('NO') ```
0
115
A
Party
PROGRAMMING
900
[ "dfs and similar", "graphs", "trees" ]
null
null
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed?
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Print a single integer denoting the minimum number of groups that will be formed in the party.
[ "5\n-1\n1\n2\n1\n-1\n" ]
[ "3\n" ]
For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5
500
[ { "input": "5\n-1\n1\n2\n1\n-1", "output": "3" }, { "input": "4\n-1\n1\n2\n3", "output": "4" }, { "input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11", "output": "4" }, { "input": "6\n-1\n-1\n2\n3\n1\n1", "output": "3" }, { "input": "3\n-1\n1\n1", "output": "2" }, { "input": "1\n-1", "output": "1" }, { "input": "2\n2\n-1", "output": "2" }, { "input": "2\n-1\n-1", "output": "1" }, { "input": "3\n2\n-1\n1", "output": "3" }, { "input": "3\n-1\n-1\n-1", "output": "1" }, { "input": "5\n4\n5\n1\n-1\n4", "output": "3" }, { "input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7", "output": "4" }, { "input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4", "output": "5" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1", "output": "2" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1", "output": "1" }, { "input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11", "output": "12" }, { "input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2", "output": "11" }, { "input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6", "output": "6" }, { "input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1", "output": "3" }, { "input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1", "output": "2" }, { "input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1", "output": "3" }, { "input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8", "output": "5" }, { "input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10", "output": "6" }, { "input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1", "output": "4" }, { "input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9", "output": "4" }, { "input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8", "output": "3" }, { "input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11", "output": "7" }, { "input": "7\n7\n6\n-1\n-1\n4\n4\n-1", "output": "3" }, { "input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34", "output": "6" }, { "input": "5\n2\n3\n4\n5\n-1", "output": "5" } ]
1,681,312,130
2,147,483,647
Python 3
OK
TESTS
106
124
1,740,800
import sys from collections import defaultdict sys.setrecursionlimit(10000) def dfs(employee, depth): max_depth = depth for sub in dic[employee]: max_depth = max(max_depth, dfs(sub, depth+1)) return max_depth n = int(input()) dic = defaultdict(list) for i in range(1, n+1): manager = int(input()) if manager == -1: manager = 0 dic[manager].append(i) ans = 0 for employee in dic[0]: ans = max(ans, dfs(employee, 1)) print(ans)
Title: Party Time Limit: None seconds Memory Limit: None megabytes Problem Description: A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. Output Specification: Print a single integer denoting the minimum number of groups that will be formed in the party. Demo Input: ['5\n-1\n1\n2\n1\n-1\n'] Demo Output: ['3\n'] Note: For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5
```python import sys from collections import defaultdict sys.setrecursionlimit(10000) def dfs(employee, depth): max_depth = depth for sub in dic[employee]: max_depth = max(max_depth, dfs(sub, depth+1)) return max_depth n = int(input()) dic = defaultdict(list) for i in range(1, n+1): manager = int(input()) if manager == -1: manager = 0 dic[manager].append(i) ans = 0 for employee in dic[0]: ans = max(ans, dfs(employee, 1)) print(ans) ```
3
9
A
Die Roll
PROGRAMMING
800
[ "math", "probabilities" ]
A. Die Roll
1
64
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
[ "4 2\n" ]
[ "1/2\n" ]
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
0
[ { "input": "4 2", "output": "1/2" }, { "input": "1 1", "output": "1/1" }, { "input": "1 2", "output": "5/6" }, { "input": "1 3", "output": "2/3" }, { "input": "1 4", "output": "1/2" }, { "input": "1 5", "output": "1/3" }, { "input": "1 6", "output": "1/6" }, { "input": "2 1", "output": "5/6" }, { "input": "2 2", "output": "5/6" }, { "input": "2 3", "output": "2/3" }, { "input": "2 4", "output": "1/2" }, { "input": "2 5", "output": "1/3" }, { "input": "2 6", "output": "1/6" }, { "input": "3 1", "output": "2/3" }, { "input": "3 2", "output": "2/3" }, { "input": "3 3", "output": "2/3" }, { "input": "3 4", "output": "1/2" }, { "input": "3 5", "output": "1/3" }, { "input": "3 6", "output": "1/6" }, { "input": "4 1", "output": "1/2" }, { "input": "4 3", "output": "1/2" }, { "input": "4 4", "output": "1/2" }, { "input": "4 5", "output": "1/3" }, { "input": "4 6", "output": "1/6" }, { "input": "5 1", "output": "1/3" }, { "input": "5 2", "output": "1/3" }, { "input": "5 3", "output": "1/3" }, { "input": "5 4", "output": "1/3" }, { "input": "5 5", "output": "1/3" }, { "input": "5 6", "output": "1/6" }, { "input": "6 1", "output": "1/6" }, { "input": "6 2", "output": "1/6" }, { "input": "6 3", "output": "1/6" }, { "input": "6 4", "output": "1/6" }, { "input": "6 5", "output": "1/6" }, { "input": "6 6", "output": "1/6" } ]
1,578,539,405
2,147,483,647
Python 3
OK
TESTS
36
218
0
y, w = (int(n) for n in input().split()) d = 7 - max(y, w) x = str(d) + '/6' if d == 0: x = '0/1' elif d == 2: x = '1/3' elif d == 3: x = '1/2' elif d == 4: x = '2/3' elif d == 6: x = '1/1' print(x)
Title: Die Roll Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. Input Specification: The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. Output Specification: Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». Demo Input: ['4 2\n'] Demo Output: ['1/2\n'] Note: Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
```python y, w = (int(n) for n in input().split()) d = 7 - max(y, w) x = str(d) + '/6' if d == 0: x = '0/1' elif d == 2: x = '1/3' elif d == 3: x = '1/2' elif d == 4: x = '2/3' elif d == 6: x = '1/1' print(x) ```
3.891
1
A
Theatre Square
PROGRAMMING
1,000
[ "math" ]
A. Theatre Square
1
256
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Write the needed number of flagstones.
[ "6 6 4\n" ]
[ "4\n" ]
none
0
[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]
1,694,076,873
2,147,483,647
Python 3
OK
TESTS
20
46
0
import math def flagstones_needed(n, m, a): rows = math.ceil(n / a) cols = math.ceil(m / a) return rows * cols n, m, a = map(int, input().split()) needed_flagstones = flagstones_needed(n, m, a) print(needed_flagstones)
Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none
```python import math def flagstones_needed(n, m, a): rows = math.ceil(n / a) cols = math.ceil(m / a) return rows * cols n, m, a = map(int, input().split()) needed_flagstones = flagstones_needed(n, m, a) print(needed_flagstones) ```
3.977
858
B
Which floor?
PROGRAMMING
1,500
[ "brute force", "implementation" ]
null
null
In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat *n*?
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=100), where *n* is the number of the flat you need to restore floor for, and *m* is the number of flats in Polycarp's memory. *m* lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers *k**i*,<=*f**i* (1<=≤<=*k**i*<=≤<=100, 1<=≤<=*f**i*<=≤<=100), which means that the flat *k**i* is on the *f**i*-th floor. All values *k**i* are distinct. It is guaranteed that the given information is not self-contradictory.
Print the number of the floor in which the *n*-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor.
[ "10 3\n6 2\n2 1\n7 3\n", "8 4\n3 1\n6 2\n5 2\n2 1\n" ]
[ "4\n", "-1\n" ]
In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat.
750
[ { "input": "10 3\n6 2\n2 1\n7 3", "output": "4" }, { "input": "8 4\n3 1\n6 2\n5 2\n2 1", "output": "-1" }, { "input": "8 3\n7 2\n6 2\n1 1", "output": "2" }, { "input": "4 2\n8 3\n3 1", "output": "2" }, { "input": "11 4\n16 4\n11 3\n10 3\n15 4", "output": "3" }, { "input": "16 6\n3 1\n16 4\n10 3\n9 3\n19 5\n8 2", "output": "4" }, { "input": "1 0", "output": "1" }, { "input": "1 1\n1 1", "output": "1" }, { "input": "1 1\n1 1", "output": "1" }, { "input": "1 2\n1 1\n2 2", "output": "1" }, { "input": "2 2\n2 1\n1 1", "output": "1" }, { "input": "2 0", "output": "-1" }, { "input": "2 1\n3 3", "output": "2" }, { "input": "3 2\n1 1\n3 3", "output": "3" }, { "input": "3 3\n1 1\n3 3\n2 2", "output": "3" }, { "input": "3 0", "output": "-1" }, { "input": "1 1\n2 1", "output": "1" }, { "input": "2 2\n2 1\n1 1", "output": "1" }, { "input": "2 3\n3 2\n1 1\n2 1", "output": "1" }, { "input": "3 0", "output": "-1" }, { "input": "3 1\n1 1", "output": "-1" }, { "input": "2 2\n1 1\n3 1", "output": "1" }, { "input": "1 3\n1 1\n2 1\n3 1", "output": "1" }, { "input": "81 0", "output": "-1" }, { "input": "22 1\n73 73", "output": "22" }, { "input": "63 2\n10 10\n64 64", "output": "63" }, { "input": "88 3\n37 37\n15 15\n12 12", "output": "88" }, { "input": "29 4\n66 66\n47 47\n62 62\n2 2", "output": "29" }, { "input": "9 40\n72 72\n47 47\n63 63\n66 66\n21 21\n94 94\n28 28\n45 45\n93 93\n25 25\n100 100\n43 43\n49 49\n9 9\n74 74\n26 26\n42 42\n50 50\n2 2\n92 92\n76 76\n3 3\n78 78\n44 44\n69 69\n36 36\n65 65\n81 81\n13 13\n46 46\n24 24\n96 96\n73 73\n82 82\n68 68\n64 64\n41 41\n31 31\n29 29\n10 10", "output": "9" }, { "input": "50 70\n3 3\n80 80\n23 23\n11 11\n87 87\n7 7\n63 63\n61 61\n67 67\n53 53\n9 9\n43 43\n55 55\n27 27\n5 5\n1 1\n99 99\n65 65\n37 37\n60 60\n32 32\n38 38\n81 81\n2 2\n34 34\n17 17\n82 82\n26 26\n71 71\n4 4\n16 16\n19 19\n39 39\n51 51\n6 6\n49 49\n64 64\n83 83\n10 10\n56 56\n30 30\n76 76\n90 90\n42 42\n47 47\n91 91\n21 21\n52 52\n40 40\n77 77\n35 35\n88 88\n75 75\n95 95\n28 28\n15 15\n69 69\n22 22\n48 48\n66 66\n31 31\n98 98\n73 73\n25 25\n97 97\n18 18\n13 13\n54 54\n72 72\n29 29", "output": "50" }, { "input": "6 0", "output": "-1" }, { "input": "32 1\n9 5", "output": "16" }, { "input": "73 2\n17 9\n21 11", "output": "37" }, { "input": "6 3\n48 24\n51 26\n62 31", "output": "3" }, { "input": "43 4\n82 41\n52 26\n88 44\n41 21", "output": "22" }, { "input": "28 40\n85 43\n19 10\n71 36\n39 20\n57 29\n6 3\n15 8\n11 6\n99 50\n77 39\n79 40\n31 16\n35 18\n24 12\n54 27\n93 47\n90 45\n72 36\n63 32\n22 11\n83 42\n5 3\n12 6\n56 28\n94 47\n25 13\n41 21\n29 15\n36 18\n23 12\n1 1\n84 42\n55 28\n58 29\n9 5\n68 34\n86 43\n3 2\n48 24\n98 49", "output": "14" }, { "input": "81 70\n55 28\n85 43\n58 29\n20 10\n4 2\n47 24\n42 21\n28 14\n26 13\n38 19\n9 5\n83 42\n7 4\n72 36\n18 9\n61 31\n41 21\n64 32\n90 45\n46 23\n67 34\n2 1\n6 3\n27 14\n87 44\n39 20\n11 6\n21 11\n35 18\n48 24\n44 22\n3 2\n71 36\n62 31\n34 17\n16 8\n99 50\n57 29\n13 7\n79 40\n100 50\n53 27\n89 45\n36 18\n43 22\n92 46\n98 49\n75 38\n40 20\n97 49\n37 19\n68 34\n30 15\n96 48\n17 9\n12 6\n45 23\n65 33\n76 38\n84 42\n23 12\n91 46\n52 26\n8 4\n32 16\n77 39\n88 44\n86 43\n70 35\n51 26", "output": "41" }, { "input": "34 0", "output": "-1" }, { "input": "63 1\n94 24", "output": "16" }, { "input": "4 2\n38 10\n48 12", "output": "1" }, { "input": "37 3\n66 17\n89 23\n60 15", "output": "10" }, { "input": "71 4\n15 4\n13 4\n4 1\n70 18", "output": "18" }, { "input": "77 40\n49 13\n66 17\n73 19\n15 4\n36 9\n1 1\n41 11\n91 23\n51 13\n46 12\n39 10\n42 11\n56 14\n61 16\n70 18\n92 23\n65 17\n54 14\n97 25\n8 2\n87 22\n33 9\n28 7\n38 10\n50 13\n26 7\n7 2\n31 8\n84 21\n47 12\n27 7\n53 14\n19 5\n93 24\n29 8\n3 1\n77 20\n62 16\n9 3\n44 11", "output": "20" }, { "input": "18 70\n51 13\n55 14\n12 3\n43 11\n42 11\n95 24\n96 24\n29 8\n65 17\n71 18\n18 5\n62 16\n31 8\n100 25\n4 1\n77 20\n56 14\n24 6\n93 24\n97 25\n79 20\n40 10\n49 13\n86 22\n21 6\n46 12\n6 2\n14 4\n23 6\n20 5\n52 13\n88 22\n39 10\n70 18\n94 24\n13 4\n37 10\n41 11\n91 23\n85 22\n83 21\n89 23\n33 9\n64 16\n67 17\n57 15\n47 12\n36 9\n72 18\n81 21\n76 19\n35 9\n80 20\n34 9\n5 2\n22 6\n84 21\n63 16\n74 19\n90 23\n68 17\n98 25\n87 22\n2 1\n92 23\n50 13\n38 10\n28 7\n8 2\n60 15", "output": "5" }, { "input": "89 0", "output": "-1" }, { "input": "30 1\n3 1", "output": "-1" }, { "input": "63 2\n48 6\n17 3", "output": "8" }, { "input": "96 3\n45 6\n25 4\n35 5", "output": "12" }, { "input": "37 4\n2 1\n29 4\n27 4\n47 6", "output": "5" }, { "input": "64 40\n40 5\n92 12\n23 3\n75 10\n71 9\n2 1\n54 7\n18 3\n9 2\n74 10\n87 11\n11 2\n90 12\n30 4\n48 6\n12 2\n91 12\n60 8\n35 5\n13 2\n53 7\n46 6\n38 5\n59 8\n97 13\n32 4\n6 1\n36 5\n43 6\n83 11\n81 11\n99 13\n69 9\n10 2\n21 3\n78 10\n31 4\n27 4\n57 8\n1 1", "output": "8" }, { "input": "17 70\n63 8\n26 4\n68 9\n30 4\n61 8\n84 11\n39 5\n53 7\n4 1\n81 11\n50 7\n91 12\n59 8\n90 12\n20 3\n21 3\n83 11\n94 12\n37 5\n8 1\n49 7\n34 5\n19 3\n44 6\n74 10\n2 1\n73 10\n88 11\n43 6\n36 5\n57 8\n64 8\n76 10\n40 5\n71 9\n95 12\n15 2\n41 6\n89 12\n42 6\n96 12\n1 1\n52 7\n38 5\n45 6\n78 10\n82 11\n16 2\n48 6\n51 7\n56 7\n28 4\n87 11\n93 12\n46 6\n29 4\n97 13\n54 7\n35 5\n3 1\n79 10\n99 13\n13 2\n55 7\n100 13\n11 2\n75 10\n24 3\n33 5\n22 3", "output": "3" }, { "input": "9 0", "output": "-1" }, { "input": "50 1\n31 2", "output": "-1" }, { "input": "79 2\n11 1\n22 2", "output": "-1" }, { "input": "16 3\n100 7\n94 6\n3 1", "output": "1" }, { "input": "58 4\n73 5\n52 4\n69 5\n3 1", "output": "4" }, { "input": "25 40\n70 5\n28 2\n60 4\n54 4\n33 3\n21 2\n51 4\n20 2\n44 3\n79 5\n65 5\n1 1\n52 4\n23 2\n38 3\n92 6\n63 4\n3 1\n91 6\n5 1\n64 4\n34 3\n25 2\n97 7\n89 6\n61 4\n71 5\n88 6\n29 2\n56 4\n45 3\n6 1\n53 4\n57 4\n90 6\n76 5\n8 1\n46 3\n73 5\n87 6", "output": "2" }, { "input": "78 70\n89 6\n52 4\n87 6\n99 7\n3 1\n25 2\n46 3\n78 5\n35 3\n68 5\n85 6\n23 2\n60 4\n88 6\n17 2\n8 1\n15 1\n67 5\n95 6\n59 4\n94 6\n31 2\n4 1\n16 1\n10 1\n97 7\n42 3\n2 1\n24 2\n34 3\n37 3\n70 5\n18 2\n41 3\n48 3\n58 4\n20 2\n38 3\n72 5\n50 4\n49 4\n40 3\n61 4\n6 1\n45 3\n28 2\n13 1\n27 2\n96 6\n56 4\n91 6\n77 5\n12 1\n11 1\n53 4\n76 5\n74 5\n82 6\n55 4\n80 5\n14 1\n44 3\n7 1\n83 6\n79 5\n92 6\n66 5\n36 3\n73 5\n100 7", "output": "5" }, { "input": "95 0", "output": "-1" }, { "input": "33 1\n30 1", "output": "-1" }, { "input": "62 2\n14 1\n15 1", "output": "-1" }, { "input": "3 3\n6 1\n25 1\n38 2", "output": "1" }, { "input": "44 4\n72 3\n80 3\n15 1\n36 2", "output": "2" }, { "input": "34 40\n25 1\n28 1\n78 3\n5 1\n13 1\n75 3\n15 1\n67 3\n57 2\n23 1\n26 1\n61 2\n22 1\n48 2\n85 3\n24 1\n82 3\n83 3\n53 2\n38 2\n19 1\n33 2\n69 3\n17 1\n79 3\n54 2\n77 3\n97 4\n20 1\n35 2\n14 1\n18 1\n71 3\n21 1\n36 2\n56 2\n44 2\n63 2\n72 3\n32 1", "output": "2" }, { "input": "83 70\n79 3\n49 2\n2 1\n44 2\n38 2\n77 3\n86 3\n31 1\n83 3\n82 3\n35 2\n7 1\n78 3\n23 1\n39 2\n58 2\n1 1\n87 3\n72 3\n20 1\n48 2\n14 1\n13 1\n6 1\n70 3\n55 2\n52 2\n25 1\n11 1\n61 2\n76 3\n95 3\n32 1\n66 3\n29 1\n9 1\n5 1\n3 1\n88 3\n59 2\n96 3\n10 1\n63 2\n40 2\n42 2\n34 2\n43 2\n19 1\n89 3\n94 3\n24 1\n98 4\n12 1\n30 1\n69 3\n17 1\n50 2\n8 1\n93 3\n16 1\n97 4\n54 2\n71 3\n18 1\n33 2\n80 3\n15 1\n99 4\n75 3\n4 1", "output": "3" }, { "input": "2 0", "output": "-1" }, { "input": "36 1\n96 1", "output": "1" }, { "input": "73 2\n34 1\n4 1", "output": "-1" }, { "input": "6 3\n37 1\n22 1\n70 1", "output": "1" }, { "input": "47 4\n66 1\n57 1\n85 1\n47 1", "output": "1" }, { "input": "9 40\n73 1\n21 1\n37 1\n87 1\n33 1\n69 1\n49 1\n19 1\n35 1\n93 1\n71 1\n43 1\n79 1\n85 1\n29 1\n72 1\n76 1\n47 1\n17 1\n67 1\n95 1\n41 1\n54 1\n88 1\n42 1\n80 1\n98 1\n96 1\n10 1\n24 1\n78 1\n18 1\n3 1\n91 1\n2 1\n15 1\n5 1\n60 1\n36 1\n46 1", "output": "1" }, { "input": "63 70\n82 1\n53 1\n57 1\n46 1\n97 1\n19 1\n36 1\n90 1\n23 1\n88 1\n68 1\n45 1\n2 1\n70 1\n86 1\n8 1\n83 1\n40 1\n99 1\n42 1\n32 1\n52 1\n81 1\n50 1\n77 1\n37 1\n54 1\n75 1\n4 1\n49 1\n73 1\n22 1\n21 1\n98 1\n18 1\n51 1\n14 1\n76 1\n92 1\n80 1\n78 1\n33 1\n79 1\n89 1\n67 1\n9 1\n44 1\n60 1\n64 1\n55 1\n29 1\n100 1\n16 1\n87 1\n10 1\n12 1\n25 1\n85 1\n30 1\n63 1\n39 1\n38 1\n31 1\n5 1\n26 1\n91 1\n43 1\n72 1\n48 1\n94 1", "output": "1" }, { "input": "2 0", "output": "-1" } ]
1,505,657,223
3,723
Python 3
WRONG_ANSWER
PRETESTS
2
46
0
n, m = map(int, input().split()) v = [] for i in range(m): v.append(list(map(int, input().split()))) ans = [] for i in range(1, 100): o = True for p in v: if (p[0] - 1) // i != p[1] - 1: o = False break if o: ans.append(i) if len(ans) == 1: print((n - 1) // ans[0] + 1) else: print(-1)
Title: Which floor? Time Limit: None seconds Memory Limit: None megabytes Problem Description: In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat *n*? Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=100), where *n* is the number of the flat you need to restore floor for, and *m* is the number of flats in Polycarp's memory. *m* lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers *k**i*,<=*f**i* (1<=≤<=*k**i*<=≤<=100, 1<=≤<=*f**i*<=≤<=100), which means that the flat *k**i* is on the *f**i*-th floor. All values *k**i* are distinct. It is guaranteed that the given information is not self-contradictory. Output Specification: Print the number of the floor in which the *n*-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Demo Input: ['10 3\n6 2\n2 1\n7 3\n', '8 4\n3 1\n6 2\n5 2\n2 1\n'] Demo Output: ['4\n', '-1\n'] Note: In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat.
```python n, m = map(int, input().split()) v = [] for i in range(m): v.append(list(map(int, input().split()))) ans = [] for i in range(1, 100): o = True for p in v: if (p[0] - 1) // i != p[1] - 1: o = False break if o: ans.append(i) if len(ans) == 1: print((n - 1) // ans[0] + 1) else: print(-1) ```
0
433
B
Kuriyama Mirai's Stones
PROGRAMMING
1,200
[ "dp", "implementation", "sortings" ]
null
null
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions: 1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones. The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
[ "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n", "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n" ]
[ "24\n9\n28\n", "10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n" ]
Please note that the answers to the questions may overflow 32-bit integer type.
1,500
[ { "input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6", "output": "24\n9\n28" }, { "input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2", "output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5" }, { "input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1 1 3\n1 1 3\n2 1 4\n1 1 2", "output": "5\n7\n5\n5\n2\n7\n7\n13\n4" }, { "input": "18\n26 46 56 18 78 88 86 93 13 77 21 84 59 61 5 74 72 52\n25\n1 10 10\n1 9 13\n2 13 17\n1 8 14\n2 2 6\n1 12 16\n2 15 17\n2 3 6\n1 3 13\n2 8 9\n2 17 17\n1 17 17\n2 5 10\n2 1 18\n1 4 16\n1 1 13\n1 1 8\n2 7 11\n2 6 12\n1 5 9\n1 4 5\n2 7 15\n1 8 8\n1 8 14\n1 3 7", "output": "77\n254\n413\n408\n124\n283\n258\n111\n673\n115\n88\n72\n300\n1009\n757\n745\n491\n300\n420\n358\n96\n613\n93\n408\n326" }, { "input": "56\n43 100 44 66 65 11 26 75 96 77 5 15 75 96 11 44 11 97 75 53 33 26 32 33 90 26 68 72 5 44 53 26 33 88 68 25 84 21 25 92 1 84 21 66 94 35 76 51 11 95 67 4 61 3 34 18\n27\n1 20 38\n1 11 46\n2 42 53\n1 8 11\n2 11 42\n2 35 39\n2 37 41\n1 48 51\n1 32 51\n1 36 40\n1 31 56\n1 18 38\n2 9 51\n1 7 48\n1 15 52\n1 27 31\n2 5 19\n2 35 50\n1 31 34\n1 2 7\n2 15 33\n2 46 47\n1 26 28\n2 3 29\n1 23 45\n2 29 55\n1 14 29", "output": "880\n1727\n1026\n253\n1429\n335\n350\n224\n1063\n247\n1236\n1052\n2215\n2128\n1840\n242\n278\n1223\n200\n312\n722\n168\n166\n662\n1151\n2028\n772" }, { "input": "18\n38 93 48 14 69 85 26 47 71 11 57 9 38 65 72 78 52 47\n38\n2 10 12\n1 6 18\n2 2 2\n1 3 15\n2 1 16\n2 5 13\n1 9 17\n1 2 15\n2 5 17\n1 15 15\n2 4 11\n2 3 4\n2 2 5\n2 1 17\n2 6 16\n2 8 16\n2 8 14\n1 9 12\n2 8 13\n2 1 14\n2 5 13\n1 2 3\n1 9 14\n2 12 15\n2 3 3\n2 9 13\n2 4 12\n2 11 14\n2 6 16\n1 8 14\n1 12 15\n2 3 4\n1 3 5\n2 4 14\n1 6 6\n2 7 14\n2 7 18\n1 8 12", "output": "174\n658\n11\n612\n742\n461\n453\n705\n767\n72\n353\n40\n89\n827\n644\n559\n409\n148\n338\n592\n461\n141\n251\n277\n14\n291\n418\n262\n644\n298\n184\n40\n131\n558\n85\n456\n784\n195" }, { "input": "1\n2\n10\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1", "output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2" }, { "input": "2\n1 5\n8\n2 1 2\n1 1 1\n1 1 2\n1 1 1\n2 2 2\n2 1 2\n1 1 1\n1 2 2", "output": "6\n1\n6\n1\n5\n6\n1\n5" }, { "input": "8\n2 6 4 6 8 4 7 7\n9\n2 6 8\n1 2 3\n2 3 4\n2 1 2\n2 4 5\n1 3 8\n2 4 4\n2 2 4\n1 2 4", "output": "22\n10\n10\n6\n12\n36\n6\n14\n16" }, { "input": "4\n1 1 2 8\n9\n1 2 4\n1 2 4\n1 1 2\n2 3 4\n2 1 1\n1 2 3\n2 3 4\n1 1 4\n1 1 4", "output": "11\n11\n2\n10\n1\n3\n10\n12\n12" }, { "input": "5\n8 8 1 10 7\n5\n1 2 2\n1 1 4\n1 3 3\n1 2 2\n1 2 5", "output": "8\n27\n1\n8\n26" }, { "input": "10\n10 3 3 2 1 2 7 1 1 1\n4\n2 10 10\n2 6 6\n1 5 10\n1 6 8", "output": "10\n2\n13\n10" }, { "input": "6\n6 6 3 8 5 4\n3\n2 2 3\n2 1 5\n1 1 5", "output": "9\n24\n28" }, { "input": "1\n1\n2\n1 1 1\n2 1 1", "output": "1\n1" } ]
1,685,551,355
2,147,483,647
Python 3
OK
TESTS
46
857
18,329,600
# Online Python - IDE, Editor, Compiler, Interpreter n=int(input()) sum1=[] sum2=[] orig=input().split(' ') orig=[int(x) for x in orig] sum1.append(orig[0]) for i in range(1,n): sum1.append(sum1[i-1]+orig[i]) orig.sort() sum2.append(orig[0]) for i in range(1,n): sum2.append(sum2[i-1]+orig[i]) q=int(input()) for _ in range(q): query=input().split(' ') query=[int(x) for x in query] l=query[1] -1 r=query[2] -1 if(query[0]==1): if(l==0): print(sum1[r]) else: print(sum1[r]-sum1[l-1]) else: if(l==0): print(sum2[r]) else: print(sum2[r]-sum2[l-1])
Title: Kuriyama Mirai's Stones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions: 1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones. The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. Output Specification: Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. Demo Input: ['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n'] Demo Output: ['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n'] Note: Please note that the answers to the questions may overflow 32-bit integer type.
```python # Online Python - IDE, Editor, Compiler, Interpreter n=int(input()) sum1=[] sum2=[] orig=input().split(' ') orig=[int(x) for x in orig] sum1.append(orig[0]) for i in range(1,n): sum1.append(sum1[i-1]+orig[i]) orig.sort() sum2.append(orig[0]) for i in range(1,n): sum2.append(sum2[i-1]+orig[i]) q=int(input()) for _ in range(q): query=input().split(' ') query=[int(x) for x in query] l=query[1] -1 r=query[2] -1 if(query[0]==1): if(l==0): print(sum1[r]) else: print(sum1[r]-sum1[l-1]) else: if(l==0): print(sum2[r]) else: print(sum2[r]-sum2[l-1]) ```
3
312
A
Whose sentence is it?
