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157
|
B
|
Trace
|
PROGRAMMING
| 1,000
|
[
"geometry",
"sortings"
] | null | null |
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
|
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different.
|
Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.
|
[
"1\n1\n",
"3\n1 4 2\n"
] |
[
"3.1415926536\n",
"40.8407044967\n"
] |
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 1<sup class="upper-index">2</sup> = π.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 4<sup class="upper-index">2</sup> - π × 2<sup class="upper-index">2</sup>) + π × 1<sup class="upper-index">2</sup> = π × 12 + π = 13π
| 1,000
|
[
{
"input": "1\n1",
"output": "3.1415926536"
},
{
"input": "3\n1 4 2",
"output": "40.8407044967"
},
{
"input": "4\n4 1 3 2",
"output": "31.4159265359"
},
{
"input": "4\n100 10 2 1",
"output": "31111.1920484997"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "172.7875959474"
},
{
"input": "1\n1000",
"output": "3141592.6535897931"
},
{
"input": "8\n8 1 7 2 6 3 5 4",
"output": "113.0973355292"
},
{
"input": "100\n1000 999 998 997 996 995 994 993 992 991 990 989 988 987 986 985 984 983 982 981 980 979 978 977 976 975 974 973 972 971 970 969 968 967 966 965 964 963 962 961 960 959 958 957 956 955 954 953 952 951 950 949 948 947 946 945 944 943 942 941 940 939 938 937 936 935 934 933 932 931 930 929 928 927 926 925 924 923 922 921 920 919 918 917 916 915 914 913 912 911 910 909 908 907 906 905 904 903 902 901",
"output": "298608.3817237098"
},
{
"input": "6\n109 683 214 392 678 10",
"output": "397266.9574170437"
},
{
"input": "2\n151 400",
"output": "431023.3704798660"
},
{
"input": "6\n258 877 696 425 663 934",
"output": "823521.3902487604"
},
{
"input": "9\n635 707 108 234 52 180 910 203 782",
"output": "1100144.9065826489"
},
{
"input": "8\n885 879 891 428 522 176 135 983",
"output": "895488.9947571954"
},
{
"input": "3\n269 918 721",
"output": "1241695.6467754442"
},
{
"input": "7\n920 570 681 428 866 935 795",
"output": "1469640.1849419588"
},
{
"input": "2\n517 331",
"output": "495517.1260654109"
},
{
"input": "2\n457 898",
"output": "1877274.3981158488"
},
{
"input": "8\n872 704 973 612 183 274 739 253",
"output": "1780774.0965755312"
},
{
"input": "74\n652 446 173 457 760 847 670 25 196 775 998 279 656 809 883 148 969 884 792 502 641 800 663 938 362 339 545 608 107 184 834 666 149 458 864 72 199 658 618 987 126 723 806 643 689 958 626 904 944 415 427 498 628 331 636 261 281 276 478 220 513 595 510 384 354 561 469 462 799 449 747 109 903 456",
"output": "1510006.5089479341"
},
{
"input": "76\n986 504 673 158 87 332 124 218 714 235 212 122 878 370 938 81 686 323 386 348 410 468 875 107 50 960 82 834 234 663 651 422 794 633 294 771 945 607 146 913 950 858 297 88 882 725 247 872 645 749 799 987 115 394 380 382 971 429 593 426 652 353 351 233 868 598 889 116 71 376 916 464 414 976 138 903",
"output": "1528494.7817143100"
},
{
"input": "70\n12 347 748 962 514 686 192 159 990 4 10 788 602 542 946 215 523 727 799 717 955 796 529 465 897 103 181 515 495 153 710 179 747 145 16 585 943 998 923 708 156 399 770 547 775 285 9 68 713 722 570 143 913 416 663 624 925 218 64 237 797 138 942 213 188 818 780 840 480 758",
"output": "1741821.4892636713"
},
{
"input": "26\n656 508 45 189 561 366 96 486 547 386 703 570 780 689 264 26 11 74 466 76 421 48 982 886 215 650",
"output": "1818821.9252031571"
},
{
"input": "52\n270 658 808 249 293 707 700 78 791 167 92 772 807 502 830 991 945 102 968 376 556 578 326 980 688 368 280 853 646 256 666 638 424 737 321 996 925 405 199 680 953 541 716 481 727 143 577 919 892 355 346 298",
"output": "1272941.9273080483"
},
{
"input": "77\n482 532 200 748 692 697 171 863 586 547 301 149 326 812 147 698 303 691 527 805 681 387 619 947 598 453 167 799 840 508 893 688 643 974 998 341 804 230 538 669 271 404 477 759 943 596 949 235 880 160 151 660 832 82 969 539 708 889 258 81 224 655 790 144 462 582 646 256 445 52 456 920 67 819 631 484 534",
"output": "2045673.1891262225"
},
{
"input": "27\n167 464 924 575 775 97 944 390 297 315 668 296 533 829 851 406 702 366 848 512 71 197 321 900 544 529 116",
"output": "1573959.9105970615"
},
{
"input": "38\n488 830 887 566 720 267 583 102 65 200 884 220 263 858 510 481 316 804 754 568 412 166 374 869 356 977 145 421 500 58 664 252 745 70 381 927 670 772",
"output": "1479184.3434235646"
},
{
"input": "64\n591 387 732 260 840 397 563 136 571 876 831 953 799 493 579 13 559 872 53 678 256 232 969 993 847 14 837 365 547 997 604 199 834 529 306 443 739 49 19 276 343 835 904 588 900 870 439 576 975 955 518 117 131 347 800 83 432 882 869 709 32 950 314 450",
"output": "1258248.6984672088"
},
{
"input": "37\n280 281 169 68 249 389 977 101 360 43 448 447 368 496 125 507 747 392 338 270 916 150 929 428 118 266 589 470 774 852 263 644 187 817 808 58 637",
"output": "1495219.0323274869"
},
{
"input": "97\n768 569 306 968 437 779 227 561 412 60 44 807 234 645 169 858 580 396 343 145 842 723 416 80 456 247 81 150 297 116 760 964 312 558 101 850 549 650 299 868 121 435 579 705 118 424 302 812 970 397 659 565 916 183 933 459 6 593 518 717 326 305 744 470 75 981 824 221 294 324 194 293 251 446 481 215 338 861 528 829 921 945 540 89 450 178 24 460 990 392 148 219 934 615 932 340 937",
"output": "1577239.7333274092"
},
{
"input": "94\n145 703 874 425 277 652 239 496 458 658 339 842 564 699 893 352 625 980 432 121 798 872 499 859 850 721 414 825 543 843 304 111 342 45 219 311 50 748 465 902 781 822 504 985 919 656 280 310 917 438 464 527 491 713 906 329 635 777 223 810 501 535 156 252 806 112 971 719 103 443 165 98 579 554 244 996 221 560 301 51 977 422 314 858 528 772 448 626 185 194 536 66 577 677",
"output": "1624269.3753516484"
},
{
"input": "97\n976 166 649 81 611 927 480 231 998 711 874 91 969 521 531 414 993 790 317 981 9 261 437 332 173 573 904 777 882 990 658 878 965 64 870 896 271 732 431 53 761 943 418 602 708 949 930 130 512 240 363 458 673 319 131 784 224 48 919 126 208 212 911 59 677 535 450 273 479 423 79 807 336 18 72 290 724 28 123 605 287 228 350 897 250 392 885 655 746 417 643 114 813 378 355 635 905",
"output": "1615601.7212203942"
},
{
"input": "91\n493 996 842 9 748 178 1 807 841 519 796 998 84 670 778 143 707 208 165 893 154 943 336 150 761 881 434 112 833 55 412 682 552 945 758 189 209 600 354 325 440 844 410 20 136 665 88 791 688 17 539 821 133 236 94 606 483 446 429 60 960 476 915 134 137 852 754 908 276 482 117 252 297 903 981 203 829 811 471 135 188 667 710 393 370 302 874 872 551 457 692",
"output": "1806742.5014501044"
},
{
"input": "95\n936 736 17 967 229 607 589 291 242 244 29 698 800 566 630 667 90 416 11 94 812 838 668 520 678 111 490 823 199 973 681 676 683 721 262 896 682 713 402 691 874 44 95 704 56 322 822 887 639 433 406 35 988 61 176 496 501 947 440 384 372 959 577 370 754 802 1 945 427 116 746 408 308 391 397 730 493 183 203 871 831 862 461 565 310 344 504 378 785 137 279 123 475 138 415",
"output": "1611115.5269110680"
},
{
"input": "90\n643 197 42 218 582 27 66 704 195 445 641 675 285 639 503 686 242 327 57 955 848 287 819 992 756 749 363 48 648 736 580 117 752 921 923 372 114 313 202 337 64 497 399 25 883 331 24 871 917 8 517 486 323 529 325 92 891 406 864 402 263 773 931 253 625 31 17 271 140 131 232 586 893 525 846 54 294 562 600 801 214 55 768 683 389 738 314 284 328 804",
"output": "1569819.2914796301"
},
{
"input": "98\n29 211 984 75 333 96 840 21 352 168 332 433 130 944 215 210 620 442 363 877 91 491 513 955 53 82 351 19 998 706 702 738 770 453 344 117 893 590 723 662 757 16 87 546 312 669 568 931 224 374 927 225 751 962 651 587 361 250 256 240 282 600 95 64 384 589 813 783 39 918 412 648 506 283 886 926 443 173 946 241 310 33 622 565 261 360 547 339 943 367 354 25 479 743 385 485 896 741",
"output": "2042921.1539616778"
},
{
"input": "93\n957 395 826 67 185 4 455 880 683 654 463 84 258 878 553 592 124 585 9 133 20 609 43 452 725 125 801 537 700 685 771 155 566 376 19 690 383 352 174 208 177 416 304 1000 533 481 87 509 358 233 681 22 507 659 36 859 952 259 138 271 594 779 576 782 119 69 608 758 283 616 640 523 710 751 34 106 774 92 874 568 864 660 998 992 474 679 180 409 15 297 990 689 501",
"output": "1310703.8710041976"
},
{
"input": "97\n70 611 20 30 904 636 583 262 255 501 604 660 212 128 199 138 545 576 506 528 12 410 77 888 783 972 431 188 338 485 148 793 907 678 281 922 976 680 252 724 253 920 177 361 721 798 960 572 99 622 712 466 608 49 612 345 266 751 63 594 40 695 532 789 520 930 825 929 48 59 405 135 109 735 508 186 495 772 375 587 201 324 447 610 230 947 855 318 856 956 313 810 931 175 668 183 688",
"output": "1686117.9099228707"
},
{
"input": "96\n292 235 391 180 840 172 218 997 166 287 329 20 886 325 400 471 182 356 448 337 417 319 58 106 366 764 393 614 90 831 924 314 667 532 64 874 3 434 350 352 733 795 78 640 967 63 47 879 635 272 145 569 468 792 153 761 770 878 281 467 209 208 298 37 700 18 334 93 5 750 412 779 523 517 360 649 447 328 311 653 57 578 767 460 647 663 50 670 151 13 511 580 625 907 227 89",
"output": "1419726.5608617242"
},
{
"input": "100\n469 399 735 925 62 153 707 723 819 529 200 624 57 708 245 384 889 11 639 638 260 419 8 142 403 298 204 169 887 388 241 983 885 267 643 943 417 237 452 562 6 839 149 742 832 896 100 831 712 754 679 743 135 222 445 680 210 955 220 63 960 487 514 824 481 584 441 997 795 290 10 45 510 678 844 503 407 945 850 84 858 934 500 320 936 663 736 592 161 670 606 465 864 969 293 863 868 393 899 744",
"output": "1556458.0979239127"
},
{
"input": "100\n321 200 758 415 190 710 920 992 873 898 814 259 359 66 971 210 838 545 663 652 684 277 36 756 963 459 335 484 462 982 532 423 131 703 307 229 391 938 253 847 542 975 635 928 220 980 222 567 557 181 366 824 900 180 107 979 112 564 525 413 300 422 876 615 737 343 902 8 654 628 469 913 967 785 893 314 909 215 912 262 20 709 363 915 997 954 986 454 596 124 74 159 660 550 787 418 895 786 293 50",
"output": "1775109.8050211088"
},
{
"input": "100\n859 113 290 762 701 63 188 431 810 485 671 673 99 658 194 227 511 435 941 212 551 124 89 222 42 321 657 815 898 171 216 482 707 567 724 491 414 942 820 351 48 653 685 312 586 24 20 627 602 498 533 173 463 262 621 466 119 299 580 964 510 987 40 698 521 998 847 651 746 215 808 563 785 837 631 772 404 923 682 244 232 214 390 350 968 771 517 900 70 543 934 554 681 368 642 575 891 728 478 317",
"output": "1447969.4788174964"
},
{
"input": "100\n941 283 349 457 52 837 299 284 796 305 893 624 101 972 738 204 121 70 17 704 836 791 95 111 162 952 472 724 733 580 878 177 705 804 11 211 463 417 288 409 410 485 896 755 921 267 164 656 505 765 539 439 535 19 991 689 220 474 114 944 884 144 926 849 486 566 117 35 749 499 797 303 362 905 690 890 976 66 590 183 234 683 39 297 769 787 376 541 571 759 495 200 261 352 73 493 831 442 273 339",
"output": "1597889.4218394549"
},
{
"input": "100\n110 868 147 888 291 282 916 542 917 337 235 595 498 621 814 249 261 713 445 666 142 275 319 49 900 543 333 606 487 670 620 769 852 258 230 867 603 491 714 839 879 37 21 74 590 190 397 506 499 967 152 862 200 107 145 23 614 167 857 475 79 598 270 906 626 116 89 59 651 846 723 201 50 699 522 408 634 132 226 414 824 764 513 586 526 238 556 162 667 173 684 415 527 743 10 442 552 274 350 822",
"output": "1567230.6191330721"
},
{
"input": "100\n901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "298608.3817237098"
}
] | 1,594,041,795
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 248
| 20,172,800
|
n = int(input())
pie = 3.1415926536
arr = list(map(int, input().split()))
ans = 0.0
arr.sort()
for i in range(n):
if i % 2 == 0:
ans += arr[i] * arr[i] * pie
else:
ans -= arr[i] * arr[i] * pie
print(ans)
|
Title: Trace
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different.
Output Specification:
Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.
Demo Input:
['1\n1\n', '3\n1 4 2\n']
Demo Output:
['3.1415926536\n', '40.8407044967\n']
Note:
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 1<sup class="upper-index">2</sup> = π.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 4<sup class="upper-index">2</sup> - π × 2<sup class="upper-index">2</sup>) + π × 1<sup class="upper-index">2</sup> = π × 12 + π = 13π
|
```python
n = int(input())
pie = 3.1415926536
arr = list(map(int, input().split()))
ans = 0.0
arr.sort()
for i in range(n):
if i % 2 == 0:
ans += arr[i] * arr[i] * pie
else:
ans -= arr[i] * arr[i] * pie
print(ans)
```
| 0
|
|
797
|
B
|
Odd sum
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence.
|
The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum.
|
Print sum of resulting subseqeuence.
|
[
"4\n-2 2 -3 1\n",
"3\n2 -5 -3\n"
] |
[
"3\n",
"-1\n"
] |
In the first example sum of the second and the fourth elements is 3.