PROGRAMMING
1,100
[ "implementation", "strings" ]
null
null
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG&gt;.&lt; I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
[ "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n" ]
[ "Freda's\nOMG&gt;.&lt; I don't know!\nOMG&gt;.&lt; I don't know!\nRainbow's\nOMG&gt;.&lt; I don't know!\n" ]
none
500
[ { "input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .", "output": "Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!" }, { "input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iapelala.\nCFPlbUgObrXLejPNu.F\nZSUfvisiHyrIMjMlala.\nmiao. lala.\nd,IWSeumytrVlala.", "output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's" }, { "input": "10\nmiao.,taUvXPVlala.\nmiao.txEeId.X_lala.\nLZIeAEd JaeBVlala.\ncKPIsWpwIlala.\nfYp.eSvn,g\nKMx,nFEslala.\nmiao.QtMyxYqiajjuM\nDutxNkCqywgcnCYskcd\ngFLKACjeqfD\n,Ss UmY.wJvcX", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nmiao.Plala.\nDVm,VYslala.\nmiao.rlala.\nmiao.,KQNL.fO_.QRc\nUBLCKEUePlala.\nIouS.Alala.\nmiao.lala.\nmiao.rlala.\nEJZwRJeKlala.\nmiao.Olala.", "output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nRainbow's\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!" }, { "input": "10\nmiao.grFTpju.jCLRnZ\ng.pVHYA_Usnm\nlloWONolcMFElala.\nAW,n.JJkOTe.Nd\n.bP.HvKlala.\nGziqPGQa,lala.\nmiao.,QkOCH.vFlala.\n.PUtOwImvUsoeh \nmiao.Z,KIds.R\nmiao.,_MDzoaAiJlala.", "output": "Rainbow's\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!" }, { "input": "10\nmiao.xWfjV\nHFVrGCDQXyZ,Sbm\nLMDS.xVkTCAY.vm\nmiao.lLBglala.\nnl,jRPyClala.\nFYnHoXlala.\nmiao. oxaHE\n.WTrw_mNpOQCa\nHOk..wHYoyMhl\nQX,XpMuPIROM", "output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nJBQqiXlala.\npUNUWQRiMPCXv\nAiLnfNHWznwkC.lala.\nmiao.Dl_Oy\nxJJJkVkdfOzQBH_SmKh\nfgD_IHvdHiorE,W\nmiao.usBKixglala.\nwCpqPUzEtD\nmiao.rlala.\nmiao.JylcGvWlala.", "output": "Freda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nmiao..FLhPl_Wjslala.\nmiao. tdEGtfdJlala.\nGAzEUlala.\nKCcmOa .aKBlZyYsdu.V\nmiao.lala.\njKylnM,FXK\nmiao.GBWqjGH.v\nmiao.RefxS Cni.\nOxaaEihuHQR_s,\nmiao.a,Axtlala.", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nNo.I_aTXlala.\nmiao.JKSCoRZS\nnOBMIlala.\nmiao.nlala.\nmiao._xqxoHIIlala.\nmiao.NJPy SWyiUDWc\nmiao.cCnahFaqqj.Xqp\nnreSMDeXPPYAQxI,W\nAktPajWimdd_qRn\nmiao.QHwKCYlala.", "output": "Freda's\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\n \n,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ \n \nmiao.miao.miao.\nlala.lala.lala.\nlala.miao.\nmiaolala. \nmiao.lala\nmiaolala_\n,.._ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nduClyjMIPsEuWmx_Ce.byVoizYlTM,sF\nuZHsNip_,Mwtg,FZjM_LzPC,_pSvEOyTHfAOvoZXvxCZdgYDTCDdCAoSVZWyxXGcLgWlala.\nEGtJFPAvTEcqjkhaGxdduaQ_rmUzF.WaU, EIuX B,aVzFFpFrxpwADXuayRD azDfj \n_tJqYzXyqc.,u.F,mUYukveBPWnPq,f,dJnPHuBazdnbRHfzwNUdRbheAIjcoaPcnLvocrzcioxCapb R\n.YUBeb_zmwUt.QQuUdQIiOXtqshcsycEe,HLytHlala.\ndJndLqGBHt.GfpN.BgvsbXoLh_DIzAJOtFDmLSCYEztvPcS_GHPxivzV,NPMmSAtfk.Mg.w,A UcCt_lCD.csEzyJJBYtSMkzqiA\nmiao.qlala.\nmiao.FmDlY\nmiao.UQI.aJmnanNvRLskuVaMybDMsOlala.\nmiao.lala.", "output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nmiao.vyscfysAtWcPkpFHdwZqAQ,UPPcjhKQTlala.\nmiao.KESqus DybUuYFoWVpo..LWZh.UqEdUsTHFlKfzqkThAUPklala.\nUNoE vfZIAdxkiWKhsHPfsqRPTNQoHgAxooVLYxRzugHjo jaEHWQFF\nCCmdIwr.UkoiYWK.Z,,ZesMpISTXNgnpYnJaWquCyL,gO\n.JvOayhXK_bgoYbfAtnXg\nbvdSzRrXoGxVgWvdXnsjEnEfxDzIQo_aZVGDGrzwuAMtzVAHioMBx_DHuTxyieGbGuSRNUojOREqxBBxvCgqAOMzwIWT\nMBuaWduZmRaOGyIPzWOsBVeqtDrblAbXxmM_uRfqMvnVlLEuhVKlhidN_aigiXyq,ZEDqQAx\nmiao.wCHVCuVKNePKmIUFLL_lala.\nmiao.iAqstXHUv\n pMO yvPkNtnNwmUCao W,wW.OvIMVaEeVYHmqaniWq.ivlala.", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's" }, { "input": "10\nmiao.\nmiao.jrwLBCpNaDCjyoK.PFzbwWU.h.. wfQquG_P..lala.\nmiao.LGlYdKjw__.Chlala.\nW.wtr qG KDOHj.xWxPbXIXjD_,GJZDaAZ,JBHphsjWJwSKcZAIAi\nmiao.pHsGAZQDWPJQwKC.zHjJituLgp.eUrzObTI.wrpect.FMUJqu,Zuslala.\nmiao.YVlOpXccUA_YU igbsbZbhOVwyYTyOjnWqgiTmxwAuFa.flCHn.,MtVbqxZQl_BGHXWkwijGjuL, ,ezyNlala.\nmiao.xCrVSz.aMv UOSOroDlQxWeBmlWe.FA.ZfUmviMlala.\nxebAlala.\nmiao.qVSxqf vOTlala.\nD.oBUwsLQRgXAoNkQJhQN.w.oMhuvtujnmiwgQYMfjlNTSHh .lSKgI.OEp", "output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nZXXzYlTiQU\nkXE.DdcbOojSaSgjMcFBPubKHefEVAzbi,PDFgSZIz,lala.\nxEfrTCjKhhwBC.UNmJXgTGUdkQeVDlala.\nLfaEw.jvMmuOBWtfoiJNtDIlQAVWNU,xWK_efBBtfkM\nqtBpqKZMWZMX_NKrUAEKYyQcLZWQlqbM\nmiao.PrJEbUtInremuaKRItqXOrfQEjQcAak VQ\nMpGCq awvQaHRvDr uvtVMKsvZI\nmiao.A.RVGu.szCEp.pXQJwL EuTltlN.WradoTvWHJyhcNSoulala.\nmiao.rzlUHzUdxtDRpWRuc,QZwEBfsKKGHMLGtFymPPQdptLFlzZ_ORWqrlfOrlntuDkpXEvz.CxwAsFYUvpnOnFWG\nmiao.VXUoNBwlgBwcna_n.CgAAcKKUuiVA.doOJKHpMdwNwlHAcLpdfN.Awa SthrlEWpUcuOonUTxIQNszYcHDXxnhArrM..A", "output": "OMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's" }, { "input": "10\nmiao.qbxBFzrjtWv.yOk\nDBgi,loApO AACrGnwssCHN\nmiao.LV.wbQEE_V.BSAtdTIHTQOJVJ_nGOthbL,nJvQ.UeWFpsa.GGsK_Uv,HQxHS,AN_bkrolala.\nmiao.tBEqk rIQuByGKhfq_iP.BW,nySZEfrfySEcqnnIzxC,lrjIiivbxlkoVXJFiegGFRn NO,txGPhVBcv.CVhMmNO zlala.\nmiao.aBZWDWxk.wkR ,NyCzGxJnJDqBZpetdUPAmmBZDXl_Tbflala.\nmiao. XN,uMwWm. VqloYr..jTLszlala.\n.rshcgfZ.eZOdMu_RMh\nmiao.ahiwpECEe.lala.\nLeoUSroTekQAMSO__M L_ZEeRD_tUihYvQETFB,RzJmFtFiKrU\nBtygQG_OoFEFBL.KsVWTYbtqtalXoStFCZ RINHda.NuLmlkRB.vAQJFvelbsfoJ.T,M sJn", "output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nYoYBCcaqhXLfvKKf.UYMODTHyPZlala.\ncxgWn J.Q\nmiao.nwH.IHntgKYDhdsjU DMTHXEVRyeJP ZaAecCIBJXuv.YjhEmtbjvjKnK.U,oc,x\nmiao.EcQ.FDtRJgmpAzxhq.RwXBLxjyC,IeMqaFoheMPFCGWBcwUAFnbiwlbz_fcsEGPfJaeryCtFocBNEWTlala.\nmiao.W\nmiao. ZQpIeyCXJSnFgAIzu.THfrmyoogYWQzFqblala.\nmiao.ifzdCwnTDcxpvdr OTC.YqPv.MKDp..utICtAsbfYyGlala.\nmiao.\nmiao.tS.U.wH.s,CxORZJsBAHLi,fXeoDJWVBH\nrcUMpeupOVRKrcIRAvU.rP kgUEfoeXcrFPQOBYG.BNvAQPg.XHMWizhLpZNljXc .LQmVXCi", "output": "Freda's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!" }, { "input": "10\nlala.\nmiao.milalala.lmmialamiao.la.o.iao.a.ao.\nmialala.o.\nmiao.millala.allala.amiao..miao.miao.lala.ao.miammiao.iao.o.\nmiao.miaomiao..\nlalmiao.amiao..\nmiao.lala.lamiamiaolala..o.lalala.miao..\nmlala.iao.lalamiao..\nlmlala.iao.alalamiao.lmialala.lala.miao.o.alala..lala..lalmiaomiao..lalmiao.a.lalamiao..miao.alala..\nlalllamiao.la.lala.alamiao.lalalala.lala..miao.lamiao.la.lallalamiao..a..a.", "output": "Freda's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nlalllala.ala.lala.a.mmimiao.aomiao.lllala.ala.amiao.la.mialalala.la.o..imiao.miao.amlala.iao.o.\nmilala.alllala.ala.amiao.lamiao..o.\nlala.lalalala..lalalala..\nlala.miao.\nmimiao.ao.lala.\nlalmiao.amlala.iamialala.o.o..\nlalammlala.iaolammiao.imiao.ao.la..iao..\nmiao.mialala.omiao..mlala.iaolala..\nmiamiao.o.llallala.ala.la.miao.ala.miao.mimialmiao.ala.o.alala.miaomiao..olala..\nmialala.lamiao.la.lala.miao.ollala.allala.ala.lmiaommiao.imiao.ao.lallallala.a.miao.a..a..", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "10\nlamiao.lamiao.mimiao.ao..\nllala.almiaomiao..lala.miao.a.\nlalala.lala.lalala.lala.lalala..la.\nlalala.la.miao.\nmiao.llalallala.miao.a.la.lala.almimiao.ao.a.\nmiao.mlala.ilala.aomilala.ao..\nmilala.ao.\nmmiao.iao.\nmlala.lala.ialalmiao.alala..olala.lala..\nlala.lmlalalalalala...iao.milala.aommiao.iao..alamiaolallala.milala.ao.a..lalalmiao.a..llala.amiao..", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "4\n \n miao. \n lala. \n ", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" }, { "input": "1\nMiao.", "output": "OMG>.< I don't know!" }, { "input": "1\nm", "output": "OMG>.< I don't know!" }, { "input": "1\nMiAo.sdsdlala.", "output": "Freda's" }, { "input": "1\nLalA.", "output": "OMG>.< I don't know!" }, { "input": "3\n.\nm\nl", "output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!" } ]
1,450,624,792
2,147,483,647
Python 3
OK
TESTS
36
77
0
def lala(s): f1 = 0 f2 = 0 l = len(s) if s[0:5]=="miao.": f1 = 1 if s[len(s)-5:] =="lala.": f2 = 1 if f1==1 and f2==0: return 1 elif f1==0 and f2==1 : return 2 else: return 0 n = int(input()) for i in range(0,n,1): s = input() if lala(s)==1: print("Rainbow's") elif lala(s)==2: print("Freda's") else : print("OMG>.< I don't know!")
Title: Whose sentence is it? Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. Output Specification: For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG&gt;.&lt; I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions. Demo Input: ['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n'] Demo Output: ["Freda's\nOMG&gt;.&lt; I don't know!\nOMG&gt;.&lt; I don't know!\nRainbow's\nOMG&gt;.&lt; I don't know!\n"] Note: none
```python def lala(s): f1 = 0 f2 = 0 l = len(s) if s[0:5]=="miao.": f1 = 1 if s[len(s)-5:] =="lala.": f2 = 1 if f1==1 and f2==0: return 1 elif f1==0 and f2==1 : return 2 else: return 0 n = int(input()) for i in range(0,n,1): s = input() if lala(s)==1: print("Rainbow's") elif lala(s)==2: print("Freda's") else : print("OMG>.< I don't know!") ```
3
1,000
B
Light It Up
PROGRAMMING
1,500
[ "greedy" ]
null
null
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$.
Print the only integer — maximum possible total time when the lamp is lit.
[ "3 10\n4 6 7\n", "2 12\n1 10\n", "2 7\n3 4\n" ]
[ "8\n", "9\n", "6\n" ]
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
0
[ { "input": "3 10\n4 6 7", "output": "8" }, { "input": "2 12\n1 10", "output": "9" }, { "input": "2 7\n3 4", "output": "6" }, { "input": "1 2\n1", "output": "1" }, { "input": "5 10\n1 3 5 6 8", "output": "6" }, { "input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "1 10\n1", "output": "9" }, { "input": "1 10000000\n1", "output": "9999999" }, { "input": "1 8\n1", "output": "7" }, { "input": "7 17\n1 5 9 10 11 14 16", "output": "9" }, { "input": "4 17\n1 5 9 10", "output": "12" }, { "input": "5 12\n1 2 3 4 5", "output": "9" }, { "input": "2 1000000000\n100 111", "output": "999999989" }, { "input": "3 90591\n90579 90580 90581", "output": "90589" }, { "input": "5 16\n1 2 3 4 5", "output": "13" }, { "input": "6 981231233\n1 6 15 30 130 1875", "output": "981229468" }, { "input": "1 3\n1", "output": "2" }, { "input": "3 12\n4 7 11", "output": "8" }, { "input": "1 100000\n1", "output": "99999" }, { "input": "2 5\n1 3", "output": "3" }, { "input": "1 1000000000\n1", "output": "999999999" }, { "input": "1 1000000000\n999999999", "output": "999999999" }, { "input": "3 4\n1 2 3", "output": "2" }, { "input": "3 5\n1 2 3", "output": "3" }, { "input": "1 4\n2", "output": "3" }, { "input": "3 5\n1 3 4", "output": "3" }, { "input": "9 20\n5 9 11 12 14 15 16 17 19", "output": "12" }, { "input": "2 4\n1 2", "output": "3" }, { "input": "1 10\n2", "output": "9" }, { "input": "2 6\n2 4", "output": "4" }, { "input": "2 4\n1 3", "output": "2" }, { "input": "3 6\n1 2 4", "output": "4" }, { "input": "7 9863\n65 96 97 98 101 112 1115", "output": "9819" } ]
1,700,556,913
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
8
78
13,209,600
# -*- coding: utf-8 -*- """ Created on Tue Nov 21 16:33:10 2023 @author: GaoMingze 2300011427 """ n,M=map(int,input().split()) num=[0]+list(map(int,input().split()))+[M] on=[] ans=0 for i in range(1,n+2,2): ans+=num[i]-num[i-1] for i in range(2,n+2,2): s=0 if num[i]-num[i-1]>1: s+=num[i-1]-num[i-2] ans=max(ans,M+s-ans-1) print(ans)
Title: Light It Up Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up. Input Specification: First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$. Output Specification: Print the only integer — maximum possible total time when the lamp is lit. Demo Input: ['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n'] Demo Output: ['8\n', '9\n', '6\n'] Note: In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
```python # -*- coding: utf-8 -*- """ Created on Tue Nov 21 16:33:10 2023 @author: GaoMingze 2300011427 """ n,M=map(int,input().split()) num=[0]+list(map(int,input().split()))+[M] on=[] ans=0 for i in range(1,n+2,2): ans+=num[i]-num[i-1] for i in range(2,n+2,2): s=0 if num[i]-num[i-1]>1: s+=num[i-1]-num[i-2] ans=max(ans,M+s-ans-1) print(ans) ```
0
313
A
Ilya and Bank Account
PROGRAMMING
900
[ "implementation", "number theory" ]
null
null
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money. Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance. Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
In a single line print an integer — the maximum state of the bank account that Ilya can get.
[ "2230\n", "-10\n", "-100003\n" ]
[ "2230\n", "0\n", "-10000\n" ]
In the first test sample Ilya doesn't profit from using the present. In the second test sample you can delete digit 1 and get the state of the account equal to 0.
500
[ { "input": "2230", "output": "2230" }, { "input": "-10", "output": "0" }, { "input": "-100003", "output": "-10000" }, { "input": "544883178", "output": "544883178" }, { "input": "-847251738", "output": "-84725173" }, { "input": "423654797", "output": "423654797" }, { "input": "-623563697", "output": "-62356367" }, { "input": "645894116", "output": "645894116" }, { "input": "-384381709", "output": "-38438170" }, { "input": "437587210", "output": "437587210" }, { "input": "-297534606", "output": "-29753460" }, { "input": "891773002", "output": "891773002" }, { "input": "-56712976", "output": "-5671296" }, { "input": "963662765", "output": "963662765" }, { "input": "-272656295", "output": "-27265625" }, { "input": "383441522", "output": "383441522" }, { "input": "-477665112", "output": "-47766511" }, { "input": "791725034", "output": "791725034" }, { "input": "-812168727", "output": "-81216872" }, { "input": "528894922", "output": "528894922" }, { "input": "-479977172", "output": "-47997712" }, { "input": "568044564", "output": "568044564" }, { "input": "-392784794", "output": "-39278474" }, { "input": "925596634", "output": "925596634" }, { "input": "-836078769", "output": "-83607876" }, { "input": "71036059", "output": "71036059" }, { "input": "-337396162", "output": "-33739612" }, { "input": "87129297", "output": "87129297" }, { "input": "-648171877", "output": "-64817187" }, { "input": "20218400", "output": "20218400" }, { "input": "10", "output": "10" }, { "input": "1000000000", "output": "1000000000" }, { "input": "-1000000000", "output": "-100000000" }, { "input": "-102", "output": "-10" }, { "input": "-120", "output": "-10" }, { "input": "-20", "output": "0" }, { "input": "-15", "output": "-1" }, { "input": "-5575533", "output": "-557553" }, { "input": "-50", "output": "0" }, { "input": "-31", "output": "-1" }, { "input": "-55", "output": "-5" }, { "input": "-81", "output": "-1" }, { "input": "-99", "output": "-9" }, { "input": "-23", "output": "-2" }, { "input": "-41", "output": "-1" }, { "input": "-24", "output": "-2" }, { "input": "46", "output": "46" } ]
1,696,252,871
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
1
92
0
n = int(input()) max_balance = n if n < 0: last_digit_removed = n // 10 digit_before_last_removed = n // 100 * 10 + n % 10 max_balance = max(n, last_digit_removed, digit_before_last_removed) print(max_balance)
Title: Ilya and Bank Account Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money. Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance. Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift. Input Specification: The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account. Output Specification: In a single line print an integer — the maximum state of the bank account that Ilya can get. Demo Input: ['2230\n', '-10\n', '-100003\n'] Demo Output: ['2230\n', '0\n', '-10000\n'] Note: In the first test sample Ilya doesn't profit from using the present. In the second test sample you can delete digit 1 and get the state of the account equal to 0.
```python n = int(input()) max_balance = n if n < 0: last_digit_removed = n // 10 digit_before_last_removed = n // 100 * 10 + n % 10 max_balance = max(n, last_digit_removed, digit_before_last_removed) print(max_balance) ```
0
653
A
Bear and Three Balls
PROGRAMMING
900
[ "brute force", "implementation", "sortings" ]
null
null
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above.
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
[ "4\n18 55 16 17\n", "6\n40 41 43 44 44 44\n", "8\n5 972 3 4 1 4 970 971\n" ]
[ "YES\n", "NO\n", "YES\n" ]
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
500
[ { "input": "4\n18 55 16 17", "output": "YES" }, { "input": "6\n40 41 43 44 44 44", "output": "NO" }, { "input": "8\n5 972 3 4 1 4 970 971", "output": "YES" }, { "input": "3\n959 747 656", "output": "NO" }, { "input": "4\n1 2 2 3", "output": "YES" }, { "input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543", "output": "NO" }, { "input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869", "output": "YES" }, { "input": "3\n408 410 409", "output": "YES" }, { "input": "3\n903 902 904", "output": "YES" }, { "input": "3\n399 400 398", "output": "YES" }, { "input": "3\n450 448 449", "output": "YES" }, { "input": "3\n390 389 388", "output": "YES" }, { "input": "3\n438 439 440", "output": "YES" }, { "input": "11\n488 688 490 94 564 615 641 170 489 517 669", "output": "YES" }, { "input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954", "output": "YES" }, { "input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318", "output": "YES" }, { "input": "6\n10 79 306 334 304 305", "output": "YES" }, { "input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365", "output": "YES" }, { "input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73", "output": "YES" }, { "input": "11\n325 325 324 324 324 325 325 324 324 324 324", "output": "NO" }, { "input": "7\n517 517 518 517 518 518 518", "output": "NO" }, { "input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710", "output": "NO" }, { "input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29", "output": "NO" }, { "input": "7\n880 880 514 536 881 881 879", "output": "YES" }, { "input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375", "output": "YES" }, { "input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404", "output": "YES" }, { "input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987", "output": "YES" }, { "input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116", "output": "NO" }, { "input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995", "output": "NO" }, { "input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22", "output": "YES" }, { "input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952", "output": "YES" }, { "input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911", "output": "NO" }, { "input": "3\n3 1 2", "output": "YES" }, { "input": "3\n500 999 1000", "output": "NO" }, { "input": "10\n101 102 104 105 107 109 110 112 113 115", "output": "NO" }, { "input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "50\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "3\n1000 999 998", "output": "YES" }, { "input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456", "output": "NO" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3", "output": "YES" }, { "input": "3\n999 999 1000", "output": "NO" }, { "input": "9\n2 4 5 13 25 100 200 300 400", "output": "NO" }, { "input": "9\n1 1 1 2 2 2 3 3 3", "output": "YES" }, { "input": "3\n1 1 2", "output": "NO" }, { "input": "3\n998 999 1000", "output": "YES" }, { "input": "12\n1 1 1 1 1 1 1 1 1 2 2 4", "output": "NO" }, { "input": "4\n4 3 4 5", "output": "YES" }, { "input": "6\n1 1 1 2 2 2", "output": "NO" }, { "input": "3\n2 3 2", "output": "NO" }, { "input": "5\n10 5 6 3 2", "output": "NO" }, { "input": "3\n1 2 1", "output": "NO" }, { "input": "3\n1 2 3", "output": "YES" }, { "input": "4\n998 999 1000 1000", "output": "YES" }, { "input": "5\n2 3 9 9 4", "output": "YES" }, { "input": "4\n1 2 4 4", "output": "NO" }, { "input": "3\n1 1 1", "output": "NO" }, { "input": "3\n2 2 3", "output": "NO" }, { "input": "7\n1 2 2 2 4 5 6", "output": "YES" }, { "input": "5\n1 3 10 3 10", "output": "NO" }, { "input": "3\n1 2 2", "output": "NO" }, { "input": "4\n1000 1000 999 998", "output": "YES" }, { "input": "3\n5 3 7", "output": "NO" }, { "input": "6\n1 1 2 2 3 3", "output": "YES" }, { "input": "9\n6 6 6 5 5 5 4 4 4", "output": "YES" }, { "input": "7\n5 6 6 6 7 7 7", "output": "YES" }, { "input": "5\n2 3 3 3 4", "output": "YES" }, { "input": "5\n2 1 2 1 3", "output": "YES" }, { "input": "3\n1 2 7", "output": "NO" }, { "input": "3\n1000 1000 1000", "output": "NO" }, { "input": "5\n1 100 2 100 3", "output": "YES" }, { "input": "5\n5 4 6 5 5", "output": "YES" }, { "input": "12\n1 1 1 1 2 2 2 2 3 3 3 3", "output": "YES" }, { "input": "5\n9 9 1 2 3", "output": "YES" }, { "input": "6\n1 2 3 1 2 3", "output": "YES" }, { "input": "7\n1 1 1 1 2 3 3", "output": "YES" }, { "input": "3\n13 13 13", "output": "NO" }, { "input": "3\n42 42 42", "output": "NO" }, { "input": "8\n1 1 1 1 2 2 2 2", "output": "NO" }, { "input": "6\n1 1 1 1 2 3", "output": "YES" }, { "input": "6\n1 1 2 2 6 6", "output": "NO" }, { "input": "6\n1 2 5 5 5 5", "output": "NO" }, { "input": "9\n1 2 3 1 2 3 1 2 3", "output": "YES" }, { "input": "4\n1 2 1 100", "output": "NO" }, { "input": "5\n1 1 2 2 3", "output": "YES" } ]
1,594,046,181
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
93
6,656,000
n=int(input()) m=sorted(map(int,input().split())) flag=0 for i in range(n-2): if m[i+2]-m[i]==2 and m[i+1]-m[i]==1: flag=1 break else: flag=0 print(['NO','YES'][flag==1])
Title: Bear and Three Balls Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above. Input Specification: The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball. Output Specification: Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes). Demo Input: ['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
```python n=int(input()) m=sorted(map(int,input().split())) flag=0 for i in range(n-2): if m[i+2]-m[i]==2 and m[i+1]-m[i]==1: flag=1 break else: flag=0 print(['NO','YES'][flag==1]) ```
0
721
A
One-dimensional Japanese Crossword
PROGRAMMING
800
[ "implementation" ]
null
null
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew.
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
[ "3\nBBW\n", "5\nBWBWB\n", "4\nWWWW\n", "4\nBBBB\n", "13\nWBBBBWWBWBBBW\n" ]
[ "1\n2 ", "3\n1 1 1 ", "0\n", "1\n4 ", "3\n4 1 3 " ]
The last sample case correspond to the picture in the statement.
500
[ { "input": "3\nBBW", "output": "1\n2 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "4\nWWWW", "output": "0" }, { "input": "4\nBBBB", "output": "1\n4 " }, { "input": "13\nWBBBBWWBWBBBW", "output": "3\n4 1 3 " }, { "input": "1\nB", "output": "1\n1 " }, { "input": "2\nBB", "output": "1\n2 " }, { "input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB", "output": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "1\nW", "output": "0" }, { "input": "2\nWW", "output": "0" }, { "input": "2\nWB", "output": "1\n1 " }, { "input": "2\nBW", "output": "1\n1 " }, { "input": "3\nBBB", "output": "1\n3 " }, { "input": "3\nBWB", "output": "2\n1 1 " }, { "input": "3\nWBB", "output": "1\n2 " }, { "input": "3\nWWB", "output": "1\n1 " }, { "input": "3\nWBW", "output": "1\n1 " }, { "input": "3\nBWW", "output": "1\n1 " }, { "input": "3\nWWW", "output": "0" }, { "input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB", "output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 " }, { "input": "5\nBBBWB", "output": "2\n3 1 " }, { "input": "5\nBWWWB", "output": "2\n1 1 " }, { "input": "5\nWWWWB", "output": "1\n1 " }, { "input": "5\nBWWWW", "output": "1\n1 " }, { "input": "5\nBBBWW", "output": "1\n3 " }, { "input": "5\nWWBBB", "output": "1\n3 " }, { "input": "10\nBBBBBWWBBB", "output": "2\n5 3 " }, { "input": "10\nBBBBWBBWBB", "output": "3\n4 2 2 " }, { "input": "20\nBBBBBWWBWBBWBWWBWBBB", "output": "6\n5 1 2 1 1 3 " }, { "input": "20\nBBBWWWWBBWWWBWBWWBBB", "output": "5\n3 2 1 1 3 " }, { "input": "20\nBBBBBBBBWBBBWBWBWBBB", "output": "5\n8 3 1 1 3 " }, { "input": "20\nBBBWBWBWWWBBWWWWBWBB", "output": "6\n3 1 1 2 1 2 " }, { "input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB", "output": "5\n6 1 1 1 12 " }, { "input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB", "output": "9\n5 1 2 1 1 4 1 1 8 " }, { "input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB", "output": "7\n11 1 1 8 1 1 6 " }, { "input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW", "output": "0" }, { "input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW", "output": "9\n5 1 1 1 1 2 2 1 1 " }, { "input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB", "output": "6\n1 1 1 1 1 5 " }, { "input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB", "output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB", "output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n100 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB", "output": "11\n20 1 1 1 2 1 1 1 1 1 20 " }, { "input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB", "output": "11\n4 1 1 1 1 2 1 1 1 1 6 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "10\nWWBWWWBWBB", "output": "3\n1 1 2 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n17 31 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n24 42 " }, { "input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB", "output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 " }, { "input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB", "output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 " }, { "input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB", "output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 " }, { "input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB", "output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 " }, { "input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB", "output": "11\n1 5 3 4 3 1 1 2 1 2 4 " } ]
1,619,504,107
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
62
0
A = int(input()) B = str(input()) Groupcount = 0 Bcount = 0 Bgroup = "" for i in range(A): if B[i] == 'B': if i == 0 or B[i-1] == 'W': Groupcount = Groupcount + 1 Bcount = Bcount +1 elif B[i] == 'W': Bgroup += str(Bcount) + " " Bcount = 0 if B[-1] != 'W': Bgroup += str(Bcount) + "" print(Groupcount) print(Bgroup)
Title: One-dimensional Japanese Crossword Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). Output Specification: The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. Demo Input: ['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n'] Demo Output: ['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 '] Note: The last sample case correspond to the picture in the statement.
```python A = int(input()) B = str(input()) Groupcount = 0 Bcount = 0 Bgroup = "" for i in range(A): if B[i] == 'B': if i == 0 or B[i-1] == 'W': Groupcount = Groupcount + 1 Bcount = Bcount +1 elif B[i] == 'W': Bgroup += str(Bcount) + " " Bcount = 0 if B[-1] != 'W': Bgroup += str(Bcount) + "" print(Groupcount) print(Bgroup) ```
0
200
B
Drinks
PROGRAMMING
800
[ "implementation", "math" ]
null
null
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent. One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has. Find the volume fraction of orange juice in the final drink.
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
[ "3\n50 50 100\n", "4\n0 25 50 75\n" ]
[ "66.666666666667\n", "37.500000000000\n" ]
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
500
[ { "input": "3\n50 50 100", "output": "66.666666666667" }, { "input": "4\n0 25 50 75", "output": "37.500000000000" }, { "input": "3\n0 1 8", "output": "3.000000000000" }, { "input": "5\n96 89 93 95 70", "output": "88.600000000000" }, { "input": "7\n62 41 78 4 38 39 75", "output": "48.142857142857" }, { "input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22", "output": "11.615384615385" }, { "input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18", "output": "12.761904761905" }, { "input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84", "output": "69.538461538462" }, { "input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94", "output": "91.551724137931" }, { "input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100", "output": "99.515151515152" }, { "input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2", "output": "8.147058823529" }, { "input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93", "output": "91.921052631579" }, { "input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100", "output": "97.019230769231" }, { "input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6", "output": "50.965517241379" }, { "input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10", "output": "7.505882352941" }, { "input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6", "output": "10.418918918919" }, { "input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41", "output": "20.928571428571" }, { "input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96", "output": "49.191919191919" }, { "input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92", "output": "73.484848484848" }, { "input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6", "output": "7.640000000000" }, { "input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100", "output": "77.660000000000" }, { "input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81", "output": "91.480000000000" }, { "input": "1\n0", "output": "0.000000000000" }, { "input": "1\n100", "output": "100.000000000000" }, { "input": "1\n78", "output": "78.000000000000" }, { "input": "2\n0 100", "output": "50.000000000000" }, { "input": "2\n100 100", "output": "100.000000000000" }, { "input": "5\n0 0 0 0 1", "output": "0.200000000000" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99", "output": "99.990000000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1", "output": "0.010000000000" }, { "input": "5\n100 100 100 100 100", "output": "100.000000000000" } ]
1,683,519,327
2,147,483,647
Python 3
OK
TESTS
31
92
0
n = int(input()) percentages = list(map(int, input().split())) total_percent = sum(percentages) avg_percent = total_percent / n print(avg_percent)
Title: Drinks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent. One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has. Find the volume fraction of orange juice in the final drink. Input Specification: The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. Output Specification: Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. Demo Input: ['3\n50 50 100\n', '4\n0 25 50 75\n'] Demo Output: ['66.666666666667\n', '37.500000000000\n'] Note: Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
```python n = int(input()) percentages = list(map(int, input().split())) total_percent = sum(percentages) avg_percent = total_percent / n print(avg_percent) ```
3
862
B
Mahmoud and Ehab and the bipartiteness
PROGRAMMING
1,300
[ "dfs and similar", "graphs", "trees" ]
null
null
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees. A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below. Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add? A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105). The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree. It's guaranteed that the given graph is a tree.
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
[ "3\n1 2\n1 3\n", "5\n1 2\n2 3\n3 4\n4 5\n" ]
[ "0\n", "2\n" ]
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)) Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph) In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0. In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).
1,000
[ { "input": "3\n1 2\n1 3", "output": "0" }, { "input": "5\n1 2\n2 3\n3 4\n4 5", "output": "2" }, { "input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6", "output": "16" }, { "input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4", "output": "16" }, { "input": "10\n2 6\n3 7\n8 4\n4 10\n6 9\n9 7\n3 10\n1 2\n5 8", "output": "16" }, { "input": "10\n6 9\n9 7\n9 4\n10 9\n9 1\n9 8\n9 2\n9 5\n3 9", "output": "0" }, { "input": "2\n1 2", "output": "0" } ]
1,506,783,544
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
16
561
10,956,800
n = int(input()) c = [0, 0] g = [[] for _ in range(n)] def col(i, p, cl): c[cl] += 1 for j in range(len(g[i])): if g[i][j] != p: col(g[i][j], i, abs(cl - 1)) for i in range(n - 1): inp = input().split(' ') f = int(inp[0]) - 1 t = int(inp[1]) - 1 g[f].append(t) g[t].append(f) col(0, -1, 1) print(c[0] * c[1] - (n - 1))
Title: Mahmoud and Ehab and the bipartiteness Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees. A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below. Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add? A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same . Input Specification: The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105). The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree. It's guaranteed that the given graph is a tree. Output Specification: Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions. Demo Input: ['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n'] Demo Output: ['0\n', '2\n'] Note: Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)) Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph) In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0. In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).
```python n = int(input()) c = [0, 0] g = [[] for _ in range(n)] def col(i, p, cl): c[cl] += 1 for j in range(len(g[i])): if g[i][j] != p: col(g[i][j], i, abs(cl - 1)) for i in range(n - 1): inp = input().split(' ') f = int(inp[0]) - 1 t = int(inp[1]) - 1 g[f].append(t) g[t].append(f) col(0, -1, 1) print(c[0] * c[1] - (n - 1)) ```
-1
991
D
Bishwock
PROGRAMMING
1,500
[ "dp", "greedy" ]
null
null
Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns.
The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$.
Output a single integer — the maximum amount of bishwocks that can be placed onto the given board.
[ "00\n00\n", "00X00X0XXX0\n0XXX0X00X00\n", "0X0X0\n0X0X0\n", "0XXX0\n00000\n" ]
[ "1", "4", "0", "2" ]
none
1,500
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"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX00XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "0" }, { "input": "0000000000000000000000000000000000000000000000000000000000X0000000000000000000000000000000000000X000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "66" }, { "input": "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000XX\n000000000000000000000000000000000X00000000000000000X000000000000000000000000000000000000000000000000", "output": "65" }, { "input": "0000X00X000000X0000X00X00X0000000000X0000000X000X00000X0X000XXX00000000XX0XX000000000000X00000000000\n000000000XX000000X00000X00X00X00000000000000000X0X000XX0000000000000X0X00X0000X0000X000000X0000000XX", "output": "49" }, { "input": "0000000000000000000000000000000000X0000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "65" }, { "input": "00000000000000000000000000X000000000000000000000000000000000000000000X00000X0000X000000000000000000\n000X0000000000X000000000000000000000X0000000000X0X0000000000000000000X00000000000000000000000000000", "output": "62" }, { "input": "000X00XX0XX0X00X0XX0XXXX00XXX0X00000000XXX0XXXXXXX0X00X00XX00X0XXX00000XXXX0XX00X00XXX00X0X0XXXX000\nXXXXX000X0XXX000XXXXX0XX0000XX0XXX0XXX000XXX00XXXXX00X00XX0000X0XX0XX0XX000X0XX00X00XX00XX00X00XX0X", "output": "16" }, { "input": "X0X0XXXX0XXXXXXXXXX00XXXXXXXXXXXXXXXXXX0XXXXX0XX0X00X0XXXXXXX0X0XXXXXXXXX0X0X0XX0XX0000XXXX00XXX0XX\nXX0XXXXXXX0X0X00XXXX0X000X0XXXXXX0XXX0X0XXXX0XXXXXXXXXXXXXX00XX00XXX0XXXXXXXXXX00XXXX0XXXX0XXXXXXXX", "output": "4" }, { "input": "000\n000", "output": "2" }, { "input": "000000000000000000000\n000000000000000000000", "output": "14" }, { "input": "00000000000000000000000000000000000000\n00000000000000000000000000000000000000", "output": "25" }, { "input": "0000\n00X0", "output": "2" }, { "input": "000\n00X", "output": "1" }, { "input": "X000\nX000", "output": "2" }, { "input": "X0000X000XX00000000000000000000XX000X0000X00X0X00XX000000000000X0000X0X0XX00\n00X00XX00000000X0000000X000X000X0X0X00X000X0X0X0000X0000000X000XX000XX000X00", "output": "33" }, { "input": "0X00\n0000", "output": "2" }, { "input": "0X0\n000", "output": "1" }, { "input": "000000\n000000", "output": "4" }, { "input": "X00X0\n00000", "output": "2" }, { "input": "000\n0X0", "output": "1" }, { "input": "X000\n00X0", "output": "2" }, { "input": "X0000\n00000", "output": "3" }, { "input": "X000X\nX000X", "output": "2" }, { "input": "X0000X\n00000X", "output": "3" }, { "input": "000000000000\n000000000000", "output": "8" }, { "input": "00000\n0000X", "output": "3" } ]
1,675,207,609
2,147,483,647
PyPy 3-64
OK
TESTS
72
62
0
a = input() b = input() v2, oa, ob, o2 = 0,0,0,0 if a[0] == b[0] == '0': v2, oa, ob, o2 = 0,-1,-1,-1 elif a[0] == '0': v2, oa, ob, o2 = -1,-1,0,-1 elif b[0] == '0': v2, oa, ob, o2 = -1,0,-1,-1 else: v2, oa, ob, o2 = -1,-1,-1,0 for i in range(1, len(a)): nv2 = max(v2, oa, ob, o2) if a[i] == 'X' or b[i] == 'X': nv2 = -1 no2 = -1 if a[i] == '0' == b[i]: no2 = max(v2, oa, ob) + 1 elif a[i] == '0' or b[i] == '0': if v2 > -1: no2 = v2 + 1 else: no2 = max(v2, oa, ob, o2) noa = -1 if a[i] == b[i] == '0': if v2 > -1: noa = v2 + 1 elif a[i] == '0': noa = -1 elif b[i] == '0': noa = max(v2, oa, ob, o2) nob = -1 if a[i] == b[i] == '0': if v2 > -1: nob = v2 + 1 elif b[i] == '0': nob = -1 elif a[i] == '0': nob = max(v2, oa, ob, o2) v2, oa, ob, o2 = nv2, noa, nob, no2 print(max(v2, oa, ob, o2, 0))
Title: Bishwock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns. Input Specification: The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$. Output Specification: Output a single integer — the maximum amount of bishwocks that can be placed onto the given board. Demo Input: ['00\n00\n', '00X00X0XXX0\n0XXX0X00X00\n', '0X0X0\n0X0X0\n', '0XXX0\n00000\n'] Demo Output: ['1', '4', '0', '2'] Note: none
```python a = input() b = input() v2, oa, ob, o2 = 0,0,0,0 if a[0] == b[0] == '0': v2, oa, ob, o2 = 0,-1,-1,-1 elif a[0] == '0': v2, oa, ob, o2 = -1,-1,0,-1 elif b[0] == '0': v2, oa, ob, o2 = -1,0,-1,-1 else: v2, oa, ob, o2 = -1,-1,-1,0 for i in range(1, len(a)): nv2 = max(v2, oa, ob, o2) if a[i] == 'X' or b[i] == 'X': nv2 = -1 no2 = -1 if a[i] == '0' == b[i]: no2 = max(v2, oa, ob) + 1 elif a[i] == '0' or b[i] == '0': if v2 > -1: no2 = v2 + 1 else: no2 = max(v2, oa, ob, o2) noa = -1 if a[i] == b[i] == '0': if v2 > -1: noa = v2 + 1 elif a[i] == '0': noa = -1 elif b[i] == '0': noa = max(v2, oa, ob, o2) nob = -1 if a[i] == b[i] == '0': if v2 > -1: nob = v2 + 1 elif b[i] == '0': nob = -1 elif a[i] == '0': nob = max(v2, oa, ob, o2) v2, oa, ob, o2 = nv2, noa, nob, no2 print(max(v2, oa, ob, o2, 0)) ```
3
401
C
Team
PROGRAMMING
1,400
[ "constructive algorithms", "greedy", "implementation" ]
null
null
Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork. For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that: - there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one. Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=106) — the number of cards containing number 0; *m* (1<=≤<=*m*<=≤<=106) — the number of cards containing number 1.
In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.
[ "1 2\n", "4 8\n", "4 10\n", "1 5\n" ]
[ "101\n", "110110110101\n", "11011011011011\n", "-1\n" ]
none
1,500
[ { "input": "1 2", "output": "101" }, { "input": "4 8", "output": "110110110101" }, { "input": "4 10", "output": "11011011011011" }, { "input": "1 5", "output": "-1" }, { "input": "3 4", "output": "1010101" }, { "input": "3 10", "output": "-1" }, { "input": "74 99", "output": "11011011011011011011011011011011011011011011011011011011011011011011011010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101" }, { "input": "19 30", "output": "1101101101101101101101101101101010101010101010101" }, { "input": "33 77", "output": "-1" }, { "input": "3830 6966", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "1000000 1000000", "output": "1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..." }, { "input": "1027 2030", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "4610 4609", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "3342 3339", "output": "-1" }, { "input": "7757 7755", "output": "-1" }, { "input": "10 8", "output": "-1" }, { "input": "4247 8495", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "7101 14204", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "9801 19605", "output": "-1" }, { "input": "4025 6858", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "7129 13245", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "8826 12432", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "6322 9256", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "8097 14682", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "6196 6197", "output": "1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..." }, { "input": "1709 2902", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "455 512", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101..." }, { "input": "1781 1272", "output": "-1" }, { "input": "3383 5670", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "954 1788", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "9481 15554", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "9079 100096", "output": "-1" }, { "input": "481533 676709", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "423472 564888", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "227774 373297", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "42346 51898", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "739107 1000000", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "455043 798612", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "801460 801459", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "303498 503791", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "518822 597833", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "32342 64687", "output": "-1" }, { "input": "873192 873189", "output": "-1" }, { "input": "384870 450227", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "201106 208474", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "775338 980888", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "263338 393171", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "241043 330384", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "307203 614408", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "379310 417986", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "661101 785111", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "284634 319008", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "500000 1000000", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "499999 1000000", "output": "1101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101101..." }, { "input": "3 1", "output": "-1" }, { "input": "14124 242112", "output": "-1" }, { "input": "2 1", "output": "010" }, { "input": "2 3", "output": "10101" }, { "input": "1 4", "output": "11011" }, { "input": "2 6", "output": "11011011" }, { "input": "2 5", "output": "1101101" }, { "input": "2 4", "output": "110101" }, { "input": "2 3", "output": "10101" }, { "input": "2 2", "output": "1010" }, { "input": "2 1", "output": "010" }, { "input": "1 1", "output": "10" }, { "input": "1 3", "output": "1101" }, { "input": "3 2", "output": "01010" }, { "input": "5 4", "output": "010101010" }, { "input": "4 3", "output": "0101010" }, { "input": "11 10", "output": "010101010101010101010" }, { "input": "8 7", "output": "010101010101010" }, { "input": "4 2", "output": "-1" }, { "input": "6 5", "output": "01010101010" }, { "input": "3 7", "output": "1101101101" }, { "input": "5 1", "output": "-1" }, { "input": "10 1", "output": "-1" }, { "input": "4 9", "output": "1101101101101" }, { "input": "6 4", "output": "-1" }, { "input": "12 10", "output": "-1" }, { "input": "4 100", "output": "-1" } ]
1,589,354,080
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
0
187
22,220,800
from math import * import numpy as np def arrange(n, m): div = ceil(m/(n+1)) if div >= 3 or (m < n and n >= 2): print(-1) return num1 = m num0 = n ans = np.asarray([]) while num0 > 0: div = ceil(num1/(num0+1)) ans = np.append(ans,'1'*div) num1 -= div ans = np.append(ans,'0') num0 -= 1 ans = np.append(ans,'1'*num1) res = '' for ele in ans: res += ele print(res) def main(): n, m = list(map(int, input().split())) arrange(n, m) if __name__ == "__main__": main()
Title: Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork. For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that: - there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one. Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way. Input Specification: The first line contains two integers: *n* (1<=≤<=*n*<=≤<=106) — the number of cards containing number 0; *m* (1<=≤<=*m*<=≤<=106) — the number of cards containing number 1. Output Specification: In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1. Demo Input: ['1 2\n', '4 8\n', '4 10\n', '1 5\n'] Demo Output: ['101\n', '110110110101\n', '11011011011011\n', '-1\n'] Note: none
```python from math import * import numpy as np def arrange(n, m): div = ceil(m/(n+1)) if div >= 3 or (m < n and n >= 2): print(-1) return num1 = m num0 = n ans = np.asarray([]) while num0 > 0: div = ceil(num1/(num0+1)) ans = np.append(ans,'1'*div) num1 -= div ans = np.append(ans,'0') num0 -= 1 ans = np.append(ans,'1'*num1) res = '' for ele in ans: res += ele print(res) def main(): n, m = list(map(int, input().split())) arrange(n, m) if __name__ == "__main__": main() ```
-1
20
C
Dijkstra?
PROGRAMMING
1,900
[ "graphs", "shortest paths" ]
C. Dijkstra?
1
64
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge. It is possible that the graph has loops and multiple edges between pair of vertices.
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
[ "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n" ]
[ "1 4 3 5 ", "1 4 3 5 " ]
none
1,500
[ { "input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "output": "1 4 3 5 " }, { "input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "output": "1 4 3 5 " }, { "input": "2 1\n1 2 1", "output": "1 2 " }, { "input": "3 1\n1 2 1", "output": "-1" }, { "input": "3 3\n1 2 1\n1 3 2\n2 3 1", "output": "1 3 " }, { "input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173", "output": "1 5 8 7 3 10 " }, { "input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157", "output": "1 8 10 " }, { "input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135", "output": "1 9 5 8 10 " }, { "input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242", "output": "1 4 6 10 " } ]
1,619,319,864
2,147,483,647
PyPy 3
TIME_LIMIT_EXCEEDED
TESTS
19
1,000
5,324,800
def solver(n,m,graph): d = [float("inf")] * n used = [False] * n d[0] = 0 parent = [-1] * n for i in range(n): v = -1 for j in range(n): if not used[j] and (v == -1 or d[j] < d[v]): v =j used[v] = True for to,w in graph[v]: if d[v] + w < d[to]: d[to] = d[v] + w parent[to] = v if d[n-1] == float("inf"): print(-1) return else: u = n-1 ans = [] while u != -1: ans.append(u+1) u = parent[u] ans.reverse() print(*ans) n,m = [int(x) for x in input().split()] graph = [[] for _ in range(n)] for _ in range(m): a,b,w = [int(x) for x in input().split()] graph[a-1].append((b-1,w)) graph[b-1].append((a-1,w)) solver(n,m,graph)
Title: Dijkstra? Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. Input Specification: The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge. It is possible that the graph has loops and multiple edges between pair of vertices. Output Specification: Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. Demo Input: ['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n'] Demo Output: ['1 4 3 5 ', '1 4 3 5 '] Note: none
```python def solver(n,m,graph): d = [float("inf")] * n used = [False] * n d[0] = 0 parent = [-1] * n for i in range(n): v = -1 for j in range(n): if not used[j] and (v == -1 or d[j] < d[v]): v =j used[v] = True for to,w in graph[v]: if d[v] + w < d[to]: d[to] = d[v] + w parent[to] = v if d[n-1] == float("inf"): print(-1) return else: u = n-1 ans = [] while u != -1: ans.append(u+1) u = parent[u] ans.reverse() print(*ans) n,m = [int(x) for x in input().split()] graph = [[] for _ in range(n)] for _ in range(m): a,b,w = [int(x) for x in input().split()] graph[a-1].append((b-1,w)) graph[b-1].append((a-1,w)) solver(n,m,graph) ```
0
731
A
Night at the Museum
PROGRAMMING
800
[ "implementation", "strings" ]
null
null
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition. Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture: After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'. Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
[ "zeus\n", "map\n", "ares\n" ]
[ "18\n", "35\n", "34\n" ]
To print the string from the first sample it would be optimal to perform the following sequence of rotations: 1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
500
[ { "input": "zeus", "output": "18" }, { "input": "map", "output": "35" }, { "input": "ares", "output": "34" }, { "input": "l", "output": "11" }, { "input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv", "output": "99" }, { "input": "gngvi", "output": "44" }, { "input": "aaaaa", "output": "0" }, { "input": "a", "output": "0" }, { "input": "z", "output": "1" }, { "input": "vyadeehhikklnoqrs", "output": "28" }, { "input": "jjiihhhhgggfedcccbazyxx", "output": "21" }, { "input": "fyyptqqxuciqvwdewyppjdzur", "output": "117" }, { "input": "fqcnzmzmbobmancqcoalzmanaobpdse", "output": "368" }, { "input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza", "output": "8" }, { "input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy", "output": "644" }, { "input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss", "output": "8" }, { "input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl", "output": "421" }, { "input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa", "output": "84" }, { "input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco", "output": "666" }, { "input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww", "output": "22" }, { "input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh", "output": "643" }, { "input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib", "output": "245" }, { "input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro", "output": "468" }, { "input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned", "output": "523" }, { "input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna", "output": "130" }, { "input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh", "output": "163" }, { "input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb", "output": "155" }, { "input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp", "output": "57" }, { "input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs", "output": "1236" }, { "input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx", "output": "49" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "0" }, { "input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt", "output": "331" }, { "input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte", "output": "692" }, { "input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh", "output": "1293" }, { "input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg", "output": "16" }, { "input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple", "output": "616" }, { "input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl", "output": "605" }, { "input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud", "output": "549" }, { "input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore", "output": "688" }, { "input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc", "output": "604" }, { "input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa", "output": "572" }, { "input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp", "output": "609" }, { "input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl", "output": "223" }, { "input": "aa", "output": "0" } ]
1,684,203,632
2,147,483,647
PyPy 3-64
OK
TESTS
44
124
9,625,600
import math from sys import stdin from collections import Counter, defaultdict, deque, namedtuple from bisect import bisect_right, bisect_left from typing import List, DefaultDict from itertools import permutations def readarray(typ): return list(map(typ, stdin.readline().split())) def readint(): return int(input()) def getRight(char: str, embosser: List[str], index: int) -> int: distance = 0 while embosser[index] != char: index += 1 distance += 1 if index == len(embosser): index = 0 return distance def getLeft(char: str, embosser: List[str], index: int) -> int: distance = 0 while embosser[index] != char: index -= 1 distance += 1 if index == -1: index = len(embosser)-1 return distance embosser = [char for char in "abcdefghijklmnopqrstuvwxyz"] s = input() index = 0 ans = 0 for char in s: rightDistance = getRight(char, embosser, index) leftDistance = getLeft(char, embosser, index) index = embosser.index(char) ans += min(rightDistance, leftDistance) print(ans)
Title: Night at the Museum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition. Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture: After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'. Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. Input Specification: The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. Output Specification: Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. Demo Input: ['zeus\n', 'map\n', 'ares\n'] Demo Output: ['18\n', '35\n', '34\n'] Note: To print the string from the first sample it would be optimal to perform the following sequence of rotations: 1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
```python import math from sys import stdin from collections import Counter, defaultdict, deque, namedtuple from bisect import bisect_right, bisect_left from typing import List, DefaultDict from itertools import permutations def readarray(typ): return list(map(typ, stdin.readline().split())) def readint(): return int(input()) def getRight(char: str, embosser: List[str], index: int) -> int: distance = 0 while embosser[index] != char: index += 1 distance += 1 if index == len(embosser): index = 0 return distance def getLeft(char: str, embosser: List[str], index: int) -> int: distance = 0 while embosser[index] != char: index -= 1 distance += 1 if index == -1: index = len(embosser)-1 return distance embosser = [char for char in "abcdefghijklmnopqrstuvwxyz"] s = input() index = 0 ans = 0 for char in s: rightDistance = getRight(char, embosser, index) leftDistance = getLeft(char, embosser, index) index = embosser.index(char) ans += min(rightDistance, leftDistance) print(ans) ```
3
55
C
Pie or die
PROGRAMMING
1,900
[ "games" ]
C. Pie or die
2
256
Volodya and Vlad play the following game. There are *k* pies at the cells of *n*<=<=×<=<=*m* board. Each turn Volodya moves one pie to the neighbouring (by side) cell. If the pie lies at the border of the board then Volodya can move it outside the board, get the pie and win. After Volodya's move, Vlad bans some edge at the border of the board of length 1 (between two knots of the board) so that Volodya is not able to move the pie outside the board through this edge anymore. The question is: will Volodya win this game? We suppose both players follow the optimal strategy.
First line contains 3 integers, separated by space: 1<=≤<=*n*,<=*m*<=≤<=100 — dimensions of the board and 0<=≤<=*k*<=≤<=100 — the number of pies. Each of the next *k* lines contains 2 integers, separated by space: 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* — coordinates of the corresponding pie. There could be more than one pie at a cell.
Output only one word: "YES" — if Volodya wins, "NO" — otherwise.