| 0
|
[
{
"input": "4\n-2 2 -3 1",
"output": "3"
},
{
"input": "3\n2 -5 -3",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "15\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046",
"output": "53507"
},
{
"input": "2\n-5439 -6705",
"output": "-5439"
},
{
"input": "2\n2850 6843",
"output": "9693"
},
{
"input": "2\n144 9001",
"output": "9145"
},
{
"input": "10\n7535 -819 2389 4933 5495 4887 -5181 -9355 7955 5757",
"output": "38951"
},
{
"input": "10\n-9169 -1574 3580 -8579 -7177 -3216 7490 3470 3465 -1197",
"output": "18005"
},
{
"input": "10\n941 7724 2220 -4704 -8374 -8249 7606 9502 612 -9097",
"output": "28605"
},
{
"input": "10\n4836 -2331 -3456 2312 -1574 3134 -670 -204 512 -5504",
"output": "8463"
},
{
"input": "10\n1184 5136 1654 3254 6576 6900 6468 327 179 7114",
"output": "38613"
},
{
"input": "10\n-2152 -1776 -1810 -9046 -6090 -2324 -8716 -6103 -787 -812",
"output": "-787"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "5\n5 5 5 3 -1",
"output": "17"
},
{
"input": "5\n-1 -2 5 3 0",
"output": "7"
},
{
"input": "5\n-3 -2 5 -1 3",
"output": "7"
},
{
"input": "3\n-2 2 -1",
"output": "1"
},
{
"input": "5\n5 0 7 -2 3",
"output": "15"
},
{
"input": "2\n-2 -5",
"output": "-5"
},
{
"input": "3\n-1 -3 0",
"output": "-1"
},
{
"input": "5\n2 -1 0 -3 -2",
"output": "1"
},
{
"input": "4\n2 3 0 5",
"output": "7"
},
{
"input": "5\n-5 3 -2 2 5",
"output": "7"
},
{
"input": "59\n8593 5929 3016 -859 4366 -6842 8435 -3910 -2458 -8503 -3612 -9793 -5360 -9791 -362 -7180 727 -6245 -8869 -7316 8214 -7944 7098 3788 -5436 -6626 -1131 -2410 -5647 -7981 263 -5879 8786 709 6489 5316 -4039 4909 -4340 7979 -89 9844 -906 172 -7674 -3371 -6828 9505 3284 5895 3646 6680 -1255 3635 -9547 -5104 -1435 -7222 2244",
"output": "129433"
},
{
"input": "17\n-6170 2363 6202 -9142 7889 779 2843 -5089 2313 -3952 1843 5171 462 -3673 5098 -2519 9565",
"output": "43749"
},
{
"input": "26\n-8668 9705 1798 -1766 9644 3688 8654 -3077 -5462 2274 6739 2732 3635 -4745 -9144 -9175 -7488 -2010 1637 1118 8987 1597 -2873 -5153 -8062 146",
"output": "60757"
},
{
"input": "51\n8237 -7239 -3545 -6059 -5110 4066 -4148 -7641 -5797 -994 963 1144 -2785 -8765 -1216 5410 1508 -6312 -6313 -680 -7657 4579 -6898 7379 2015 -5087 -5417 -6092 3819 -9101 989 -8380 9161 -7519 -9314 -3838 7160 5180 567 -1606 -3842 -9665 -2266 1296 -8417 -3976 7436 -2075 -441 -4565 3313",
"output": "73781"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n-2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "5"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n0 -1",
"output": "-1"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "2\n3 0",
"output": "3"
},
{
"input": "2\n0 -1",
"output": "-1"
},
{
"input": "3\n-3 1 -1",
"output": "1"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n1 3 1",
"output": "5"
},
{
"input": "3\n-1 0 1",
"output": "1"
},
{
"input": "3\n-3 -3 -2",
"output": "-3"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n-2 -2 1",
"output": "1"
},
{
"input": "4\n0 -1 -3 -4",
"output": "-1"
},
{
"input": "4\n5 3 2 1",
"output": "11"
},
{
"input": "4\n-1 -2 4 -2",
"output": "3"
},
{
"input": "4\n-1 -3 0 -3",
"output": "-1"
},
{
"input": "4\n1 -4 -3 -4",
"output": "1"
},
{
"input": "4\n5 3 3 4",
"output": "15"
},
{
"input": "4\n-1 -3 -1 2",
"output": "1"
},
{
"input": "4\n3 2 -1 -4",
"output": "5"
},
{
"input": "5\n-5 -4 -3 -5 2",
"output": "-1"
},
{
"input": "5\n5 5 1 2 -2",
"output": "13"
},
{
"input": "5\n-2 -1 -5 -1 4",
"output": "3"
},
{
"input": "5\n-5 -5 -4 4 0",
"output": "-1"
},
{
"input": "5\n2 -3 -1 -4 -5",
"output": "1"
},
{
"input": "5\n4 3 4 2 3",
"output": "13"
},
{
"input": "5\n0 -2 -5 3 3",
"output": "3"
},
{
"input": "5\n4 -2 -2 -3 0",
"output": "1"
},
{
"input": "6\n6 7 -1 1 5 -1",
"output": "19"
},
{
"input": "6\n-1 7 2 -3 -4 -5",
"output": "9"
},
{
"input": "6\n0 -1 -3 -5 2 -6",
"output": "1"
},
{
"input": "6\n4 -1 0 3 6 1",
"output": "13"
},
{
"input": "6\n5 3 3 4 4 -3",
"output": "19"
},
{
"input": "6\n0 -3 5 -4 5 -4",
"output": "7"
},
{
"input": "6\n-5 -3 1 -1 -5 -3",
"output": "1"
},
{
"input": "6\n-2 1 3 -2 7 4",
"output": "15"
},
{
"input": "7\n0 7 6 2 7 0 6",
"output": "21"
},
{
"input": "7\n6 -6 -1 -5 7 1 7",
"output": "21"
},
{
"input": "7\n2 3 -5 0 -4 0 -4",
"output": "5"
},
{
"input": "7\n-6 3 -3 -1 -6 -6 -5",
"output": "3"
},
{
"input": "7\n7 6 3 2 4 2 0",
"output": "21"
},
{
"input": "7\n-2 3 -3 4 4 0 -1",
"output": "11"
},
{
"input": "7\n-5 -7 4 0 5 -3 -5",
"output": "9"
},
{
"input": "7\n-3 -5 -4 1 3 -4 -7",
"output": "3"
},
{
"input": "8\n5 2 4 5 7 -2 7 3",
"output": "33"
},
{
"input": "8\n-8 -3 -1 3 -8 -4 -4 4",
"output": "7"
},
{
"input": "8\n-6 -7 -7 -5 -4 -9 -2 -7",
"output": "-5"
},
{
"input": "8\n8 7 6 8 3 4 8 -2",
"output": "41"
},
{
"input": "8\n6 7 0 -6 6 5 4 7",
"output": "35"
},
{
"input": "8\n0 -7 -5 -5 5 -1 -8 -7",
"output": "5"
},
{
"input": "8\n1 -6 -5 7 -3 -4 2 -2",
"output": "9"
},
{
"input": "8\n1 -8 -6 -6 -6 -7 -5 -1",
"output": "1"
},
{
"input": "9\n-3 -1 4 4 8 -8 -5 9 -2",
"output": "25"
},
{
"input": "9\n-9 -1 3 -2 -7 2 -9 -1 -4",
"output": "5"
},
{
"input": "9\n-6 -9 -3 -8 -5 2 -6 0 -5",
"output": "-1"
},
{
"input": "9\n5 4 3 3 6 7 8 5 9",
"output": "47"
},
{
"input": "9\n5 3 9 1 5 2 -3 7 0",
"output": "31"
},
{
"input": "9\n-3 -9 -1 -7 5 6 -4 -6 -6",
"output": "11"
},
{
"input": "9\n-6 -5 6 -5 -2 0 1 2 -9",
"output": "9"
},
{
"input": "9\n8 3 6 1 -3 5 2 9 1",
"output": "35"
},
{
"input": "10\n-6 -4 -7 -1 -9 -10 -10 1 0 -3",
"output": "1"
},
{
"input": "10\n-2 -10 -5 -6 -10 -3 -6 -3 -8 -8",
"output": "-3"
},
{
"input": "10\n8 5 9 2 3 3 -6 1 -1 8",
"output": "39"
},
{
"input": "10\n2 10 -7 6 -1 -1 7 -9 -4 -6",
"output": "25"
},
{
"input": "10\n-10 -2 -2 -1 -10 -7 1 0 -4 -5",
"output": "1"
},
{
"input": "10\n4 3 10 -2 -1 0 10 6 7 0",
"output": "39"
},
{
"input": "10\n-2 6 6 5 0 10 6 7 -1 1",
"output": "41"
},
{
"input": "10\n-10 2 8 -6 -1 -5 1 -10 -10 -1",
"output": "11"
}
] | 1,634,565,465
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 126
| 93
| 7,270,400
|
def solution(n, a):
oddMin = 10**4
isOdd = False
sum = 0
for i in range(n):
if (a[i] > 0):
sum = sum+a[i]
if (a[i]%2 !=0):
isOdd = True
if (oddMin > abs(a[i])):
oddMin = abs(a[i])
if (isOdd == False):
return -1
if (sum%2==0):
sum = sum - oddMin
return sum
n = int(input())
a = list(map(int, input().split()))
print(solution(n, a))
|
Title: Odd sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence.
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum.
Output Specification:
Print sum of resulting subseqeuence.
Demo Input:
['4\n-2 2 -3 1\n', '3\n2 -5 -3\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first example sum of the second and the fourth elements is 3.
|
```python
def solution(n, a):
oddMin = 10**4
isOdd = False
sum = 0
for i in range(n):
if (a[i] > 0):
sum = sum+a[i]
if (a[i]%2 !=0):
isOdd = True
if (oddMin > abs(a[i])):
oddMin = abs(a[i])
if (isOdd == False):
return -1
if (sum%2==0):
sum = sum - oddMin
return sum
n = int(input())
a = list(map(int, input().split()))
print(solution(n, a))
```
| 3
|
|
189
|
A
|
Cut Ribbon
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp"
] | null | null |
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
|
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
|
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
|
[
"5 5 3 2\n",
"7 5 5 2\n"
] |
[
"2\n",
"2\n"
] |
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
| 500
|
[
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
{
"input": "10 3 4 5",
"output": "3"
},
{
"input": "100 23 15 50",
"output": "2"
},
{
"input": "3119 3515 1021 7",
"output": "11"
},
{
"input": "918 102 1327 1733",
"output": "9"
},
{
"input": "3164 42 430 1309",
"output": "15"
},
{
"input": "3043 317 1141 2438",
"output": "7"
},
{
"input": "26 1 772 2683",
"output": "26"
},
{
"input": "370 2 1 15",
"output": "370"
},
{
"input": "734 12 6 2",
"output": "367"
},
{
"input": "418 18 14 17",
"output": "29"
},
{
"input": "18 16 28 9",
"output": "2"
},
{
"input": "14 6 2 17",
"output": "7"
},
{
"input": "29 27 18 2",
"output": "2"
},
{
"input": "29 12 7 10",
"output": "3"
},
{
"input": "27 23 4 3",
"output": "9"
},
{
"input": "5 14 5 2",
"output": "1"
},
{
"input": "5 17 26 5",
"output": "1"
},
{
"input": "9 1 10 3",
"output": "9"
},
{
"input": "2 19 15 1",
"output": "2"
},
{
"input": "4 6 4 9",
"output": "1"
},
{
"input": "10 6 2 9",
"output": "5"
},
{
"input": "2 2 9 6",
"output": "1"
},
{
"input": "6 2 4 1",
"output": "6"
},
{
"input": "27 24 5 27",
"output": "1"
},
{
"input": "2683 83 26 2709",
"output": "101"
},
{
"input": "728 412 789 158",
"output": "3"
},
{
"input": "3964 4 2916 176",
"output": "991"
},
{
"input": "3399 2035 2 3334",
"output": "683"
},
{
"input": "3455 244 3301 3",
"output": "991"
},
{
"input": "595 2263 3625 1",
"output": "595"
},
{
"input": "4000 1 1 1",
"output": "4000"
},
{
"input": "3999 2 2 3999",
"output": "1"
},
{
"input": "25 6 8 11",
"output": "3"
},
{
"input": "4000 500 1000 2000",
"output": "8"
},
{
"input": "53 10 11 23",
"output": "5"
},
{
"input": "100 100 1 1",
"output": "100"
},
{
"input": "17 3 4 10",
"output": "5"
},
{
"input": "413 101 102 105",
"output": "4"
},
{
"input": "490 4 49 50",
"output": "111"
},
{
"input": "3999 2 3 3",
"output": "1999"
},
{
"input": "8 3 8 4",
"output": "2"
},
{
"input": "5 1 3 3",
"output": "5"
},
{
"input": "100 3 17 22",
"output": "27"
},
{
"input": "4000 2 3 4",
"output": "2000"
},
{
"input": "4000 3 3 5",
"output": "1332"
},
{
"input": "13 4 6 7",
"output": "2"
},
{
"input": "4000 5 2 2",
"output": "2000"
},
{
"input": "3999 2 2 3",
"output": "1999"
},
{
"input": "4000 33 7 3333",
"output": "564"
},
{
"input": "60 33 20 9",
"output": "4"
},
{
"input": "100 9 11 99",
"output": "10"
},
{
"input": "2009 6 8 9",
"output": "334"
}
] | 1,676,099,968
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 46
| 0
|
def cut_ribbon(n, a, b, c):
dp = [0] + [-1] * n
for i in range(1, n+1):
if i >= a and dp[i-a] != -1:
dp[i] = max(dp[i], dp[i-a]+1)
if i >= b and dp[i-b] != -1:
dp[i] = max(dp[i], dp[i-b]+1)
if i >= c and dp[i-c] != -1:
dp[i] = max(dp[i], dp[i-c]+1)
return dp[n]
n, a, b, c = map(int, input().split())
print(cut_ribbon(n, a, b, c))
|
Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
|
```python
def cut_ribbon(n, a, b, c):
dp = [0] + [-1] * n
for i in range(1, n+1):
if i >= a and dp[i-a] != -1:
dp[i] = max(dp[i], dp[i-a]+1)
if i >= b and dp[i-b] != -1:
dp[i] = max(dp[i], dp[i-b]+1)
if i >= c and dp[i-c] != -1:
dp[i] = max(dp[i], dp[i-c]+1)
return dp[n]
n, a, b, c = map(int, input().split())
print(cut_ribbon(n, a, b, c))
```
| 3
|
|
342
|
A
|
Xenia and Divisors
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
|
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
|
[
"6\n1 1 1 2 2 2\n",
"6\n2 2 1 1 4 6\n"
] |
[
"-1\n",
"1 2 4\n1 2 6\n"
] |
none
| 500
|
[
{
"input": "6\n1 1 1 2 2 2",
"output": "-1"
},
{
"input": "6\n2 2 1 1 4 6",
"output": "1 2 4\n1 2 6"
},
{
"input": "3\n1 2 3",
"output": "-1"
},
{
"input": "3\n7 5 7",
"output": "-1"
},
{
"input": "3\n1 3 4",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 3 6 6 3 1 3 1 6",
"output": "1 3 6\n1 3 6\n1 3 6"
},
{
"input": "6\n1 2 4 1 3 5",
"output": "-1"
},
{
"input": "3\n1 3 7",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 2 4 1 2 4 1 3 6",
"output": "1 2 4\n1 2 4\n1 3 6"
},
{
"input": "12\n3 6 1 1 3 6 1 1 2 6 2 6",
"output": "1 3 6\n1 3 6\n1 2 6\n1 2 6"
},
{
"input": "9\n1 1 1 4 4 4 6 2 2",
"output": "-1"
},
{
"input": "9\n1 2 4 6 3 1 3 1 5",
"output": "-1"
},
{
"input": "15\n2 1 2 1 3 6 1 2 1 6 1 3 4 6 4",
"output": "1 2 4\n1 2 4\n1 3 6\n1 3 6\n1 2 6"
},
{
"input": "3\n2 3 6",
"output": "-1"
},
{
"input": "3\n2 4 6",
"output": "-1"
},
{
"input": "3\n2 5 6",
"output": "-1"
},
{
"input": "3\n2 4 7",
"output": "-1"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "-1"
},
{
"input": "3\n7 7 7",
"output": "-1"
},
{
"input": "6\n1 2 4 7 7 7",
"output": "-1"
},
{
"input": "6\n1 1 2 6 6 6",
"output": "-1"
},
{
"input": "9\n1 1 1 3 3 2 4 4 6",
"output": "-1"
},
{
"input": "6\n1 2 4 5 5 5",
"output": "-1"
},
{
"input": "15\n1 1 1 1 1 2 2 2 2 4 4 6 6 6 6",
"output": "-1"
},
{
"input": "6\n1 1 5 5 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 5 6 7",
"output": "-1"
},
{
"input": "6\n1 1 4 4 7 7",
"output": "-1"
},
{
"input": "24\n1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 4 4 4 6 6 6",
"output": "-1"
},
{
"input": "3\n1 7 6",
"output": "-1"
},
{
"input": "6\n1 1 2 4 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 7 7 7 7 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 6 5 5",
"output": "-1"
}
] | 1,650,505,179
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 512,000
|
def main():
n = int(input())
vals = [int(i) for i in input().split()]
counts = {}
for v in vals:
if v not in counts:
counts[v] = 1
else:
counts[v] += 1
groups = []
unique = sorted(list(set(vals)), reverse=True)
for i in range(0, n//3):
group = []
for j in unique:
if counts[j] != 0:
if len(group) == 0:
group.append(j)
counts[j] -= 1
elif j == 0 or group[-1] % j == 0:
group.append(j)
counts[j] -= 1
if len(group) == 3:
break
groups.append(group[::-1])
for group in groups:
print(" ".join([str(i) for i in group]))
if __name__ == "__main__":
main()
|
Title: Xenia and Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
Output Specification:
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
Demo Input:
['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n']
Demo Output:
['-1\n', '1 2 4\n1 2 6\n']
Note:
none
|
```python
def main():
n = int(input())
vals = [int(i) for i in input().split()]
counts = {}
for v in vals:
if v not in counts:
counts[v] = 1
else:
counts[v] += 1
groups = []
unique = sorted(list(set(vals)), reverse=True)
for i in range(0, n//3):
group = []
for j in unique:
if counts[j] != 0:
if len(group) == 0:
group.append(j)
counts[j] -= 1
elif j == 0 or group[-1] % j == 0:
group.append(j)
counts[j] -= 1
if len(group) == 3:
break
groups.append(group[::-1])
for group in groups:
print(" ".join([str(i) for i in group]))
if __name__ == "__main__":
main()
```
| 0
|
|
496
|
B
|
Secret Combination
|
PROGRAMMING
| 1,500
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display.
|
Print a single line containing *n* digits — the desired state of the display containing the smallest possible number.
|
[
"3\n579\n",
"4\n2014\n"
] |
[
"024\n",
"0142\n"
] |
none
| 1,000
|
[
{
"input": "3\n579",
"output": "024"
},
{
"input": "4\n2014",
"output": "0142"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n039",
"output": "014"
},
{
"input": "4\n4444",
"output": "0000"
},
{
"input": "5\n46802",
"output": "02468"
},
{
"input": "10\n4447444444",
"output": "0000000003"
},
{
"input": "10\n5810438174",
"output": "0147609473"
},
{
"input": "30\n027027027027027027027027027027",
"output": "027027027027027027027027027027"
},
{
"input": "50\n41012516454101251645410125164541012516454101251645",
"output": "01076781720107678172010767817201076781720107678172"
},
{
"input": "72\n464553044645330446455304464553064645530445455304464553044645530446455304",
"output": "001011960020119600201196002011960020119600201996002011960020119620201196"
},
{
"input": "100\n2144315253572020279108092911160072328496568665545836825277616363478721946398140227406814602154768031",
"output": "0005996121738545755443472571416650525236761083528703911639570359104365792010332041424619191680979818"
},
{
"input": "200\n79025531557298703099245700860027432585447902553155729870309924570086002743258544790255315572987030992457008600274325854479025531557298703099245700860027432585447902553155729870309924570086002743258544",
"output": "00274325854479025531557298703099245700860027432585447902553155729870309924570086002743258544790255315572987030992457008600274325854479025531557298703099245700860027432585447902553155729870309924570086"
},
{
"input": "100\n6669666666666666666866266666666666666666666666666666666666666666626666666666666966666766665667666656",
"output": "0000000000000000000000000000000000000000006000000000000030000010000900100009000030000000000000002006"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,608,981,273
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 28
| 624
| 10,137,600
|
n=int(input())
l=[int(x) for x in input()]
l=l+l
g=list(l)
ans='9'*10000
while True:
for x in range(0, len(l)-n):
ans=min(ans, ''.join(str(i) for i in l[x:x+n]))
l=[(x+1)%10 for x in l]
if l==g:
break
print(ans)
|
Title: Secret Combination
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display.
Output Specification:
Print a single line containing *n* digits — the desired state of the display containing the smallest possible number.
Demo Input:
['3\n579\n', '4\n2014\n']
Demo Output:
['024\n', '0142\n']
Note:
none
|
```python
n=int(input())
l=[int(x) for x in input()]
l=l+l
g=list(l)
ans='9'*10000
while True:
for x in range(0, len(l)-n):
ans=min(ans, ''.join(str(i) for i in l[x:x+n]))
l=[(x+1)%10 for x in l]
if l==g:
break
print(ans)
```
| 3
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,547,968,714
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 186
| 0
|
row, col = [*map(int, input().split())]
map = []
missing_vertices = None
for map_piece in range(row):
map_line = input()
map.append(map_line)
if '*' in map_line:
if map_line.count('*') == 1:
missing_vertices = map_piece
row = missing_vertices
col = 0 if map[row][0] != '*' else len(map) - 1
print(str(row + 1) + " " + str(col + 1))
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
row, col = [*map(int, input().split())]
map = []
missing_vertices = None
for map_piece in range(row):
map_line = input()
map.append(map_line)
if '*' in map_line:
if map_line.count('*') == 1:
missing_vertices = map_piece
row = missing_vertices
col = 0 if map[row][0] != '*' else len(map) - 1
print(str(row + 1) + " " + str(col + 1))
```
| 0
|
|
496
|
A
|
Minimum Difficulty
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
|
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
|
Print a single number — the minimum difficulty of the track after removing a single hold.
|
[
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] |
[
"5\n",
"2\n",
"4\n"
] |
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
| 500
|
[
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9 16 25 36 49 64 81 100",
"output": "19"
},
{
"input": "10\n300 315 325 338 350 365 379 391 404 416",
"output": "23"
},
{
"input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112",
"output": "2"
},
{
"input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500",
"output": "310"
},
{
"input": "3\n159 282 405",
"output": "246"
},
{
"input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741",
"output": "241"
},
{
"input": "10\n218 300 388 448 535 629 680 740 836 925",
"output": "111"
},
{
"input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996",
"output": "20"
},
{
"input": "100\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 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "901"
},
{
"input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978",
"output": "605"
},
{
"input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999",
"output": "803"
},
{
"input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932",
"output": "17"
},
{
"input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000",
"output": "54"
},
{
"input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000",
"output": "74"
}
] | 1,549,964,431
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 108
| 0
|
input()
X = list(map(int, input().split()))
MIN = X[-1] - X[0]
for i in range(1, len(X) - 1):
Y = X[:i] + X[i + 1:]
MAX = Y[1] - Y[0]
for j in range(len(Y)):
MAX = max(MAX, Y[j] - Y[j - 1])
MIN = min(MIN, MAX)
print(MIN)
|
Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
|
```python
input()
X = list(map(int, input().split()))
MIN = X[-1] - X[0]
for i in range(1, len(X) - 1):
Y = X[:i] + X[i + 1:]
MAX = Y[1] - Y[0]
for j in range(len(Y)):
MAX = max(MAX, Y[j] - Y[j - 1])
MIN = min(MIN, MAX)
print(MIN)
```
| 3
|
|
622
|
B
|
The Time
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
You are given the current time in 24-hour format hh:mm. Find and print the time after *a* minutes.
Note that you should find only the time after *a* minutes, see the examples to clarify the problem statement.
You can read more about 24-hour format here [https://en.wikipedia.org/wiki/24-hour_clock](https://en.wikipedia.org/wiki/24-hour_clock).
|
The first line contains the current time in the format hh:mm (0<=≤<=*hh*<=<<=24,<=0<=≤<=*mm*<=<<=60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).
The second line contains integer *a* (0<=≤<=*a*<=≤<=104) — the number of the minutes passed.
|
The only line should contain the time after *a* minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).
See the examples to check the input/output format.
|
[
"23:59\n10\n",
"20:20\n121\n",
"10:10\n0\n"
] |
[
"00:09\n",
"22:21\n",
"10:10\n"
] |
none
| 0
|
[
{
"input": "23:59\n10",
"output": "00:09"
},
{
"input": "20:20\n121",
"output": "22:21"
},
{
"input": "10:10\n0",
"output": "10:10"
},
{
"input": "12:34\n10000",
"output": "11:14"
},
{
"input": "00:00\n10000",
"output": "22:40"
},
{
"input": "00:00\n1440",
"output": "00:00"
},
{
"input": "23:59\n8640",
"output": "23:59"
},
{
"input": "10:01\n0",
"output": "10:01"
},
{
"input": "04:05\n0",
"output": "04:05"
},
{
"input": "02:59\n1",
"output": "03:00"
},
{
"input": "05:15\n10",
"output": "05:25"
},
{
"input": "03:10\n20",
"output": "03:30"
},
{
"input": "09:11\n0",
"output": "09:11"
},
{
"input": "19:00\n0",
"output": "19:00"
},
{
"input": "23:59\n1",
"output": "00:00"
},
{
"input": "11:59\n1",
"output": "12:00"
},
{
"input": "19:34\n566",
"output": "05:00"
},
{
"input": "00:01\n59",
"output": "01:00"
},
{
"input": "03:30\n0",
"output": "03:30"
},
{
"input": "22:30\n30",
"output": "23:00"
},
{
"input": "22:50\n70",
"output": "00:00"
},
{
"input": "05:12\n0",
"output": "05:12"
},
{
"input": "09:20\n40",
"output": "10:00"
},
{
"input": "15:04\n36",
"output": "15:40"
},
{
"input": "05:37\n23",
"output": "06:00"
},
{
"input": "23:59\n59",
"output": "00:58"
},
{
"input": "21:09\n9997",
"output": "19:46"
},
{
"input": "11:00\n1",
"output": "11:01"
},
{
"input": "20:01\n2699",
"output": "17:00"
},
{
"input": "01:00\n59",
"output": "01:59"
},
{
"input": "07:09\n6538",
"output": "20:07"
},
{
"input": "00:00\n10",
"output": "00:10"
},
{
"input": "02:09\n2074",
"output": "12:43"
},
{
"input": "01:10\n1",
"output": "01:11"
},
{
"input": "23:01\n59",
"output": "00:00"
},
{
"input": "08:50\n20",
"output": "09:10"
},
{
"input": "13:18\n5121",
"output": "02:39"
},
{
"input": "18:31\n2677",
"output": "15:08"
},
{
"input": "14:17\n108",
"output": "16:05"
},
{
"input": "02:45\n5617",
"output": "00:22"
},
{
"input": "00:00\n3600",
"output": "12:00"
},
{
"input": "19:01\n59",
"output": "20:00"
},
{
"input": "19:02\n59",
"output": "20:01"
},
{
"input": "14:00\n2880",
"output": "14:00"
},
{
"input": "01:10\n44",
"output": "01:54"
},
{
"input": "02:01\n59",
"output": "03:00"
},
{
"input": "07:02\n121",
"output": "09:03"
},
{
"input": "10:00\n61",
"output": "11:01"
},
{
"input": "23:59\n61",
"output": "01:00"
},
{
"input": "00:00\n0",
"output": "00:00"
},
{
"input": "23:59\n121",
"output": "02:00"
},
{
"input": "00:00\n60",
"output": "01:00"
},
{
"input": "15:52\n60",
"output": "16:52"
},
{
"input": "00:39\n6525",
"output": "13:24"
},
{
"input": "00:30\n30",
"output": "01:00"
},
{
"input": "00:59\n1",
"output": "01:00"
},
{
"input": "00:55\n4321",
"output": "00:56"
},
{
"input": "10:05\n1",
"output": "10:06"
},
{
"input": "23:00\n60",
"output": "00:00"
}
] | 1,589,648,109
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 59
| 156
| 20,172,800
|
a= input()
t=int(input())
hr= int(a[:a.index(':')])
mi = int(a[a.index(':')+1:])
v = mi+t
nm = str(v%60)
hr +=v//60
hw = str(hr%24)
if len(nm)==1:
nm= '0'+nm
if len(hw)==1:
hw = '0'+hw
print(hw+':'+nm)
|
Title: The Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the current time in 24-hour format hh:mm. Find and print the time after *a* minutes.
Note that you should find only the time after *a* minutes, see the examples to clarify the problem statement.
You can read more about 24-hour format here [https://en.wikipedia.org/wiki/24-hour_clock](https://en.wikipedia.org/wiki/24-hour_clock).
Input Specification:
The first line contains the current time in the format hh:mm (0<=≤<=*hh*<=<<=24,<=0<=≤<=*mm*<=<<=60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).
The second line contains integer *a* (0<=≤<=*a*<=≤<=104) — the number of the minutes passed.
Output Specification:
The only line should contain the time after *a* minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).
See the examples to check the input/output format.
Demo Input:
['23:59\n10\n', '20:20\n121\n', '10:10\n0\n']
Demo Output:
['00:09\n', '22:21\n', '10:10\n']
Note:
none
|
```python
a= input()
t=int(input())
hr= int(a[:a.index(':')])
mi = int(a[a.index(':')+1:])
v = mi+t
nm = str(v%60)
hr +=v//60
hw = str(hr%24)
if len(nm)==1:
nm= '0'+nm
if len(hw)==1:
hw = '0'+hw
print(hw+':'+nm)
```
| 3
|
|
812
|
C
|
Sagheer and Nubian Market
|
PROGRAMMING
| 1,500
|
[
"binary search",
"sortings"
] | null | null |
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
|
The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs.
|
On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
|
[
"3 11\n2 3 5\n",
"4 100\n1 2 5 6\n",
"1 7\n7\n"
] |
[
"2 11\n",
"4 54\n",
"0 0\n"
] |
In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
| 1,500
|
[
{
"input": "3 11\n2 3 5",
"output": "2 11"
},
{
"input": "4 100\n1 2 5 6",
"output": "4 54"
},
{
"input": "1 7\n7",
"output": "0 0"
},
{
"input": "1 7\n5",
"output": "1 6"
},
{
"input": "1 1\n1",
"output": "0 0"
},
{
"input": "4 33\n4 3 2 1",
"output": "3 27"
},
{
"input": "86 96\n89 48 14 55 5 35 7 79 49 70 74 18 64 63 35 93 63 97 90 77 33 11 100 75 60 99 54 38 3 6 55 1 7 64 56 90 21 76 35 16 61 78 38 78 93 21 89 1 58 53 34 77 56 37 46 59 30 5 85 1 52 87 84 99 97 9 15 66 29 60 17 16 59 23 88 93 32 2 98 89 63 42 9 86 70 80",
"output": "3 71"
},
{
"input": "9 2727\n73 41 68 90 51 7 20 48 69",
"output": "9 872"
},
{
"input": "35 792600\n61 11 82 29 3 50 65 60 62 86 83 78 15 82 7 77 38 87 100 12 93 86 96 79 14 58 60 47 94 39 36 23 69 93 18",
"output": "35 24043"
},
{
"input": "63 47677090\n53 4 59 68 6 12 47 63 28 93 9 53 61 63 53 70 77 63 49 76 70 23 4 40 4 34 24 70 42 83 84 95 11 46 38 83 26 85 34 29 67 96 3 62 97 7 42 65 49 45 50 54 81 74 83 59 10 87 95 87 89 27 3",
"output": "63 130272"
},
{
"input": "88 631662736\n93 75 25 7 6 55 92 23 22 32 4 48 61 29 91 79 16 18 18 9 66 9 57 62 3 81 48 16 21 90 93 58 30 8 31 47 44 70 34 85 52 71 58 42 99 53 43 54 96 26 6 13 38 4 13 60 1 48 32 100 52 8 27 99 66 34 98 45 19 50 37 59 31 56 58 70 61 14 100 66 74 85 64 57 92 89 7 92",
"output": "88 348883"
},
{
"input": "12 12\n1232 1848 2048 4694 5121 3735 9968 4687 2040 6033 5839 2507",
"output": "0 0"
},
{
"input": "37 5271\n368 6194 4856 8534 944 4953 2085 5350 788 7772 9786 1321 4310 4453 7078 9912 5799 4066 5471 5079 5161 9773 1300 5474 1202 1353 9499 9694 9020 6332 595 7619 1271 7430 1199 3127 8867",
"output": "5 4252"
},
{
"input": "65 958484\n9597 1867 5346 637 6115 5833 3318 6059 4430 9169 8155 7895 3534 7962 9900 9495 5694 3461 5370 1945 1724 9264 3475 618 3421 551 8359 6889 1843 6716 9216 2356 1592 6265 2945 6496 4947 2840 9057 6141 887 4823 4004 8027 1993 1391 796 7059 5500 4369 4012 4983 6495 8990 3633 5439 421 1129 6970 8796 7826 1200 8741 6555 5037",
"output": "65 468998"
},
{
"input": "90 61394040\n2480 6212 4506 829 8191 797 5336 6722 3178 1007 5849 3061 3588 6684 5983 5452 7654 5321 660 2569 2809 2179 679 4858 6887 2580 6880 6120 4159 5542 4999 8703 2386 8221 7046 1229 1662 4542 7089 3548 4298 1973 1854 2473 5507 241 359 5248 7907 5201 9624 4596 1723 2622 4800 4716 693 961 7402 9004 7994 8048 6590 5866 7502 3304 4331 5218 6906 1016 5342 6644 2205 5823 8525 4839 1914 2651 3940 7751 3489 4178 7234 6640 7602 9765 8559 7819 5827 163",
"output": "90 795634"
},
{
"input": "14 891190480\n1424 3077 9632 6506 4568 9650 5534 1085 6934 9340 2867 367 7075 618",
"output": "14 70147"
},
{
"input": "39 43\n22166 81842 15513 80979 39645 60168 96994 13493 12904 79871 49910 45356 93691 51829 18226 34288 11525 41944 40433 67295 30123 1081 55623 22279 75814 82316 2963 39329 38223 8445 43202 61912 15122 86367 37200 68113 57194 38541 49641",
"output": "0 0"
},
{
"input": "67 8824\n75515 67590 86373 34191 3446 27408 31581 24727 40005 23718 39738 30960 4786 51040 32590 80454 14335 47173 20079 41204 67289 58347 88969 88396 37681 43963 13886 85690 12259 14732 42036 62620 15011 41890 20150 59469 62104 30136 47163 19790 25699 27453 36151 52914 52684 20503 78622 81082 94500 55756 94030 54764 72763 37830 13210 64559 53600 87998 80472 19001 83769 79700 88794 10161 99980 95184 74439",
"output": "2 8268"
},
{
"input": "16 56532535\n84567 85265 99012 62115 58908 10120 27355 90226 17191 35588 43807 6327 89157 71156 16253 22387",
"output": "16 821610"
},
{
"input": "42 818723640\n57019 99450 43192 25131 80678 41786 52490 34454 51620 56132 41773 94622 23949 11597 96944 51420 44164 85727 72959 26622 1838 36302 99335 35572 92734 10651 22293 97601 38634 24066 81870 10904 5825 71968 2377 61198 3104 4191 97851 99920 65295 1385",
"output": "42 2094569"
}
] | 1,573,371,330
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 109
| 0
|
n,S = list(map(int, input().split()))
arr = list(map(int, input().split()))
c = []
m=n
for i in range(n):
sum=0
for j in range(m):
sum+= arr[j] + (j+1)*m
if sum<=S:
c.append((m,sum))
m-=1
if len(c)==0:
print(0,0)
else:
print(*max(c),sep=" ")
|
Title: Sagheer and Nubian Market
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
Input Specification:
The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs.
Output Specification:
On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
Demo Input:
['3 11\n2 3 5\n', '4 100\n1 2 5 6\n', '1 7\n7\n']
Demo Output:
['2 11\n', '4 54\n', '0 0\n']
Note:
In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
|
```python
n,S = list(map(int, input().split()))
arr = list(map(int, input().split()))
c = []
m=n
for i in range(n):
sum=0
for j in range(m):
sum+= arr[j] + (j+1)*m
if sum<=S:
c.append((m,sum))
m-=1
if len(c)==0:
print(0,0)
else:
print(*max(c),sep=" ")
```
| 0
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,678,383,247
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
a,b,c = [int(i) for i in input().split()]
print(min((a*2+b*2), a+b+c))
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
a,b,c = [int(i) for i in input().split()]
print(min((a*2+b*2), a+b+c))
```
| 0
|
|
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"
}
] | 1,591,032,927
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 218
| 307,200
|
n =int(input())
lst = [int(i) for i in input().split()]
d = dict()
for i in lst:
if i not in d:
d[i] = 1
else:
d[i]+=1
if n%2 == 0:
for i in d:
if d[i] >= n//2+1:
print("NO")
break
else:
print("YES")
else:
for i in d:
if d[i] >= n//2+2:
print("NO")
break
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
n =int(input())
lst = [int(i) for i in input().split()]
d = dict()
for i in lst:
if i not in d:
d[i] = 1
else:
d[i]+=1
if n%2 == 0:
for i in d:
if d[i] >= n//2+1:
print("NO")
break
else:
print("YES")
else:
for i in d:
if d[i] >= n//2+2:
print("NO")
break
else:
print("YES")
```
| 3
|
|
342
|
A
|
Xenia and Divisors
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
|
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
|
[
"6\n1 1 1 2 2 2\n",
"6\n2 2 1 1 4 6\n"
] |
[
"-1\n",
"1 2 4\n1 2 6\n"
] |
none
| 500
|
[
{
"input": "6\n1 1 1 2 2 2",
"output": "-1"
},
{
"input": "6\n2 2 1 1 4 6",
"output": "1 2 4\n1 2 6"
},
{
"input": "3\n1 2 3",
"output": "-1"
},
{
"input": "3\n7 5 7",
"output": "-1"
},
{
"input": "3\n1 3 4",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 3 6 6 3 1 3 1 6",
"output": "1 3 6\n1 3 6\n1 3 6"
},
{
"input": "6\n1 2 4 1 3 5",
"output": "-1"
},
{
"input": "3\n1 3 7",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 2 4 1 2 4 1 3 6",
"output": "1 2 4\n1 2 4\n1 3 6"
},
{
"input": "12\n3 6 1 1 3 6 1 1 2 6 2 6",
"output": "1 3 6\n1 3 6\n1 2 6\n1 2 6"
},
{
"input": "9\n1 1 1 4 4 4 6 2 2",
"output": "-1"
},
{
"input": "9\n1 2 4 6 3 1 3 1 5",
"output": "-1"
},
{
"input": "15\n2 1 2 1 3 6 1 2 1 6 1 3 4 6 4",
"output": "1 2 4\n1 2 4\n1 3 6\n1 3 6\n1 2 6"
},
{
"input": "3\n2 3 6",
"output": "-1"
},
{
"input": "3\n2 4 6",
"output": "-1"
},
{
"input": "3\n2 5 6",
"output": "-1"
},
{
"input": "3\n2 4 7",
"output": "-1"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "-1"
},
{
"input": "3\n7 7 7",
"output": "-1"
},
{
"input": "6\n1 2 4 7 7 7",
"output": "-1"
},
{
"input": "6\n1 1 2 6 6 6",
"output": "-1"
},
{
"input": "9\n1 1 1 3 3 2 4 4 6",
"output": "-1"
},
{
"input": "6\n1 2 4 5 5 5",
"output": "-1"
},
{
"input": "15\n1 1 1 1 1 2 2 2 2 4 4 6 6 6 6",
"output": "-1"
},
{
"input": "6\n1 1 5 5 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 5 6 7",
"output": "-1"
},
{
"input": "6\n1 1 4 4 7 7",
"output": "-1"
},
{
"input": "24\n1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 4 4 4 6 6 6",
"output": "-1"
},
{
"input": "3\n1 7 6",
"output": "-1"
},
{
"input": "6\n1 1 2 4 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 7 7 7 7 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 6 5 5",
"output": "-1"
}
] | 1,657,102,540
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 23
| 1,000
| 4,812,800
|
n = int(input())
a = sorted(list(map(int, input().split())))
m = n // 3
g = [[0 for j in range(3)] for i in range(m)]
index = [0] * m
b = False
for x in a:
b = False
for i in range(m):
j = index[i]
if j == 0 or (j > 0 and j < 3 and x > g[i][j - 1] and x % g[i][j - 1] == 0):
g[i][j] = x
index[i] += 1
b = True
break
if not b:
print(-1)
break
if b:
for v in g:
for e in v:
print(e, end=" ")
print()
|
Title: Xenia and Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
Output Specification:
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
Demo Input:
['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n']
Demo Output:
['-1\n', '1 2 4\n1 2 6\n']
Note:
none
|
```python
n = int(input())
a = sorted(list(map(int, input().split())))
m = n // 3
g = [[0 for j in range(3)] for i in range(m)]
index = [0] * m
b = False
for x in a:
b = False
for i in range(m):
j = index[i]
if j == 0 or (j > 0 and j < 3 and x > g[i][j - 1] and x % g[i][j - 1] == 0):
g[i][j] = x
index[i] += 1
b = True
break
if not b:
print(-1)
break
if b:
for v in g:
for e in v:
print(e, end=" ")
print()
```
| 0
|
|
445
|
A
|
DZY Loves Chessboard
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"implementation"
] | null | null |
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
|
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
|
[
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] |
[
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] |
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
| 500
|
[
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,629,898,924
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 6,963,200
|
def solution(board, n, m):
def frwd(i,j):
if j<m-1:
return i, j+1
return i+1, 0
def helper(i,j):
print(i,j)
if i>=n:
return True
ii, jj = frwd(i,j)
if board[i][j]=="-":
return helper(ii, jj)
elif i==0 and j==0:
board[i][j] = "B"
elif i==0:
if board[i][j-1]=="B":
board[i][j] = "W"
else:
board[i][j] = "B"
elif i==n-1 and j==0:
if board[i-1][j]=="B":
board[i][j] = "W"
else:
board[i][j] = "B"
else:
if (board[i-1][j]=="B" or board[i-1][j]=="-") and (board[i][j-1]=="B" or board[i][j-1]=="-"):
board[i][j] = "W"
elif (board[i-1][j]=="W" or board[i-1][j]=="-") and (board[i][j-1]=="W" or board[i][j-1]=="-"):
board[i][j] = "B"
else:
return False
if helper(ii, jj):
return True
if board[i][j]=="B":
board[i][j]="W"
else:
board[i][j]="B"
return helper(ii,jj)
helper(0,0)
return board
n, m = map(int, input().split())
board = []
for i in range(n):
board.append([j for j in input()])
board = solution(board, n, m)
for i in range(n):
row = "".join(board[i])
print(row)
|
Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
|
```python
def solution(board, n, m):
def frwd(i,j):
if j<m-1:
return i, j+1
return i+1, 0
def helper(i,j):
print(i,j)
if i>=n:
return True
ii, jj = frwd(i,j)
if board[i][j]=="-":
return helper(ii, jj)
elif i==0 and j==0:
board[i][j] = "B"
elif i==0:
if board[i][j-1]=="B":
board[i][j] = "W"
else:
board[i][j] = "B"
elif i==n-1 and j==0:
if board[i-1][j]=="B":
board[i][j] = "W"
else:
board[i][j] = "B"
else:
if (board[i-1][j]=="B" or board[i-1][j]=="-") and (board[i][j-1]=="B" or board[i][j-1]=="-"):
board[i][j] = "W"
elif (board[i-1][j]=="W" or board[i-1][j]=="-") and (board[i][j-1]=="W" or board[i][j-1]=="-"):
board[i][j] = "B"
else:
return False
if helper(ii, jj):
return True
if board[i][j]=="B":
board[i][j]="W"
else:
board[i][j]="B"
return helper(ii,jj)
helper(0,0)
return board
n, m = map(int, input().split())
board = []
for i in range(n):
board.append([j for j in input()])
board = solution(board, n, m)
for i in range(n):
row = "".join(board[i])
print(row)
```
| 0
|
|
45
|
I
|
TCMCF+++
|
PROGRAMMING
| 1,400
|
[
"greedy"
] |
I. TCMCF+++
|
2
|
256
|
Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest *n* problems were suggested and every problem had a cost — a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of points of a contestant was equal to the product of the costs of all the problems he/she had completed. If a person didn't solve anything, then he/she didn't even appear in final standings and wasn't considered as participant. Vasya understood that to get the maximal number of points it is not always useful to solve all the problems. Unfortunately, he understood it only after the contest was finished. Now he asks you to help him: find out what problems he had to solve to earn the maximal number of points.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of the suggested problems. The next line contains *n* space-separated integers *c**i* (<=-<=100<=≤<=*c**i*<=≤<=100) — the cost of the *i*-th task. The tasks' costs may coinсide.