[ "2 2 1\n1 2\n", "3 4 0\n", "100 50 2\n50 25\n50 25\n" ]
[ "YES", "NO", "NO" ]
none
1,500
[ { "input": "2 2 1\n1 2", "output": "YES" }, { "input": "3 4 0", "output": "NO" }, { "input": "100 50 2\n50 25\n50 25", "output": "NO" }, { "input": "20 20 4\n10 10\n10 10\n10 10\n10 10", "output": "NO" }, { "input": "15 15 1\n8 8", "output": "NO" }, { "input": "8 8 2\n4 4\n5 5", "output": "YES" }, { "input": "100 100 2\n50 96\n51 96", "output": "YES" }, { "input": "100 100 2\n50 95\n51 95", "output": "NO" }, { "input": "20 20 1\n16 10", "output": "YES" }, { "input": "20 20 4\n15 9\n15 10\n15 11\n15 12", "output": "NO" }, { "input": "11 11 1\n6 6", "output": "NO" }, { "input": "11 11 1\n6 5", "output": "YES" }, { "input": "35 13 20\n13 8\n19 8\n24 7\n20 6\n23 7\n23 6\n30 7\n29 7\n7 7\n6 8\n9 8\n29 6\n20 7\n25 6\n19 6\n23 8\n26 6\n12 6\n15 7\n6 8", "output": "NO" }, { "input": "50 17 27\n17 8\n19 6\n25 8\n30 10\n22 10\n30 9\n25 8\n27 6\n19 7\n29 11\n39 8\n31 8\n39 8\n40 7\n11 8\n30 11\n32 8\n31 11\n36 12\n10 11\n32 8\n8 7\n7 12\n17 11\n27 7\n8 8\n23 12", "output": "NO" }, { "input": "24 29 22\n16 6\n14 22\n7 15\n11 17\n12 22\n10 13\n12 22\n12 13\n6 16\n12 21\n11 11\n9 13\n18 22\n7 20\n13 6\n6 14\n17 10\n9 13\n7 23\n14 11\n7 22\n8 12", "output": "NO" }, { "input": "32 45 3\n12 30\n27 9\n14 27", "output": "NO" }, { "input": "35 15 63\n6 6\n14 9\n7 6\n25 6\n25 8\n13 9\n18 7\n20 8\n30 10\n25 10\n7 7\n18 8\n11 10\n12 6\n8 8\n6 9\n21 9\n27 10\n28 8\n28 9\n7 9\n28 9\n10 10\n29 10\n25 8\n28 7\n22 6\n13 9\n14 7\n23 9\n20 8\n28 10\n22 7\n12 8\n13 7\n27 9\n17 8\n10 8\n19 10\n6 10\n26 6\n19 8\n28 9\n15 9\n14 7\n25 10\n17 8\n21 8\n29 6\n7 6\n16 10\n7 10\n25 7\n9 9\n30 9\n23 8\n28 8\n7 10\n12 6\n20 9\n24 8\n6 6\n26 7", "output": "NO" }, { "input": "41 50 37\n21 24\n20 32\n10 12\n35 7\n8 19\n30 22\n21 11\n35 12\n7 8\n16 10\n13 39\n6 43\n31 12\n16 14\n25 32\n27 21\n6 34\n22 26\n7 41\n18 13\n24 19\n9 44\n36 21\n17 16\n36 24\n6 31\n19 20\n12 19\n27 36\n6 31\n11 13\n19 9\n20 12\n25 25\n18 27\n17 36\n8 16", "output": "NO" }, { "input": "96 95 31\n14 23\n70 47\n11 77\n53 66\n63 87\n3 14\n57 44\n65 69\n80 74\n49 6\n57 86\n75 8\n2 32\n75 21\n14 51\n56 46\n77 6\n17 89\n87 3\n21 18\n70 67\n47 64\n13 47\n61 33\n56 30\n28 2\n65 18\n17 90\n44 77\n54 26\n32 70", "output": "YES" }, { "input": "80 51 47\n67 41\n74 7\n68 41\n6 2\n19 38\n37 28\n65 4\n6 25\n39 11\n19 34\n47 36\n62 26\n27 44\n70 45\n24 33\n41 2\n13 10\n3 17\n78 35\n53 46\n62 47\n33 17\n17 49\n2 3\n47 38\n72 35\n4 8\n32 21\n52 43\n67 12\n28 22\n53 34\n36 11\n45 45\n32 12\n5 11\n6 3\n55 21\n73 4\n55 21\n36 13\n48 18\n19 8\n70 24\n43 45\n59 50\n58 7", "output": "YES" }, { "input": "25 92 38\n21 36\n20 18\n9 29\n18 77\n10 58\n10 39\n5 3\n21 51\n11 78\n16 32\n24 71\n15 17\n23 23\n25 59\n18 57\n11 2\n16 35\n1 47\n20 59\n19 54\n11 55\n4 33\n15 41\n17 18\n16 67\n4 15\n5 23\n3 24\n20 70\n5 87\n11 1\n23 66\n21 83\n2 32\n17 22\n2 26\n16 42\n24 15", "output": "YES" }, { "input": "67 41 68\n35 16\n66 14\n1 15\n43 6\n26 17\n30 13\n42 11\n32 20\n66 14\n15 35\n35 6\n12 11\n25 9\n39 37\n31 14\n52 11\n4 32\n17 14\n32 1\n58 31\n30 20\n7 23\n13 3\n27 25\n60 27\n56 39\n60 39\n11 5\n33 14\n29 12\n13 34\n30 16\n25 16\n64 25\n47 6\n33 36\n14 40\n19 38\n57 34\n67 8\n10 13\n7 36\n22 24\n6 33\n23 40\n13 19\n65 6\n14 37\n37 21\n27 12\n41 36\n60 15\n27 11\n23 33\n67 40\n45 39\n1 41\n50 21\n28 38\n20 24\n41 34\n43 35\n51 5\n59 37\n27 4\n28 17\n63 20\n1 9", "output": "YES" }, { "input": "14 95 49\n11 48\n9 12\n1 18\n7 54\n11 20\n9 82\n12 1\n12 84\n1 13\n2 13\n12 57\n13 15\n12 18\n9 47\n13 14\n10 14\n13 94\n7 46\n14 14\n6 46\n7 95\n9 29\n13 15\n6 76\n8 60\n6 27\n9 63\n5 39\n5 70\n10 59\n5 75\n3 19\n9 32\n13 59\n5 13\n4 5\n13 80\n10 62\n13 65\n5 25\n4 81\n7 12\n10 94\n8 55\n7 61\n11 58\n7 77\n12 14\n12 47", "output": "YES" }, { "input": "15 96 22\n4 7\n7 40\n13 30\n8 53\n6 78\n5 9\n15 35\n3 13\n5 31\n2 9\n13 50\n11 17\n4 2\n10 91\n11 74\n14 49\n8 30\n10 66\n12 44\n6 19\n9 62\n15 50", "output": "YES" }, { "input": "19 19 50\n11 16\n4 11\n5 12\n19 19\n7 16\n15 10\n8 17\n8 1\n11 10\n5 19\n5 14\n17 6\n12 15\n18 17\n17 14\n10 5\n15 11\n8 8\n5 8\n18 18\n7 11\n8 4\n11 9\n6 16\n1 15\n19 13\n5 12\n10 10\n4 19\n12 4\n8 14\n19 9\n7 1\n19 11\n15 8\n4 19\n19 9\n6 7\n15 7\n2 16\n12 9\n3 18\n17 10\n3 5\n11 7\n12 6\n4 15\n19 4\n17 15\n3 10", "output": "YES" }, { "input": "93 40 43\n14 15\n58 9\n72 15\n40 40\n46 20\n17 26\n31 26\n91 36\n24 28\n32 27\n51 10\n2 35\n73 7\n6 33\n59 21\n59 39\n33 8\n22 21\n77 20\n30 38\n76 35\n40 6\n48 31\n67 29\n30 24\n6 16\n39 27\n24 29\n14 16\n5 25\n76 14\n61 25\n85 13\n60 9\n80 7\n49 19\n35 20\n90 31\n57 40\n67 27\n3 27\n21 16\n21 38", "output": "YES" }, { "input": "70 50 62\n31 22\n41 21\n31 47\n2 46\n22 8\n6 4\n45 32\n40 29\n10 11\n62 40\n70 26\n48 25\n13 44\n53 22\n3 8\n41 19\n13 8\n21 41\n66 20\n34 34\n41 48\n9 35\n23 26\n29 30\n39 27\n58 11\n35 2\n67 3\n59 23\n41 10\n54 9\n10 18\n23 44\n5 2\n37 30\n31 24\n2 21\n2 36\n34 5\n59 44\n7 4\n23 22\n47 27\n14 50\n54 50\n6 4\n64 1\n29 5\n5 37\n60 50\n58 45\n70 4\n4 46\n68 43\n62 34\n15 12\n16 2\n70 21\n59 8\n13 27\n25 41\n13 20", "output": "YES" }, { "input": "61 96 15\n27 36\n19 64\n27 53\n59 63\n48 56\n55 30\n10 23\n6 79\n32 74\n7 51\n29 65\n60 16\n43 74\n40 80\n14 31", "output": "YES" }, { "input": "87 50 62\n34 31\n42 21\n2 23\n20 25\n57 39\n46 26\n59 46\n29 33\n32 35\n79 41\n54 19\n65 7\n41 6\n40 23\n8 41\n2 31\n56 5\n37 33\n63 23\n79 4\n85 27\n53 38\n58 21\n16 11\n15 46\n33 39\n38 6\n27 41\n6 15\n25 47\n58 16\n28 50\n43 38\n48 20\n5 48\n31 6\n8 18\n40 10\n32 29\n44 20\n42 46\n63 21\n18 10\n28 49\n66 26\n64 28\n73 23\n16 29\n48 12\n23 21\n84 14\n10 45\n75 37\n80 3\n75 24\n31 25\n8 42\n67 22\n80 45\n8 31\n16 28\n49 34", "output": "YES" }, { "input": "23 100 53\n16 63\n16 31\n8 31\n4 86\n8 43\n8 27\n21 6\n13 49\n11 54\n5 86\n1 41\n19 14\n2 98\n15 76\n6 25\n6 57\n2 45\n6 98\n10 27\n16 74\n22 72\n22 13\n22 20\n15 63\n18 17\n14 32\n14 32\n2 28\n7 46\n23 16\n20 64\n18 17\n3 69\n22 77\n2 98\n11 20\n22 17\n21 8\n19 77\n19 13\n18 25\n9 24\n18 83\n19 27\n7 37\n16 19\n9 60\n11 70\n3 30\n4 84\n9 54\n22 33\n3 22", "output": "YES" }, { "input": "36 89 27\n21 66\n3 60\n11 32\n10 81\n30 31\n27 62\n11 81\n24 41\n30 6\n13 45\n34 86\n26 46\n9 62\n8 86\n17 56\n4 86\n25 36\n23 72\n18 55\n18 87\n22 67\n18 12\n19 75\n21 60\n16 49\n33 63\n26 12", "output": "YES" }, { "input": "93 93 50\n7 5\n73 91\n66 55\n12 24\n82 46\n38 49\n86 72\n51 69\n17 73\n9 85\n86 69\n65 2\n40 88\n92 26\n45 80\n74 45\n4 55\n57 93\n80 70\n49 69\n29 46\n67 38\n46 12\n16 87\n62 3\n79 62\n29 45\n58 30\n48 4\n76 73\n14 68\n31 8\n49 85\n73 78\n18 7\n87 56\n82 54\n52 73\n29 71\n87 74\n75 84\n45 28\n47 57\n44 53\n21 5\n86 5\n57 51\n45 9\n93 8\n82 43", "output": "YES" }, { "input": "11 38 21\n2 21\n2 28\n7 19\n9 18\n7 25\n8 4\n3 23\n2 32\n5 34\n10 36\n8 21\n4 6\n6 6\n4 35\n8 34\n10 18\n11 4\n10 2\n10 13\n4 37\n2 29", "output": "YES" }, { "input": "26 11 59\n13 6\n18 6\n12 6\n18 6\n21 6\n19 6\n12 6\n7 6\n6 6\n16 6\n7 6\n9 6\n19 6\n19 6\n15 6\n16 6\n16 6\n18 6\n17 6\n8 6\n13 6\n18 6\n11 6\n21 6\n9 6\n19 6\n20 6\n8 6\n20 6\n14 6\n11 6\n18 6\n7 6\n16 6\n19 6\n6 6\n6 6\n7 6\n13 6\n9 6\n16 6\n9 6\n15 6\n12 6\n17 6\n16 6\n9 6\n11 6\n10 6\n16 6\n14 6\n15 6\n7 6\n20 6\n7 6\n8 6\n17 6\n14 6\n14 6", "output": "NO" }, { "input": "30 84 35\n20 60\n23 21\n14 24\n24 72\n13 76\n25 35\n11 64\n15 57\n9 55\n14 66\n10 24\n13 68\n11 8\n19 43\n11 14\n16 26\n11 22\n10 26\n15 66\n17 65\n21 34\n7 61\n24 64\n18 16\n22 18\n12 9\n10 40\n8 24\n16 52\n10 9\n7 17\n21 78\n18 75\n10 45\n16 29", "output": "NO" }, { "input": "100 77 53\n62 72\n23 51\n42 8\n66 33\n62 16\n28 53\n72 54\n71 34\n30 26\n91 28\n27 37\n81 47\n22 40\n42 23\n92 46\n36 37\n86 70\n62 22\n20 9\n46 36\n86 67\n46 61\n33 30\n68 49\n44 57\n34 7\n89 36\n48 39\n47 62\n76 56\n22 41\n7 52\n16 8\n70 50\n52 27\n27 17\n44 30\n66 44\n62 10\n95 37\n94 39\n91 68\n12 49\n85 55\n63 28\n64 15\n75 31\n93 26\n53 51\n53 55\n66 65\n38 36\n40 15", "output": "NO" }, { "input": "66 94 26\n11 75\n46 72\n55 74\n34 10\n33 84\n25 11\n13 23\n27 73\n45 22\n54 34\n53 63\n28 8\n57 46\n26 78\n52 46\n32 38\n22 55\n17 71\n56 18\n9 60\n31 54\n6 84\n59 57\n60 81\n51 49\n41 77", "output": "NO" }, { "input": "68 100 18\n17 85\n10 77\n59 55\n29 46\n25 74\n55 11\n37 16\n57 61\n26 11\n11 88\n19 18\n28 38\n32 12\n36 49\n32 6\n57 45\n30 6\n59 95", "output": "NO" }, { "input": "28 61 4\n12 18\n21 31\n14 52\n6 36", "output": "NO" }, { "input": "11 73 1\n4 67", "output": "YES" }, { "input": "11 79 0", "output": "NO" }, { "input": "11 23 1\n11 9", "output": "YES" }, { "input": "25 11 0", "output": "NO" }, { "input": "39 11 1\n18 3", "output": "YES" }, { "input": "69 11 0", "output": "NO" }, { "input": "18 15 45\n6 7\n7 14\n12 3\n17 1\n15 3\n7 11\n9 3\n7 11\n15 4\n8 1\n12 2\n17 7\n14 15\n2 9\n12 4\n14 9\n18 8\n2 2\n17 1\n7 9\n2 4\n16 1\n12 7\n17 10\n4 1\n18 13\n10 13\n9 12\n14 1\n1 6\n3 10\n6 2\n15 3\n4 8\n14 6\n5 14\n8 11\n8 13\n6 7\n16 9\n2 7\n17 14\n17 11\n7 9\n15 8", "output": "YES" }, { "input": "16 18 70\n14 17\n16 8\n14 1\n7 1\n5 3\n7 5\n15 15\n15 2\n8 17\n12 12\n8 7\n10 16\n16 6\n14 7\n2 7\n12 4\n1 9\n6 9\n1 10\n10 13\n7 11\n2 2\n9 5\n3 10\n14 7\n4 5\n2 7\n7 16\n5 7\n7 14\n14 6\n10 16\n8 1\n4 14\n3 15\n8 11\n3 16\n12 1\n10 12\n13 3\n14 17\n5 5\n6 8\n13 10\n11 13\n3 5\n15 7\n10 3\n6 12\n13 15\n7 5\n3 8\n7 18\n6 7\n15 1\n9 6\n6 17\n11 2\n2 17\n7 16\n6 6\n2 18\n2 10\n5 16\n7 17\n3 8\n15 2\n11 11\n5 13\n16 1", "output": "YES" }, { "input": "14 20 68\n6 7\n2 15\n4 6\n10 18\n6 9\n14 14\n5 18\n9 15\n5 15\n2 9\n9 13\n10 17\n4 2\n12 12\n6 19\n7 13\n10 11\n1 1\n3 16\n7 6\n8 16\n10 17\n1 13\n12 11\n13 13\n2 20\n14 12\n11 18\n10 8\n12 4\n13 7\n13 11\n1 1\n10 6\n14 17\n1 2\n11 5\n6 12\n13 2\n4 3\n8 19\n12 8\n8 7\n5 1\n2 10\n11 10\n12 19\n2 10\n8 4\n12 13\n3 15\n8 8\n5 9\n14 15\n5 19\n7 7\n1 16\n6 12\n11 18\n5 13\n1 12\n10 14\n4 5\n2 8\n3 20\n14 7\n6 3\n4 18", "output": "YES" }, { "input": "19 13 83\n5 2\n12 11\n5 6\n3 11\n17 8\n10 8\n3 10\n9 10\n16 3\n15 12\n14 2\n11 8\n18 6\n15 10\n11 12\n2 1\n15 3\n16 3\n1 7\n15 7\n2 9\n11 13\n18 9\n4 7\n13 4\n7 4\n3 1\n14 8\n4 5\n5 7\n8 3\n17 2\n18 2\n16 3\n10 12\n6 2\n3 6\n5 2\n10 3\n18 9\n14 3\n3 6\n6 5\n12 8\n7 12\n2 11\n6 6\n18 6\n14 4\n3 10\n3 2\n13 3\n12 9\n2 10\n15 6\n1 5\n9 12\n6 12\n4 6\n18 3\n7 2\n9 13\n3 10\n19 13\n6 7\n5 1\n4 10\n12 13\n8 12\n15 1\n4 3\n3 8\n4 8\n3 7\n4 13\n8 7\n7 13\n2 8\n14 6\n12 1\n16 8\n9 4\n5 8", "output": "YES" }, { "input": "13 19 1\n6 10", "output": "NO" }, { "input": "14 17 0", "output": "NO" }, { "input": "20 19 5\n7 14\n14 12\n7 12\n15 9\n12 6", "output": "NO" }, { "input": "17 15 3\n10 7\n12 6\n8 6", "output": "NO" }, { "input": "14 17 4\n9 9\n8 7\n8 12\n7 9", "output": "NO" }, { "input": "15 11 0", "output": "NO" }, { "input": "14 16 4\n6 11\n6 8\n8 6\n6 7", "output": "NO" }, { "input": "16 16 0", "output": "NO" }, { "input": "19 20 2\n10 14\n8 11", "output": "NO" }, { "input": "13 15 1\n7 10", "output": "NO" }, { "input": "11 100 4\n6 10\n6 20\n6 30\n6 80", "output": "NO" }, { "input": "100 11 2\n40 6\n70 6", "output": "NO" }, { "input": "100 11 5\n20 6\n30 6\n43 7\n78 6\n89 6", "output": "YES" }, { "input": "20 20 5\n10 6\n6 8\n16 11\n11 11\n7 15", "output": "YES" }, { "input": "30 30 5\n7 15\n24 11\n15 15\n8 24\n9 6", "output": "NO" } ]
1,679,894,487
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
#include<bits/stdc++.h> int main(){int n,m,k,x,y;scanf("%d%d%d",&n,&m,&k);while(k--){scanf("%d%d",&x,&y);if(x<6||x>n-5||y<6||y>m-5)return 0&puts("YES");}puts("NO");}
Title: Pie or die Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Volodya and Vlad play the following game. There are *k* pies at the cells of *n*<=<=×<=<=*m* board. Each turn Volodya moves one pie to the neighbouring (by side) cell. If the pie lies at the border of the board then Volodya can move it outside the board, get the pie and win. After Volodya's move, Vlad bans some edge at the border of the board of length 1 (between two knots of the board) so that Volodya is not able to move the pie outside the board through this edge anymore. The question is: will Volodya win this game? We suppose both players follow the optimal strategy. Input Specification: First line contains 3 integers, separated by space: 1<=≤<=*n*,<=*m*<=≤<=100 — dimensions of the board and 0<=≤<=*k*<=≤<=100 — the number of pies. Each of the next *k* lines contains 2 integers, separated by space: 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* — coordinates of the corresponding pie. There could be more than one pie at a cell. Output Specification: Output only one word: "YES" — if Volodya wins, "NO" — otherwise. Demo Input: ['2 2 1\n1 2\n', '3 4 0\n', '100 50 2\n50 25\n50 25\n'] Demo Output: ['YES', 'NO', 'NO'] Note: none
```python #include<bits/stdc++.h> int main(){int n,m,k,x,y;scanf("%d%d%d",&n,&m,&k);while(k--){scanf("%d%d",&x,&y);if(x<6||x>n-5||y<6||y>m-5)return 0&puts("YES");}puts("NO");} ```
-1
254
A
Cards with Numbers
PROGRAMMING
1,200
[ "constructive algorithms", "sortings" ]
null
null
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs. Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
[ "3\n20 30 10 30 20 10\n", "1\n1 2\n" ]
[ "4 2\n1 5\n6 3\n", "-1" ]
none
500
[ { "input": "3\n20 30 10 30 20 10", "output": "4 2\n1 5\n6 3" }, { "input": "1\n1 2", "output": "-1" }, { "input": "5\n2 2 2 2 2 1 2 2 1 2", "output": "2 1\n3 4\n7 5\n6 9\n10 8" }, { "input": "5\n2 1 2 2 1 1 1 1 1 2", "output": "3 1\n2 5\n7 6\n8 9\n10 4" }, { "input": "5\n1 2 2 2 1 2 2 1 2 1", "output": "3 2\n1 5\n6 4\n7 9\n10 8" }, { "input": "5\n3 3 1 1 1 3 2 3 1 2", "output": "2 1\n3 4\n8 6\n5 9\n10 7" }, { "input": "5\n1 1 3 1 3 3 3 1 1 1", "output": "2 1\n3 5\n7 6\n4 8\n10 9" }, { "input": "5\n3 1 1 1 2 3 3 3 2 1", "output": "3 2\n1 6\n8 7\n5 9\n10 4" }, { "input": "5\n3 3 2 2 3 3 1 3 1 3", "output": "2 1\n3 4\n6 5\n7 9\n10 8" }, { "input": "5\n4 1 3 1 4 1 2 2 3 1", "output": "4 2\n1 5\n8 7\n3 9\n10 6" }, { "input": "100\n8 6 7 8 7 9 1 7 3 3 5 8 7 8 5 4 8 4 8 1 2 8 3 7 8 7 6 5 7 9 6 10 7 6 7 8 6 8 9 5 1 5 6 1 4 8 4 8 7 2 6 2 6 6 2 8 2 8 7 1 5 4 4 6 4 9 7 5 1 8 1 3 9 2 3 2 4 7 6 10 5 3 4 10 8 9 6 7 2 7 10 1 8 10 4 1 1 1 2 7 5 4 9 10 6 8 3 1 10 9 9 6 1 5 8 6 6 3 3 4 10 10 8 9 7 10 9 3 7 6 3 2 10 8 5 8 5 5 5 10 8 5 7 6 10 7 7 9 10 10 9 9 3 6 5 6 8 1 9 8 2 4 8 8 6 8 10 2 3 5 2 6 8 4 8 6 4 5 10 8 1 10 5 2 5 6 8 2 6 8 1 3 4 5 7 5 6 9 2 8", "output": "4 1\n3 5\n10 9\n8 13\n14 12\n11 15\n18 16\n17 19\n20 7\n22 25\n26 24\n2 27\n30 6\n29 33\n34 31\n36 38\n40 28\n37 43\n44 41\n45 47\n48 46\n35 49\n50 21\n51 53\n55 52\n56 58\n61 42\n62 63\n64 54\n39 66\n67 59\n60 69\n72 23\n57 74\n77 65\n32 80\n81 68\n75 82\n85 70\n73 86\n87 79\n78 88\n89 76\n84 91\n92 71\n83 95\n97 96\n90 100\n104 94\n93 106\n108 98\n103 110\n112 105\n101 114\n117 116\n107 118\n120 102\n109 121\n123 115\n111 124\n126 122\n119 128\n129 125\n99 132\n136 134\n135 137\n139 138\n133 140\n144 130..." }, { "input": "100\n7 3 8 8 1 9 6 6 3 3 8 2 7 9 9 10 2 10 4 4 9 3 6 5 2 6 3 6 3 5 2 3 8 2 5 10 3 9 7 2 1 6 7 4 8 3 9 10 9 4 3 3 7 1 4 2 2 5 6 6 1 7 9 1 8 1 2 2 5 9 7 7 6 4 6 10 1 1 8 1 5 6 4 9 5 4 4 10 6 4 5 1 9 1 7 8 6 10 3 2 4 7 10 4 8 10 6 7 8 4 1 3 8 3 2 1 9 4 2 4 3 1 6 8 6 2 2 5 6 8 6 10 1 6 4 2 7 3 6 10 6 5 6 6 3 9 4 6 4 1 5 4 4 2 8 4 10 3 7 6 6 10 2 5 5 6 1 6 1 9 9 1 10 5 10 1 1 5 7 5 2 1 4 2 3 3 3 5 1 8 10 3 3 5 9 6 3 6 8 1", "output": "4 3\n7 8\n9 2\n1 13\n14 6\n12 17\n18 16\n19 20\n21 15\n10 22\n26 23\n27 29\n30 24\n25 31\n33 11\n32 37\n40 34\n5 41\n42 28\n39 43\n47 38\n36 48\n50 44\n46 51\n57 56\n35 58\n60 59\n54 61\n62 53\n49 63\n65 45\n64 66\n68 67\n71 72\n74 55\n73 75\n78 77\n69 81\n84 70\n83 86\n88 76\n82 89\n90 87\n85 91\n92 80\n79 96\n99 52\n95 102\n103 98\n101 104\n107 97\n105 109\n111 94\n112 114\n115 100\n93 117\n118 110\n116 122\n124 113\n123 125\n126 119\n129 131\n132 106\n120 135\n136 127\n108 137\n138 121\n134 139\n142 128..." }, { "input": "100\n6 3 6 8 8 4 3 7 10 3 1 3 9 5 10 10 6 7 6 6 2 3 8 8 7 6 4 9 6 7 4 4 10 4 7 3 2 7 10 8 6 7 9 1 3 5 3 7 9 1 1 7 1 1 7 7 8 3 2 7 4 8 7 8 10 3 1 7 2 7 9 8 8 8 5 2 8 1 2 7 8 7 8 8 5 10 10 4 9 10 8 7 8 8 7 7 3 6 4 3 4 8 10 8 6 3 7 1 8 6 3 3 7 10 3 9 3 5 10 9 9 2 8 7 2 3 2 1 10 9 6 2 8 7 2 2 5 3 10 6 7 2 1 1 5 10 7 5 4 9 7 7 8 1 1 3 3 7 10 5 9 8 6 8 2 2 1 7 8 9 6 2 2 6 2 9 10 2 10 9 6 3 3 10 6 5 3 6 6 3 6 10 8 7 4 8 6 3 4 7", "output": "-1" }, { "input": "1\n2 2", "output": "2 1" }, { "input": "2\n1 2 4 7", "output": "-1" } ]
1,453,202,167
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
30
409,600
# n = int(input()) # w, h = map(int, input().split()) # to_zero_based = lambda x: int(x) - 1 from collections import defaultdict def main(): with open('input.txt') as fin: n = int(fin.readline()) nums = list(map(int, fin.readline().split())) distincts = defaultdict(list) for i, num in enumerate(nums): count = distincts[num] count.append(i + 1) if len(count) > 2: print(-1) return if len(distincts) != n: print(-1) return s = '' for i1, i2 in distincts.values(): s += i1 + ' ' + i2 with open('output.txt', 'w') as fout: fout.write(s) main()
Title: Cards with Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces. Output Specification: If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs. Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them. Demo Input: ['3\n20 30 10 30 20 10\n', '1\n1 2\n'] Demo Output: ['4 2\n1 5\n6 3\n', '-1'] Note: none
```python # n = int(input()) # w, h = map(int, input().split()) # to_zero_based = lambda x: int(x) - 1 from collections import defaultdict def main(): with open('input.txt') as fin: n = int(fin.readline()) nums = list(map(int, fin.readline().split())) distincts = defaultdict(list) for i, num in enumerate(nums): count = distincts[num] count.append(i + 1) if len(count) > 2: print(-1) return if len(distincts) != n: print(-1) return s = '' for i1, i2 in distincts.values(): s += i1 + ' ' + i2 with open('output.txt', 'w') as fout: fout.write(s) main() ```
-1
980
B
Marlin
PROGRAMMING
1,600
[ "constructive algorithms" ]
null
null
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$. The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells. A person can move from one cell to another if those cells are not occupied by hotels and share a side. Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO". If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
[ "7 2\n", "5 3\n" ]
[ "YES\n.......\n.#.....\n.#.....\n.......\n", "YES\n.....\n.###.\n.....\n.....\n" ]
none
1,000
[ { "input": "7 2", "output": "YES\n.......\n.#.....\n.#.....\n......." }, { "input": "5 3", "output": "YES\n.....\n.###.\n.....\n....." }, { "input": "3 2", "output": "YES\n...\n.#.\n.#.\n..." }, { "input": "3 0", "output": "YES\n...\n...\n...\n..." }, { "input": "49 1", "output": "YES\n.................................................\n........................#........................\n.................................................\n................................................." }, { "input": "9 4", "output": "YES\n.........\n.##......\n.##......\n........." }, { "input": "9 5", "output": "YES\n.........\n.#.#.....\n.###.....\n........." }, { "input": "99 193", "output": "YES\n...................................................................................................\n.###############################################################################################.#.\n.#################################################################################################.\n..................................................................................................." }, { "input": "99 14", "output": "YES\n...................................................................................................\n.#######...........................................................................................\n.#######...........................................................................................\n..................................................................................................." }, { "input": "57 15", "output": "YES\n.........................................................\n.######.#................................................\n.########................................................\n........................................................." }, { "input": "99 3", "output": "YES\n...................................................................................................\n................................................###................................................\n...................................................................................................\n..................................................................................................." }, { "input": "3 1", "output": "YES\n...\n.#.\n...\n..." }, { "input": "9 9", "output": "YES\n.........\n.###.#...\n.#####...\n........." }, { "input": "67 9", "output": "YES\n...................................................................\n.###.#.............................................................\n.#####.............................................................\n..................................................................." }, { "input": "99 99", "output": "YES\n...................................................................................................\n.################################################.#................................................\n.##################################################................................................\n..................................................................................................." }, { "input": "31 32", "output": "YES\n...............................\n.################..............\n.################..............\n..............................." }, { "input": "5 1", "output": "YES\n.....\n..#..\n.....\n....." }, { "input": "5 2", "output": "YES\n.....\n.#...\n.#...\n....." }, { "input": "5 4", "output": "YES\n.....\n.##..\n.##..\n....." }, { "input": "5 6", "output": "YES\n.....\n.###.\n.###.\n....." }, { "input": "5 5", "output": "YES\n.....\n.#.#.\n.###.\n....." }, { "input": "7 9", "output": "YES\n.......\n.###.#.\n.#####.\n......." }, { "input": "7 10", "output": "YES\n.......\n.#####.\n.#####.\n......." }, { "input": "19 12", "output": "YES\n...................\n.######............\n.######............\n..................." }, { "input": "19 3", "output": "YES\n...................\n........###........\n...................\n..................." }, { "input": "37 14", "output": "YES\n.....................................\n.#######.............................\n.#######.............................\n....................................." }, { "input": "37 15", "output": "YES\n.....................................\n.######.#............................\n.########............................\n....................................." }, { "input": "37 37", "output": "YES\n.....................................\n.#################.#.................\n.###################.................\n....................................." }, { "input": "37 36", "output": "YES\n.....................................\n.##################..................\n.##################..................\n....................................." }, { "input": "37 35", "output": "YES\n.....................................\n.################.#..................\n.##################..................\n....................................." }, { "input": "37 34", "output": "YES\n.....................................\n.#################...................\n.#################...................\n....................................." }, { "input": "37 38", "output": "YES\n.....................................\n.###################.................\n.###################.................\n....................................." }, { "input": "37 39", "output": "YES\n.....................................\n.##################.#................\n.####################................\n....................................." }, { "input": "37 40", "output": "YES\n.....................................\n.####################................\n.####################................\n....................................." }, { "input": "5 0", "output": "YES\n.....\n.....\n.....\n....." }, { "input": "67 1", "output": "YES\n...................................................................\n.................................#.................................\n...................................................................\n..................................................................." }, { "input": "37 19", "output": "YES\n.....................................\n.########.#..........................\n.##########..........................\n....................................." }, { "input": "77 7", "output": "YES\n.............................................................................\n.##.#........................................................................\n.####........................................................................\n............................................................................." }, { "input": "33 47", "output": "YES\n.................................\n.######################.#........\n.########################........\n................................." }, { "input": "33 48", "output": "YES\n.................................\n.########################........\n.########################........\n................................." }, { "input": "23 40", "output": "YES\n.......................\n.####################..\n.####################..\n......................." }, { "input": "23 39", "output": "YES\n.......................\n.##################.#..\n.####################..\n......................." }, { "input": "49 3", "output": "YES\n.................................................\n.......................###.......................\n.................................................\n................................................." }, { "input": "99 1", "output": "YES\n...................................................................................................\n.................................................#.................................................\n...................................................................................................\n..................................................................................................." }, { "input": "77 0", "output": "YES\n.............................................................................\n.............................................................................\n.............................................................................\n............................................................................." }, { "input": "99 0", "output": "YES\n...................................................................................................\n...................................................................................................\n...................................................................................................\n..................................................................................................." }, { "input": "99 5", "output": "YES\n...................................................................................................\n.#.#...............................................................................................\n.###...............................................................................................\n..................................................................................................." }, { "input": "99 4", "output": "YES\n...................................................................................................\n.##................................................................................................\n.##................................................................................................\n..................................................................................................." }, { "input": "99 20", "output": "YES\n...................................................................................................\n.##########........................................................................................\n.##########........................................................................................\n..................................................................................................." }, { "input": "99 194", "output": "YES\n...................................................................................................\n.#################################################################################################.\n.#################################################################################################.\n..................................................................................................." }, { "input": "99 192", "output": "YES\n...................................................................................................\n.################################################################################################..\n.################################################################################################..\n..................................................................................................." }, { "input": "99 190", "output": "YES\n...................................................................................................\n.###############################################################################################...\n.###############################################################################################...\n..................................................................................................." }, { "input": "99 189", "output": "YES\n...................................................................................................\n.#############################################################################################.#...\n.###############################################################################################...\n..................................................................................................." }, { "input": "99 177", "output": "YES\n...................................................................................................\n.#######################################################################################.#.........\n.#########################################################################################.........\n..................................................................................................." }, { "input": "99 154", "output": "YES\n...................................................................................................\n.#############################################################################.....................\n.#############################################################################.....................\n..................................................................................................." }, { "input": "99 127", "output": "YES\n...................................................................................................\n.##############################################################.#..................................\n.################################################################..................................\n..................................................................................................." }, { "input": "99 55", "output": "YES\n...................................................................................................\n.##########################.#......................................................................\n.############################......................................................................\n..................................................................................................." }, { "input": "99 40", "output": "YES\n...................................................................................................\n.####################..............................................................................\n.####################..............................................................................\n..................................................................................................." }, { "input": "97 190", "output": "YES\n.................................................................................................\n.###############################################################################################.\n.###############################################################################################.\n................................................................................................." }, { "input": "97 100", "output": "YES\n.................................................................................................\n.##################################################..............................................\n.##################################################..............................................\n................................................................................................." }, { "input": "97 111", "output": "YES\n.................................................................................................\n.######################################################.#........................................\n.########################################################........................................\n................................................................................................." }, { "input": "97 64", "output": "YES\n.................................................................................................\n.################################................................................................\n.################################................................................................\n................................................................................................." }, { "input": "97 77", "output": "YES\n.................................................................................................\n.#####################################.#.........................................................\n.#######################################.........................................................\n................................................................................................." }, { "input": "91 77", "output": "YES\n...........................................................................................\n.#####################################.#...................................................\n.#######################################...................................................\n..........................................................................................." }, { "input": "91 128", "output": "YES\n...........................................................................................\n.################################################################..........................\n.################################################################..........................\n..........................................................................................." }, { "input": "91 113", "output": "YES\n...........................................................................................\n.#######################################################.#.................................\n.#########################################################.................................\n..........................................................................................." }, { "input": "55 55", "output": "YES\n.......................................................\n.##########################.#..........................\n.############################..........................\n......................................................." }, { "input": "43 34", "output": "YES\n...........................................\n.#################.........................\n.#################.........................\n..........................................." }, { "input": "13 21", "output": "YES\n.............\n.#########.#.\n.###########.\n............." }, { "input": "27 50", "output": "YES\n...........................\n.#########################.\n.#########################.\n..........................." }, { "input": "27 49", "output": "YES\n...........................\n.#######################.#.\n.#########################.\n..........................." }, { "input": "27 48", "output": "YES\n...........................\n.########################..\n.########################..\n..........................." }, { "input": "27 40", "output": "YES\n...........................\n.####################......\n.####################......\n..........................." }, { "input": "87 80", "output": "YES\n.......................................................................................\n.########################################..............................................\n.########################################..............................................\n......................................................................................." }, { "input": "69 17", "output": "YES\n.....................................................................\n.#######.#...........................................................\n.#########...........................................................\n....................................................................." }, { "input": "39 73", "output": "YES\n.......................................\n.###################################.#.\n.#####################################.\n......................................." }, { "input": "39 72", "output": "YES\n.......................................\n.####################################..\n.####################################..\n......................................." }, { "input": "33 57", "output": "YES\n.................................\n.###########################.#...\n.#############################...\n................................." }, { "input": "33 54", "output": "YES\n.................................\n.###########################.....\n.###########################.....\n................................." }, { "input": "99 2", "output": "YES\n...................................................................................................\n.#.................................................................................................\n.#.................................................................................................\n..................................................................................................." }, { "input": "99 7", "output": "YES\n...................................................................................................\n.##.#..............................................................................................\n.####..............................................................................................\n..................................................................................................." }, { "input": "99 37", "output": "YES\n...................................................................................................\n.#################.#...............................................................................\n.###################...............................................................................\n..................................................................................................." }, { "input": "57 67", "output": "YES\n.........................................................\n.################################.#......................\n.##################################......................\n........................................................." }, { "input": "57 34", "output": "YES\n.........................................................\n.#################.......................................\n.#################.......................................\n........................................................." }, { "input": "73 17", "output": "YES\n.........................................................................\n.#######.#...............................................................\n.#########...............................................................\n........................................................................." }, { "input": "73 43", "output": "YES\n.........................................................................\n.####################.#..................................................\n.######################..................................................\n........................................................................." }, { "input": "7 3", "output": "YES\n.......\n..###..\n.......\n......." }, { "input": "7 7", "output": "YES\n.......\n.##.#..\n.####..\n......." } ]
1,525,807,146
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
62
6,963,200
n, k = [int(i) for i in input().split(" ")] print('YES') A = [['.' for i in range(n)] for j in range(4)] for i in range(1,3): for j in range(1,n-1): if k>0: A[i][j] = '#' k-=1 else: break for i in A: print(*i, sep='')
Title: Marlin Time Limit: None seconds Memory Limit: None megabytes Problem Description: The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$. The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells. A person can move from one cell to another if those cells are not occupied by hotels and share a side. Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond? Input Specification: The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively. Output Specification: Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO". If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not. Demo Input: ['7 2\n', '5 3\n'] Demo Output: ['YES\n.......\n.#.....\n.#.....\n.......\n', 'YES\n.....\n.###.\n.....\n.....\n'] Note: none
```python n, k = [int(i) for i in input().split(" ")] print('YES') A = [['.' for i in range(n)] for j in range(4)] for i in range(1,3): for j in range(1,n-1): if k>0: A[i][j] = '#' k-=1 else: break for i in A: print(*i, sep='') ```
0
265
A
Colorful Stones (Simplified Edition)
PROGRAMMING
800
[ "implementation" ]
null
null
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively. Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times. Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move. You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction. Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Print the final 1-based position of Liss in a single line.
[ "RGB\nRRR\n", "RRRBGBRBBB\nBBBRR\n", "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n" ]
[ "2\n", "3\n", "15\n" ]
none
500
[ { "input": "RGB\nRRR", "output": "2" }, { "input": "RRRBGBRBBB\nBBBRR", "output": "3" }, { "input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB", "output": "15" }, { "input": "G\nRRBBRBRRBR", "output": "1" }, { "input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB", "output": "1" }, { "input": "RRGGBRGRBG\nBRRGGBBGGR", "output": "7" }, { "input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG", "output": "15" }, { "input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR", "output": "8" }, { "input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR", "output": "16" }, { "input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR", "output": "18" }, { "input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG", "output": "23" }, { "input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR", "output": "14" }, { "input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG", "output": "12" }, { "input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB", "output": "12" }, { "input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR", "output": "2" }, { "input": "R\nB", "output": "1" }, { "input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "50" }, { "input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG", "output": "1" }, { "input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG", "output": "1" } ]
1,647,958,509
2,147,483,647
Python 3
OK
TESTS
19
92
0
stones = list(input()) instruction = list(input()) pos = 0 for i in range(len(instruction)): if stones[pos] == instruction[i]: pos+=1 print(pos+1)
Title: Colorful Stones (Simplified Edition) Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively. Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times. Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move. You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction. Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. Input Specification: The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. Output Specification: Print the final 1-based position of Liss in a single line. Demo Input: ['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n'] Demo Output: ['2\n', '3\n', '15\n'] Note: none
```python stones = list(input()) instruction = list(input()) pos = 0 for i in range(len(instruction)): if stones[pos] == instruction[i]: pos+=1 print(pos+1) ```
3
579
A
Raising Bacteria
PROGRAMMING
1,000
[ "bitmasks" ]
null
null
You are a lover of bacteria. You want to raise some bacteria in a box. Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment. What is the minimum number of bacteria you need to put into the box across those days?
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
The only line containing one integer: the answer.
[ "5\n", "8\n" ]
[ "2\n", "1\n" ]
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2. For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
250
[ { "input": "5", "output": "2" }, { "input": "8", "output": "1" }, { "input": "536870911", "output": "29" }, { "input": "1", "output": "1" }, { "input": "343000816", "output": "14" }, { "input": "559980448", "output": "12" }, { "input": "697681824", "output": "14" }, { "input": "41313494", "output": "14" }, { "input": "673935585", "output": "13" }, { "input": "117422204", "output": "19" }, { "input": "954746654", "output": "15" }, { "input": "536838144", "output": "14" }, { "input": "1000000000", "output": "13" }, { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "4", "output": "1" }, { "input": "6", "output": "2" }, { "input": "7", "output": "3" }, { "input": "9", "output": "2" }, { "input": "10", "output": "2" }, { "input": "999999999", "output": "21" }, { "input": "999999998", "output": "20" }, { "input": "999999997", "output": "20" }, { "input": "999999996", "output": "19" }, { "input": "999999995", "output": "20" }, { "input": "999999994", "output": "19" }, { "input": "999999993", "output": "19" }, { "input": "999999992", "output": "18" }, { "input": "999999991", "output": "20" }, { "input": "999999990", "output": "19" }, { "input": "536870910", "output": "28" }, { "input": "536870912", "output": "1" }, { "input": "536870913", "output": "2" }, { "input": "536870914", "output": "2" }, { "input": "22", "output": "3" }, { "input": "14", "output": "3" } ]
1,670,863,488
2,147,483,647
Python 3
OK
TESTS
36
46
0
from sys import stdin def Binary(n): string = "" while n != 0: string += str(n%2) n //= 2 return string[::-1] def main(): n, total = int(stdin.readline().strip()), 0 for i in Binary(n): total += (1 if i == "1" else 0) print(total) main()
Title: Raising Bacteria Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are a lover of bacteria. You want to raise some bacteria in a box. Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment. What is the minimum number of bacteria you need to put into the box across those days? Input Specification: The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). Output Specification: The only line containing one integer: the answer. Demo Input: ['5\n', '8\n'] Demo Output: ['2\n', '1\n'] Note: For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2. For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
```python from sys import stdin def Binary(n): string = "" while n != 0: string += str(n%2) n //= 2 return string[::-1] def main(): n, total = int(stdin.readline().strip()), 0 for i in Binary(n): total += (1 if i == "1" else 0) print(total) main() ```
3
805
A
Fake NP
PROGRAMMING
1,000
[ "greedy", "math" ]
null
null
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem.
The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).
Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them.
[ "19 29\n", "3 6\n" ]
[ "2\n", "3\n" ]
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
500
[ { "input": "19 29", "output": "2" }, { "input": "3 6", "output": "2" }, { "input": "39 91", "output": "2" }, { "input": "76 134", "output": "2" }, { "input": "93 95", "output": "2" }, { "input": "17 35", "output": "2" }, { "input": "94 95", "output": "2" }, { "input": "51 52", "output": "2" }, { "input": "47 52", "output": "2" }, { "input": "38 98", "output": "2" }, { "input": "30 37", "output": "2" }, { "input": "56 92", "output": "2" }, { "input": "900000000 1000000000", "output": "2" }, { "input": "37622224 162971117", "output": "2" }, { "input": "760632746 850720703", "output": "2" }, { "input": "908580370 968054552", "output": "2" }, { "input": "951594860 953554446", "output": "2" }, { "input": "347877978 913527175", "output": "2" }, { "input": "620769961 988145114", "output": "2" }, { "input": "820844234 892579936", "output": "2" }, { "input": "741254764 741254768", "output": "2" }, { "input": "80270976 80270977", "output": "2" }, { "input": "392602363 392602367", "output": "2" }, { "input": "519002744 519002744", "output": "519002744" }, { "input": "331900277 331900277", "output": "331900277" }, { "input": "419873015 419873018", "output": "2" }, { "input": "349533413 349533413", "output": "349533413" }, { "input": "28829775 28829776", "output": "2" }, { "input": "568814539 568814539", "output": "568814539" }, { "input": "720270740 720270743", "output": "2" }, { "input": "871232720 871232722", "output": "2" }, { "input": "305693653 305693653", "output": "305693653" }, { "input": "634097178 634097179", "output": "2" }, { "input": "450868287 450868290", "output": "2" }, { "input": "252662256 252662260", "output": "2" }, { "input": "575062045 575062049", "output": "2" }, { "input": "273072892 273072894", "output": "2" }, { "input": "770439256 770439256", "output": "770439256" }, { "input": "2 1000000000", "output": "2" }, { "input": "6 8", "output": "2" }, { "input": "2 879190747", "output": "2" }, { "input": "5 5", "output": "5" }, { "input": "999999937 999999937", "output": "999999937" }, { "input": "3 3", "output": "3" }, { "input": "5 100", "output": "2" }, { "input": "2 2", "output": "2" }, { "input": "3 18", "output": "2" }, { "input": "7 7", "output": "7" }, { "input": "39916801 39916801", "output": "39916801" }, { "input": "3 8", "output": "2" }, { "input": "13 13", "output": "13" }, { "input": "4 8", "output": "2" }, { "input": "3 12", "output": "2" }, { "input": "6 12", "output": "2" }, { "input": "999999103 999999103", "output": "999999103" }, { "input": "100000007 100000007", "output": "100000007" }, { "input": "3 99", "output": "2" }, { "input": "999999733 999999733", "output": "999999733" }, { "input": "5 10", "output": "2" }, { "input": "982451653 982451653", "output": "982451653" }, { "input": "999900001 1000000000", "output": "2" }, { "input": "999727999 999727999", "output": "999727999" }, { "input": "2 999999999", "output": "2" }, { "input": "242 244", "output": "2" }, { "input": "3 10", "output": "2" }, { "input": "15 27", "output": "2" }, { "input": "998244353 998244353", "output": "998244353" }, { "input": "5 15", "output": "2" }, { "input": "999999797 999999797", "output": "999999797" }, { "input": "2 3", "output": "2" }, { "input": "999999929 999999929", "output": "999999929" }, { "input": "3 111111", "output": "2" }, { "input": "12 18", "output": "2" }, { "input": "479001599 479001599", "output": "479001599" }, { "input": "10000019 10000019", "output": "10000019" }, { "input": "715827883 715827883", "output": "715827883" }, { "input": "999992977 999992977", "output": "999992977" }, { "input": "11 11", "output": "11" }, { "input": "29 29", "output": "29" }, { "input": "1000003 1000003", "output": "1000003" }, { "input": "6 15", "output": "2" }, { "input": "1200007 1200007", "output": "1200007" }, { "input": "3 1000000000", "output": "2" }, { "input": "990000023 990000023", "output": "990000023" }, { "input": "1717 1717", "output": "1717" }, { "input": "141650963 141650963", "output": "141650963" }, { "input": "1002523 1002523", "output": "1002523" }, { "input": "900000011 900000011", "output": "900000011" }, { "input": "104729 104729", "output": "104729" }, { "input": "4 12", "output": "2" }, { "input": "100003 100003", "output": "100003" }, { "input": "17 17", "output": "17" }, { "input": "10 100", "output": "2" } ]
1,660,576,779
2,147,483,647
Python 3
OK
TESTS
93
61
409,600
import math from re import L import sys,bisect from collections import deque,OrderedDict,defaultdict import heapq from collections import Counter def inp(): return sys.stdin.readline().rstrip() def mpp(): return map(int,inp().split()) def lis(): return list(mpp()) def yn(n): if n: return "YES" else: return "NO" def rt(i,n): return 0<=i<n def fn(arr,n,m): for i in range(n): for j in range(m): if arr[i][j]=="S": if rt(i-1,n): if arr[i-1][j]!="W" and arr[i-1][j]!="S": arr[i-1][j]="D" elif arr[i-1][j]=="W": print("NO") return if rt(i+1,n) : if arr[i+1][j]!="W" and arr[i+1][j]!="S": arr[i+1][j]="D" elif arr[i+1][j]=="W": print("NO") return if rt(j-1,m) : if arr[i][j-1]!="W" and arr[i][j-1]!="S": arr[i][j-1]="D" elif arr[i][j-1]=="W": print("NO") return if rt(j+1,m): if arr[i][j+1]!="W" and arr[i][j+1]!="S": arr[i][j+1]="D" elif arr[i][j+1]=="W": print("NO") return print("YES") for i in range(n): print("".join(arr[i])) return def main(): n,m=mpp() if n==m: print(n) else: print(2) if __name__=="__main__": main()
Title: Fake NP Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem. Input Specification: The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109). Output Specification: Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them. Demo Input: ['19 29\n', '3 6\n'] Demo Output: ['2\n', '3\n'] Note: Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
```python import math from re import L import sys,bisect from collections import deque,OrderedDict,defaultdict import heapq from collections import Counter def inp(): return sys.stdin.readline().rstrip() def mpp(): return map(int,inp().split()) def lis(): return list(mpp()) def yn(n): if n: return "YES" else: return "NO" def rt(i,n): return 0<=i<n def fn(arr,n,m): for i in range(n): for j in range(m): if arr[i][j]=="S": if rt(i-1,n): if arr[i-1][j]!="W" and arr[i-1][j]!="S": arr[i-1][j]="D" elif arr[i-1][j]=="W": print("NO") return if rt(i+1,n) : if arr[i+1][j]!="W" and arr[i+1][j]!="S": arr[i+1][j]="D" elif arr[i+1][j]=="W": print("NO") return if rt(j-1,m) : if arr[i][j-1]!="W" and arr[i][j-1]!="S": arr[i][j-1]="D" elif arr[i][j-1]=="W": print("NO") return if rt(j+1,m): if arr[i][j+1]!="W" and arr[i][j+1]!="S": arr[i][j+1]="D" elif arr[i][j+1]=="W": print("NO") return print("YES") for i in range(n): print("".join(arr[i])) return def main(): n,m=mpp() if n==m: print(n) else: print(2) if __name__=="__main__": main() ```
3
0
none
none
none
0
[ "none" ]
null
null
You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get? Definitions of subsegment and array splitting are given in notes.
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to. The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109).
Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments.
[ "5 2\n1 2 3 4 5\n", "5 1\n-4 -5 -3 -2 -1\n" ]
[ "5\n", "-5\n" ]
A subsegment [*l*,  *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>,  *a*<sub class="lower-index">*l* + 1</sub>,  ...,  *a*<sub class="lower-index">*r*</sub>. Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* &gt; 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>). In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result. In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is *min*( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.
0
[ { "input": "5 2\n1 2 3 4 5", "output": "5" }, { "input": "5 1\n-4 -5 -3 -2 -1", "output": "-5" }, { "input": "10 2\n10 9 1 -9 -7 -9 3 8 -10 5", "output": "10" }, { "input": "10 4\n-8 -1 2 -3 9 -8 4 -3 5 9", "output": "9" }, { "input": "1 1\n504262064", "output": "504262064" }, { "input": "3 3\n-54481850 -878017339 -486296116", "output": "-54481850" }, { "input": "2 2\n-333653905 224013643", "output": "224013643" }, { "input": "14 2\n-14 84 44 46 -75 -75 77 -49 44 -82 -74 -51 -9 -50", "output": "-14" }, { "input": "88 71\n-497 -488 182 104 40 183 201 282 -384 44 -29 494 224 -80 -491 -197 157 130 -52 233 -426 252 -61 -51 203 -50 195 -442 -38 385 232 -243 -49 163 340 -200 406 -254 -29 227 -194 193 487 -325 230 146 421 158 20 447 -97 479 493 -130 164 -471 -198 -330 -152 359 -554 319 544 -444 235 281 -467 337 -385 227 -366 -210 266 69 -261 525 526 -234 -355 177 109 275 -301 7 -41 553 -284 540", "output": "553" }, { "input": "39 1\n676941771 -923780377 -163050076 -230110947 -208029500 329620771 13954060 158950156 -252501602 926390671 -678745080 -921892226 -100127643 610420285 602175224 -839193819 471391946 910035173 777969600 -736144413 -489685522 60986249 830784148 278642552 -375298304 197973611 -354482364 187294011 636628282 25350767 636184407 -550869740 53830680 -42049274 -451383278 900048257 93225803 877923341 -279506435", "output": "-923780377" }, { "input": "3 2\n1 5 3", "output": "3" }, { "input": "5 2\n1 2 5 4 3", "output": "3" }, { "input": "3 2\n1 3 2", "output": "2" }, { "input": "3 2\n1 3 1", "output": "1" }, { "input": "5 3\n-2 -2 -2 -2 -2", "output": "-2" }, { "input": "5 2\n1 2 3 5 4", "output": "4" }, { "input": "5 2\n1 1 11 1 1", "output": "1" }, { "input": "3 3\n3 8 4", "output": "8" }, { "input": "6 3\n4 3 1 5 6 2", "output": "6" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "5 2\n2 5 4 3 1", "output": "2" }, { "input": "5 2\n-1 1 5 4 3", "output": "3" }, { "input": "5 2\n5 2 1 9 3", "output": "5" }, { "input": "2 1\n1000000000 1000000000", "output": "1000000000" }, { "input": "1 1\n1000000000", "output": "1000000000" }, { "input": "5 2\n1 5 3 4 1", "output": "1" }, { "input": "3 2\n-1000000000 -1000000000 -1000000000", "output": "-1000000000" }, { "input": "2 2\n5 2", "output": "5" }, { "input": "7 3\n1 1 1 10 1 1 1", "output": "10" }, { "input": "9 3\n1 2 1 1 5 1 1 1 2", "output": "5" }, { "input": "9 3\n2 2 2 2 9 2 2 2 2", "output": "9" }, { "input": "3 3\n-1000000000 -1000000000 -1000000000", "output": "-1000000000" } ]
1,508,085,490
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
31
0
x=input().split() y=input().split() a=[] z=[] for i in y: a.append(int(i)) if int(x[1])>2: print(max(a)) if int(x[1])==1: print(min(a)) if int(x[1])==2: for e in range(a): z.append(min(a[:e])) z.append(min(a[e:])) print(max(z))
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get? Definitions of subsegment and array splitting are given in notes. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to. The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109). Output Specification: Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments. Demo Input: ['5 2\n1 2 3 4 5\n', '5 1\n-4 -5 -3 -2 -1\n'] Demo Output: ['5\n', '-5\n'] Note: A subsegment [*l*,  *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>,  *a*<sub class="lower-index">*l* + 1</sub>,  ...,  *a*<sub class="lower-index">*r*</sub>. Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* &gt; 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>). In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result. In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is *min*( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.
```python x=input().split() y=input().split() a=[] z=[] for i in y: a.append(int(i)) if int(x[1])>2: print(max(a)) if int(x[1])==1: print(min(a)) if int(x[1])==2: for e in range(a): z.append(min(a[:e])) z.append(min(a[e:])) print(max(z)) ```
-1
0
none
none
none
0
[ "none" ]
null
null
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis. Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct. Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task. Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7. Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct.
Print a single integer — the required sum modulo 109<=+<=7.
[ "2\n4 7\n", "3\n4 3 1\n" ]
[ "3\n", "9\n" ]
There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3. There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
0
[ { "input": "2\n4 7", "output": "3" }, { "input": "3\n4 3 1", "output": "9" }, { "input": "20\n8 11 13 19 21 34 36 44 57 58 61 63 76 78 79 81 85 86 90 95", "output": "83396599" }, { "input": "20\n1 8 9 12 15 17 18 24 30 33 36 41 53 54 59 62 64 66 72 73", "output": "68059140" }, { "input": "20\n2 6 8 9 20 23 27 36 43 49 63 65 70 71 85 87 89 91 94 97", "output": "92743989" }, { "input": "1\n78091781", "output": "0" }, { "input": "2\n1000000000 1", "output": "999999999" }, { "input": "3\n999999998 999999999 999999992", "output": "21" }, { "input": "3\n465343471 465343474 465343473", "output": "9" }, { "input": "10\n10 3 6 2 1 9 8 4 5 7", "output": "7181" }, { "input": "10\n756734546 756734524 756734550 756734529 756734553 756734538 756734541 756734536 756734579 756734537", "output": "36489" }, { "input": "10\n877105545 939360757 849826701 845946140 803128820 926787996 967305000 904694971 921301848 971203310", "output": "861364152" }, { "input": "5\n4 7 13 17 18", "output": "270" }, { "input": "5\n20 17 13 7 2", "output": "330" }, { "input": "5\n3 17 2 5 4", "output": "237" }, { "input": "5\n999999980 999999985 999999986 999999990 999999992", "output": "210" }, { "input": "5\n1000000000 999999988 999999982 999999981 999999980", "output": "342" }, { "input": "5\n999999984 999999997 999999994 999999991 999999982", "output": "285" }, { "input": "1\n2", "output": "0" }, { "input": "5\n9 10 7 4 5", "output": "114" } ]
1,689,193,355
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
62
0
print("_RANDOM_GUESS_1689193354.922567")# 1689193354.9226089
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis. Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct. Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task. Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7. Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct. Output Specification: Print a single integer — the required sum modulo 109<=+<=7. Demo Input: ['2\n4 7\n', '3\n4 3 1\n'] Demo Output: ['3\n', '9\n'] Note: There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3. There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
```python print("_RANDOM_GUESS_1689193354.922567")# 1689193354.9226089 ```
0
323
A
Black-and-White Cube
PROGRAMMING
1,600
[ "combinatorics", "constructive algorithms" ]
null
null
You are given a cube of size *k*<=×<=*k*<=×<=*k*, which consists of unit cubes. Two unit cubes are considered neighbouring, if they have common face. Your task is to paint each of *k*3 unit cubes one of two colours (black or white), so that the following conditions must be satisfied: - each white cube has exactly 2 neighbouring cubes of white color; - each black cube has exactly 2 neighbouring cubes of black color.
The first line contains integer *k* (1<=≤<=*k*<=≤<=100), which is size of the cube.
Print -1 if there is no solution. Otherwise, print the required painting of the cube consequently by layers. Print a *k*<=×<=*k* matrix in the first *k* lines, showing how the first layer of the cube should be painted. In the following *k* lines print a *k*<=×<=*k* matrix — the way the second layer should be painted. And so on to the last *k*-th layer. Note that orientation of the cube in the space does not matter. Mark a white unit cube with symbol "w" and a black one with "b". Use the format of output data, given in the test samples. You may print extra empty lines, they will be ignored.