|
Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them.
|
[
"5\n1 2 -3 3 3\n",
"13\n100 100 100 100 100 100 100 100 100 100 100 100 100\n",
"4\n-2 -2 -2 -2\n"
] |
[
"3 1 2 3 \n",
"100 100 100 100 100 100 100 100 100 100 100 100 100 \n",
"-2 -2 -2 -2 \n"
] |
none
| 0
|
[
{
"input": "5\n1 2 -3 3 3",
"output": "3 1 2 3 "
},
{
"input": "13\n100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "4\n-2 -2 -2 -2",
"output": "-2 -2 -2 -2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "1\n-1",
"output": "-1 "
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1 1 "
},
{
"input": "2\n1 -1",
"output": "1 "
},
{
"input": "2\n-1 1",
"output": "1 "
},
{
"input": "2\n-1 -1",
"output": "-1 -1 "
},
{
"input": "2\n1 0",
"output": "1 "
},
{
"input": "2\n0 1",
"output": "1 "
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "2\n-1 0",
"output": "0"
},
{
"input": "2\n0 -1",
"output": "0"
},
{
"input": "1\n13",
"output": "13 "
},
{
"input": "1\n-13",
"output": "-13 "
},
{
"input": "1\n100",
"output": "100 "
},
{
"input": "1\n-100",
"output": "-100 "
},
{
"input": "2\n100 100",
"output": "100 100 "
},
{
"input": "2\n100 -100",
"output": "100 "
},
{
"input": "2\n-100 100",
"output": "100 "
},
{
"input": "2\n100 0",
"output": "100 "
},
{
"input": "2\n0 100",
"output": "100 "
},
{
"input": "2\n0 -100",
"output": "0"
},
{
"input": "2\n-100 0",
"output": "0"
},
{
"input": "1\n3",
"output": "3 "
},
{
"input": "2\n0 -1",
"output": "0"
},
{
"input": "2\n-1 2",
"output": "2 "
},
{
"input": "2\n2 2",
"output": "2 2 "
},
{
"input": "2\n-1 -2",
"output": "-1 -2 "
},
{
"input": "2\n-2 -1",
"output": "-1 -2 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "2\n0 -2",
"output": "0"
},
{
"input": "2\n-2 -1",
"output": "-1 -2 "
},
{
"input": "3\n0 -2 -1",
"output": "-1 -2 "
},
{
"input": "3\n2 1 -1",
"output": "2 1 "
},
{
"input": "3\n0 1 2",
"output": "2 1 "
},
{
"input": "3\n-2 2 2",
"output": "2 2 "
},
{
"input": "3\n1 -1 2",
"output": "2 1 "
},
{
"input": "3\n-2 0 2",
"output": "2 "
},
{
"input": "3\n1 0 2",
"output": "2 1 "
},
{
"input": "3\n-1 2 2",
"output": "2 2 "
},
{
"input": "4\n0 0 2 -2",
"output": "2 "
},
{
"input": "4\n1 0 -1 2",
"output": "2 1 "
},
{
"input": "4\n-2 0 -2 0",
"output": "-2 -2 "
},
{
"input": "4\n2 2 1 -1",
"output": "2 1 2 "
},
{
"input": "4\n-1 2 0 -2",
"output": "2 -1 -2 "
},
{
"input": "4\n1 2 -2 1",
"output": "2 1 1 "
},
{
"input": "4\n-2 -1 2 2",
"output": "2 -1 2 -2 "
},
{
"input": "4\n-1 -1 -2 0",
"output": "-1 -2 "
},
{
"input": "10\n-10 5 9 -10 2 -7 10 10 6 -9",
"output": "10 -10 -9 -7 2 5 6 9 10 -10 "
},
{
"input": "10\n9 10 9 10 10 10 1 5 10 5",
"output": "10 5 5 9 9 10 10 10 10 1 "
},
{
"input": "10\n-3 -9 -10 -10 -9 -8 -9 -9 -8 -9",
"output": "-3 -10 -9 -9 -9 -9 -9 -8 -8 -10 "
},
{
"input": "10\n-5 -5 4 1 -8 -3 -9 -2 4 4",
"output": "4 -8 -5 -5 -3 -2 1 4 4 -9 "
},
{
"input": "100\n-74 11 -35 -39 31 -39 43 43 2 -78 -17 -16 -70 41 -96 -70 -89 48 -98 -44 47 -92 49 20 47 -23 -19 -24 7 -79 16 18 0 29 -43 -98 27 50 -65 -50 44 -66 -64 -34 -77 -38 22 18 8 30 -62 -37 -3 -80 -94 15 -50 -61 6 -97 35 24 -19 -79 -47 -4 4 38 -37 -51 -31 -24 -3 -3 -94 -99 -87 -35 48 -57 16 -2 6 -13 -5 -60 -39 -61 -42 7 -14 -4 -99 -32 31 17 -84 13 -72 -37",
"output": "50 -99 -98 -98 -97 -96 -94 -94 -92 -89 -87 -84 -80 -79 -79 -78 -77 -74 -72 -70 -70 -66 -65 -64 -62 -61 -61 -60 -57 -51 -50 -50 -47 -44 -43 -42 -39 -39 -39 -38 -37 -37 -37 -35 -35 -34 -32 -31 -24 -24 -23 -19 -19 -17 -16 -14 -13 -5 -4 -4 -3 -3 -3 -2 2 4 6 6 7 7 8 11 13 15 16 16 17 18 18 20 22 24 27 29 30 31 31 35 38 41 43 43 44 47 47 48 48 49 -99 "
},
{
"input": "100\n8 -63 12 -31 48 1 11 7 -18 -25 -3 11 -23 44 31 33 -10 44 46 -62 29 5 -4 -35 -1 0 -20 34 -18 -46 -9 46 41 42 -69 18 24 48 50 13 -24 -19 37 -21 8 50 30 24 -48 36 -42 -43 31 50 -17 -29 -27 9 50 47 36 -50 6 -51 12 49 -1 -15 37 -44 -19 46 27 5 -37 17 19 39 11 42 44 43 -48 -1 31 -80 -30 -35 -19 50 35 -56 -1 20 50 -13 27 39 -20 -15",
"output": "50 -69 -63 -62 -56 -51 -50 -48 -48 -46 -44 -43 -42 -37 -35 -35 -31 -30 -29 -27 -25 -24 -23 -21 -20 -20 -19 -19 -19 -18 -18 -17 -15 -15 -13 -10 -9 -4 -3 -1 -1 -1 1 5 5 6 7 8 8 9 11 11 11 12 12 13 17 18 19 20 24 24 27 27 29 30 31 31 31 33 34 35 36 36 37 37 39 39 41 42 42 43 44 44 44 46 46 46 47 48 48 49 50 50 50 50 50 -80 "
},
{
"input": "100\n-39 -43 43 18 -21 -24 -8 -29 -32 -12 50 35 5 1 42 -24 44 37 6 25 -39 17 32 -67 49 -19 -19 50 45 17 -48 17 -11 -16 47 44 29 -29 30 1 50 -4 35 -18 0 -9 -14 31 49 34 -25 36 37 9 3 26 43 25 40 24 48 21 -18 50 -10 26 39 29 45 1 48 34 14 -48 17 0 50 33 -4 31 31 -63 10 26 22 8 50 -15 1 -1 -15 13 -47 25 3 45 22 -5 -16 32",
"output": "50 -63 -48 -48 -47 -43 -39 -39 -32 -29 -29 -25 -24 -24 -21 -19 -19 -18 -18 -16 -16 -15 -15 -14 -12 -11 -10 -9 -8 -5 -4 -4 1 1 1 1 3 3 5 6 8 9 10 13 14 17 17 17 17 18 21 22 22 24 25 25 25 26 26 26 29 29 30 31 31 31 32 32 33 34 34 35 35 36 37 37 39 40 42 43 43 44 44 45 45 45 47 48 48 49 49 50 50 50 50 50 -67 "
},
{
"input": "100\n43 17 39 -15 9 24 28 21 42 -56 9 12 -53 -30 -1 26 39 44 50 46 47 22 29 11 30 42 27 34 31 31 46 7 33 47 48 2 44 -19 33 32 22 23 39 34 -8 1 -18 33 43 45 47 39 -15 44 50 32 42 42 46 -13 28 35 31 -31 13 30 -10 0 9 50 24 38 24 -48 20 43 13 46 26 36 43 32 48 5 -5 39 37 41 -4 -6 -9 32 42 -3 31 37 39 48 26 38",
"output": "50 -53 -48 -31 -30 -19 -18 -15 -15 -13 -10 -9 -8 -6 -5 -4 -3 -1 1 2 5 7 9 9 9 11 12 13 13 17 20 21 22 22 23 24 24 24 26 26 26 27 28 28 29 30 30 31 31 31 31 32 32 32 32 33 33 33 34 34 35 36 37 37 38 38 39 39 39 39 39 39 41 42 42 42 42 42 43 43 43 43 44 44 44 45 46 46 46 46 47 47 47 48 48 48 50 50 -56 "
},
{
"input": "100\n35 41 38 39 46 -1 19 42 34 22 0 -23 48 24 41 12 11 4 4 35 35 2 9 33 50 30 15 21 44 47 47 27 31 24 40 14 22 26 45 1 35 31 13 8 48 50 31 36 26 26 48 41 6 -19 17 17 16 3 38 42 41 35 19 31 15 -48 43 6 -32 -18 -2 41 44 29 11 46 43 48 -12 34 30 -10 -7 44 47 24 44 32 36 29 15 25 25 -19 26 46 36 37 -10 45",
"output": "50 -32 -23 -19 -19 -18 -12 -10 -10 -7 -2 -1 1 2 3 4 4 6 6 8 9 11 11 12 13 14 15 15 15 16 17 17 19 19 21 22 22 24 24 24 25 25 26 26 26 26 27 29 29 30 30 31 31 31 31 32 33 34 34 35 35 35 35 35 36 36 36 37 38 38 39 40 41 41 41 41 41 42 42 43 43 44 44 44 44 45 45 46 46 46 47 47 47 48 48 48 48 50 -48 "
},
{
"input": "100\n42 -14 90 5 0 62 14 36 -76 -94 69 25 -2 40 -49 62 -38 0 -96 49 -24 -92 55 18 22 42 -25 72 -52 47 78 98 80 -27 -64 -4 -38 -93 -15 40 -78 -49 -49 21 50 -13 34 0 78 84 55 -95 -52 -3 -46 -49 53 23 -49 -98 -1 47 48 -93 25 37 -71 -23 74 -58 -39 -43 -14 -57 98 -6 9 -56 88 -88 7 71 -60 95 -9 15 55 63 -75 -29 -90 -38 -61 -97 9 -40 89 92 -37 50",
"output": "98 -97 -96 -95 -94 -93 -93 -92 -90 -88 -78 -76 -75 -71 -64 -61 -60 -58 -57 -56 -52 -52 -49 -49 -49 -49 -49 -46 -43 -40 -39 -38 -38 -38 -37 -29 -27 -25 -24 -23 -15 -14 -14 -13 -9 -6 -4 -3 -2 -1 5 7 9 9 14 15 18 21 22 23 25 25 34 36 37 40 40 42 42 47 47 48 49 50 50 53 55 55 55 62 62 63 69 71 72 74 78 78 80 84 88 89 90 92 95 98 -98 "
},
{
"input": "100\n-60 98 -34 30 48 69 -50 70 3 85 67 73 -23 64 31 98 57 84 54 81 24 37 41 -29 73 -6 3 62 -23 86 67 -8 79 38 60 64 -65 78 81 95 98 100 38 -46 -6 -4 14 18 58 95 94 57 21 66 8 26 89 99 74 46 69 75 97 54 29 79 1 -90 67 61 24 62 78 -1 96 82 23 87 9 87 2 -50 -26 30 74 52 -28 39 69 67 6 56 74 93 13 -22 23 97 70 -45",
"output": "100 -65 -60 -50 -50 -46 -45 -34 -29 -28 -26 -23 -23 -22 -8 -6 -6 -4 1 2 3 3 6 8 9 13 14 18 21 23 23 24 24 26 29 30 30 31 37 38 38 39 41 46 48 52 54 54 56 57 57 58 60 61 62 62 64 64 66 67 67 67 67 69 69 69 70 70 73 73 74 74 74 75 78 78 79 79 81 81 82 84 85 86 87 87 89 93 94 95 95 96 97 97 98 98 98 99 -90 "
},
{
"input": "100\n16 -41 -52 -100 -74 -57 -57 -49 63 -71 -26 -96 -50 -49 -57 -66 -27 -32 -99 6 -24 14 -79 -57 -82 -81 17 -54 -47 3 -66 -100 2 -35 -18 -83 12 46 -37 -19 -1 30 -93 -59 57 -69 -43 -83 -91 -28 -67 -17 -18 13 -35 57 -59 -85 33 -77 -74 92 -58 -82 -59 42 29 -54 -50 -89 -39 68 -64 -86 37 -73 -68 -85 -51 -25 -31 -10 -70 32 1 -64 -47 27 -86 -7 24 -55 -73 -88 21 -3 61 -61 -44 -24",
"output": "92 -100 -99 -96 -93 -91 -89 -88 -86 -86 -85 -85 -83 -83 -82 -82 -81 -79 -77 -74 -74 -73 -73 -71 -70 -69 -68 -67 -66 -66 -64 -64 -61 -59 -59 -59 -58 -57 -57 -57 -57 -55 -54 -54 -52 -51 -50 -50 -49 -49 -47 -47 -44 -43 -41 -39 -37 -35 -35 -32 -31 -28 -27 -26 -25 -24 -24 -19 -18 -18 -17 -10 -7 -3 1 2 3 6 12 13 14 16 17 21 24 27 29 30 32 33 37 42 46 57 57 61 63 68 -100 "
},
{
"input": "100\n99 100 100 99 99 96 100 100 100 98 98 99 100 99 98 98 97 99 99 100 94 100 99 98 97 100 98 99 100 99 97 91 99 95 95 97 99 99 100 98 100 99 99 100 99 100 100 93 96 96 93 99 99 99 100 96 100 97 92 100 100 100 97 100 100 99 98 98 95 97 96 92 97 100 100 90 98 100 100 99 100 98 99 99 100 97 94 99 100 100 99 99 99 100 99 100 96 99 98 100",
"output": "100 91 92 92 93 93 94 94 95 95 95 96 96 96 96 96 96 97 97 97 97 97 97 97 97 97 98 98 98 98 98 98 98 98 98 98 98 98 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 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 90 "
},
{
"input": "100\n-93 -100 -99 -100 -99 -100 -99 -99 -99 -100 -99 -99 -100 -98 -98 -100 -99 -100 -95 -100 -100 -99 -98 -100 -96 -99 -99 -100 -97 -98 -100 -98 -98 -93 -100 -100 -99 -100 -100 -98 -100 -98 -100 -98 -97 -99 -98 -99 -99 -99 -100 -100 -100 -100 -98 -93 -100 -100 -97 -100 -100 -100 -98 -99 -100 -95 -98 -99 -100 -100 -99 -99 -100 -98 -100 -100 -100 -100 -96 -100 -93 -97 -100 -100 -99 -93 -92 -98 -97 -94 -100 -100 -99 -98 -100 -100 -100 -99 -99 -100",
"output": "-92 -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 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -99 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -98 -97 -97 -97 -97 -97 -96 -96 -95 -95 -94 -93 -93 -93 -93 -93 -100 "
}
] | 1,593,458,609
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 280
| 20,172,800
|
a=int(input())
b=list(map(int,input().split()))
positivep=[]
negativep=[]
zerop=[]
for i in range(len(b)):
if b[i]>0:
positivep.append(b[i])
elif b[i]<0:
negativep.append(b[i])
else:
zerop.append(b[i])
ans=''
if len(negativep)%2==0:
for i in range(len(negativep)):
ans+=str(negativep[i])+' '
for j in range(len(positivep)):
ans+=str(positivep[j])+' '
print(ans)
else:
for j in range(len(positivep)):
ans+=str(positivep[j])+' '
print(ans)
|
Title: TCMCF+++
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest *n* problems were suggested and every problem had a cost — a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of points of a contestant was equal to the product of the costs of all the problems he/she had completed. If a person didn't solve anything, then he/she didn't even appear in final standings and wasn't considered as participant. Vasya understood that to get the maximal number of points it is not always useful to solve all the problems. Unfortunately, he understood it only after the contest was finished. Now he asks you to help him: find out what problems he had to solve to earn the maximal number of points.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of the suggested problems. The next line contains *n* space-separated integers *c**i* (<=-<=100<=≤<=*c**i*<=≤<=100) — the cost of the *i*-th task. The tasks' costs may coinсide.
Output Specification:
Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them.
Demo Input:
['5\n1 2 -3 3 3\n', '13\n100 100 100 100 100 100 100 100 100 100 100 100 100\n', '4\n-2 -2 -2 -2\n']
Demo Output:
['3 1 2 3 \n', '100 100 100 100 100 100 100 100 100 100 100 100 100 \n', '-2 -2 -2 -2 \n']
Note:
none
|
```python
a=int(input())
b=list(map(int,input().split()))
positivep=[]
negativep=[]
zerop=[]
for i in range(len(b)):
if b[i]>0:
positivep.append(b[i])
elif b[i]<0:
negativep.append(b[i])
else:
zerop.append(b[i])
ans=''
if len(negativep)%2==0:
for i in range(len(negativep)):
ans+=str(negativep[i])+' '
for j in range(len(positivep)):
ans+=str(positivep[j])+' '
print(ans)
else:
for j in range(len(positivep)):
ans+=str(positivep[j])+' '
print(ans)
```
| 0
|
510
|
D
|
Fox And Jumping
|
PROGRAMMING
| 1,900
|
[
"bitmasks",
"brute force",
"dp",
"math"
] | null | null |
Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0.
There are also *n* cards, each card has 2 attributes: length *l**i* and cost *c**i*. If she pays *c**i* dollars then she can apply *i*-th card. After applying *i*-th card she becomes able to make jumps of length *l**i*, i. e. from cell *x* to cell (*x*<=-<=*l**i*) or cell (*x*<=+<=*l**i*).
She wants to be able to jump to any cell on the tape (possibly, visiting some intermediate cells). For achieving this goal, she wants to buy some cards, paying as little money as possible.
If this is possible, calculate the minimal cost.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=300), number of cards.
The second line contains *n* numbers *l**i* (1<=≤<=*l**i*<=≤<=109), the jump lengths of cards.