[ "1\n", "2\n" ]
[ "-1\n", "bb\nww\n\nbb\nww\n" ]
none
500
[ { "input": "1", "output": "-1" }, { "input": "2", "output": "bb\nww\n\nbb\nww" }, { "input": "3", "output": "-1" }, { "input": "4", "output": "bbbb\nbwwb\nbwwb\nbbbb\n\nwwww\nwbbw\nwbbw\nwwww\n\nbbbb\nbwwb\nbwwb\nbbbb\n\nwwww\nwbbw\nwbbw\nwwww" }, { "input": "5", "output": "-1" }, { "input": "6", "output": "bbbbbb\nbwwwwb\nbwbbwb\nbwbbwb\nbwwwwb\nbbbbbb\n\nwwwwww\nwbbbbw\nwbwwbw\nwbwwbw\nwbbbbw\nwwwwww\n\nbbbbbb\nbwwwwb\nbwbbwb\nbwbbwb\nbwwwwb\nbbbbbb\n\nwwwwww\nwbbbbw\nwbwwbw\nwbwwbw\nwbbbbw\nwwwwww\n\nbbbbbb\nbwwwwb\nbwbbwb\nbwbbwb\nbwwwwb\nbbbbbb\n\nwwwwww\nwbbbbw\nwbwwbw\nwbwwbw\nwbbbbw\nwwwwww" }, { "input": "7", "output": "-1" }, { "input": "8", "output": "bbbbbbbb\nbwwwwwwb\nbwbbbbwb\nbwbwwbwb\nbwbwwbwb\nbwbbbbwb\nbwwwwwwb\nbbbbbbbb\n\nwwwwwwww\nwbbbbbbw\nwbwwwwbw\nwbwbbwbw\nwbwbbwbw\nwbwwwwbw\nwbbbbbbw\nwwwwwwww\n\nbbbbbbbb\nbwwwwwwb\nbwbbbbwb\nbwbwwbwb\nbwbwwbwb\nbwbbbbwb\nbwwwwwwb\nbbbbbbbb\n\nwwwwwwww\nwbbbbbbw\nwbwwwwbw\nwbwbbwbw\nwbwbbwbw\nwbwwwwbw\nwbbbbbbw\nwwwwwwww\n\nbbbbbbbb\nbwwwwwwb\nbwbbbbwb\nbwbwwbwb\nbwbwwbwb\nbwbbbbwb\nbwwwwwwb\nbbbbbbbb\n\nwwwwwwww\nwbbbbbbw\nwbwwwwbw\nwbwbbwbw\nwbwbbwbw\nwbwwwwbw\nwbbbbbbw\nwwwwwwww\n\nbbbbbbbb\nbwwwwwwb\n..." }, { "input": "9", "output": "-1" }, { "input": "10", "output": "bbbbbbbbbb\nbwwwwwwwwb\nbwbbbbbbwb\nbwbwwwwbwb\nbwbwbbwbwb\nbwbwbbwbwb\nbwbwwwwbwb\nbwbbbbbbwb\nbwwwwwwwwb\nbbbbbbbbbb\n\nwwwwwwwwww\nwbbbbbbbbw\nwbwwwwwwbw\nwbwbbbbwbw\nwbwbwwbwbw\nwbwbwwbwbw\nwbwbbbbwbw\nwbwwwwwwbw\nwbbbbbbbbw\nwwwwwwwwww\n\nbbbbbbbbbb\nbwwwwwwwwb\nbwbbbbbbwb\nbwbwwwwbwb\nbwbwbbwbwb\nbwbwbbwbwb\nbwbwwwwbwb\nbwbbbbbbwb\nbwwwwwwwwb\nbbbbbbbbbb\n\nwwwwwwwwww\nwbbbbbbbbw\nwbwwwwwwbw\nwbwbbbbwbw\nwbwbwwbwbw\nwbwbwwbwbw\nwbwbbbbwbw\nwbwwwwwwbw\nwbbbbbbbbw\nwwwwwwwwww\n\nbbbbbbbbbb\nbwwwwwwwwb\n..." }, { "input": "11", "output": "-1" }, { "input": "12", "output": 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"bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwbwb\nbwbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "67", "output": "-1" }, { "input": "68", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwbwb\nbwbwbwbwwwwwwwwwwwwww..." }, { "input": "69", "output": "-1" }, { "input": "70", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwbwb\nbwbwbwb..." }, { "input": "71", "output": "-1" }, { "input": "72", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbw..." }, { "input": "73", "output": "-1" }, { "input": "74", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb..." }, { "input": "75", "output": "-1" }, { "input": "76", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb..." }, { "input": "77", "output": "-1" }, { "input": "78", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbb..." }, { "input": "79", "output": "-1" }, { "input": "80", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbb..." }, { "input": "81", "output": "-1" }, { "input": "82", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwb..." }, { "input": "83", "output": "-1" }, { "input": "84", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbw..." }, { "input": "85", "output": "-1" }, { "input": "86", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "87", "output": "-1" }, { "input": "88", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "89", "output": "-1" }, { "input": "90", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "91", "output": "-1" }, { "input": "92", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "93", "output": "-1" }, { "input": "94", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwww..." }, { "input": "95", "output": "-1" }, { "input": "96", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwww..." }, { "input": "97", "output": "-1" }, { "input": "98", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwww..." }, { "input": "99", "output": "-1" }, { "input": "100", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nb..." }, { "input": "40", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\nbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwb\nbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwb\nbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwb\nbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwb\nbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwb\nbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbbbbbwbwbwb\nbwbwbwbwwwwwwwwwwwwwwwwwwwwwwwwwwbwbwbwb\nbwbwbwbwbbbbbbbbbbbbbbbbbbbbbbbbwbwbwbwb\nbwbwbwbwbwwwwwwwwwwwwwwwwwwwwwwbwbwbwbwb\nbwbwbwbwbwbbbbbbbbbbbbbbbbbbbbwbwbwbwbwb\nbwbwbwbwbwbwwwwwwwwwwwwwwwwwwbwbwbwbwbwb\nbwbwbwb..." } ]
1,590,389,759
2,147,483,647
PyPy 3
OK
TESTS
101
249
3,993,600
n=int(input()) if(n%2==1): print(-1) else: for i in range(n): for j in range(n): s = '' if (i // 2 + j)&1: s = 'wwbb' * (n//4) if n % 4 != 0: s += 'ww' else: s = 'bbww' * (n//4) if n % 4 != 0: s += 'bb' print(s)
Title: Black-and-White Cube Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a cube of size *k*<=×<=*k*<=×<=*k*, which consists of unit cubes. Two unit cubes are considered neighbouring, if they have common face. Your task is to paint each of *k*3 unit cubes one of two colours (black or white), so that the following conditions must be satisfied: - each white cube has exactly 2 neighbouring cubes of white color; - each black cube has exactly 2 neighbouring cubes of black color. Input Specification: The first line contains integer *k* (1<=≤<=*k*<=≤<=100), which is size of the cube. Output Specification: Print -1 if there is no solution. Otherwise, print the required painting of the cube consequently by layers. Print a *k*<=×<=*k* matrix in the first *k* lines, showing how the first layer of the cube should be painted. In the following *k* lines print a *k*<=×<=*k* matrix — the way the second layer should be painted. And so on to the last *k*-th layer. Note that orientation of the cube in the space does not matter. Mark a white unit cube with symbol "w" and a black one with "b". Use the format of output data, given in the test samples. You may print extra empty lines, they will be ignored. Demo Input: ['1\n', '2\n'] Demo Output: ['-1\n', 'bb\nww\n\nbb\nww\n'] Note: none
```python n=int(input()) if(n%2==1): print(-1) else: for i in range(n): for j in range(n): s = '' if (i // 2 + j)&1: s = 'wwbb' * (n//4) if n % 4 != 0: s += 'ww' else: s = 'bbww' * (n//4) if n % 4 != 0: s += 'bb' print(s) ```
3
500
A
New Year Transportation
PROGRAMMING
1,000
[ "dfs and similar", "graphs", "implementation" ]
null
null
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells. So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals. Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to. The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
[ "8 4\n1 2 1 2 1 2 1\n", "8 5\n1 2 1 2 1 1 1\n" ]
[ "YES\n", "NO\n" ]
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4. In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
500
[ { "input": "8 4\n1 2 1 2 1 2 1", "output": "YES" }, { "input": "8 5\n1 2 1 2 1 1 1", "output": "NO" }, { "input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1", "output": "YES" }, { "input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1", "output": "YES" }, { "input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1", "output": "NO" }, { "input": "10 3\n8 3 5 4 2 3 2 2 1", "output": "NO" }, { "input": "10 9\n8 3 5 4 2 3 2 2 1", "output": "YES" }, { "input": "3 2\n1 1", "output": "YES" }, { "input": "3 2\n2 1", "output": "NO" }, { "input": "4 2\n2 1 1", "output": "NO" }, { "input": "4 4\n2 2 1", "output": "YES" }, { "input": "8 8\n1 2 1 2 1 2 1", "output": "YES" }, { "input": "3 3\n1 1", "output": "YES" }, { "input": "8 8\n1 2 1 2 1 1 1", "output": "YES" }, { "input": "3 3\n2 1", "output": "YES" }, { "input": "4 4\n1 1 1", "output": "YES" }, { "input": "8 8\n1 1 1 1 1 1 1", "output": "YES" }, { "input": "5 5\n1 1 1 1", "output": "YES" } ]
1,672,206,305
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
30
0
x,a=[int(i) for i in input().split()] b=[int(i) for i in input().split()] x=0 y=0 if a==1 or a==2: print("YES") else: for i in b: x=x+i if x==a: print("YES") y=1 break elif x>a: print("NO") break if y==0: print("NO")
Title: New Year Transportation Time Limit: None seconds Memory Limit: None megabytes Problem Description: New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells. So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals. Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. Input Specification: The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to. The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. Output Specification: If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". Demo Input: ['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4. In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
```python x,a=[int(i) for i in input().split()] b=[int(i) for i in input().split()] x=0 y=0 if a==1 or a==2: print("YES") else: for i in b: x=x+i if x==a: print("YES") y=1 break elif x>a: print("NO") break if y==0: print("NO") ```
0
875
A
Classroom Watch
PROGRAMMING
1,200
[ "brute force", "math" ]
null
null
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system. Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
In the first line print one integer *k* — number of different values of *x* satisfying the condition. In next *k* lines print these values in ascending order.
[ "21\n", "20\n" ]
[ "1\n15\n", "0\n" ]
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21. In the second test case there are no such *x*.
500
[ { "input": "21", "output": "1\n15" }, { "input": "20", "output": "0" }, { "input": "1", "output": "0" }, { "input": "2", "output": "1\n1" }, { "input": "3", "output": "0" }, { "input": "100000001", "output": "2\n99999937\n100000000" }, { "input": "1000000000", "output": "1\n999999932" }, { "input": "999999979", "output": "2\n999999899\n999999908" }, { "input": "9", "output": "0" }, { "input": "10", "output": "1\n5" }, { "input": "11", "output": "1\n10" }, { "input": "39", "output": "1\n33" }, { "input": "66", "output": "1\n60" }, { "input": "75", "output": "0" }, { "input": "100", "output": "1\n86" }, { "input": "101", "output": "2\n91\n100" }, { "input": "2014", "output": "2\n1988\n2006" }, { "input": "999999994", "output": "0" } ]
1,508,316,158
2,147,483,647
Python 3
OK
TESTS
18
62
5,529,600
n = int(input()) def digits(v): s = 0 while v > 0: s += v % 10 v //= 10 return s a = [] for x in range(max(0, n - 81), n): if x + digits(x) == n: a.append(x) print(len(a)) if len(a) > 0: print(" ".join(map(str, a)))
Title: Classroom Watch Time Limit: None seconds Memory Limit: None megabytes Problem Description: Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system. Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=109). Output Specification: In the first line print one integer *k* — number of different values of *x* satisfying the condition. In next *k* lines print these values in ascending order. Demo Input: ['21\n', '20\n'] Demo Output: ['1\n15\n', '0\n'] Note: In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21. In the second test case there are no such *x*.
```python n = int(input()) def digits(v): s = 0 while v > 0: s += v % 10 v //= 10 return s a = [] for x in range(max(0, n - 81), n): if x + digits(x) == n: a.append(x) print(len(a)) if len(a) > 0: print(" ".join(map(str, a))) ```
3
237
D
T-decomposition
PROGRAMMING
2,000
[ "dfs and similar", "graphs", "greedy", "trees" ]
null
null
You've got a undirected tree *s*, consisting of *n* nodes. Your task is to build an optimal T-decomposition for it. Let's define a T-decomposition as follows. Let's denote the set of all nodes *s* as *v*. Let's consider an undirected tree *t*, whose nodes are some non-empty subsets of *v*, we'll call them *x**i* . The tree *t* is a T-decomposition of *s*, if the following conditions holds: 1. the union of all *x**i* equals *v*; 1. for any edge (*a*,<=*b*) of tree *s* exists the tree node *t*, containing both *a* and *b*; 1. if the nodes of the tree *t* *x**i* and *x**j* contain the node *a* of the tree *s*, then all nodes of the tree *t*, lying on the path from *x**i* to *x**j* also contain node *a*. So this condition is equivalent to the following: all nodes of the tree *t*, that contain node *a* of the tree *s*, form a connected subtree of tree *t*. There are obviously many distinct trees *t*, that are T-decompositions of the tree *s*. For example, a T-decomposition is a tree that consists of a single node, equal to set *v*. Let's define the cardinality of node *x**i* as the number of nodes in tree *s*, containing in the node. Let's choose the node with the maximum cardinality in *t*. Let's assume that its cardinality equals *w*. Then the weight of T-decomposition *t* is value *w*. The optimal T-decomposition is the one with the minimum weight. Your task is to find the optimal T-decomposition of the given tree *s* that has the minimum number of nodes.
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105), that denotes the number of nodes in tree *s*. Each of the following *n*<=-<=1 lines contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), denoting that the nodes of tree *s* with indices *a**i* and *b**i* are connected by an edge. Consider the nodes of tree *s* indexed from 1 to *n*. It is guaranteed that *s* is a tree.
In the first line print a single integer *m* that denotes the number of nodes in the required T-decomposition. Then print *m* lines, containing descriptions of the T-decomposition nodes. In the *i*-th (1<=≤<=*i*<=≤<=*m*) of them print the description of node *x**i* of the T-decomposition. The description of each node *x**i* should start from an integer *k**i*, that represents the number of nodes of the initial tree *s*, that are contained in the node *x**i*. Then you should print *k**i* distinct space-separated integers — the numbers of nodes from *s*, contained in *x**i*, in arbitrary order. Then print *m*<=-<=1 lines, each consisting two integers *p**i*,<=*q**i* (1<=≤<=*p**i*,<=*q**i*<=≤<=*m*; *p**i*<=≠<=*q**i*). The pair of integers *p**i*,<=*q**i* means there is an edge between nodes *x**p**i* and *x**q**i* of T-decomposition. The printed T-decomposition should be the optimal T-decomposition for the given tree *s* and have the minimum possible number of nodes among all optimal T-decompositions. If there are multiple optimal T-decompositions with the minimum number of nodes, print any of them.
[ "2\n1 2\n", "3\n1 2\n2 3\n", "4\n2 1\n3 1\n4 1\n" ]
[ "1\n2 1 2\n", "2\n2 1 2\n2 2 3\n1 2\n", "3\n2 2 1\n2 3 1\n2 4 1\n1 2\n2 3\n" ]
none
2,000
[ { "input": "2\n1 2", "output": "1\n2 1 2" }, { "input": "3\n1 2\n2 3", "output": "2\n2 1 2\n2 2 3\n1 2" }, { "input": "4\n2 1\n3 1\n4 1", "output": "3\n2 2 1\n2 3 1\n2 4 1\n1 2\n2 3" }, { "input": "6\n2 5\n4 3\n4 2\n4 6\n3 1", "output": "5\n2 2 5\n2 4 3\n2 4 2\n2 4 6\n2 3 1\n2 3\n1 3\n3 4\n2 5" }, { "input": "6\n3 6\n4 2\n3 4\n3 1\n6 5", "output": "5\n2 3 6\n2 4 2\n2 3 4\n2 3 1\n2 6 5\n1 3\n2 3\n3 4\n1 5" }, { "input": "6\n3 2\n6 5\n1 3\n1 4\n6 1", "output": "5\n2 3 2\n2 6 5\n2 1 3\n2 1 4\n2 6 1\n1 3\n3 4\n2 5\n4 5" }, { "input": "6\n5 3\n4 2\n5 6\n6 1\n5 4", "output": "5\n2 5 3\n2 4 2\n2 5 6\n2 6 1\n2 5 4\n1 3\n3 4\n3 5\n2 5" }, { "input": "5\n1 2\n1 3\n4 5\n4 1", "output": "4\n2 1 2\n2 1 3\n2 4 5\n2 4 1\n1 2\n3 4\n2 4" }, { "input": "5\n1 5\n5 3\n2 4\n4 1", "output": "4\n2 1 5\n2 5 3\n2 2 4\n2 4 1\n1 2\n3 4\n1 4" }, { "input": "5\n1 5\n5 2\n1 4\n1 3", "output": "4\n2 1 5\n2 5 2\n2 1 4\n2 1 3\n1 2\n1 3\n3 4" }, { "input": "4\n1 4\n3 1\n3 2", "output": "3\n2 1 4\n2 3 1\n2 3 2\n1 2\n2 3" }, { "input": "20\n19 2\n19 18\n20 9\n20 10\n18 4\n17 5\n17 13\n11 17\n20 3\n11 1\n18 7\n11 20\n20 16\n5 15\n19 6\n11 14\n20 8\n17 12\n11 19", "output": "19\n2 19 2\n2 19 18\n2 20 9\n2 20 10\n2 18 4\n2 17 5\n2 17 13\n2 11 17\n2 20 3\n2 11 1\n2 18 7\n2 11 20\n2 20 16\n2 5 15\n2 19 6\n2 11 14\n2 20 8\n2 17 12\n2 11 19\n1 2\n3 4\n2 5\n6 7\n7 8\n4 9\n8 10\n5 11\n10 12\n9 12\n12 13\n6 14\n2 15\n12 16\n13 17\n8 18\n16 19\n15 19" }, { "input": "21\n2 8\n9 15\n7 5\n14 6\n19 7\n9 1\n2 10\n16 14\n16 17\n19 2\n2 12\n19 11\n16 18\n2 13\n19 9\n19 16\n1 20\n14 21\n1 3\n2 4", "output": "20\n2 2 8\n2 9 15\n2 7 5\n2 14 6\n2 19 7\n2 9 1\n2 2 10\n2 16 14\n2 16 17\n2 19 2\n2 2 12\n2 19 11\n2 16 18\n2 2 13\n2 19 9\n2 19 16\n2 1 20\n2 14 21\n2 1 3\n2 2 4\n3 5\n2 6\n1 7\n4 8\n8 9\n5 10\n7 10\n10 11\n10 12\n9 13\n11 14\n12 15\n6 15\n15 16\n13 16\n6 17\n8 18\n17 19\n14 20" }, { "input": "22\n10 11\n8 10\n8 15\n3 17\n8 20\n15 5\n10 1\n10 13\n11 9\n19 3\n9 14\n5 7\n19 2\n8 18\n11 4\n15 22\n15 19\n15 6\n8 12\n17 21\n13 16", "output": "21\n2 10 11\n2 8 10\n2 8 15\n2 3 17\n2 8 20\n2 15 5\n2 10 1\n2 10 13\n2 11 9\n2 19 3\n2 9 14\n2 5 7\n2 19 2\n2 8 18\n2 11 4\n2 15 22\n2 15 19\n2 15 6\n2 8 12\n2 17 21\n2 13 16\n1 2\n2 3\n3 5\n3 6\n2 7\n7 8\n1 9\n4 10\n9 11\n6 12\n10 13\n5 14\n9 15\n6 16\n16 17\n13 17\n17 18\n14 19\n4 20\n8 21" }, { "input": "23\n10 19\n11 2\n15 18\n8 14\n15 7\n23 6\n21 5\n14 1\n10 13\n8 23\n19 16\n12 3\n8 10\n8 21\n14 11\n6 22\n7 8\n4 15\n9 12\n15 9\n1 20\n11 17", "output": "22\n2 10 19\n2 11 2\n2 15 18\n2 8 14\n2 15 7\n2 23 6\n2 21 5\n2 14 1\n2 10 13\n2 8 23\n2 19 16\n2 12 3\n2 8 10\n2 8 21\n2 14 11\n2 6 22\n2 7 8\n2 4 15\n2 9 12\n2 15 9\n2 1 20\n2 11 17\n3 5\n4 8\n1 9\n4 10\n6 10\n1 11\n10 13\n9 13\n13 14\n7 14\n8 15\n2 15\n6 16\n5 17\n14 17\n5 18\n12 19\n18 20\n19 20\n8 21\n15 22" }, { "input": "24\n12 22\n4 12\n11 9\n14 19\n20 3\n16 24\n3 14\n14 23\n15 8\n8 20\n1 11\n1 7\n11 13\n2 15\n3 10\n16 5\n14 21\n6 2\n11 16\n24 17\n8 1\n3 4\n7 18", "output": "23\n2 12 22\n2 4 12\n2 11 9\n2 14 19\n2 20 3\n2 16 24\n2 3 14\n2 14 23\n2 15 8\n2 8 20\n2 1 11\n2 1 7\n2 11 13\n2 2 15\n2 3 10\n2 16 5\n2 14 21\n2 6 2\n2 11 16\n2 24 17\n2 8 1\n2 3 4\n2 7 18\n1 2\n5 7\n4 7\n7 8\n9 10\n5 10\n3 11\n11 12\n11 13\n9 14\n7 15\n6 16\n8 17\n14 18\n13 19\n16 19\n6 20\n10 21\n12 21\n15 22\n2 22\n12 23" }, { "input": "25\n13 1\n4 17\n15 25\n3 21\n1 6\n1 9\n12 15\n13 4\n24 19\n22 24\n8 20\n4 11\n11 14\n17 16\n15 7\n23 3\n22 13\n3 5\n6 10\n16 18\n24 23\n10 2\n9 8\n7 22", "output": "24\n2 13 1\n2 4 17\n2 15 25\n2 3 21\n2 1 6\n2 1 9\n2 12 15\n2 13 4\n2 24 19\n2 22 24\n2 8 20\n2 4 11\n2 11 14\n2 17 16\n2 15 7\n2 23 3\n2 22 13\n2 3 5\n2 6 10\n2 16 18\n2 24 23\n2 10 2\n2 9 8\n2 7 22\n1 5\n5 6\n3 7\n1 8\n2 8\n9 10\n8 12\n12 13\n2 14\n7 15\n4 16\n10 17\n8 17\n16 18\n5 19\n14 20\n10 21\n16 21\n19 22\n6 23\n11 23\n15 24\n17 24" } ]
1,616,104,890
2,147,483,647
PyPy 3
TIME_LIMIT_EXCEEDED
TESTS
8
2,000
23,859,200
N = 10**5 + 5 g = [[] for _ in range(N)] p = [0]*N def dfs(cur, par): p[cur] = par for nxt in g[cur]: if nxt != par: dfs(nxt, cur) n = int(input()) for i in range(n-1): a, b = map(int, input().split()) g[a].append(b) g[b].append(a) dfs(1, 0) print(n-1) for i in range(2, n+1): print(2, str(i), str(p[i])) for i in range(len(g[1])-1): print(str(g[1][i]-1), str(g[1][i+1]-1)) for i in range(2, n+1): for c in g[i]: if c != p[i]: print(str(i-1), str(c-1))
Title: T-decomposition Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a undirected tree *s*, consisting of *n* nodes. Your task is to build an optimal T-decomposition for it. Let's define a T-decomposition as follows. Let's denote the set of all nodes *s* as *v*. Let's consider an undirected tree *t*, whose nodes are some non-empty subsets of *v*, we'll call them *x**i* . The tree *t* is a T-decomposition of *s*, if the following conditions holds: 1. the union of all *x**i* equals *v*; 1. for any edge (*a*,<=*b*) of tree *s* exists the tree node *t*, containing both *a* and *b*; 1. if the nodes of the tree *t* *x**i* and *x**j* contain the node *a* of the tree *s*, then all nodes of the tree *t*, lying on the path from *x**i* to *x**j* also contain node *a*. So this condition is equivalent to the following: all nodes of the tree *t*, that contain node *a* of the tree *s*, form a connected subtree of tree *t*. There are obviously many distinct trees *t*, that are T-decompositions of the tree *s*. For example, a T-decomposition is a tree that consists of a single node, equal to set *v*. Let's define the cardinality of node *x**i* as the number of nodes in tree *s*, containing in the node. Let's choose the node with the maximum cardinality in *t*. Let's assume that its cardinality equals *w*. Then the weight of T-decomposition *t* is value *w*. The optimal T-decomposition is the one with the minimum weight. Your task is to find the optimal T-decomposition of the given tree *s* that has the minimum number of nodes. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105), that denotes the number of nodes in tree *s*. Each of the following *n*<=-<=1 lines contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), denoting that the nodes of tree *s* with indices *a**i* and *b**i* are connected by an edge. Consider the nodes of tree *s* indexed from 1 to *n*. It is guaranteed that *s* is a tree. Output Specification: In the first line print a single integer *m* that denotes the number of nodes in the required T-decomposition. Then print *m* lines, containing descriptions of the T-decomposition nodes. In the *i*-th (1<=≤<=*i*<=≤<=*m*) of them print the description of node *x**i* of the T-decomposition. The description of each node *x**i* should start from an integer *k**i*, that represents the number of nodes of the initial tree *s*, that are contained in the node *x**i*. Then you should print *k**i* distinct space-separated integers — the numbers of nodes from *s*, contained in *x**i*, in arbitrary order. Then print *m*<=-<=1 lines, each consisting two integers *p**i*,<=*q**i* (1<=≤<=*p**i*,<=*q**i*<=≤<=*m*; *p**i*<=≠<=*q**i*). The pair of integers *p**i*,<=*q**i* means there is an edge between nodes *x**p**i* and *x**q**i* of T-decomposition. The printed T-decomposition should be the optimal T-decomposition for the given tree *s* and have the minimum possible number of nodes among all optimal T-decompositions. If there are multiple optimal T-decompositions with the minimum number of nodes, print any of them. Demo Input: ['2\n1 2\n', '3\n1 2\n2 3\n', '4\n2 1\n3 1\n4 1\n'] Demo Output: ['1\n2 1 2\n', '2\n2 1 2\n2 2 3\n1 2\n', '3\n2 2 1\n2 3 1\n2 4 1\n1 2\n2 3\n'] Note: none
```python N = 10**5 + 5 g = [[] for _ in range(N)] p = [0]*N def dfs(cur, par): p[cur] = par for nxt in g[cur]: if nxt != par: dfs(nxt, cur) n = int(input()) for i in range(n-1): a, b = map(int, input().split()) g[a].append(b) g[b].append(a) dfs(1, 0) print(n-1) for i in range(2, n+1): print(2, str(i), str(p[i])) for i in range(len(g[1])-1): print(str(g[1][i]-1), str(g[1][i+1]-1)) for i in range(2, n+1): for c in g[i]: if c != p[i]: print(str(i-1), str(c-1)) ```
0
478
C
Table Decorations
PROGRAMMING
1,800
[ "greedy" ]
null
null
You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color? Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner.
The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.
Print a single integer *t* — the maximum number of tables that can be decorated in the required manner.
[ "5 4 3\n", "1 1 1\n", "2 3 3\n" ]
[ "4\n", "1\n", "2\n" ]
In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.
1,500
[ { "input": "5 4 3", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 3 3", "output": "2" }, { "input": "0 1 0", "output": "0" }, { "input": "0 3 3", "output": "2" }, { "input": "4 0 4", "output": "2" }, { "input": "1000000000 1000000000 1000000000", "output": "1000000000" }, { "input": "100 99 56", "output": "85" }, { "input": "1000 1000 1002", "output": "1000" }, { "input": "0 1 1000000000", "output": "1" }, { "input": "500000000 1000000000 500000000", "output": "666666666" }, { "input": "1000000000 2000000000 1000000000", "output": "1333333333" }, { "input": "2000000000 2000000000 2000000000", "output": "2000000000" }, { "input": "0 0 0", "output": "0" }, { "input": "1 2000000000 1000000000", "output": "1000000000" }, { "input": "1585222789 1889821127 2000000000", "output": "1825014638" }, { "input": "10000 7500 7500", "output": "8333" }, { "input": "150000 75000 75000", "output": "100000" }, { "input": "999288131 55884921 109298382", "output": "165183303" }, { "input": "100500 100500 3", "output": "67001" }, { "input": "1463615122 1988383731 837331500", "output": "1429776784" }, { "input": "1938 8999 1882", "output": "3820" }, { "input": "45 33 76", "output": "51" }, { "input": "100000 1 2", "output": "3" }, { "input": "198488 50 18", "output": "68" }, { "input": "82728372 939848 100139442", "output": "61269220" }, { "input": "99 5747 5298", "output": "3714" }, { "input": "3 5 2", "output": "3" }, { "input": "7511 7512 7513", "output": "7512" }, { "input": "1234567890 123456789 987654321", "output": "781893000" }, { "input": "500000000 2000000000 500000000", "output": "1000000000" }, { "input": "500000002 2000000000 500000001", "output": "1000000001" }, { "input": "520000000 1000000033 501000000", "output": "673666677" }, { "input": "10000 1000 100000", "output": "11000" }, { "input": "2000000000 500000000 499999999", "output": "999999999" }, { "input": "1999999999 500000000 500000000", "output": "999999999" }, { "input": "1 1 9", "output": "2" }, { "input": "3 0 0", "output": "0" }, { "input": "6 1 1", "output": "2" }, { "input": "2000000000 1999999999 1999999999", "output": "1999999999" }, { "input": "3 4 9", "output": "5" }, { "input": "3 3 6", "output": "4" } ]
1,656,985,297
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
9
61
0
from math import floor r, g, b = list(map(int, input().split())) s = r + g + b s = floor(s / 3) print(min(g + b, s))
Title: Table Decorations Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color? Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner. Input Specification: The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space. Output Specification: Print a single integer *t* — the maximum number of tables that can be decorated in the required manner. Demo Input: ['5 4 3\n', '1 1 1\n', '2 3 3\n'] Demo Output: ['4\n', '1\n', '2\n'] Note: In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.
```python from math import floor r, g, b = list(map(int, input().split())) s = r + g + b s = floor(s / 3) print(min(g + b, s)) ```
0
893
C
Rumor
PROGRAMMING
1,300
[ "dfs and similar", "graphs", "greedy" ]
null
null
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely.
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
[ "5 2\n2 5 3 4 8\n1 4\n4 5\n", "10 0\n1 2 3 4 5 6 7 8 9 10\n", "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n" ]
[ "10\n", "55\n", "15\n" ]
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
0
[ { "input": "5 2\n2 5 3 4 8\n1 4\n4 5", "output": "10" }, { "input": "10 0\n1 2 3 4 5 6 7 8 9 10", "output": "55" }, { "input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10", "output": "15" }, { "input": "1 0\n0", "output": "0" }, { "input": "1 0\n1000000000", "output": "1000000000" }, { "input": "2 0\n0 0", "output": "0" }, { "input": "2 0\n1000000000 0", "output": "1000000000" }, { "input": "2 0\n0 1000000000", "output": "1000000000" }, { "input": "2 0\n1000000000 1000000000", "output": "2000000000" }, { "input": "2 1\n0 0\n1 2", "output": "0" } ]
1,610,863,604
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
3
358
18,124,800
from collections import defaultdict import sys sys.setrecursionlimit(15000) def dfs(v,graph,visited,mini,l): visited[v]=True for i in graph[v]: if(visited[i]==False): mini=min(mini,l[i]) dfs(i,graph,visited,mini,l) return mini ver,edg=map(int,input().split()) l=list(map(int,input().split())) graph=defaultdict(list) zp=0 for i in range(edg): u,v=map(int,input().split()) graph[u].append(v) graph[v].append(u) if(u==0 or v==0): zp=1 if(zp==1): ver=ver else: l=[0]+l ver+=1 visited=[False]*(ver) summ=0 for i in range(len(visited)): if(visited[i]==False): mini=l[i] summ+=dfs(i,graph,visited,mini,l) print(summ)
Title: Rumor Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely. Input Specification: The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. Output Specification: Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. Demo Input: ['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n'] Demo Output: ['10\n', '55\n', '15\n'] Note: In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
```python from collections import defaultdict import sys sys.setrecursionlimit(15000) def dfs(v,graph,visited,mini,l): visited[v]=True for i in graph[v]: if(visited[i]==False): mini=min(mini,l[i]) dfs(i,graph,visited,mini,l) return mini ver,edg=map(int,input().split()) l=list(map(int,input().split())) graph=defaultdict(list) zp=0 for i in range(edg): u,v=map(int,input().split()) graph[u].append(v) graph[v].append(u) if(u==0 or v==0): zp=1 if(zp==1): ver=ver else: l=[0]+l ver+=1 visited=[False]*(ver) summ=0 for i in range(len(visited)): if(visited[i]==False): mini=l[i] summ+=dfs(i,graph,visited,mini,l) print(summ) ```
-1
938
C
Constructing Tests
PROGRAMMING
1,700
[ "binary search", "brute force", "constructive algorithms" ]
null
null
Let's denote a *m*-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size *m*<=×<=*m* of this matrix contains at least one zero. Consider the following problem: You are given two integers *n* and *m*. You have to construct an *m*-free square matrix of size *n*<=×<=*n* such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix. You don't have to solve this problem. Instead, you have to construct a few tests for it. You will be given *t* numbers *x*1, *x*2, ..., *x**t*. For every , find two integers *n**i* and *m**i* (*n**i*<=≥<=*m**i*) such that the answer for the aforementioned problem is exactly *x**i* if we set *n*<==<=*n**i* and *m*<==<=*m**i*.