The third line contains *n* numbers *c**i* (1<=≤<=*c**i*<=≤<=105), the costs of cards.
|
If it is impossible to buy some cards and become able to jump to any cell, output -1. Otherwise output the minimal cost of buying such set of cards.
|
[
"3\n100 99 9900\n1 1 1\n",
"5\n10 20 30 40 50\n1 1 1 1 1\n",
"7\n15015 10010 6006 4290 2730 2310 1\n1 1 1 1 1 1 10\n",
"8\n4264 4921 6321 6984 2316 8432 6120 1026\n4264 4921 6321 6984 2316 8432 6120 1026\n"
] |
[
"2\n",
"-1\n",
"6\n",
"7237\n"
] |
In first sample test, buying one card is not enough: for example, if you buy a card with length 100, you can't jump to any cell whose index is not a multiple of 100. The best way is to buy first and second card, that will make you be able to jump to any cell.
In the second sample test, even if you buy all cards, you can't jump to any cell whose index is not a multiple of 10, so you should output -1.
| 2,000
|
[
{
"input": "3\n100 99 9900\n1 1 1",
"output": "2"
},
{
"input": "5\n10 20 30 40 50\n1 1 1 1 1",
"output": "-1"
},
{
"input": "7\n15015 10010 6006 4290 2730 2310 1\n1 1 1 1 1 1 10",
"output": "6"
},
{
"input": "8\n4264 4921 6321 6984 2316 8432 6120 1026\n4264 4921 6321 6984 2316 8432 6120 1026",
"output": "7237"
},
{
"input": "6\n1 2 4 8 16 32\n32 16 8 4 2 1",
"output": "32"
},
{
"input": "1\n1\n1",
"output": "1"
},
{
"input": "1\n2\n2",
"output": "-1"
},
{
"input": "8\n2 3 5 7 11 13 17 19\n4 8 7 1 5 2 6 3",
"output": "3"
},
{
"input": "1\n1000000000\n100000",
"output": "-1"
},
{
"input": "2\n1000000000 999999999\n100000 100000",
"output": "200000"
},
{
"input": "39\n692835 4849845 22610 1995 19019 114 6270 15 85085 27170 1365 1155 7410 238 3135 546 373065 715 110 969 15 10374 2730 19019 85 65 5187 26 3233230 1122 399 1122 53295 910 110 12597 16302 125970 67830\n4197 6490 2652 99457 65400 96257 33631 23456 14319 22288 16179 74656 89713 31503 45895 31777 64534 27989 60861 69846 44586 87185 96589 62279 62478 6180 26977 12112 9975 72933 73239 65856 98253 18875 55266 55867 36397 40743 47977",
"output": "18961"
},
{
"input": "35\n512 268435456 8 128 134217728 8192 33554432 33554432 536870912 512 65536 1048576 32768 512 524288 1024 536870912 536870912 16 32 33554432 134217728 2 16 16777216 8192 262144 65536 33554432 128 4096 2097152 33554432 2097152 2\n36157 67877 79710 63062 12683 36255 61053 83828 93590 74236 5281 28143 7350 45953 96803 15998 11240 45207 63010 74076 85227 83498 68320 77288 48100 51373 87843 70054 28986 25365 98581 11195 43674 75769 22053",
"output": "-1"
}
] | 1,691,494,168
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1691494167.9766536")# 1691494167.9766698
|
Title: Fox And Jumping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0.
There are also *n* cards, each card has 2 attributes: length *l**i* and cost *c**i*. If she pays *c**i* dollars then she can apply *i*-th card. After applying *i*-th card she becomes able to make jumps of length *l**i*, i. e. from cell *x* to cell (*x*<=-<=*l**i*) or cell (*x*<=+<=*l**i*).
She wants to be able to jump to any cell on the tape (possibly, visiting some intermediate cells). For achieving this goal, she wants to buy some cards, paying as little money as possible.
If this is possible, calculate the minimal cost.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=300), number of cards.
The second line contains *n* numbers *l**i* (1<=≤<=*l**i*<=≤<=109), the jump lengths of cards.
The third line contains *n* numbers *c**i* (1<=≤<=*c**i*<=≤<=105), the costs of cards.
Output Specification:
If it is impossible to buy some cards and become able to jump to any cell, output -1. Otherwise output the minimal cost of buying such set of cards.
Demo Input:
['3\n100 99 9900\n1 1 1\n', '5\n10 20 30 40 50\n1 1 1 1 1\n', '7\n15015 10010 6006 4290 2730 2310 1\n1 1 1 1 1 1 10\n', '8\n4264 4921 6321 6984 2316 8432 6120 1026\n4264 4921 6321 6984 2316 8432 6120 1026\n']
Demo Output:
['2\n', '-1\n', '6\n', '7237\n']
Note:
In first sample test, buying one card is not enough: for example, if you buy a card with length 100, you can't jump to any cell whose index is not a multiple of 100. The best way is to buy first and second card, that will make you be able to jump to any cell.
In the second sample test, even if you buy all cards, you can't jump to any cell whose index is not a multiple of 10, so you should output -1.
|
```python
print("_RANDOM_GUESS_1691494167.9766536")# 1691494167.9766698
```
| 0
|
|
816
|
B
|
Karen and Coffee
|
PROGRAMMING
| 1,400
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
|
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
[
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] |
[
"3\n3\n0\n4\n",
"0\n"
] |
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
| 1,000
|
[
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,528,988,791
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 13
| 2,074
| 28,364,800
|
from sys import stdin,stdout
n,k,q=map(int,input().split())
a=[[] for i in range(200001)];b=[0]*200001
for i in range(1,n+1):
c,d=(int(x) for x in stdin.readline().split())
a[c].append(d);a[d].append(c)
c=0
for i in range(1,200001):
r=0
for j in a[i]:
if j>i:
c+=1
elif i==j:
c+=1;r-=1
else:
r-=1
b[i]=c
c+=r
c=0;d=[0]*200001
for i in range(200001):
if b[i]>=k:
c+=1
d[i]=c
for i in range(q):
c,r=(int(x) for x in stdin.readline().split())
stdout.write(str(d[r]-d[c-1])+'\n')
|
Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
|
```python
from sys import stdin,stdout
n,k,q=map(int,input().split())
a=[[] for i in range(200001)];b=[0]*200001
for i in range(1,n+1):
c,d=(int(x) for x in stdin.readline().split())
a[c].append(d);a[d].append(c)
c=0
for i in range(1,200001):
r=0
for j in a[i]:
if j>i:
c+=1
elif i==j:
c+=1;r-=1
else:
r-=1
b[i]=c
c+=r
c=0;d=[0]*200001
for i in range(200001):
if b[i]>=k:
c+=1
d[i]=c
for i in range(q):
c,r=(int(x) for x in stdin.readline().split())
stdout.write(str(d[r]-d[c-1])+'\n')
```
| 0
|
|
556
|
A
|
Case of the Zeros and Ones
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
|
Output the minimum length of the string that may remain after applying the described operations several times.
|
[
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] |
[
"0\n",
"1\n",
"6\n"
] |
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 250
|
[
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,679,007,730
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 11
| 1,000
| 716,800
|
numero = int(input())
lista = str(input())
while "01" in lista or "10" in lista:
if("01" in lista):
lista = lista.replace("01", "")
if("10" in lista):
lista = lista.replace("10", "")
print(len(lista))
|
Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
numero = int(input())
lista = str(input())
while "01" in lista or "10" in lista:
if("01" in lista):
lista = lista.replace("01", "")
if("10" in lista):
lista = lista.replace("10", "")
print(len(lista))
```
| 0
|
|
909
|
C
|
Python Indentation
|
PROGRAMMING
| 1,800
|
[
"dp"
] | null | null |
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
|
The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
|
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7.
|
[
"4\ns\nf\nf\ns\n",
"4\nf\ns\nf\ns\n"
] |
[
"1\n",
"2\n"
] |
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or
| 1,500
|
[
{
"input": "4\ns\nf\nf\ns",
"output": "1"
},
{
"input": "4\nf\ns\nf\ns",
"output": "2"
},
{
"input": "156\nf\ns\nf\ns\nf\ns\ns\ns\ns\nf\ns\ns\nf\nf\ns\nf\nf\nf\nf\ns\ns\ns\nf\ns\ns\nf\nf\nf\nf\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\nf\nf\nf\ns\ns\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\ns\ns\ns\nf\ns\ns\nf\ns\nf\nf\ns\ns\ns\nf\nf\nf\nf\ns\ns\nf\nf\nf\nf\nf\nf\nf\ns\nf\ns\ns\ns\nf\nf\ns\ns\ns\ns\ns\nf\nf\nf\nf\ns\nf\nf\ns\nf\ns\ns\nf\nf\nf\ns\ns\ns\nf\ns\ns\nf\ns\nf\nf\nf\ns\nf\nf\ns\ns\nf\ns\nf\nf\ns\ns\ns\ns\nf\ns\nf\nf\ns\ns\nf\nf\nf\ns\ns\nf\nf\nf\ns\nf\ns\nf\nf\ns",
"output": "666443222"
},
{
"input": "4\nf\nf\ns\ns",
"output": "3"
},
{
"input": "2\nf\ns",
"output": "1"
},
{
"input": "1\ns",
"output": "1"
},
{
"input": "3\nf\nf\ns",
"output": "1"
},
{
"input": "2\ns\ns",
"output": "1"
},
{
"input": "156\ns\nf\ns\ns\ns\ns\nf\ns\ns\ns\nf\nf\ns\nf\nf\ns\nf\nf\nf\ns\nf\nf\ns\nf\nf\ns\ns\nf\nf\ns\nf\nf\nf\nf\nf\ns\ns\nf\ns\nf\nf\nf\ns\nf\nf\nf\ns\ns\ns\nf\ns\ns\nf\nf\ns\ns\nf\ns\nf\nf\ns\nf\nf\nf\ns\ns\nf\nf\ns\nf\ns\ns\ns\ns\ns\ns\ns\nf\ns\nf\nf\nf\ns\ns\ns\ns\nf\nf\ns\nf\nf\ns\ns\nf\ns\nf\ns\ns\nf\nf\nf\nf\nf\ns\nf\ns\ns\nf\nf\ns\nf\nf\ns\ns\ns\nf\ns\ns\ns\ns\nf\nf\ns\nf\nf\nf\nf\ns\nf\ns\ns\nf\nf\ns\nf\ns\nf\nf\nf\nf\ns\ns\nf\nf\nf\nf\ns\nf\ns\nf\ns\ns\ns\nf\nf\ns",
"output": "479461584"
},
{
"input": "66\ns\nf\ns\ns\nf\ns\ns\ns\ns\nf\ns\ns\nf\nf\ns\ns\nf\ns\ns\nf\ns\ns\nf\nf\ns\ns\nf\nf\ns\ns\nf\ns\ns\ns\ns\nf\nf\ns\ns\nf\nf\ns\ns\nf\ns\ns\nf\ns\ns\nf\ns\ns\nf\nf\ns\nf\ns\ns\nf\nf\ns\nf\ns\nf\nf\ns",
"output": "392847498"
}
] | 1,603,961,529
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 171
| 409,600
|
n = int(input())
l = []
for _ in range(n):
l.append(input().strip())
tb = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
tb[i][i] = 0 if l[i]=='f' else 1
for k in range(1,n):
for i in range(n-k):
# i to i+k
if l[i+k]=='f':
tb[i][i+k]= 0
elif l[i]=='f':
c = 0
for j in range(1,k):
c+= tb[i+1][i+j]*tb[i+j+1][i+k]
tb[i][i+k] = c+tb[i+1][i+k]
elif l[i]=='s':
tb[i][i+k] = tb[i+1][i+k]
#print(i,i+k)
#print(c)
print(tb[0][n-1])
|
Title: Python Indentation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input Specification:
The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output Specification:
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7.
Demo Input:
['4\ns\nf\nf\ns\n', '4\nf\ns\nf\ns\n']
Demo Output:
['1\n', '2\n']
Note:
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or
|
```python
n = int(input())
l = []
for _ in range(n):
l.append(input().strip())
tb = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
tb[i][i] = 0 if l[i]=='f' else 1
for k in range(1,n):
for i in range(n-k):
# i to i+k
if l[i+k]=='f':
tb[i][i+k]= 0
elif l[i]=='f':
c = 0
for j in range(1,k):
c+= tb[i+1][i+j]*tb[i+j+1][i+k]
tb[i][i+k] = c+tb[i+1][i+k]
elif l[i]=='s':
tb[i][i+k] = tb[i+1][i+k]
#print(i,i+k)
#print(c)
print(tb[0][n-1])
```
| 0
|
|
784
|
A
|
Numbers Joke
|
PROGRAMMING
| 1,700
|
[
"*special"
] | null | null |
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output a single integer.
|
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
|
Output a single integer.
|
[
"3\n"
] |
[
"27\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "27"
},
{
"input": "1",
"output": "4"
},
{
"input": "2",
"output": "22"
},
{
"input": "4",
"output": "58"
},
{
"input": "5",
"output": "85"
},
{
"input": "6",
"output": "94"
},
{
"input": "7",
"output": "121"
},
{
"input": "8",
"output": "166"
},
{
"input": "9",
"output": "202"
},
{
"input": "10",
"output": "265"
},
{
"input": "11",
"output": "274"
},
{
"input": "12",
"output": "319"
},
{
"input": "13",
"output": "346"
},
{
"input": "14",
"output": "355"
},
{
"input": "15",
"output": "378"
},
{
"input": "16",
"output": "382"
},
{
"input": "17",
"output": "391"
},
{
"input": "18",
"output": "438"
},
{
"input": "19",
"output": "454"
},
{
"input": "20",
"output": "483"
},
{
"input": "21",
"output": "517"
},
{
"input": "22",
"output": "526"
},
{
"input": "23",
"output": "535"
},
{
"input": "24",
"output": "562"
},
{
"input": "25",
"output": "576"
},
{
"input": "26",
"output": "588"
},
{
"input": "27",
"output": "627"
},
{
"input": "28",
"output": "634"
},
{
"input": "29",
"output": "636"
},
{
"input": "30",
"output": "645"
}
] | 1,491,245,518
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 5,529,600
|
print((int(input()))**3)
|
Title: Numbers Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output a single integer.
Input Specification:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output Specification:
Output a single integer.
Demo Input:
['3\n']
Demo Output:
['27\n']
Note:
none
|
```python
print((int(input()))**3)
```
| 0
|
|
816
|
B
|
Karen and Coffee
|
PROGRAMMING
| 1,400
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
|
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
[
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] |
[
"3\n3\n0\n4\n",
"0\n"
] |
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
| 1,000
|
[
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,564,936,775
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 2,500
| 819,200
|
n,k,q = map(int,input().split())
countArray = [0]*200001
for i in range(n):
l,r = map(int,input().split())
for j in range(l,r+1):
countArray[j] += 1
for i in range(q):
a,b = map(int,input().split())
count = 0
for j in range(a,b+1):
if countArray[j] >= k:
count += 1
print(count)
|
Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
|
```python
n,k,q = map(int,input().split())
countArray = [0]*200001
for i in range(n):
l,r = map(int,input().split())
for j in range(l,r+1):
countArray[j] += 1
for i in range(q):
a,b = map(int,input().split())
count = 0
for j in range(a,b+1):
if countArray[j] >= k:
count += 1
print(count)
```
| 0
|
|
967
|
A
|
Mind the Gap
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
|
The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order.
|
Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
|
[
"6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n",
"16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n",
"3 17\n0 30\n1 0\n12 0\n"
] |
[
"6 1\n",
"24 50\n",
"0 0\n"
] |
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing.
| 500
|
[
{
"input": "6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40",
"output": "6 1"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "24 50"
},
{
"input": "3 17\n0 30\n1 0\n12 0",
"output": "0 0"
},
{
"input": "24 60\n0 21\n2 21\n2 46\n3 17\n4 15\n5 43\n6 41\n7 50\n8 21\n9 8\n10 31\n10 45\n12 30\n14 8\n14 29\n14 32\n14 52\n15 16\n16 7\n16 52\n18 44\n20 25\n21 13\n22 7",
"output": "23 8"
},
{
"input": "20 60\n0 9\n0 19\n0 57\n2 42\n3 46\n3 47\n5 46\n8 1\n9 28\n9 41\n10 54\n12 52\n13 0\n14 49\n17 28\n17 39\n19 34\n20 52\n21 35\n23 22",
"output": "6 47"
},
{
"input": "57 20\n0 2\n0 31\n1 9\n1 42\n1 58\n2 4\n2 35\n2 49\n3 20\n3 46\n4 23\n4 52\n5 5\n5 39\n6 7\n6 48\n6 59\n7 8\n7 35\n8 10\n8 46\n8 53\n9 19\n9 33\n9 43\n10 18\n10 42\n11 0\n11 26\n12 3\n12 5\n12 30\n13 1\n13 38\n14 13\n14 54\n15 31\n16 5\n16 44\n17 18\n17 30\n17 58\n18 10\n18 34\n19 13\n19 49\n19 50\n19 59\n20 17\n20 23\n20 40\n21 18\n21 57\n22 31\n22 42\n22 56\n23 37",
"output": "23 58"
},
{
"input": "66 20\n0 16\n0 45\n0 58\n1 6\n1 19\n2 7\n2 9\n3 9\n3 25\n3 57\n4 38\n4 58\n5 21\n5 40\n6 16\n6 19\n6 58\n7 6\n7 26\n7 51\n8 13\n8 36\n8 55\n9 1\n9 15\n9 33\n10 12\n10 37\n11 15\n11 34\n12 8\n12 37\n12 55\n13 26\n14 0\n14 34\n14 36\n14 48\n15 23\n15 29\n15 43\n16 8\n16 41\n16 45\n17 5\n17 7\n17 15\n17 29\n17 46\n18 12\n18 19\n18 38\n18 57\n19 32\n19 58\n20 5\n20 40\n20 44\n20 50\n21 18\n21 49\n22 18\n22 47\n23 1\n23 38\n23 50",
"output": "1 40"
},
{
"input": "1 1\n0 0",
"output": "0 2"
},
{
"input": "10 1\n0 2\n0 4\n0 5\n0 8\n0 9\n0 11\n0 13\n0 16\n0 19\n0 21",
"output": "0 0"
},
{
"input": "10 1\n0 2\n0 5\n0 8\n0 11\n0 15\n0 17\n0 25\n0 28\n0 29\n0 32",
"output": "0 0"
},
{
"input": "15 20\n0 47\n2 24\n4 19\n4 34\n5 46\n8 15\n9 8\n10 28\n17 47\n17 52\n18 32\n19 50\n20 46\n20 50\n23 21",
"output": "0 0"
},
{
"input": "1 5\n1 0",
"output": "0 0"
},
{
"input": "24 60\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "1 30\n0 29",
"output": "1 0"
},
{
"input": "1 2\n3 0",
"output": "0 0"
},
{
"input": "16 60\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "25 0"
},
{
"input": "1 5\n0 6",
"output": "0 0"
},
{
"input": "2 60\n0 59\n23 59",
"output": "2 0"
},
{
"input": "1 58\n0 1",
"output": "1 0"
},
{
"input": "25 60\n0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "2 3\n0 3\n0 30",
"output": "0 7"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 9",
"output": "24 0"
},
{
"input": "1 60\n2 0",
"output": "0 0"
},
{
"input": "2 60\n0 0\n5 0",
"output": "1 1"
},
{
"input": "1 30\n0 31",
"output": "0 0"
},
{
"input": "2 60\n0 59\n3 1",
"output": "2 0"
},
{
"input": "2 60\n0 59\n5 0",
"output": "2 0"
},
{
"input": "1 59\n0 0",
"output": "1 0"
},
{
"input": "3 25\n0 0\n1 0\n2 0",
"output": "0 26"
},
{
"input": "1 2\n2 3",
"output": "0 0"
}
] | 1,525,008,622
| 922
|
Python 3
|
OK
|
TESTS
| 34
| 108
| 7,065,600
|
n, s = map(int, input().split())
a = []
ind = True
for i in range(n):
x, y = map(int, input().split())
time = (60 * x) + y
if len(a) == 0 and time >= s + 1 and ind:
print(0, 0)
ind = False
elif len(a) > 0 and ind:
last = a[len(a) - 1]
if time - last >= (2 * s) + 2:
y = last + s + 1
print(y // 60, y % 60)
ind = False
a.append(time)
#print(a)
if ind:
y = a[len(a) - 1] + s + 1
#print(y)
print(y // 60, y % 60)
|
Title: Mind the Gap
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input Specification:
The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order.