The first line contains one integer *t* (1<=≤<=*t*<=≤<=100) — the number of tests you have to construct. Then *t* lines follow, *i*-th line containing one integer *x**i* (0<=≤<=*x**i*<=≤<=109). Note that in hacks you have to set *t*<==<=1.
For each test you have to construct, output two positive numbers *n**i* and *m**i* (1<=≤<=*m**i*<=≤<=*n**i*<=≤<=109) such that the maximum number of 1's in a *m**i*-free *n**i*<=×<=*n**i* matrix is exactly *x**i*. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer <=-<=1.
[ "3\n21\n0\n1\n" ]
[ "5 2\n1 1\n-1\n" ]
none
0
[ { "input": "3\n21\n0\n1", "output": "5 2\n1 1\n-1" }, { "input": "1\n420441920", "output": "-1" }, { "input": "1\n4", "output": "-1" }, { "input": "1\n297540", "output": "546 22" }, { "input": "1\n9", "output": "-1" }, { "input": "1\n144", "output": "-1" }, { "input": "1\n16", "output": "-1" }, { "input": "1\n25", "output": "-1" }, { "input": "1\n999944", "output": "-1" }, { "input": "1\n6", "output": "-1" }, { "input": "1\n14", "output": "-1" }, { "input": "1\n81", "output": "-1" }, { "input": "1\n2", "output": "-1" }, { "input": "1\n36", "output": "-1" }, { "input": "1\n2925", "output": "-1" }, { "input": "1\n5704", "output": "77 5" }, { "input": "1\n4104", "output": "-1" }, { "input": "1\n1980", "output": "-1" }, { "input": "1\n10", "output": "-1" }, { "input": "1\n4860", "output": "72 4" }, { "input": "1\n2601", "output": "-1" }, { "input": "1\n28", "output": "-1" }, { "input": "1\n56", "output": "-1" } ]
1,690,911,768
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
30
0
print("_RANDOM_GUESS_1690911768.8700302")# 1690911768.8700488
Title: Constructing Tests Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's denote a *m*-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size *m*<=×<=*m* of this matrix contains at least one zero. Consider the following problem: You are given two integers *n* and *m*. You have to construct an *m*-free square matrix of size *n*<=×<=*n* such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix. You don't have to solve this problem. Instead, you have to construct a few tests for it. You will be given *t* numbers *x*1, *x*2, ..., *x**t*. For every , find two integers *n**i* and *m**i* (*n**i*<=≥<=*m**i*) such that the answer for the aforementioned problem is exactly *x**i* if we set *n*<==<=*n**i* and *m*<==<=*m**i*. Input Specification: The first line contains one integer *t* (1<=≤<=*t*<=≤<=100) — the number of tests you have to construct. Then *t* lines follow, *i*-th line containing one integer *x**i* (0<=≤<=*x**i*<=≤<=109). Note that in hacks you have to set *t*<==<=1. Output Specification: For each test you have to construct, output two positive numbers *n**i* and *m**i* (1<=≤<=*m**i*<=≤<=*n**i*<=≤<=109) such that the maximum number of 1's in a *m**i*-free *n**i*<=×<=*n**i* matrix is exactly *x**i*. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer <=-<=1. Demo Input: ['3\n21\n0\n1\n'] Demo Output: ['5 2\n1 1\n-1\n'] Note: none
```python print("_RANDOM_GUESS_1690911768.8700302")# 1690911768.8700488 ```
0
248
B
Chilly Willy
PROGRAMMING
1,400
[ "math", "number theory" ]
null
null
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros.
A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
[ "1\n", "5\n" ]
[ "-1\n", "10080" ]
none
1,000
[ { "input": "1", "output": "-1" }, { "input": "5", "output": "10080" }, { "input": "6", "output": "100170" }, { "input": "4", "output": "1050" }, { "input": "15", "output": "100000000000110" }, { "input": "16", "output": "1000000000000050" }, { "input": "17", "output": "10000000000000080" }, { "input": "7", "output": "1000020" }, { "input": "120", "output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000170" }, { "input": "8", "output": "10000200" }, { "input": "3", "output": "210" }, { "input": "2", "output": "-1" }, { "input": "9", "output": "100000110" }, { "input": "10", "output": "1000000050" }, { "input": "11", "output": "10000000080" }, { "input": "12", "output": "100000000170" }, { "input": "13", "output": "1000000000020" }, { "input": "14", "output": "10000000000200" }, { "input": "100000", "output": 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"18", "output": "100000000000000170" }, { "input": "19", "output": "1000000000000000020" }, { "input": "20", "output": "10000000000000000200" }, { "input": "21", "output": "100000000000000000110" }, { "input": "22", "output": "1000000000000000000050" }, { "input": "23", "output": "10000000000000000000080" }, { "input": "24", "output": "100000000000000000000170" }, { "input": "25", "output": "1000000000000000000000020" }, { "input": "31", "output": "1000000000000000000000000000020" }, { "input": "33", "output": "100000000000000000000000000000110" }, { "input": "65", "output": "10000000000000000000000000000000000000000000000000000000000000080" }, { "input": "2345", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "5522", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "8824", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "9003", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "88888", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "77777", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "66666", "output": 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"1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "34532", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "27324", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "45332", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "1000", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "12398", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." } ]
1,695,631,633
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
e = int(input("Enter: ")) if e < 3: print("-1") else: x = 10**(e - 1) f = x % 210 print(x - f + 210)
Title: Chilly Willy Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros. Input Specification: A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105). Output Specification: Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist. Demo Input: ['1\n', '5\n'] Demo Output: ['-1\n', '10080'] Note: none
```python e = int(input("Enter: ")) if e < 3: print("-1") else: x = 10**(e - 1) f = x % 210 print(x - f + 210) ```
0
217
A
Ice Skating
PROGRAMMING
1,200
[ "brute force", "dfs and similar", "dsu", "graphs" ]
null
null
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created. We assume that Bajtek can only heap up snow drifts at integer coordinates.
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift. Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
[ "2\n2 1\n1 2\n", "2\n2 1\n4 1\n" ]
[ "1\n", "0\n" ]
none
500
[ { "input": "2\n2 1\n1 2", "output": "1" }, { "input": "2\n2 1\n4 1", "output": "0" }, { "input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459", "output": "21" }, { "input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815", "output": "16" }, { "input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663", "output": "10" }, { "input": "1\n321 88", "output": "0" }, { "input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689", "output": "7" }, { "input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510", "output": "6" }, { "input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888", "output": "5" }, { "input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802", "output": "13" }, { "input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662", "output": "11" }, { "input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786", "output": "38" }, { "input": "3\n175 201\n907 909\n388 360", "output": "2" }, { "input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86", "output": "6" }, { "input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820", "output": "16" }, { "input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901", "output": "31" }, { "input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210", "output": "29" }, { "input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894", "output": "29" }, { "input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823", "output": "13" }, { "input": "5\n175 158\n16 2\n397 381\n668 686\n957 945", "output": "4" }, { "input": "5\n312 284\n490 509\n730 747\n504 497\n782 793", "output": "4" }, { "input": "2\n802 903\n476 348", "output": "1" }, { "input": "4\n325 343\n425 442\n785 798\n275 270", "output": "3" }, { "input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653", "output": "25" }, { "input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124", "output": "34" }, { "input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255", "output": "29" }, { "input": "5\n664 666\n951 941\n739 742\n844 842\n2 2", "output": "4" }, { "input": "3\n939 867\n411 427\n757 708", "output": "2" }, { "input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636", "output": "34" }, { "input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195", "output": "28" }, { "input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488", "output": "22" }, { "input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235", "output": "8" }, { "input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621", "output": "22" }, { "input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461", "output": "22" }, { "input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924", "output": "6" }, { "input": "3\n1 1\n2 1\n2 2", "output": "0" }, { "input": "4\n1 1\n1 2\n2 1\n2 2", "output": "0" }, { "input": "5\n1 1\n1 2\n2 2\n3 1\n3 3", "output": "0" }, { "input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3", "output": "0" }, { "input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10", "output": "1" }, { "input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11", "output": "1" }, { "input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11", "output": "3" }, { "input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25", "output": "7" }, { "input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28", "output": "5" }, { "input": "3\n1 2\n2 1\n2 2", "output": "0" }, { "input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1", "output": "0" }, { "input": "4\n1 1\n1 2\n2 1\n2 2", "output": "0" }, { "input": "3\n1 1\n2 2\n1 2", "output": "0" }, { "input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4", "output": "1" }, { "input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100", "output": "6" }, { "input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4", "output": "0" }, { "input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3", "output": "0" }, { "input": "4\n3 1\n2 1\n2 2\n1 2", "output": "0" }, { "input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3", "output": "0" }, { "input": "3\n1 2\n1 3\n1 4", "output": "0" }, { "input": "4\n1 1\n2 2\n1 2\n2 1", "output": "0" }, { "input": "4\n1 3\n2 1\n3 2\n3 1", "output": "1" }, { "input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4", "output": "0" }, { "input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700", "output": "2" }, { "input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9", "output": "1" }, { "input": "4\n1 1\n2 2\n3 2\n3 1", "output": "0" }, { "input": "3\n1 2\n3 4\n3 2", "output": "0" }, { "input": "3\n1 1\n1 2\n2 2", "output": "0" }, { "input": "4\n5 5\n5 4\n6 3\n6 4", "output": "0" }, { "input": "3\n1 1\n2 2\n2 1", "output": "0" } ]
1,584,555,263
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
218
409,600
from collections import Counter n = int(input()) List = [] for i in range(n): a,b = map(int,input().split()) List.append(a) List.append(b) Dict = Counter(List) if(len(Dict.keys())==n+1): print(0) else: print(1)
Title: Ice Skating Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created. We assume that Bajtek can only heap up snow drifts at integer coordinates. Input Specification: The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift. Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. Output Specification: Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. Demo Input: ['2\n2 1\n1 2\n', '2\n2 1\n4 1\n'] Demo Output: ['1\n', '0\n'] Note: none
```python from collections import Counter n = int(input()) List = [] for i in range(n): a,b = map(int,input().split()) List.append(a) List.append(b) Dict = Counter(List) if(len(Dict.keys())==n+1): print(0) else: print(1) ```
0
835
A
Key races
PROGRAMMING
800
[ "math" ]
null
null
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds. If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows: 1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it. The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw. Given the length of the text and the information about participants, determine the result of the game.
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
[ "5 1 2 1 2\n", "3 3 1 1 1\n", "4 5 3 1 5\n" ]
[ "First\n", "Second\n", "Friendship\n" ]
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins. In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins. In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
500
[ { "input": "5 1 2 1 2", "output": "First" }, { "input": "3 3 1 1 1", "output": "Second" }, { "input": "4 5 3 1 5", "output": "Friendship" }, { "input": "1000 1000 1000 1000 1000", "output": "Friendship" }, { "input": "1 1 1 1 1", "output": "Friendship" }, { "input": "8 8 1 1 1", "output": "Second" }, { "input": "15 14 32 65 28", "output": "First" }, { "input": "894 197 325 232 902", "output": "First" }, { "input": "1 2 8 8 5", "output": "Friendship" }, { "input": "37 261 207 1 1000", "output": "Friendship" }, { "input": "29 344 406 900 1", "output": "Friendship" }, { "input": "1 2 8 9 8", "output": "First" }, { "input": "2 9 8 8 9", "output": "Friendship" }, { "input": "213 480 811 134 745", "output": "First" }, { "input": "2 313 856 964 421", "output": "Friendship" }, { "input": "1 10 2 6 10", "output": "Friendship" }, { "input": "2 7 6 2 3", "output": "Friendship" }, { "input": "637 324 69 612 998", "output": "Second" }, { "input": "13 849 819 723 918", "output": "Friendship" }, { "input": "9 5 7 8 7", "output": "First" }, { "input": "6 5 7 10 4", "output": "Friendship" }, { "input": "61 464 623 89 548", "output": "First" }, { "input": "641 31 29 161 802", "output": "Friendship" }, { "input": "3 3 1 6 9", "output": "Friendship" }, { "input": "2 3 9 8 2", "output": "Friendship" }, { "input": "485 117 368 567 609", "output": "First" }, { "input": "4 202 512 995 375", "output": "Friendship" }, { "input": "424 41 41 909 909", "output": "Friendship" }, { "input": "884 913 263 641 265", "output": "Second" }, { "input": "12 462 8 311 327", "output": "Second" }, { "input": "436 306 266 493 580", "output": "Second" }, { "input": "69 1 2 1 2", "output": "First" } ]
1,587,005,677
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
#include<bits/stdc++.h> using namespace std; int main(){ int s,v1,v2,t1,t2,f,se; cin >>s >> v1 >> v2 >> t1 >> t2; f = v1*s + 2*t1; se = v2*s + 2*t2; if(f > se) cout << "Second" << endl; else if ( f == se) cout << "Friendship" << endl; else cout << "First" << endl; return 0; }
Title: Key races Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds. If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows: 1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it. The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw. Given the length of the text and the information about participants, determine the result of the game. Input Specification: The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. Output Specification: If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". Demo Input: ['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n'] Demo Output: ['First\n', 'Second\n', 'Friendship\n'] Note: In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins. In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins. In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
```python #include<bits/stdc++.h> using namespace std; int main(){ int s,v1,v2,t1,t2,f,se; cin >>s >> v1 >> v2 >> t1 >> t2; f = v1*s + 2*t1; se = v2*s + 2*t2; if(f > se) cout << "Second" << endl; else if ( f == se) cout << "Friendship" << endl; else cout << "First" << endl; return 0; } ```
-1
0
none
none
none
0
[ "none" ]
null
null
Innokenty is a president of a new football league in Byteland. The first task he should do is to assign short names to all clubs to be shown on TV next to the score. Of course, the short names should be distinct, and Innokenty wants that all short names consist of three letters. Each club's full name consist of two words: the team's name and the hometown's name, for example, "DINAMO BYTECITY". Innokenty doesn't want to assign strange short names, so he wants to choose such short names for each club that: 1. the short name is the same as three first letters of the team's name, for example, for the mentioned club it is "DIN", 1. or, the first two letters of the short name should be the same as the first two letters of the team's name, while the third letter is the same as the first letter in the hometown's name. For the mentioned club it is "DIB". Apart from this, there is a rule that if for some club *x* the second option of short name is chosen, then there should be no club, for which the first option is chosen which is the same as the first option for the club *x*. For example, if the above mentioned club has short name "DIB", then no club for which the first option is chosen can have short name equal to "DIN". However, it is possible that some club have short name "DIN", where "DI" are the first two letters of the team's name, and "N" is the first letter of hometown's name. Of course, no two teams can have the same short name. Help Innokenty to choose a short name for each of the teams. If this is impossible, report that. If there are multiple answer, any of them will suit Innokenty. If for some team the two options of short name are equal, then Innokenty will formally think that only one of these options is chosen.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of clubs in the league. Each of the next *n* lines contains two words — the team's name and the hometown's name for some club. Both team's name and hometown's name consist of uppercase English letters and have length at least 3 and at most 20.
It it is not possible to choose short names and satisfy all constraints, print a single line "NO". Otherwise, in the first line print "YES". Then print *n* lines, in each line print the chosen short name for the corresponding club. Print the clubs in the same order as they appeared in input. If there are multiple answers, print any of them.
[ "2\nDINAMO BYTECITY\nFOOTBALL MOSCOW\n", "2\nDINAMO BYTECITY\nDINAMO BITECITY\n", "3\nPLAYFOOTBALL MOSCOW\nPLAYVOLLEYBALL SPB\nGOGO TECHNOCUP\n", "3\nABC DEF\nABC EFG\nABD OOO\n" ]
[ "YES\nDIN\nFOO\n", "NO\n", "YES\nPLM\nPLS\nGOG\n", "YES\nABD\nABE\nABO\n" ]
In the first sample Innokenty can choose first option for both clubs. In the second example it is not possible to choose short names, because it is not possible that one club has first option, and the other has second option if the first options are equal for both clubs. In the third example Innokenty can choose the second options for the first two clubs, and the first option for the third club. In the fourth example note that it is possible that the chosen short name for some club *x* is the same as the first option of another club *y* if the first options of *x* and *y* are different.
0
[ { "input": "2\nDINAMO BYTECITY\nFOOTBALL MOSCOW", "output": "YES\nDIN\nFOO" }, { "input": "2\nDINAMO BYTECITY\nDINAMO BITECITY", "output": "NO" }, { "input": "3\nPLAYFOOTBALL MOSCOW\nPLAYVOLLEYBALL SPB\nGOGO TECHNOCUP", "output": "YES\nPLM\nPLS\nGOG" }, { "input": "3\nABC DEF\nABC EFG\nABD OOO", "output": "YES\nABD\nABE\nABO" }, { "input": "3\nABC DEF\nABC EFG\nABC EEEEE", "output": "NO" }, { "input": "3\nABC DEF\nABC EFG\nABD CABA", "output": "YES\nABD\nABE\nABC" }, { "input": "3\nABC DEF\nABC EFG\nABD EABA", "output": "NO" }, { "input": "1\nAAA AAA", "output": "YES\nAAA" }, { "input": "1\nAAAAAAAAAAAAAAAAAAAA ZZZZZZZZZZZZZZZZZZZZ", "output": "YES\nAAA" }, { "input": "5\nADAC BABC\nABB DCB\nABB BCDC\nDBAC BAC\nDBBC DBC", "output": "YES\nADA\nABD\nABB\nDBA\nDBB" }, { "input": "5\nIAH HJIE\nIAH FJK\nIAH BIAA\nIAH AFG\nIAH DEFF", "output": "YES\nIAH\nIAF\nIAB\nIAA\nIAD" }, { "input": "10\nIJGDI KHB\nHBI CKKCG\nFHE GCAA\nEDCGH HHICE\nGFH AIHD\nHED KIK\nDCK BCFIJ\nFFIHE FDB\nJGB AKKI\nIJD CAG", "output": "YES\nIJG\nHBI\nFHE\nEDC\nGFH\nHED\nDCK\nFFI\nJGB\nIJD" }, { "input": "10\nEDBG IGGAC\nEDBG GIKAG\nEDBG IKGEI\nEDBG AJEG\nEDBG HAD\nEDBG ACKK\nEDBG FEDE\nEDBG DAB\nEDBG CCJBD\nEDBG KKGFB", "output": "NO" }, { "input": "10\nADE GBH\nJJDGJ AAF\nJJDGJ BBKG\nADE FKH\nADE CIA\nAIE JCBJ\nAIE BBJB\nEBAK JDB\nJJDGJ IDBG\nCEJE FIG", "output": "YES\nADG\nJJA\nJJB\nADF\nADC\nAIJ\nAIB\nEBA\nJJI\nCEJ" }, { "input": "4\nABA DEF\nABB DEF\nABC DEF\nABE DEF", "output": "YES\nABA\nABB\nABC\nABE" }, { "input": "2\nABC CCC\nABE CCC", "output": "YES\nABC\nABE" }, { "input": "2\nABS SSS\nABD SSD", "output": "YES\nABS\nABD" } ]
1,689,423,249
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
46
0
print("_RANDOM_GUESS_1689423249.2576973")# 1689423249.2577193
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Innokenty is a president of a new football league in Byteland. The first task he should do is to assign short names to all clubs to be shown on TV next to the score. Of course, the short names should be distinct, and Innokenty wants that all short names consist of three letters. Each club's full name consist of two words: the team's name and the hometown's name, for example, "DINAMO BYTECITY". Innokenty doesn't want to assign strange short names, so he wants to choose such short names for each club that: 1. the short name is the same as three first letters of the team's name, for example, for the mentioned club it is "DIN", 1. or, the first two letters of the short name should be the same as the first two letters of the team's name, while the third letter is the same as the first letter in the hometown's name. For the mentioned club it is "DIB". Apart from this, there is a rule that if for some club *x* the second option of short name is chosen, then there should be no club, for which the first option is chosen which is the same as the first option for the club *x*. For example, if the above mentioned club has short name "DIB", then no club for which the first option is chosen can have short name equal to "DIN". However, it is possible that some club have short name "DIN", where "DI" are the first two letters of the team's name, and "N" is the first letter of hometown's name. Of course, no two teams can have the same short name. Help Innokenty to choose a short name for each of the teams. If this is impossible, report that. If there are multiple answer, any of them will suit Innokenty. If for some team the two options of short name are equal, then Innokenty will formally think that only one of these options is chosen. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of clubs in the league. Each of the next *n* lines contains two words — the team's name and the hometown's name for some club. Both team's name and hometown's name consist of uppercase English letters and have length at least 3 and at most 20. Output Specification: It it is not possible to choose short names and satisfy all constraints, print a single line "NO". Otherwise, in the first line print "YES". Then print *n* lines, in each line print the chosen short name for the corresponding club. Print the clubs in the same order as they appeared in input. If there are multiple answers, print any of them. Demo Input: ['2\nDINAMO BYTECITY\nFOOTBALL MOSCOW\n', '2\nDINAMO BYTECITY\nDINAMO BITECITY\n', '3\nPLAYFOOTBALL MOSCOW\nPLAYVOLLEYBALL SPB\nGOGO TECHNOCUP\n', '3\nABC DEF\nABC EFG\nABD OOO\n'] Demo Output: ['YES\nDIN\nFOO\n', 'NO\n', 'YES\nPLM\nPLS\nGOG\n', 'YES\nABD\nABE\nABO\n'] Note: In the first sample Innokenty can choose first option for both clubs. In the second example it is not possible to choose short names, because it is not possible that one club has first option, and the other has second option if the first options are equal for both clubs. In the third example Innokenty can choose the second options for the first two clubs, and the first option for the third club. In the fourth example note that it is possible that the chosen short name for some club *x* is the same as the first option of another club *y* if the first options of *x* and *y* are different.
```python print("_RANDOM_GUESS_1689423249.2576973")# 1689423249.2577193 ```
0
1,003
A
Polycarp's Pockets
PROGRAMMING
800
[ "implementation" ]
null
null
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
[ "6\n1 2 4 3 3 2\n", "1\n100\n" ]
[ "2\n", "1\n" ]
none
0
[ { "input": "6\n1 2 4 3 3 2", "output": "2" }, { "input": "1\n100", "output": "1" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" }, { "input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71", "output": "51" }, { "input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29", "output": "1" }, { "input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24", "output": "2" }, { "input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6", "output": "10" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "99" }, { "input": "7\n1 2 3 3 3 1 2", "output": "3" }, { "input": "5\n1 2 3 4 5", "output": "1" }, { "input": "7\n1 2 3 4 5 6 7", "output": "1" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "1" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "1" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "1" }, { "input": "3\n2 1 1", "output": "2" }, { "input": "11\n1 2 3 4 5 6 7 8 9 1 1", "output": "3" }, { "input": "12\n1 2 1 1 1 1 1 1 1 1 1 1", "output": "11" }, { "input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "14" }, { "input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15" }, { "input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "16" }, { "input": "3\n1 1 1", "output": "3" }, { "input": "3\n1 2 3", "output": "1" }, { "input": "10\n1 1 1 1 2 2 1 1 9 10", "output": "6" }, { "input": "2\n1 1", "output": "2" }, { "input": "56\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "56" }, { "input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92", "output": "6" }, { "input": "10\n1 2 2 3 3 3 4 4 4 4", "output": "4" }, { "input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21", "output": "3" }, { "input": "5\n5 5 5 5 1", "output": "4" }, { "input": "6\n2 3 2 5 2 6", "output": "3" }, { "input": "3\n58 59 58", "output": "2" }, { "input": "9\n1 2 3 4 5 6 7 8 8", "output": "2" }, { "input": "97\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "97" }, { "input": "3\n95 95 4", "output": "2" }, { "input": "3\n2 2 5", "output": "2" } ]
1,592,632,521
2,147,483,647
Python 3
OK
TESTS
37
109
0
n = input() arr = list(map(int, input().split())) arrS = set(arr) c=0 for x in arrS: y = arr.count(x) if y>c: c=y print(c)
Title: Polycarp's Pockets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. Input Specification: The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. Output Specification: Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. Demo Input: ['6\n1 2 4 3 3 2\n', '1\n100\n'] Demo Output: ['2\n', '1\n'] Note: none
```python n = input() arr = list(map(int, input().split())) arrS = set(arr) c=0 for x in arrS: y = arr.count(x) if y>c: c=y print(c) ```
3
244
A
Dividing Orange
PROGRAMMING
900
[ "implementation" ]
null
null
One day Ms Swan bought an orange in a shop. The orange consisted of *n*·*k* segments, numbered with integers from 1 to *n*·*k*. There were *k* children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the *i*-th (1<=≤<=*i*<=≤<=*k*) child wrote the number *a**i* (1<=≤<=*a**i*<=≤<=*n*·*k*). All numbers *a**i* accidentally turned out to be different. Now the children wonder, how to divide the orange so as to meet these conditions: - each child gets exactly *n* orange segments; - the *i*-th child gets the segment with number *a**i* for sure; - no segment goes to two children simultaneously. Help the children, divide the orange and fulfill the requirements, described above.
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=30). The second line contains *k* space-separated integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*·*k*), where *a**i* is the number of the orange segment that the *i*-th child would like to get. It is guaranteed that all numbers *a**i* are distinct.
Print exactly *n*·*k* distinct integers. The first *n* integers represent the indexes of the segments the first child will get, the second *n* integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces. You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.