Output Specification:
Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Demo Input:
['6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n', '16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n', '3 17\n0 30\n1 0\n12 0\n']
Demo Output:
['6 1\n', '24 50\n', '0 0\n']
Note:
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing.
|
```python
n, s = map(int, input().split())
a = []
ind = True
for i in range(n):
x, y = map(int, input().split())
time = (60 * x) + y
if len(a) == 0 and time >= s + 1 and ind:
print(0, 0)
ind = False
elif len(a) > 0 and ind:
last = a[len(a) - 1]
if time - last >= (2 * s) + 2:
y = last + s + 1
print(y // 60, y % 60)
ind = False
a.append(time)
#print(a)
if ind:
y = a[len(a) - 1] + s + 1
#print(y)
print(y // 60, y % 60)
```
| 3
|
|
611
|
A
|
New Year and Days
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
|
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
|
Print one integer — the number of candies Limak will save in the year 2016.
|
[
"4 of week\n",
"30 of month\n"
] |
[
"52\n",
"11\n"
] |
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
| 500
|
[
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output": "52"
},
{
"input": "2 of week",
"output": "52"
},
{
"input": "3 of week",
"output": "52"
},
{
"input": "5 of week",
"output": "53"
},
{
"input": "7 of week",
"output": "52"
},
{
"input": "1 of month",
"output": "12"
},
{
"input": "2 of month",
"output": "12"
},
{
"input": "3 of month",
"output": "12"
},
{
"input": "4 of month",
"output": "12"
},
{
"input": "5 of month",
"output": "12"
},
{
"input": "6 of month",
"output": "12"
},
{
"input": "7 of month",
"output": "12"
},
{
"input": "8 of month",
"output": "12"
},
{
"input": "9 of month",
"output": "12"
},
{
"input": "10 of month",
"output": "12"
},
{
"input": "11 of month",
"output": "12"
},
{
"input": "12 of month",
"output": "12"
},
{
"input": "13 of month",
"output": "12"
},
{
"input": "14 of month",
"output": "12"
},
{
"input": "15 of month",
"output": "12"
},
{
"input": "16 of month",
"output": "12"
},
{
"input": "18 of month",
"output": "12"
},
{
"input": "19 of month",
"output": "12"
},
{
"input": "20 of month",
"output": "12"
},
{
"input": "21 of month",
"output": "12"
},
{
"input": "22 of month",
"output": "12"
},
{
"input": "23 of month",
"output": "12"
},
{
"input": "24 of month",
"output": "12"
},
{
"input": "25 of month",
"output": "12"
},
{
"input": "26 of month",
"output": "12"
},
{
"input": "27 of month",
"output": "12"
},
{
"input": "28 of month",
"output": "12"
},
{
"input": "29 of month",
"output": "12"
}
] | 1,460,792,064
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
S=input().split()
a=int(S[0])
S1=S[2]
k=0
if S1=='month' :
if a>=29 :
print(11)
else :
print(12)
else :
i=5
j=365
while a!=i :
i=i+1
j=j-1
if i==8 :
i=1
while j>0 :
j=j-7
k=k+1
print(k)
|
Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
|
```python
S=input().split()
a=int(S[0])
S1=S[2]
k=0
if S1=='month' :
if a>=29 :
print(11)
else :
print(12)
else :
i=5
j=365
while a!=i :
i=i+1
j=j-1
if i==8 :
i=1
while j>0 :
j=j-7
k=k+1
print(k)
```
| 0
|
|
909
|
B
|
Segments
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"math"
] | null | null |
You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them.
You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis.
Find the minimal number of layers you have to use for the given *N*.
|
The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100).
|
Output a single integer - the minimal number of layers required to draw the segments for the given *N*.
|
[
"2\n",
"3\n",
"4\n"
] |
[
"2\n",
"4\n",
"6\n"
] |
As an example, here are the segments and their optimal arrangement into layers for *N* = 4.
| 1,000
|
[
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "4"
},
{
"input": "4",
"output": "6"
},
{
"input": "21",
"output": "121"
},
{
"input": "100",
"output": "2550"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "9"
},
{
"input": "6",
"output": "12"
},
{
"input": "7",
"output": "16"
},
{
"input": "8",
"output": "20"
},
{
"input": "9",
"output": "25"
},
{
"input": "10",
"output": "30"
},
{
"input": "11",
"output": "36"
},
{
"input": "12",
"output": "42"
},
{
"input": "13",
"output": "49"
},
{
"input": "14",
"output": "56"
},
{
"input": "15",
"output": "64"
},
{
"input": "16",
"output": "72"
},
{
"input": "17",
"output": "81"
},
{
"input": "18",
"output": "90"
},
{
"input": "19",
"output": "100"
},
{
"input": "20",
"output": "110"
},
{
"input": "22",
"output": "132"
},
{
"input": "23",
"output": "144"
},
{
"input": "24",
"output": "156"
},
{
"input": "25",
"output": "169"
},
{
"input": "26",
"output": "182"
},
{
"input": "27",
"output": "196"
},
{
"input": "28",
"output": "210"
},
{
"input": "29",
"output": "225"
},
{
"input": "30",
"output": "240"
},
{
"input": "31",
"output": "256"
},
{
"input": "32",
"output": "272"
},
{
"input": "33",
"output": "289"
},
{
"input": "34",
"output": "306"
},
{
"input": "35",
"output": "324"
},
{
"input": "36",
"output": "342"
},
{
"input": "37",
"output": "361"
},
{
"input": "38",
"output": "380"
},
{
"input": "39",
"output": "400"
},
{
"input": "40",
"output": "420"
},
{
"input": "41",
"output": "441"
},
{
"input": "42",
"output": "462"
},
{
"input": "43",
"output": "484"
},
{
"input": "44",
"output": "506"
},
{
"input": "45",
"output": "529"
},
{
"input": "46",
"output": "552"
},
{
"input": "47",
"output": "576"
},
{
"input": "48",
"output": "600"
},
{
"input": "49",
"output": "625"
},
{
"input": "50",
"output": "650"
},
{
"input": "51",
"output": "676"
},
{
"input": "52",
"output": "702"
},
{
"input": "53",
"output": "729"
},
{
"input": "54",
"output": "756"
},
{
"input": "55",
"output": "784"
},
{
"input": "56",
"output": "812"
},
{
"input": "57",
"output": "841"
},
{
"input": "58",
"output": "870"
},
{
"input": "59",
"output": "900"
},
{
"input": "60",
"output": "930"
},
{
"input": "61",
"output": "961"
},
{
"input": "62",
"output": "992"
},
{
"input": "63",
"output": "1024"
},
{
"input": "64",
"output": "1056"
},
{
"input": "65",
"output": "1089"
},
{
"input": "66",
"output": "1122"
},
{
"input": "67",
"output": "1156"
},
{
"input": "68",
"output": "1190"
},
{
"input": "69",
"output": "1225"
},
{
"input": "70",
"output": "1260"
},
{
"input": "71",
"output": "1296"
},
{
"input": "72",
"output": "1332"
},
{
"input": "73",
"output": "1369"
},
{
"input": "74",
"output": "1406"
},
{
"input": "75",
"output": "1444"
},
{
"input": "76",
"output": "1482"
},
{
"input": "77",
"output": "1521"
},
{
"input": "78",
"output": "1560"
},
{
"input": "79",
"output": "1600"
},
{
"input": "80",
"output": "1640"
},
{
"input": "81",
"output": "1681"
},
{
"input": "82",
"output": "1722"
},
{
"input": "83",
"output": "1764"
},
{
"input": "84",
"output": "1806"
},
{
"input": "85",
"output": "1849"
},
{
"input": "86",
"output": "1892"
},
{
"input": "87",
"output": "1936"
},
{
"input": "88",
"output": "1980"
},
{
"input": "89",
"output": "2025"
},
{
"input": "90",
"output": "2070"
},
{
"input": "91",
"output": "2116"
},
{
"input": "92",
"output": "2162"
},
{
"input": "93",
"output": "2209"
},
{
"input": "94",
"output": "2256"
},
{
"input": "95",
"output": "2304"
},
{
"input": "96",
"output": "2352"
},
{
"input": "97",
"output": "2401"
},
{
"input": "98",
"output": "2450"
},
{
"input": "99",
"output": "2500"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "9"
}
] | 1,620,867,682
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 77
| 6,758,400
|
n=int(input())
print((n+n%2)*((n+2)//2)//2)
|
Title: Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them.
You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis.
Find the minimal number of layers you have to use for the given *N*.
Input Specification:
The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100).
Output Specification:
Output a single integer - the minimal number of layers required to draw the segments for the given *N*.
Demo Input:
['2\n', '3\n', '4\n']
Demo Output:
['2\n', '4\n', '6\n']
Note:
As an example, here are the segments and their optimal arrangement into layers for *N* = 4.
|
```python
n=int(input())
print((n+n%2)*((n+2)//2)//2)
```
| 3
|
|
177
|
A2
|
Good Matrix Elements
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
|
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
|
Print a single integer — the sum of good matrix elements.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] |
[
"45\n",
"17\n"
] |
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
| 70
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
"output": "756"
},
{
"input": "3\n19 7 16\n12 15 5\n15 15 5",
"output": "109"
},
{
"input": "3\n36 4 33\n11 46 32\n20 49 34",
"output": "265"
},
{
"input": "3\n79 91 74\n33 82 22\n18 28 54",
"output": "481"
},
{
"input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6",
"output": "65"
},
{
"input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6",
"output": "225"
},
{
"input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6",
"output": "495"
},
{
"input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6",
"output": "705"
},
{
"input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83",
"output": "708"
},
{
"input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "5"
},
{
"input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "3\n0 0 0\n0 0 0\n0 0 0",
"output": "0"
},
{
"input": "11\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "15\n0 2 1 2 1 0 2 0 2 2 2 2 2 0 2\n2 0 1 0 1 1 2 2 0 2 2 0 3 0 1\n3 3 2 2 2 1 2 3 2 3 1 2 3 3 2\n0 3 0 3 3 3 2 1 0 2 3 2 3 3 2\n3 0 0 2 1 2 3 1 1 1 2 2 2 1 0\n1 3 2 3 3 0 3 2 0 3 1 2 3 0 3\n2 2 3 2 0 0 2 3 0 1 2 1 1 2 1\n2 1 3 2 3 3 2 2 2 0 3 3 1 1 3\n0 2 1 3 1 2 0 0 1 0 0 2 0 0 0\n3 1 2 1 2 3 2 3 2 1 0 2 2 0 3\n0 0 1 3 3 0 1 1 0 0 2 3 0 3 0\n2 3 0 3 1 1 2 2 1 3 2 3 0 0 2\n3 0 2 2 3 3 0 2 2 2 0 1 0 0 3\n2 2 2 1 2 2 2 0 2 2 2 2 0 2 3\n2 2 3 0 0 0 0 2 1 1 1 3 2 2 0",
"output": "97"
}
] | 1,651,095,220
| 2,147,483,647
|
Python 3
|
OK
|
TESTS2
| 33
| 92
| 0
|
n=int(input())
mat1=[]
for i in range(0,n):
temp=list(map(int,input().split()))
mat1.append(temp)
sum=0
for i in range(0,n):
sum=sum+mat1[i][i]
p1=n-1
for i in range(0,n):
sum=sum+mat1[i][p1]
p1=p1-1
temp=int((n-1)//2)
for i in range(0,n):
sum=sum+mat1[temp][i]
for i in range(0,n):
sum=sum+mat1[i][temp]
result=sum-(3*mat1[temp][temp])
print(result)
|
Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
|
```python
n=int(input())
mat1=[]
for i in range(0,n):
temp=list(map(int,input().split()))
mat1.append(temp)
sum=0
for i in range(0,n):
sum=sum+mat1[i][i]
p1=n-1
for i in range(0,n):
sum=sum+mat1[i][p1]
p1=p1-1
temp=int((n-1)//2)
for i in range(0,n):
sum=sum+mat1[temp][i]
for i in range(0,n):
sum=sum+mat1[i][temp]
result=sum-(3*mat1[temp][temp])
print(result)
```
| 3
|
|
784
|
A
|
Numbers Joke
|
PROGRAMMING
| 1,700
|
[
"*special"
] | null | null |
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output a single integer.
|
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
|
Output a single integer.
|
[
"3\n"
] |
[
"27\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "27"
},
{
"input": "1",
"output": "4"
},
{
"input": "2",
"output": "22"
},
{
"input": "4",
"output": "58"
},
{
"input": "5",
"output": "85"
},
{
"input": "6",
"output": "94"
},
{
"input": "7",
"output": "121"
},
{
"input": "8",
"output": "166"
},
{
"input": "9",
"output": "202"
},
{
"input": "10",
"output": "265"
},
{
"input": "11",
"output": "274"
},
{
"input": "12",
"output": "319"
},
{
"input": "13",
"output": "346"
},
{
"input": "14",
"output": "355"
},
{
"input": "15",
"output": "378"
},
{
"input": "16",
"output": "382"
},
{
"input": "17",
"output": "391"
},
{
"input": "18",
"output": "438"
},
{
"input": "19",
"output": "454"
},
{
"input": "20",
"output": "483"
},
{
"input": "21",
"output": "517"
},
{
"input": "22",
"output": "526"
},
{
"input": "23",
"output": "535"
},
{
"input": "24",
"output": "562"
},
{
"input": "25",
"output": "576"
},
{
"input": "26",
"output": "588"
},
{
"input": "27",
"output": "627"
},
{
"input": "28",
"output": "634"
},
{
"input": "29",
"output": "636"
},
{
"input": "30",
"output": "645"
}
] | 1,522,530,554
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 77
| 7,065,600
|
#print(27) #test number 1
print(pow(int(input()),3))
|
Title: Numbers Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output a single integer.
Input Specification:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=30).
Output Specification:
Output a single integer.
Demo Input:
['3\n']
Demo Output:
['27\n']
Note:
none
|
```python
#print(27) #test number 1
print(pow(int(input()),3))
```
| 0
|
|
327
|
A
|
Flipping Game
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dp",
"implementation"
] | null | null |
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
|
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
|
[
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] |
[
"4\n",
"4\n"
] |
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
| 500
|
[
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "18"
},
{
"input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "22"
},
{
"input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1",
"output": "70"
},
{
"input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1",
"output": "60"
},
{
"input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1",
"output": "18"
},
{
"input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1",
"output": "36"
},
{
"input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "44"
},
{
"input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1",
"output": "61"
},
{
"input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0",
"output": "61"
},
{
"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 0",
"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": "99"
},
{
"input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0",
"output": "61"
},
{
"input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1",
"output": "61"
},
{
"input": "2\n1 1",
"output": "1"
}
] | 1,696,358,755
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
n = int(input())
*a, = map(int, input().split())
r, mr = 0, 0
for ai in a:
mr = max(r, mr)
r += (1, -r)[ai]
print(sum(a)+mr)
|
Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
|
```python
n = int(input())
*a, = map(int, input().split())
r, mr = 0, 0
for ai in a:
mr = max(r, mr)
r += (1, -r)[ai]
print(sum(a)+mr)
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,662,723,051
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
m,n = input().split()
m = int(m)
n = int(n)
print(m*n//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
m,n = input().split()
m = int(m)
n = int(n)
print(m*n//2)
```
| 3.977
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
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{
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},
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},
{
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},
{
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},
{
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},
{
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"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
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"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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},
{
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"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
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{
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"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
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"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
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"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
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"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
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"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
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"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
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"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
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{
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},
{
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"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
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{
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{
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{
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{
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},
{
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},
{
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"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
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"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
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"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,640,662,459
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
str1=input()
str2=input()
result=""
for i in range(len(str1)):
if str1[i]==str2[i]:
result+="0"
else:
result+="1"
print(result)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
str1=input()
str2=input()
result=""
for i in range(len(str1)):
if str1[i]==str2[i]:
result+="0"
else:
result+="1"
print(result)
```
| 3.9885
|
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"
}
] | 1,633,774,147
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 62
| 6,758,400
|
'''
Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.
'''
n=int(input())
a=list(map(int,input().split()))
c=list(map(a.count,a))
print (max(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
'''
Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.
'''
n=int(input())
a=list(map(int,input().split()))
c=list(map(a.count,a))
print (max(c))
```
| 3
|
|
118
|
A
|
String Task
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
|
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
|
Print the resulting string. It is guaranteed that this string is not empty.
|
[
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] |
[
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] |
none
| 500
|
[
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,695,956,911
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 92
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Fri Sep 29 11:00:12 2023
@author: ZHAO XUDI
"""
s = str(input()).lower()
l = list(s)
vowels = ["a","o","y","e","u","i"]
new = []
for letter in l:
if letter not in vowels:
new.append(letter)
print("."+".".join(new))
|
Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Fri Sep 29 11:00:12 2023
@author: ZHAO XUDI
"""
s = str(input()).lower()
l = list(s)
vowels = ["a","o","y","e","u","i"]
new = []
for letter in l:
if letter not in vowels:
new.append(letter)
print("."+".".join(new))
```
| 3
|
|
378
|
A
|
Playing with Dice
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
|
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
|
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
|
[
"2 5\n",
"2 4\n"
] |
[
"3 0 3\n",
"2 1 3\n"
] |
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.
| 500
|
[
{
"input": "2 5",
"output": "3 0 3"
},
{
"input": "2 4",
"output": "2 1 3"
},
{
"input": "5 3",
"output": "2 1 3"
},
{
"input": "1 6",
"output": "3 0 3"
},
{
"input": "5 1",
"output": "3 1 2"
},
{
"input": "6 3",
"output": "2 0 4"
},
{
"input": "2 3",
"output": "2 0 4"
},
{
"input": "5 6",
"output": "5 0 1"
},
{
"input": "4 4",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
},
{
"input": "6 4",
"output": "1 1 4"
},
{
"input": "1 4",
"output": "2 0 4"
},
{
"input": "5 5",
"output": "0 6 0"
},
{
"input": "4 5",
"output": "4 0 2"
},
{
"input": "4 3",
"output": "3 0 3"
},
{
"input": "1 5",
"output": "2 1 3"
},
{
"input": "6 5",
"output": "1 0 5"
},
{
"input": "2 2",
"output": "0 6 0"
},
{
"input": "1 3",
"output": "1 1 4"
},
{
"input": "3 6",
"output": "4 0 2"
},
{
"input": "3 1",
"output": "4 1 1"
},
{
"input": "3 2",
"output": "4 0 2"
},
{
"input": "3 5",
"output": "3 1 2"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "6 2",
"output": "2 1 3"
},
{
"input": "4 1",
"output": "4 0 2"
},
{
"input": "5 2",
"output": "3 0 3"
},
{
"input": "4 2",
"output": "3 1 2"
},
{
"input": "2 1",
"output": "5 0 1"
},
{
"input": "6 1",
"output": "3 0 3"
},
{
"input": "4 6",
"output": "4 1 1"
},
{
"input": "2 6",
"output": "3 1 2"
},
{
"input": "3 4",
"output": "3 0 3"
},
{
"input": "1 2",
"output": "1 0 5"
},
{
"input": "6 6",
"output": "0 6 0"
},
{
"input": "5 4",
"output": "2 0 4"
},
{
"input": "3 3",
"output": "0 6 0"
},
{
"input": "1 1",
"output": "0 6 0"
}
] | 1,634,487,919
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 61
| 0
|
a,b = map(int,input().split())
awin = 0
draw = 0
bwin = 0
for i in range(1,7):
if abs(a-i) < abs(b-i):
awin += 1
elif abs(a-i) > abs(b-i):
bwin += 1
else:
draw += 1
print(awin,draw,bwin)
|
Title: Playing with Dice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
Input Specification:
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
Output Specification:
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
Demo Input:
['2 5\n', '2 4\n']
Demo Output:
['3 0 3\n', '2 1 3\n']
Note:
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.
|
```python
a,b = map(int,input().split())
awin = 0
draw = 0
bwin = 0
for i in range(1,7):
if abs(a-i) < abs(b-i):
awin += 1
elif abs(a-i) > abs(b-i):
bwin += 1
else:
draw += 1
print(awin,draw,bwin)
```
| 3
|
|
166
|
E
|
Tetrahedron
|
PROGRAMMING
| 1,500
|
[
"dp",
"math",
"matrices"
] | null | null |
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
|
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
|
[
"2\n",
"4\n"
] |
[
"3\n",
"21\n"
] |
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*
| 1,000
|
[
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "21"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "6"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "183"
},
{
"input": "7",
"output": "546"
},
{
"input": "8",
"output": "1641"
},
{
"input": "9",
"output": "4920"
},
{
"input": "10",
"output": "14763"
},
{
"input": "15",
"output": "3587226"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "100",
"output": "721510432"
},
{
"input": "300",
"output": "327873818"
},
{
"input": "900",
"output": "295068084"
},
{
"input": "1500",
"output": "451187545"
},
{
"input": "3000",
"output": "645417275"
},
{
"input": "5000",
"output": "755610910"
},
{
"input": "10000",
"output": "723907367"
},
{
"input": "50000",
"output": "969527595"
},
{
"input": "90000",
"output": "548978368"
},
{
"input": "99999",
"output": "909741855"
},
{
"input": "100000",
"output": "729225554"
},
{
"input": "300000",
"output": "93822635"
},
{
"input": "800000",
"output": "178940616"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "4000000",
"output": "882155933"
},
{
"input": "9000000",
"output": "295060537"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "9999998",
"output": "576926295"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "31",
"output": "347990060"
}
] | 1,657,007,185
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 342
| 1,638,400
|
mod = 10**9+7
n = int(input())
# A, B, C, D = 0, 0, 0, 1
# for i in range(1, n+1):
# new_A = (B + C + D) % mod
# new_B = (A + C + D) % mod
# new_C = (A + B + D) % mod
# new_D = (A + B + C) % mod
# A, B, C, D = new_A, new_B, new_C, new_D
D, not_D = 1, 0
for i in range(1, n+1):
new_D = (not_D * 3) % mod
new_not_D = (not_D * 2 + D) % mod
D, not_D = new_D, new_not_D
print(D)
|
Title: Tetrahedron
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
Output Specification:
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '21\n']
Note:
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*
|
```python
mod = 10**9+7
n = int(input())
# A, B, C, D = 0, 0, 0, 1
# for i in range(1, n+1):
# new_A = (B + C + D) % mod
# new_B = (A + C + D) % mod
# new_C = (A + B + D) % mod
# new_D = (A + B + C) % mod
# A, B, C, D = new_A, new_B, new_C, new_D
D, not_D = 1, 0
for i in range(1, n+1):
new_D = (not_D * 3) % mod
new_not_D = (not_D * 2 + D) % mod
D, not_D = new_D, new_not_D
print(D)
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,697,053,774
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 1,000
| 0
|
input_ = input().split()
k, r = int(input_[0]), int(input_[1])
n = 1
while True:
if ((k%10)*n)%10 == r:
break
n += 1
print(n)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
input_ = input().split()
k, r = int(input_[0]), int(input_[1])
n = 1
while True:
if ((k%10)*n)%10 == r:
break
n += 1
print(n)
```
| 0
|
|
681
|
A
|
A Good Contest
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
|
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
|
[
"3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n",
"3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n"
] |
[
"YES",
"NO"
] |
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.
| 500
|
[
{
"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450",
"output": "NO"
},
{
"input": "1\nDb -3373 3591",
"output": "NO"
},
{
"input": "5\nQ2bz 960 2342\nhmX 2710 -1348\ngbAe -1969 -963\nE -160 196\npsi 2665 -3155",
"output": "NO"
},
{
"input": "9\nmwAz9lQ 1786 -1631\nnYgYFXZQfY -1849 -1775\nKU4jF -1773 -3376\nopR 3752 2931\nGl -1481 -1002\nR -1111 3778\n0i9B21DC 3650 289\nQ8L2dS0 358 -3305\ng -2662 3968",
"output": "NO"
},
{
"input": "5\nzMSBcOUf -2883 -2238\nYN -3314 -1480\nfHpuccQn06 -1433 -589\naM1NVEPQi 399 3462\n_L 2516 -3290",
"output": "NO"
},
{
"input": "1\na 2400 2401",
"output": "YES"
},
{
"input": "1\nfucker 4000 4000",
"output": "NO"
},
{
"input": "1\nJora 2400 2401",
"output": "YES"
},
{
"input": "1\nACA 2400 2420",
"output": "YES"
},
{
"input": "1\nAca 2400 2420",
"output": "YES"
},
{
"input": "1\nSub_d 2401 2402",
"output": "YES"
},
{
"input": "2\nHack 2400 2401\nDum 1243 555",
"output": "YES"
},
{
"input": "1\nXXX 2400 2500",
"output": "YES"
},
{
"input": "1\nfucker 2400 2401",
"output": "YES"
},
{
"input": "1\nX 2400 2500",
"output": "YES"
},
{
"input": "1\nvineet 2400 2401",
"output": "YES"
},
{
"input": "1\nabc 2400 2500",
"output": "YES"
},
{
"input": "1\naaaaa 2400 2401",
"output": "YES"
},
{
"input": "1\nhoge 2400 2401",
"output": "YES"
},
{
"input": "1\nInfinity 2400 2468",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2401",
"output": "YES"
},
{
"input": "1\nFuck 2400 2401",
"output": "YES"
},
{
"input": "1\nfuck 2400 2401",
"output": "YES"
},
{
"input": "3\nApplejack 2400 2401\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450",
"output": "YES"
},
{
"input": "1\nalex 2400 2401",
"output": "YES"
},
{
"input": "1\nA 2400 2401",
"output": "YES"
},
{
"input": "1\na 2400 2455",
"output": "YES"
},
{
"input": "1\nlol 2400 2401",
"output": "YES"
},
{
"input": "2\nBurunduk1 2400 2537\nBudAlNik 2084 2214",
"output": "YES"
},
{
"input": "1\naaaaaa 2400 2401",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2500",
"output": "YES"
},
{
"input": "1\nds 2400 2410",
"output": "YES"
},
{
"input": "1\nas 2400 2401",
"output": "YES"
},
{
"input": "1\nabc 2400 2401",
"output": "YES"
},
{
"input": "3\nBudAlNik 2084 2214\nsubscriber 2833 2749\nBurunduk1 2526 2537",
"output": "YES"
},
{
"input": "1\ncaonima 2400 2401",
"output": "YES"
},
{
"input": "1\narr 2400 2500",
"output": "YES"
},
{
"input": "1\nx 2400 2401",
"output": "YES"
},
{
"input": "1\narrr 2400 2500",
"output": "YES"
},
{
"input": "1\nabc 2400 2405",
"output": "YES"
},
{
"input": "3\nBurunduk1 2400 2420\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2537",
"output": "YES"
},
{
"input": "1\nHELLO 2400 2401",
"output": "YES"
},
{
"input": "1\neatmore 2400 2500",
"output": "YES"
},
{
"input": "1\nb 2400 2401",
"output": "YES"
},
{
"input": "3\nBurunduk1 2400 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "1\nApplejack 2400 2410",
"output": "YES"
},
{
"input": "1\nabacaba 2400 2451",
"output": "YES"
},
{
"input": "1\nrekt_n00b 2500 2600",
"output": "YES"
}
] | 1,672,163,954
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 60
| 93
| 0
|
n=int(input())
flag=False
for i in range(n):
info=list(input().split())
if int(info[1])>=2400 and int(info[2])>int(info[1]):
flag=True
break
if flag:
print("YES")
else:
print("NO")
|
Title: A Good Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
Output Specification:
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
Demo Input:
['3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n', '3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.
|
```python
n=int(input())
flag=False
for i in range(n):
info=list(input().split())
if int(info[1])>=2400 and int(info[2])>int(info[1]):
flag=True
break
if flag:
print("YES")
else:
print("NO")
```
| 3
|
|
681
|
A
|
A Good Contest
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
|
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
|
[
"3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n",
"3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n"
] |
[
"YES",
"NO"
] |
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.
| 500
|
[
{
"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450",
"output": "NO"
},
{
"input": "1\nDb -3373 3591",
"output": "NO"
},
{
"input": "5\nQ2bz 960 2342\nhmX 2710 -1348\ngbAe -1969 -963\nE -160 196\npsi 2665 -3155",
"output": "NO"
},
{
"input": "9\nmwAz9lQ 1786 -1631\nnYgYFXZQfY -1849 -1775\nKU4jF -1773 -3376\nopR 3752 2931\nGl -1481 -1002\nR -1111 3778\n0i9B21DC 3650 289\nQ8L2dS0 358 -3305\ng -2662 3968",
"output": "NO"
},
{
"input": "5\nzMSBcOUf -2883 -2238\nYN -3314 -1480\nfHpuccQn06 -1433 -589\naM1NVEPQi 399 3462\n_L 2516 -3290",
"output": "NO"
},
{
"input": "1\na 2400 2401",
"output": "YES"
},
{
"input": "1\nfucker 4000 4000",
"output": "NO"
},
{
"input": "1\nJora 2400 2401",
"output": "YES"
},
{
"input": "1\nACA 2400 2420",
"output": "YES"
},
{
"input": "1\nAca 2400 2420",
"output": "YES"
},
{
"input": "1\nSub_d 2401 2402",
"output": "YES"
},
{
"input": "2\nHack 2400 2401\nDum 1243 555",
"output": "YES"
},
{
"input": "1\nXXX 2400 2500",
"output": "YES"
},
{
"input": "1\nfucker 2400 2401",
"output": "YES"
},
{
"input": "1\nX 2400 2500",
"output": "YES"
},
{
"input": "1\nvineet 2400 2401",
"output": "YES"
},
{
"input": "1\nabc 2400 2500",
"output": "YES"
},
{
"input": "1\naaaaa 2400 2401",
"output": "YES"
},
{
"input": "1\nhoge 2400 2401",
"output": "YES"
},
{
"input": "1\nInfinity 2400 2468",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2401",
"output": "YES"
},
{
"input": "1\nFuck 2400 2401",
"output": "YES"
},
{
"input": "1\nfuck 2400 2401",
"output": "YES"
},
{
"input": "3\nApplejack 2400 2401\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450",
"output": "YES"
},
{
"input": "1\nalex 2400 2401",
"output": "YES"
},
{
"input": "1\nA 2400 2401",
"output": "YES"
},
{
"input": "1\na 2400 2455",
"output": "YES"
},
{
"input": "1\nlol 2400 2401",
"output": "YES"
},
{
"input": "2\nBurunduk1 2400 2537\nBudAlNik 2084 2214",
"output": "YES"
},
{
"input": "1\naaaaaa 2400 2401",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2500",
"output": "YES"
},
{
"input": "1\nds 2400 2410",
"output": "YES"
},
{
"input": "1\nas 2400 2401",
"output": "YES"
},
{
"input": "1\nabc 2400 2401",
"output": "YES"
},
{
"input": "3\nBudAlNik 2084 2214\nsubscriber 2833 2749\nBurunduk1 2526 2537",
"output": "YES"
},
{
"input": "1\ncaonima 2400 2401",
"output": "YES"
},
{
"input": "1\narr 2400 2500",
"output": "YES"
},
{
"input": "1\nx 2400 2401",
"output": "YES"
},
{
"input": "1\narrr 2400 2500",
"output": "YES"
},
{
"input": "1\nabc 2400 2405",
"output": "YES"
},
{
"input": "3\nBurunduk1 2400 2420\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "1\nBurunduk1 2400 2537",
"output": "YES"
},
{
"input": "1\nHELLO 2400 2401",
"output": "YES"
},
{
"input": "1\neatmore 2400 2500",
"output": "YES"
},
{
"input": "1\nb 2400 2401",
"output": "YES"
},
{
"input": "3\nBurunduk1 2400 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "1\nApplejack 2400 2410",
"output": "YES"
},
{
"input": "1\nabacaba 2400 2451",
"output": "YES"
},
{
"input": "1\nrekt_n00b 2500 2600",
"output": "YES"
}
] | 1,648,367,187
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 60
| 46
| 0
|
def check(b,a):
if (b>=2400 and a>=2400 and a>b):
return 1
else:
return 0
z=0
for i in range(int(input())):
name,before,after=input().split()
z+=check(int(before),int(after))
if(z>0):
print('YES')
else:
print('NO')
|
Title: A Good Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
Output Specification:
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
Demo Input:
['3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n', '3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.
|
```python
def check(b,a):
if (b>=2400 and a>=2400 and a>b):
return 1
else:
return 0
z=0
for i in range(int(input())):
name,before,after=input().split()
z+=check(int(before),int(after))
if(z>0):
print('YES')
else:
print('NO')
```
| 3
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,694,670,009
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 122
| 0
|
l=[4,7,44,47,74,77,444,447,474,477,747,774,777]
a=int(input())
c=1
for i in l:
if a % i ==0:
print("YES")
c=0
break
if c==1:
print("NO")
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
l=[4,7,44,47,74,77,444,447,474,477,747,774,777]
a=int(input())
c=1
for i in l:
if a % i ==0:
print("YES")
c=0
break
if c==1:
print("NO")
```
| 3
|
|
519
|
C
|
A and B and Team Training
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed?
|
The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session.
|
Print the maximum number of teams that can be formed.
|
[
"2 6\n",
"4 5\n"
] |
[
"2\n",
"3\n"
] |
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).
| 1,500
|
[
{
"input": "2 6",
"output": "2"
},
{
"input": "4 5",
"output": "3"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "500000 500000",
"output": "333333"
},
{
"input": "70 100",
"output": "56"
},
{
"input": "5 12525",
"output": "5"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "5 10",
"output": "5"
},
{
"input": "0 0",
"output": "0"
},
{
"input": "0 1",
"output": "0"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "0 21233",
"output": "0"
},
{
"input": "12523 0",
"output": "0"
},
{
"input": "1231 1253",
"output": "828"
},
{
"input": "500000 0",
"output": "0"
},
{
"input": "1 500000",
"output": "1"
},
{
"input": "250000 500000",
"output": "250000"
},
{
"input": "500000 250000",
"output": "250000"
},
{
"input": "33333 77777",
"output": "33333"
},
{
"input": "30900 174529",
"output": "30900"
},
{
"input": "89979 57154",
"output": "49044"
},
{
"input": "231646 398487",
"output": "210044"
},
{
"input": "332019 281112",
"output": "204377"
},
{
"input": "473686 122443",
"output": "122443"
},
{
"input": "481245 86879",
"output": "86879"
},
{
"input": "39935 123534",
"output": "39935"
},
{
"input": "10000 20000",
"output": "10000"
},
{
"input": "10000 20001",
"output": "10000"
},
{
"input": "10000 20002",
"output": "10000"
},
{
"input": "10000 20003",
"output": "10000"
},
{
"input": "10000 20004",
"output": "10000"
},
{
"input": "10001 20000",
"output": "10000"
},
{
"input": "10001 20001",
"output": "10000"
},
{
"input": "10001 20002",
"output": "10001"
},
{
"input": "10001 20003",
"output": "10001"
},
{
"input": "10001 20004",
"output": "10001"
},
{
"input": "20000 10000",
"output": "10000"
},
{
"input": "20001 10000",
"output": "10000"
},
{
"input": "20002 10000",
"output": "10000"
},
{
"input": "20003 10000",
"output": "10000"
},
{
"input": "20004 10000",
"output": "10000"
},
{
"input": "20000 10001",
"output": "10000"
},
{
"input": "20001 10001",
"output": "10000"
},
{
"input": "20002 10001",
"output": "10001"
},
{
"input": "20003 10001",
"output": "10001"
},
{
"input": "20004 10001",
"output": "10001"
},
{
"input": "10 0",
"output": "0"
},
{
"input": "0 6",
"output": "0"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "0 0",
"output": "0"
}
] | 1,680,410,264
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 52
| 62
| 1,433,600
|
n,m = map(int,input().split())
ans = 0
while n>0 and m>0:
if n >= m:
n -= 2
m -= 1
if n>=0 and m>=0:
ans += 1
else:
m -= 2
n -= 1
if n >=0 and m>=0:
ans += 1
print(ans)
|
Title: A and B and Team Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are *n* experienced members and *m* newbies on the training session. Can you calculate what maximum number of teams can be formed?
Input Specification:
The first line contains two integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=5·105) — the number of experienced participants and newbies that are present at the training session.
Output Specification:
Print the maximum number of teams that can be formed.
Demo Input:
['2 6\n', '4 5\n']
Demo Output:
['2\n', '3\n']
Note:
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).
|
```python
n,m = map(int,input().split())
ans = 0
while n>0 and m>0:
if n >= m:
n -= 2
m -= 1
if n>=0 and m>=0:
ans += 1
else:
m -= 2
n -= 1
if n >=0 and m>=0:
ans += 1
print(ans)
```
| 3
|
|
689
|
B
|
Mike and Shortcuts
|
PROGRAMMING
| 1,600
|
[
"dfs and similar",
"graphs",
"greedy",
"shortest paths"
] | null | null |
Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city.
City consists of *n* intersections numbered from 1 to *n*. Mike starts walking from his house located at the intersection number 1 and goes along some sequence of intersections. Walking from intersection number *i* to intersection *j* requires |*i*<=-<=*j*| units of energy. The total energy spent by Mike to visit a sequence of intersections *p*1<==<=1,<=*p*2,<=...,<=*p**k* is equal to units of energy.
Of course, walking would be boring if there were no shortcuts. A shortcut is a special path that allows Mike walking from one intersection to another requiring only 1 unit of energy. There are exactly *n* shortcuts in Mike's city, the *i**th* of them allows walking from intersection *i* to intersection *a**i* (*i*<=≤<=*a**i*<=≤<=*a**i*<=+<=1) (but not in the opposite direction), thus there is exactly one shortcut starting at each intersection. Formally, if Mike chooses a sequence *p*1<==<=1,<=*p*2,<=...,<=*p**k* then for each 1<=≤<=*i*<=<<=*k* satisfying *p**i*<=+<=1<==<=*a**p**i* and *a**p**i*<=≠<=*p**i* Mike will spend only 1 unit of energy instead of |*p**i*<=-<=*p**i*<=+<=1| walking from the intersection *p**i* to intersection *p**i*<=+<=1. For example, if Mike chooses a sequence *p*1<==<=1,<=*p*2<==<=*a**p*1,<=*p*3<==<=*a**p*2,<=...,<=*p**k*<==<=*a**p**k*<=-<=1, he spends exactly *k*<=-<=1 units of total energy walking around them.