[ "2 2\n4 1\n", "3 1\n2\n" ]
[ "2 4 \n1 3 \n", "3 2 1 \n" ]
none
500
[ { "input": "2 2\n4 1", "output": "2 4 \n1 3 " }, { "input": "3 1\n2", "output": "3 2 1 " }, { "input": "5 5\n25 24 23 22 21", "output": "2 3 1 25 4 \n7 6 8 5 24 \n10 12 9 23 11 \n13 15 14 16 22 \n19 21 20 17 18 " }, { "input": "1 30\n8 22 13 25 10 30 12 27 6 4 7 2 20 16 26 14 15 17 23 3 24 9 5 11 29 1 19 28 21 18", "output": "8 \n22 \n13 \n25 \n10 \n30 \n12 \n27 \n6 \n4 \n7 \n2 \n20 \n16 \n26 \n14 \n15 \n17 \n23 \n3 \n24 \n9 \n5 \n11 \n29 \n1 \n19 \n28 \n21 \n18 " }, { "input": "30 1\n29", "output": "8 20 17 12 5 26 13 2 19 22 28 16 10 4 6 11 3 25 1 27 15 9 30 24 21 18 14 23 29 7 " }, { "input": "10 10\n13 39 6 75 84 94 96 21 85 71", "output": "9 3 1 13 5 7 4 2 10 8 \n17 12 19 11 39 14 15 18 16 20 \n22 27 6 24 25 30 26 28 23 29 \n36 33 75 34 38 31 35 40 37 32 \n43 44 49 42 46 48 47 45 84 41 \n51 94 52 56 57 54 50 55 53 58 \n64 60 62 61 66 59 63 96 67 65 \n72 69 76 77 70 78 73 21 74 68 \n81 85 87 88 80 83 89 86 79 82 \n93 91 100 99 98 71 90 95 92 97 " }, { "input": "10 15\n106 109 94 50 3 143 147 10 89 145 29 28 87 126 110", "output": "9 4 1 106 6 7 5 2 11 8 \n17 13 19 12 109 14 15 18 16 20 \n21 26 94 23 24 31 25 27 22 30 \n37 34 50 35 39 32 36 40 38 33 \n43 44 49 42 46 48 47 45 3 41 \n52 143 53 57 58 55 51 56 54 59 \n65 61 63 62 67 60 64 147 68 66 \n72 70 75 76 71 77 73 10 74 69 \n80 89 84 85 79 82 86 83 78 81 \n92 90 98 97 96 145 88 93 91 95 \n100 104 105 103 102 108 99 101 29 107 \n111 114 112 116 119 118 28 113 117 115 \n128 120 122 125 129 127 87 124 123 121 \n133 136 130 134 132 131 135 126 137 138 \n142 141 144 148 146 149 110 140..." }, { "input": "15 10\n126 111 12 6 28 47 51 116 53 35", "output": "9 13 1 14 5 16 15 2 10 8 126 3 11 4 7 \n111 22 21 26 20 30 17 23 18 19 24 31 27 25 29 \n43 40 41 39 42 12 45 44 34 37 32 36 38 33 46 \n59 6 57 56 58 49 62 54 50 52 63 61 48 55 60 \n70 67 71 75 69 77 72 65 68 73 76 74 28 64 66 \n80 89 86 79 87 91 81 78 88 83 85 82 90 84 47 \n95 93 51 99 104 98 103 101 100 102 97 96 94 92 105 \n120 115 113 118 109 119 110 116 114 106 121 117 108 107 112 \n135 133 128 125 123 131 129 122 124 53 134 132 130 127 136 \n148 139 141 143 146 144 147 138 137 145 142 149 140 150 35 \n..." }, { "input": "30 30\n455 723 796 90 7 881 40 736 147 718 560 619 468 363 161 767 282 19 111 369 443 850 871 242 713 789 208 435 135 411", "output": "9 22 18 13 5 28 14 2 21 24 30 17 11 4 6 12 3 27 1 29 16 10 31 26 23 20 15 25 455 8 \n723 52 49 60 45 48 34 59 58 44 32 57 61 56 51 33 42 37 41 38 47 53 36 50 54 55 46 39 43 35 \n89 71 796 74 78 70 88 67 84 85 63 83 82 62 72 79 81 80 73 91 69 66 65 87 77 75 64 68 86 76 \n115 90 102 121 104 106 109 98 112 120 119 105 103 97 113 93 100 118 107 96 117 92 94 116 95 101 110 108 114 99 \n136 133 148 123 144 139 149 142 7 140 138 127 150 129 122 130 143 126 134 152 132 145 131 146 125 151 137 128 124 141 \n154 177..." }, { "input": "1 1\n1", "output": "1 " }, { "input": "2 1\n1", "output": "2 1 " }, { "input": "1 2\n2 1", "output": "2 \n1 " }, { "input": "1 3\n2 3 1", "output": "2 \n3 \n1 " }, { "input": "2 3\n3 2 1", "output": "4 3 \n2 5 \n1 6 " }, { "input": "3 3\n6 7 8", "output": "2 6 1 \n7 4 3 \n5 9 8 " }, { "input": "3 1\n3", "output": "2 3 1 " }, { "input": "3 2\n5 4", "output": "2 5 1 \n4 6 3 " }, { "input": "12 13\n149 22 133 146 151 64 45 88 77 126 92 134 143", "output": "8 11 1 10 5 6 4 2 9 7 149 3 \n14 13 19 12 17 16 22 20 21 23 15 18 \n133 28 34 32 31 25 30 33 24 29 26 27 \n35 42 38 40 43 46 39 41 44 146 36 37 \n56 51 48 49 50 54 53 151 57 52 47 55 \n61 58 65 68 67 59 62 66 69 63 64 60 \n80 70 75 74 76 81 45 72 78 73 79 71 \n94 85 88 83 90 87 86 89 93 82 84 91 \n99 104 98 96 103 105 102 97 77 95 101 100 \n116 109 107 111 115 113 126 108 112 110 114 106 \n127 121 125 118 120 128 123 92 119 122 117 124 \n139 132 136 130 131 140 141 134 137 138 135 129 \n150 142 144 155 154..." }, { "input": "30 29\n427 740 444 787 193 268 19 767 46 276 245 468 661 348 402 62 665 425 398 503 89 455 200 772 355 442 863 416 164", "output": "8 21 17 12 5 27 13 2 20 23 29 16 10 4 6 11 3 26 1 28 15 9 30 25 22 18 14 24 427 7 \n740 51 48 59 43 47 33 58 57 42 31 56 60 55 50 32 40 36 39 37 45 52 35 49 53 54 44 38 41 34 \n90 71 444 74 78 70 88 67 84 85 63 83 82 61 72 79 81 80 73 91 69 66 65 87 77 75 64 68 86 76 \n114 787 102 120 104 106 109 98 111 119 118 105 103 97 112 93 100 117 107 96 116 92 94 115 95 101 110 108 113 99 \n134 132 145 122 142 137 146 140 193 138 136 126 147 128 121 129 141 125 133 149 131 143 130 144 124 148 135 127 123 139 \n151 1..." }, { "input": "29 30\n173 601 360 751 194 411 708 598 236 812 855 647 100 106 59 38 822 196 529 417 606 159 384 389 300 172 544 726 702 799", "output": "8 20 17 12 5 26 13 2 19 22 28 16 10 4 6 11 3 25 1 27 15 9 7 24 21 18 14 23 173 \n47 36 37 35 45 51 49 41 31 33 29 32 46 57 52 48 54 34 55 53 56 30 601 44 43 39 40 42 50 \n77 79 84 86 64 72 75 60 76 78 81 73 80 58 82 69 70 67 83 65 68 62 360 71 61 63 85 66 74 \n90 107 751 110 105 93 98 96 95 97 116 91 109 102 115 87 99 104 114 88 92 113 94 111 101 89 103 112 108 \n140 127 144 134 118 125 141 137 119 133 128 139 124 121 130 126 120 142 136 122 132 117 194 131 129 143 138 123 135 \n147 168 163 154 174 160 146..." }, { "input": "29 29\n669 371 637 18 176 724 137 757 407 420 658 737 188 408 185 416 425 293 178 557 8 104 139 819 268 403 255 63 793", "output": "9 22 19 13 5 28 14 2 21 24 30 17 11 4 6 12 3 27 1 29 16 10 7 26 23 20 15 25 669 \n48 38 39 37 46 52 50 42 33 35 31 34 47 58 53 49 55 36 56 54 57 32 371 45 44 40 41 43 51 \n78 80 85 87 65 73 76 60 77 79 82 74 81 59 83 70 71 68 84 66 69 62 637 72 61 64 86 67 75 \n91 107 18 110 106 94 99 97 96 98 116 92 109 102 115 88 100 105 114 89 93 113 95 111 101 90 103 112 108 \n142 127 146 134 118 125 143 138 119 133 128 141 124 121 130 126 120 144 136 122 132 117 176 131 129 145 140 123 135 \n149 169 164 156 173 161 14..." }, { "input": "28 29\n771 736 590 366 135 633 68 789 193 459 137 370 216 692 730 712 537 356 752 757 796 541 804 27 431 162 196 630 684", "output": "8 20 17 12 5 26 13 2 19 22 771 16 10 4 6 11 3 25 1 28 15 9 7 24 21 18 14 23 \n34 55 49 41 54 45 33 37 35 53 29 40 30 32 43 31 36 51 736 44 39 46 38 50 48 52 47 42 \n77 65 78 73 63 56 72 590 76 62 74 57 83 69 58 80 60 79 66 59 64 82 67 70 81 61 71 75 \n107 104 92 94 106 109 84 88 86 99 98 105 366 93 103 101 89 87 95 90 100 85 91 102 97 108 110 96 \n124 125 113 123 119 120 121 134 127 132 117 129 116 130 138 111 118 131 122 139 128 114 112 126 115 136 133 135 \n141 633 142 153 160 152 149 156 166 158 161 144..." }, { "input": "29 29\n669 371 637 18 176 724 137 757 407 420 658 737 188 408 185 416 425 293 178 557 8 104 139 819 268 403 255 63 793", "output": "9 22 19 13 5 28 14 2 21 24 30 17 11 4 6 12 3 27 1 29 16 10 7 26 23 20 15 25 669 \n48 38 39 37 46 52 50 42 33 35 31 34 47 58 53 49 55 36 56 54 57 32 371 45 44 40 41 43 51 \n78 80 85 87 65 73 76 60 77 79 82 74 81 59 83 70 71 68 84 66 69 62 637 72 61 64 86 67 75 \n91 107 18 110 106 94 99 97 96 98 116 92 109 102 115 88 100 105 114 89 93 113 95 111 101 90 103 112 108 \n142 127 146 134 118 125 143 138 119 133 128 141 124 121 130 126 120 144 136 122 132 117 176 131 129 145 140 123 135 \n149 169 164 156 173 161 14..." }, { "input": "27 3\n12 77 80", "output": "8 21 18 13 5 27 14 2 20 23 12 17 10 4 6 11 3 26 1 24 16 9 7 25 22 19 15 \n43 32 46 48 51 37 41 49 77 30 40 28 34 38 44 35 31 45 52 50 47 29 36 53 42 39 33 \n62 61 78 63 81 55 70 79 67 73 58 69 59 64 80 54 56 57 68 72 65 60 71 66 74 75 76 " }, { "input": "3 27\n77 9 32 56 7 65 58 24 64 19 49 62 47 44 28 79 76 71 21 4 18 23 51 53 12 6 20", "output": "2 77 1 \n9 5 3 \n8 10 32 \n13 56 11 \n15 7 14 \n65 17 16 \n22 58 25 \n24 26 27 \n29 64 30 \n31 33 19 \n35 34 49 \n62 37 36 \n47 38 39 \n44 40 41 \n42 43 28 \n46 45 79 \n48 50 76 \n71 54 52 \n57 21 55 \n60 4 59 \n61 18 63 \n66 23 67 \n68 51 69 \n72 70 53 \n12 73 74 \n75 6 78 \n81 20 80 " }, { "input": "10 30\n165 86 241 45 144 43 95 250 28 240 42 15 295 211 48 99 199 156 206 109 100 194 229 224 57 10 220 79 44 203", "output": "8 3 1 165 5 6 4 2 9 7 \n17 12 19 11 86 13 14 18 16 20 \n21 26 241 23 24 30 25 27 22 29 \n36 33 45 34 38 31 35 39 37 32 \n46 47 53 41 50 52 51 49 144 40 \n55 43 56 61 62 59 54 60 58 63 \n69 65 67 66 71 64 68 95 72 70 \n76 74 80 81 75 82 77 250 78 73 \n85 28 90 91 84 88 92 89 83 87 \n97 94 104 103 102 240 93 98 96 101 \n106 111 112 110 108 114 105 107 42 113 \n115 118 116 120 123 122 15 117 121 119 \n131 124 126 129 132 130 295 128 127 125 \n136 139 133 137 135 134 138 211 140 141 \n146 145 147 149 148 150 4..." }, { "input": "30 10\n71 146 274 157 190 85 32 152 25 278", "output": "8 20 17 12 5 27 13 2 19 22 29 16 10 4 6 11 3 26 1 28 15 9 30 24 21 18 14 23 71 7 \n146 51 48 59 44 47 34 58 57 43 31 56 60 55 50 33 41 37 40 38 46 52 36 49 53 54 45 39 42 35 \n90 70 274 74 78 69 89 66 84 86 62 83 82 61 72 79 81 80 73 91 68 65 64 88 77 75 63 67 87 76 \n114 157 102 120 104 106 109 98 111 119 118 105 103 97 112 93 100 117 107 96 116 92 94 115 95 101 110 108 113 99 \n134 132 145 122 142 137 147 140 190 138 136 126 148 128 121 129 141 125 133 150 131 143 130 144 124 149 135 127 123 139 \n153 17..." }, { "input": "7 1\n5", "output": "2 3 1 5 6 7 4 " }, { "input": "6 1\n5", "output": "2 3 1 5 6 4 " } ]
1,555,325,533
2,147,483,647
Python 3
OK
TESTS
29
218
0
N, K = map(int, input().split()) A = list(map(int, input().split())) vals = set(range(1, N * K + 1)) answer = [[] for k in range(K)] for k in range(K): answer[k].append(A[k]) vals.remove(A[k]) k = 0 while len(vals) > 0: answer[k].append(vals.pop()) k = (k + 1) % K for k in range(K): print(" ".join([str(x) for x in answer[k]]))
Title: Dividing Orange Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Ms Swan bought an orange in a shop. The orange consisted of *n*·*k* segments, numbered with integers from 1 to *n*·*k*. There were *k* children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the *i*-th (1<=≤<=*i*<=≤<=*k*) child wrote the number *a**i* (1<=≤<=*a**i*<=≤<=*n*·*k*). All numbers *a**i* accidentally turned out to be different. Now the children wonder, how to divide the orange so as to meet these conditions: - each child gets exactly *n* orange segments; - the *i*-th child gets the segment with number *a**i* for sure; - no segment goes to two children simultaneously. Help the children, divide the orange and fulfill the requirements, described above. Input Specification: The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=30). The second line contains *k* space-separated integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*·*k*), where *a**i* is the number of the orange segment that the *i*-th child would like to get. It is guaranteed that all numbers *a**i* are distinct. Output Specification: Print exactly *n*·*k* distinct integers. The first *n* integers represent the indexes of the segments the first child will get, the second *n* integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces. You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them. Demo Input: ['2 2\n4 1\n', '3 1\n2\n'] Demo Output: ['2 4 \n1 3 \n', '3 2 1 \n'] Note: none
```python N, K = map(int, input().split()) A = list(map(int, input().split())) vals = set(range(1, N * K + 1)) answer = [[] for k in range(K)] for k in range(K): answer[k].append(A[k]) vals.remove(A[k]) k = 0 while len(vals) > 0: answer[k].append(vals.pop()) k = (k + 1) % K for k in range(K): print(" ".join([str(x) for x in answer[k]])) ```
3
96
A
Football
PROGRAMMING
900
[ "implementation", "strings" ]
A. Football
2
256
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Print "YES" if the situation is dangerous. Otherwise, print "NO".
[ "001001\n", "1000000001\n" ]
[ "NO\n", "YES\n" ]
none
500
[ { "input": "001001", "output": "NO" }, { "input": "1000000001", "output": "YES" }, { "input": "00100110111111101", "output": "YES" }, { "input": "11110111111111111", "output": "YES" }, { "input": "01", "output": "NO" }, { "input": "10100101", "output": "NO" }, { "input": "1010010100000000010", "output": "YES" }, { "input": "101010101", "output": "NO" }, { "input": "000000000100000000000110101100000", "output": "YES" }, { "input": "100001000000110101100000", "output": "NO" }, { "input": "100001000011010110000", "output": "NO" }, { "input": "010", "output": "NO" }, { "input": "10101011111111111111111111111100", "output": "YES" }, { "input": "1001101100", "output": "NO" }, { "input": "1001101010", "output": "NO" }, { "input": "1111100111", "output": "NO" }, { "input": "00110110001110001111", "output": "NO" }, { "input": "11110001001111110001", "output": "NO" }, { "input": "10001111001011111101", "output": "NO" }, { "input": "10000010100000001000110001010100001001001010011", "output": "YES" }, { "input": "01111011111010111100101100001011001010111110000010", "output": "NO" }, { "input": "00100000100100101110011001011011101110110110010100", "output": "NO" }, { "input": "10110100110001001011110101110010100010000000000100101010111110111110100011", "output": "YES" }, { "input": "00011101010101111001011011001101101011111101000010100000111000011100101011", "output": "NO" }, { "input": "01110000110100110101110100111000101101011101011110110100100111100001110111", "output": "NO" }, { "input": "11110110011000100111100111101101011111110100010101011011111101110110110111", "output": "YES" }, { "input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100", "output": "NO" }, { "input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010", "output": "NO" }, { "input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110", "output": "NO" }, { "input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000", "output": "NO" }, { "input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110", "output": "YES" }, { "input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000", "output": "NO" }, { "input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101", "output": "YES" }, { "input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101", "output": "YES" }, { "input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000", "output": "NO" }, { "input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000", "output": "YES" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "YES" }, { "input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111", "output": "YES" }, { "input": "10100101000", "output": "NO" }, { "input": "11110111011101", "output": "NO" }, { "input": "10000000", "output": "YES" }, { "input": "00000001", "output": "YES" }, { "input": "01111111", "output": "YES" }, { "input": "11111110", "output": "YES" } ]
1,697,249,440
2,147,483,647
Python 3
OK
TESTS
44
62
0
players = input() length = len(players) previous = players[0] count = 1 flag = False for i in range(1, length): if players[i] == previous: count += 1 else: previous = players[i] count = 1 if count == 7: flag = True break if flag: print("YES") else: print("NO") # 化院 荆屹然 2300011884
Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. Input Specification: The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. Output Specification: Print "YES" if the situation is dangerous. Otherwise, print "NO". Demo Input: ['001001\n', '1000000001\n'] Demo Output: ['NO\n', 'YES\n'] Note: none
```python players = input() length = len(players) previous = players[0] count = 1 flag = False for i in range(1, length): if players[i] == previous: count += 1 else: previous = players[i] count = 1 if count == 7: flag = True break if flag: print("YES") else: print("NO") # 化院 荆屹然 2300011884 ```
3.9845
950
A
Left-handers, Right-handers and Ambidexters
PROGRAMMING
800
[ "implementation", "math" ]
null
null
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
[ "1 4 2\n", "5 5 5\n", "0 2 0\n" ]
[ "6\n", "14\n", "0\n" ]
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
500
[ { "input": "1 4 2", "output": "6" }, { "input": "5 5 5", "output": "14" }, { "input": "0 2 0", "output": "0" }, { "input": "30 70 34", "output": "128" }, { "input": "89 32 24", "output": "112" }, { "input": "89 44 77", "output": "210" }, { "input": "0 0 0", "output": "0" }, { "input": "100 100 100", "output": "300" }, { "input": "1 1 1", "output": "2" }, { "input": "30 70 35", "output": "130" }, { "input": "89 44 76", "output": "208" }, { "input": "0 100 100", "output": "200" }, { "input": "100 0 100", "output": "200" }, { "input": "100 1 100", "output": "200" }, { "input": "1 100 100", "output": "200" }, { "input": "100 100 0", "output": "200" }, { "input": "100 100 1", "output": "200" }, { "input": "1 2 1", "output": "4" }, { "input": "0 0 100", "output": "100" }, { "input": "0 100 0", "output": "0" }, { "input": "100 0 0", "output": "0" }, { "input": "10 8 7", "output": "24" }, { "input": "45 47 16", "output": "108" }, { "input": "59 43 100", "output": "202" }, { "input": "34 1 30", "output": "62" }, { "input": "14 81 1", "output": "30" }, { "input": "53 96 94", "output": "242" }, { "input": "62 81 75", "output": "218" }, { "input": "21 71 97", "output": "188" }, { "input": "49 82 73", "output": "204" }, { "input": "88 19 29", "output": "96" }, { "input": "89 4 62", "output": "132" }, { "input": "58 3 65", "output": "126" }, { "input": "27 86 11", "output": "76" }, { "input": "35 19 80", "output": "134" }, { "input": "4 86 74", "output": "156" }, { "input": "32 61 89", "output": "182" }, { "input": "68 60 98", "output": "226" }, { "input": "37 89 34", "output": "142" }, { "input": "92 9 28", "output": "74" }, { "input": "79 58 98", "output": "234" }, { "input": "35 44 88", "output": "166" }, { "input": "16 24 19", "output": "58" }, { "input": "74 71 75", "output": "220" }, { "input": "83 86 99", "output": "268" }, { "input": "97 73 15", "output": "176" }, { "input": "77 76 73", "output": "226" }, { "input": "48 85 55", "output": "188" }, { "input": "1 2 2", "output": "4" }, { "input": "2 2 2", "output": "6" }, { "input": "2 1 2", "output": "4" }, { "input": "2 2 1", "output": "4" }, { "input": "3 2 1", "output": "6" }, { "input": "1 2 3", "output": "6" }, { "input": "1 3 2", "output": "6" }, { "input": "2 1 3", "output": "6" }, { "input": "2 3 1", "output": "6" }, { "input": "3 1 2", "output": "6" }, { "input": "99 99 99", "output": "296" }, { "input": "99 99 100", "output": "298" }, { "input": "99 100 99", "output": "298" }, { "input": "99 100 100", "output": "298" }, { "input": "100 99 99", "output": "298" }, { "input": "100 99 100", "output": "298" }, { "input": "100 100 99", "output": "298" }, { "input": "89 32 23", "output": "110" }, { "input": "4 5 0", "output": "8" }, { "input": "3 0 3", "output": "6" }, { "input": "0 0 2", "output": "2" }, { "input": "97 97 0", "output": "194" }, { "input": "1 4 0", "output": "2" }, { "input": "5 2 0", "output": "4" }, { "input": "0 5 10", "output": "14" }, { "input": "0 1 2", "output": "2" }, { "input": "5 2 3", "output": "10" }, { "input": "5 5 0", "output": "10" }, { "input": "0 0 10", "output": "10" }, { "input": "0 1 1", "output": "2" }, { "input": "0 0 1", "output": "0" } ]
1,666,342,953
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
31
0
l,r,a=(map(int,input().split())) if(l==0 or r==0): print(0) elif(l==r and l==a and r==a): x=l+r+a if(x%2==0): print(x) else: print(x-1); exit() else: if(l<r): l=l+a elif(r<l): r=r+a if(l<r): print(l*2) else: print(r*2)
Title: Left-handers, Right-handers and Ambidexters Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. Input Specification: The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. Output Specification: Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. Demo Input: ['1 4 2\n', '5 5 5\n', '0 2 0\n'] Demo Output: ['6\n', '14\n', '0\n'] Note: In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
```python l,r,a=(map(int,input().split())) if(l==0 or r==0): print(0) elif(l==r and l==a and r==a): x=l+r+a if(x%2==0): print(x) else: print(x-1); exit() else: if(l<r): l=l+a elif(r<l): r=r+a if(l<r): print(l*2) else: print(r*2) ```
0
141
A
Amusing Joke
PROGRAMMING
800
[ "implementation", "sortings", "strings" ]
null
null
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
[ "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n", "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n", "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n" ]
[ "YES\n", "NO\n", "NO\n" ]
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".
500
[ { "input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS", "output": "YES" }, { "input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI", "output": "NO" }, { "input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER", "output": "NO" }, { "input": "B\nA\nAB", "output": "YES" }, { "input": "ONDOL\nJNPB\nONLNJBODP", "output": "YES" }, { "input": "Y\nW\nYW", "output": "YES" }, { "input": "OI\nM\nIMO", "output": "YES" }, { "input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF", "output": "YES" }, { "input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB", "output": "NO" }, { "input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH", "output": "NO" }, { "input": "IQ\nOQ\nQOQIGGKFNHJSGCGM", "output": "NO" }, { "input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR", "output": "YES" }, { "input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY", "output": "YES" }, { "input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX", "output": "YES" }, { "input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW", "output": "YES" }, { "input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU", "output": "YES" }, { "input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK", "output": "YES" }, { "input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE", "output": "NO" }, { "input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ", "output": "NO" }, { "input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI", "output": "NO" }, { "input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT", "output": "NO" }, { "input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY", "output": "NO" }, { "input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT", "output": "NO" }, { "input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY", "output": "NO" }, { "input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI", "output": "NO" }, { "input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE", "output": "NO" }, { "input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH", "output": "NO" }, { "input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU", "output": "NO" }, { "input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG", "output": "NO" }, { "input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM", "output": "NO" }, { "input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW", "output": "NO" }, { "input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ", "output": "YES" }, { "input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS", "output": "YES" }, { "input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH", "output": "YES" }, { "input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB", "output": "YES" }, { "input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP", "output": "YES" }, { "input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR", "output": "YES" }, { "input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT", "output": "YES" }, { "input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH", "output": "YES" }, { "input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG", "output": "YES" }, { "input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF", "output": "YES" }, { "input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA", "output": "NO" }, { "input": "IXFDY\nJRMOU\nDF", "output": "NO" }, { "input": "JPSPZ\nUGCUB\nJMZZZZZZZZ", "output": "NO" }, { "input": "AC\nA\nBBA", "output": "NO" }, { "input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB", "output": "NO" }, { "input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW", "output": "NO" }, { "input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE", "output": "NO" }, { "input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH", "output": "NO" }, { "input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ", "output": "NO" }, { "input": "GIO\nPRL\nPRL", "output": "NO" }, { "input": "A\nB\nABC", "output": "NO" }, { "input": "KKK\nKKK\nZZZZZ", "output": "NO" }, { "input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA", "output": "NO" } ]
1,685,696,165
2,147,483,647
Python 3
OK
TESTS
54
92
0
n=list(input()) n1=list(input()) n2=sorted(list(input())) ll=sorted(n+n1) if ll==n2:print("YES") else:print("NO")
Title: Amusing Joke Time Limit: None seconds Memory Limit: None megabytes Problem Description: So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. Input Specification: The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. Output Specification: Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. Demo Input: ['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".
```python n=list(input()) n1=list(input()) n2=sorted(list(input())) ll=sorted(n+n1) if ll==n2:print("YES") else:print("NO") ```
3
934
A
A Compatible Pair
PROGRAMMING
1,400
[ "brute force", "games" ]
null
null
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming. Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively. Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns. Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible. You are asked to find the brightness of the chosen pair if both of them choose optimally.
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*. The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*. All the integers range from <=-<=109 to 109.
Print a single integer — the brightness of the chosen pair.
[ "2 2\n20 18\n2 14\n", "5 3\n-1 0 1 2 3\n-1 0 1\n" ]
[ "252\n", "2\n" ]
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself. In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
500
[ { "input": "2 2\n20 18\n2 14", "output": "252" }, { "input": "5 3\n-1 0 1 2 3\n-1 0 1", "output": "2" }, { "input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7", "output": "70" }, { "input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 2 6 7 9 4 3 7 2 0\n0 5 9 4 4 6 1 8 2 1 6 6 8 6 4 4 7 2 1 8 6 7 4 9 8 3 0 2 0 10 7 1 4 9 4 4 2 5 3 5 1 3 2 4 1 6 5 3 8 6", "output": "100" }, { "input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n-775179088 631683023 -974858199 -157471745 -629658630 71825477 -6235611", "output": "127184126241438168" }, { "input": "16 15\n-94580188 -713689767 -559972014 -632609438 -930348091 -567718487 -611395744 -819913097 -924009672 -427913920 -812510647 -546415480 -982072775 -693369647 -693004777 -714181162\n-772924706 -202246100 -165871667 -991426281 -490838183 209351416 134956137 -36128588 -754413937 -616596290 696201705 -201191199 967464971 -244181984 -729907974", "output": "922371547895579571" }, { "input": "12 22\n-102896616 -311161241 -67541276 -402842686 -830595520 -813834033 -44046671 -584806552 -598620444 -968935604 -303048547 -545969410\n545786451 262898403 442511997 -441241260 -479587986 -752123290 720443264 500646237 737842681 -571966572 -798463881 -477248830 89875164 410339460 -359022689 -251280099 -441455542 -538431186 -406793869 374561004 -108755237 -440143410", "output": "663200522440413120" }, { "input": "33 14\n-576562007 -218618150 -471719380 -583840778 -256368365 -68451917 -405045344 -775538133 -896830082 -439261765 -947070124 -716577019 -456110999 -689862512 -132480131 -10805271 -518903339 -196240188 -222292638 -828546042 -43887962 -161359263 -281422097 -484060534 963147664 -492377073 -154570101 -52145116 187803553 858844161 66540410 418777176 434025748\n-78301978 -319393213 -12393024 542953412 786804661 845642067 754996432 -985617475 -487171947 56142664 203173079 -268261708 -817080591 -511720682", "output": "883931400924882950" }, { "input": "15 8\n-966400308 -992207261 -302395973 -837980754 -516443826 -492405613 -378127629 -762650324 -519519776 -36132939 -286460372 -351445284 -407653342 -604960925 -523442015\n610042288 27129580 -103108347 -942517864 842060508 -588904868 614786155 37455106", "output": "910849554065102112" }, { "input": "6 30\n-524297819 -947277203 -444186475 -182837689 -385379656 -453917269\n834529938 35245081 663687669 585422565 164412867 850052113 796429008 -307345676 -127653313 426960600 211854713 -733687358 251466836 -33491050 -882811238 455544614 774581544 768447941 -241033484 441104324 -493975870 308277556 275268265 935941507 -152292053 -961509996 -740482111 -954176110 -924254634 -518710544", "output": "504117593849498724" }, { "input": "5 32\n-540510995 -841481393 -94342377 -74818927 -93445356\n686714668 -82581175 736472406 502016312 575563638 -899308712 503504178 -644271272 -437408397 385778869 -746757839 306275973 -663503743 -431116516 -418708278 -515261493 -988182324 900230931 218258353 -714420102 -241118202 294802602 -937785552 -857537498 -723195312 -690515139 -214508504 -44086454 -231621215 -418360090 -810003786 -675944617", "output": "534123411186652380" }, { "input": "32 13\n-999451897 -96946179 -524159869 -906101658 -63367320 -629803888 -968586834 -658416130 -874232857 -926556428 -749908220 -517073321 -659752288 -910152878 -786916085 -607633039 -191428642 -867952926 -873793977 -584331784 -733245792 -779809700 -554228536 -464503499 561577340 258991071 -569805979 -372655165 -106685554 -619607960 188856473 -268960803\n886429660 -587284372 911396803 -462990289 -228681210 -876239914 -822830527 -750131315 -401234943 116991909 -582713480 979631847 813552478", "output": "848714444125692276" }, { "input": "12 25\n-464030345 -914672073 -483242132 -856226270 -925135169 -353124606 -294027092 -619650850 -490724485 -240424784 -483066792 -921640365\n279850608 726838739 -431610610 242749870 -244020223 -396865433 129534799 182767854 -939698671 342579400 330027106 893561388 -263513962 643369418 276245179 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1,518,794,917
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
46
5,632,000
s=input() n=input() m=input() n=n.split() m=m.split() n=list(n) m=list(m) a=max(n) n.remove(str(a)) a=max(n) b=max(m) print(a*b)
Title: A Compatible Pair Time Limit: None seconds Memory Limit: None megabytes Problem Description: Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming. Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively. Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns. Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible. You are asked to find the brightness of the chosen pair if both of them choose optimally. Input Specification: The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*. The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*. All the integers range from <=-<=109 to 109. Output Specification: Print a single integer — the brightness of the chosen pair. Demo Input: ['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n'] Demo Output: ['252\n', '2\n'] Note: In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself. In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
```python s=input() n=input() m=input() n=n.split() m=m.split() n=list(n) m=list(m) a=max(n) n.remove(str(a)) a=max(n) b=max(m) print(a*b) ```
-1