Before going on his adventure, Mike asks you to find the minimum amount of energy required to reach each of the intersections from his home. Formally, for each 1<=≤<=*i*<=≤<=*n* Mike is interested in finding minimum possible total energy of some sequence *p*1<==<=1,<=*p*2,<=...,<=*p**k*<==<=*i*.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Mike's city intersection.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*i*<=≤<=*a**i*<=≤<=*n* , , describing shortcuts of Mike's city, allowing to walk from intersection *i* to intersection *a**i* using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from *a**i* to *i*).
|
In the only line print *n* integers *m*1,<=*m*2,<=...,<=*m**n*, where *m**i* denotes the least amount of total energy required to walk from intersection 1 to intersection *i*.
|
[
"3\n2 2 3\n",
"5\n1 2 3 4 5\n",
"7\n4 4 4 4 7 7 7\n"
] |
[
"0 1 2 \n",
"0 1 2 3 4 \n",
"0 1 2 1 2 3 3 \n"
] |
In the first sample case desired sequences are:
1: 1; *m*<sub class="lower-index">1</sub> = 0;
2: 1, 2; *m*<sub class="lower-index">2</sub> = 1;
3: 1, 3; *m*<sub class="lower-index">3</sub> = |3 - 1| = 2.
In the second sample case the sequence for any intersection 1 < *i* is always 1, *i* and *m*<sub class="lower-index">*i*</sub> = |1 - *i*|.
In the third sample case — consider the following intersection sequences:
1: 1; *m*<sub class="lower-index">1</sub> = 0;
2: 1, 2; *m*<sub class="lower-index">2</sub> = |2 - 1| = 1;
3: 1, 4, 3; *m*<sub class="lower-index">3</sub> = 1 + |4 - 3| = 2;
4: 1, 4; *m*<sub class="lower-index">4</sub> = 1;
5: 1, 4, 5; *m*<sub class="lower-index">5</sub> = 1 + |4 - 5| = 2;
6: 1, 4, 6; *m*<sub class="lower-index">6</sub> = 1 + |4 - 6| = 3;
7: 1, 4, 5, 7; *m*<sub class="lower-index">7</sub> = 1 + |4 - 5| + 1 = 3.
| 1,000
|
[
{
"input": "3\n2 2 3",
"output": "0 1 2 "
},
{
"input": "5\n1 2 3 4 5",
"output": "0 1 2 3 4 "
},
{
"input": "7\n4 4 4 4 7 7 7",
"output": "0 1 2 1 2 3 3 "
},
{
"input": "98\n17 17 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 57 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 90 90 90 90 90 90 90 90 90 90 90 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 92 95 95 95 95 95 97 98 98",
"output": "0 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 4 4 5 6 5 6 7 8 "
},
{
"input": "91\n4 6 23 23 23 23 23 28 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 47 47 47 54 54 54 54 54 54 54 58 58 58 58 58 58 69 69 69 69 69 69 69 69 69 69 69 69 70 70 70 70 70 70 70 70 70 70 71 72 72 72 73 75 77 77 77 82 82 84 84 84 84 84 85 86 87 89 89 90 91",
"output": "0 1 2 1 2 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 3 4 5 6 5 6 7 8 9 9 8 7 6 5 4 3 4 5 6 7 8 9 10 9 10 9 8 7 6 5 4 5 6 7 6 7 8 9 10 11 10 9 8 7 6 5 6 6 7 8 9 10 11 11 12 13 14 14 13 14 14 15 16 17 18 19 20 21 "
},
{
"input": "82\n1 5 11 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 39 39 39 39 39 45 45 45 45 45 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 71 71 71 71 71 71 71 73 73 75 75 76 77 79 81 81 81 82",
"output": "0 1 2 3 2 3 4 5 5 4 3 4 5 6 7 8 9 10 11 12 13 12 11 10 9 8 7 6 5 4 3 4 5 6 7 8 9 10 9 9 8 7 6 5 4 5 6 7 8 9 10 11 12 13 14 15 16 16 15 14 13 12 11 10 9 8 7 6 5 6 6 7 8 9 10 11 12 13 14 15 15 16 "
},
{
"input": "4\n2 3 3 4",
"output": "0 1 2 3 "
}
] | 1,467,829,412
| 6,512
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 3
| 46
| 0
|
n = int(input())
ai = list(map(int,input().split()))
ai2 = [1000000]*(n+1)
ai2[0] = -1
num = 1
for i in range(n):
num2 = num
for j in range(num,ai[i]):
ai2[j] = ai2[j-1] + 1
num = ai[i]+1
ai2[num-1] = min(ai2[num-1],ai2[num-2] + 1,ai2[i+1]+1)
for j in range(num-2,num2-1,-1):
ai2[j] = min(ai2[j+1] + 1,ai2[j])
for i in range(1,n+1):
print(ai2[i],end=" ")
|
Title: Mike and Shortcuts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city.
City consists of *n* intersections numbered from 1 to *n*. Mike starts walking from his house located at the intersection number 1 and goes along some sequence of intersections. Walking from intersection number *i* to intersection *j* requires |*i*<=-<=*j*| units of energy. The total energy spent by Mike to visit a sequence of intersections *p*1<==<=1,<=*p*2,<=...,<=*p**k* is equal to units of energy.
Of course, walking would be boring if there were no shortcuts. A shortcut is a special path that allows Mike walking from one intersection to another requiring only 1 unit of energy. There are exactly *n* shortcuts in Mike's city, the *i**th* of them allows walking from intersection *i* to intersection *a**i* (*i*<=≤<=*a**i*<=≤<=*a**i*<=+<=1) (but not in the opposite direction), thus there is exactly one shortcut starting at each intersection. Formally, if Mike chooses a sequence *p*1<==<=1,<=*p*2,<=...,<=*p**k* then for each 1<=≤<=*i*<=<<=*k* satisfying *p**i*<=+<=1<==<=*a**p**i* and *a**p**i*<=≠<=*p**i* Mike will spend only 1 unit of energy instead of |*p**i*<=-<=*p**i*<=+<=1| walking from the intersection *p**i* to intersection *p**i*<=+<=1. For example, if Mike chooses a sequence *p*1<==<=1,<=*p*2<==<=*a**p*1,<=*p*3<==<=*a**p*2,<=...,<=*p**k*<==<=*a**p**k*<=-<=1, he spends exactly *k*<=-<=1 units of total energy walking around them.
Before going on his adventure, Mike asks you to find the minimum amount of energy required to reach each of the intersections from his home. Formally, for each 1<=≤<=*i*<=≤<=*n* Mike is interested in finding minimum possible total energy of some sequence *p*1<==<=1,<=*p*2,<=...,<=*p**k*<==<=*i*.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Mike's city intersection.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*i*<=≤<=*a**i*<=≤<=*n* , , describing shortcuts of Mike's city, allowing to walk from intersection *i* to intersection *a**i* using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from *a**i* to *i*).
Output Specification:
In the only line print *n* integers *m*1,<=*m*2,<=...,<=*m**n*, where *m**i* denotes the least amount of total energy required to walk from intersection 1 to intersection *i*.
Demo Input:
['3\n2 2 3\n', '5\n1 2 3 4 5\n', '7\n4 4 4 4 7 7 7\n']
Demo Output:
['0 1 2 \n', '0 1 2 3 4 \n', '0 1 2 1 2 3 3 \n']
Note:
In the first sample case desired sequences are:
1: 1; *m*<sub class="lower-index">1</sub> = 0;
2: 1, 2; *m*<sub class="lower-index">2</sub> = 1;
3: 1, 3; *m*<sub class="lower-index">3</sub> = |3 - 1| = 2.
In the second sample case the sequence for any intersection 1 < *i* is always 1, *i* and *m*<sub class="lower-index">*i*</sub> = |1 - *i*|.
In the third sample case — consider the following intersection sequences:
1: 1; *m*<sub class="lower-index">1</sub> = 0;
2: 1, 2; *m*<sub class="lower-index">2</sub> = |2 - 1| = 1;
3: 1, 4, 3; *m*<sub class="lower-index">3</sub> = 1 + |4 - 3| = 2;
4: 1, 4; *m*<sub class="lower-index">4</sub> = 1;
5: 1, 4, 5; *m*<sub class="lower-index">5</sub> = 1 + |4 - 5| = 2;
6: 1, 4, 6; *m*<sub class="lower-index">6</sub> = 1 + |4 - 6| = 3;
7: 1, 4, 5, 7; *m*<sub class="lower-index">7</sub> = 1 + |4 - 5| + 1 = 3.
|
```python
n = int(input())
ai = list(map(int,input().split()))
ai2 = [1000000]*(n+1)
ai2[0] = -1
num = 1
for i in range(n):
num2 = num
for j in range(num,ai[i]):
ai2[j] = ai2[j-1] + 1
num = ai[i]+1
ai2[num-1] = min(ai2[num-1],ai2[num-2] + 1,ai2[i+1]+1)
for j in range(num-2,num2-1,-1):
ai2[j] = min(ai2[j+1] + 1,ai2[j])
for i in range(1,n+1):
print(ai2[i],end=" ")
```
| 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"
}
] | 1,699,813,762
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n,k = map(int, input().split())
lista = list(map(int, input().split()))
ganadores = 0
for i in lista:
if i > k:
ganadores += 1
print(ganadores)
|
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())
lista = list(map(int, input().split()))
ganadores = 0
for i in lista:
if i > k:
ganadores += 1
print(ganadores)
```
| 0
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,689,346,137
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 71
| 154
| 0
|
n=input()
l=[i for i in n.split('WUB')]
for i in l:
print(i+" ")
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
n=input()
l=[i for i in n.split('WUB')]
for i in l:
print(i+" ")
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
|
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
|
Print the maximum integer which can be shown on the display of Stepan's newest device.
|
[
"2\n",
"3\n"
] |
[
"1\n",
"7\n"
] |
none
| 0
|
[
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "11"
},
{
"input": "5",
"output": "71"
},
{
"input": "6",
"output": "111"
},
{
"input": "85651",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "85666",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99999",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "100000",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99998",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9",
"output": "7111"
},
{
"input": "99997",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99996",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99995",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "156",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "255",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "4568",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "5431",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "6782",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "8343",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9514",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
}
] | 1,683,909,737
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) {
BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
try {
int n = Integer.parseInt(buffer.readLine().trim());
StringBuilder answer = new StringBuilder();
if (n % 2 == 1) {
answer.append(7);
n -= 3;
}
while (n - 2 >= 0) {
answer.append(1);
n -= 2;
}
System.out.println(answer.toString());
} catch (Exception e) {
System.out.println(e.getMessage());
e.printStackTrace();
}
}
}
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
Output Specification:
Print the maximum integer which can be shown on the display of Stepan's newest device.
Demo Input:
['2\n', '3\n']
Demo Output:
['1\n', '7\n']
Note:
none
|
```python
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) {
BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
try {
int n = Integer.parseInt(buffer.readLine().trim());
StringBuilder answer = new StringBuilder();
if (n % 2 == 1) {
answer.append(7);
n -= 3;
}
while (n - 2 >= 0) {
answer.append(1);
n -= 2;
}
System.out.println(answer.toString());
} catch (Exception e) {
System.out.println(e.getMessage());
e.printStackTrace();
}
}
}
```
| -1
|
|
264
|
A
|
Escape from Stones
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"data structures",
"implementation",
"two pointers"
] | null | null |
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
|
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
|
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
|
[
"llrlr\n",
"rrlll\n",
"lrlrr\n"
] |
[
"3\n5\n4\n2\n1\n",
"1\n2\n5\n4\n3\n",
"2\n4\n5\n3\n1\n"
] |
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
| 500
|
[
{
"input": "llrlr",
"output": "3\n5\n4\n2\n1"
},
{
"input": "rrlll",
"output": "1\n2\n5\n4\n3"
},
{
"input": "lrlrr",
"output": "2\n4\n5\n3\n1"
},
{
"input": "lllrlrllrl",
"output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1"
},
{
"input": "llrlrrrlrr",
"output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1"
},
{
"input": "rlrrrllrrr",
"output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2"
},
{
"input": "lrrlrrllrrrrllllllrr",
"output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1"
},
{
"input": "rlrrrlrrrllrrllrlrll",
"output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2"
},
{
"input": "lllrrlrlrllrrrrrllrl",
"output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1"
},
{
"input": "rrrllrrrlllrlllrlrrr",
"output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4"
},
{
"input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl",
"output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3"
},
{
"input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll",
"output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1"
},
{
"input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr",
"output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1"
},
{
"input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl",
"output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1"
},
{
"input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl",
"output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1"
},
{
"input": "l",
"output": "1"
},
{
"input": "r",
"output": "1"
}
] | 1,580,327,897
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 1,668
| 38,400,000
|
s = input()
ans = [0] * len(s)
left = 0
right = len(s) - 1
for i in range(len(s)):
if s[i] == 'r':
ans[left] = i+1
left += 1
else:
ans[right] = i+1
right -= 1
print(*list(ans))
|
Title: Escape from Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
Input Specification:
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
Output Specification:
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
Demo Input:
['llrlr\n', 'rrlll\n', 'lrlrr\n']
Demo Output:
['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n']
Note:
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
|
```python
s = input()
ans = [0] * len(s)
left = 0
right = len(s) - 1
for i in range(len(s)):
if s[i] == 'r':
ans[left] = i+1
left += 1
else:
ans[right] = i+1
right -= 1
print(*list(ans))
```
| 3
|
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,619,521,001
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, i, a[2000+10], j;
while(scanf("%d",&n)!=EOF)
{
for(i = 0;i<n;i++)
{
scanf("%d",&a[i]);
}
int sum;
for(i = 0;i<n;i++)
{
sum = 0;
for(j = 0;j<n;j++)
{
if(a[j]>a[i])
{
sum+=1;
}
}
sum++;
printf("%d%c",sum, i==n-1?'\n':' ');
}
}
return 0;
}
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, i, a[2000+10], j;
while(scanf("%d",&n)!=EOF)
{
for(i = 0;i<n;i++)
{
scanf("%d",&a[i]);
}
int sum;
for(i = 0;i<n;i++)
{
sum = 0;
for(j = 0;j<n;j++)
{
if(a[j]>a[i])
{
sum+=1;
}
}
sum++;
printf("%d%c",sum, i==n-1?'\n':' ');
}
}
return 0;
}
```
| -1
|
|
611
|
A
|
New Year and Days
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
|
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
|
Print one integer — the number of candies Limak will save in the year 2016.
|
[
"4 of week\n",
"30 of month\n"
] |
[
"52\n",
"11\n"
] |
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
| 500
|
[
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output": "52"
},
{
"input": "2 of week",
"output": "52"
},
{
"input": "3 of week",
"output": "52"
},
{
"input": "5 of week",
"output": "53"
},
{
"input": "7 of week",
"output": "52"
},
{
"input": "1 of month",
"output": "12"
},
{
"input": "2 of month",
"output": "12"
},
{
"input": "3 of month",
"output": "12"
},
{
"input": "4 of month",
"output": "12"
},
{
"input": "5 of month",
"output": "12"
},
{
"input": "6 of month",
"output": "12"
},
{
"input": "7 of month",
"output": "12"
},
{
"input": "8 of month",
"output": "12"
},
{
"input": "9 of month",
"output": "12"
},
{
"input": "10 of month",
"output": "12"
},
{
"input": "11 of month",
"output": "12"
},
{
"input": "12 of month",
"output": "12"
},
{
"input": "13 of month",
"output": "12"
},
{
"input": "14 of month",
"output": "12"
},
{
"input": "15 of month",
"output": "12"
},
{
"input": "16 of month",
"output": "12"
},
{
"input": "18 of month",
"output": "12"
},
{
"input": "19 of month",
"output": "12"
},
{
"input": "20 of month",
"output": "12"
},
{
"input": "21 of month",
"output": "12"
},
{
"input": "22 of month",
"output": "12"
},
{
"input": "23 of month",
"output": "12"
},
{
"input": "24 of month",
"output": "12"
},
{
"input": "25 of month",
"output": "12"
},
{
"input": "26 of month",
"output": "12"
},
{
"input": "27 of month",
"output": "12"
},
{
"input": "28 of month",
"output": "12"
},
{
"input": "29 of month",
"output": "12"
}
] | 1,472,376,562
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 62
| 4,608,000
|
a = input().split()
b = int(a[0])
if a[2] == 'week':
if b >= 5 and b <= 6:
print(53)
else:
print(52)
else:
if b == 31:
print(7)
elif b == 30:
print(11)
else:
print(12)
|
Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
|
```python
a = input().split()
b = int(a[0])
if a[2] == 'week':
if b >= 5 and b <= 6:
print(53)
else:
print(52)
else:
if b == 31:
print(7)
elif b == 30:
print(11)
else:
print(12)
```
| 3
|
|
556
|
A
|
Case of the Zeros and Ones
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
|
Output the minimum length of the string that may remain after applying the described operations several times.
|
[
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] |
[
"0\n",
"1\n",
"6\n"
] |
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 250
|
[
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,629,773,208
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 77
| 7,475,200
|
n = int(input())
a = input()
um = a.count("1")
zero = a.count("0")
print(abs(um - zero))
|
Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n = int(input())
a = input()
um = a.count("1")
zero = a.count("0")
print(abs(um - zero))
```
| 3
|
|
6
|
D
|
Lizards and Basements 2
|
PROGRAMMING
| 2,600
|
[
"brute force",
"dp"
] |
D. Lizards and Basements 2
|
2
|
64
|
This is simplified version of the problem used on the original contest. The original problem seems to have too difiicult solution. The constraints for input data have been reduced.
Polycarp likes to play computer role-playing game «Lizards and Basements». At the moment he is playing it as a magician. At one of the last levels he has to fight the line of archers. The only spell with which he can damage them is a fire ball. If Polycarp hits the *i*-th archer with his fire ball (they are numbered from left to right), the archer loses *a* health points. At the same time the spell damages the archers adjacent to the *i*-th (if any) — they lose *b* (1<=≤<=*b*<=<<=*a*<=≤<=10) health points each.
As the extreme archers (i.e. archers numbered 1 and *n*) are very far, the fire ball cannot reach them. Polycarp can hit any other archer with his fire ball.
The amount of health points for each archer is known. An archer will be killed when this amount is less than 0. What is the minimum amount of spells Polycarp can use to kill all the enemies?
Polycarp can throw his fire ball into an archer if the latter is already killed.
|
The first line of the input contains three integers *n*,<=*a*,<=*b* (3<=≤<=*n*<=≤<=10; 1<=≤<=*b*<=<<=*a*<=≤<=10). The second line contains a sequence of *n* integers — *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=15), where *h**i* is the amount of health points the *i*-th archer has.
|
In the first line print *t* — the required minimum amount of fire balls.
In the second line print *t* numbers — indexes of the archers that Polycarp should hit to kill all the archers in *t* shots. All these numbers should be between 2 and *n*<=-<=1. Separate numbers with spaces. If there are several solutions, output any of them. Print numbers in any order.
|
[
"3 2 1\n2 2 2\n",
"4 3 1\n1 4 1 1\n"
] |
[
"3\n2 2 2 ",
"4\n2 2 3 3 "
] |
none
| 0
|
[
{
"input": "3 2 1\n2 2 2",
"output": "3\n2 2 2 "
},
{
"input": "4 3 1\n1 4 1 1",
"output": "4\n2 2 3 3 "
},
{
"input": "3 5 3\n1 2 1",
"output": "1\n2 "
},
{
"input": "3 5 3\n3 2 2",
"output": "2\n2 2 "
},
{
"input": "3 5 3\n3 2 2",
"output": "2\n2 2 "
},
{
"input": "3 5 1\n10 10 10",
"output": "11\n2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "3 5 3\n10 9 7",
"output": "4\n2 2 2 2 "
},
{
"input": "3 5 1\n1 9 10",
"output": "11\n2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "3 5 3\n10 9 7",
"output": "4\n2 2 2 2 "
},
{
"input": "3 5 2\n9 3 6",
"output": "5\n2 2 2 2 2 "
},
{
"input": "4 5 3\n2 2 2 1",
"output": "2\n2 3 "
},
{
"input": "4 5 3\n2 3 2 2",
"output": "2\n2 3 "
},
{
"input": "4 5 3\n4 2 4 2",
"output": "3\n2 2 3 "
},
{
"input": "4 5 1\n4 9 1 8",
"output": "14\n2 2 2 2 2 3 3 3 3 3 3 3 3 3 "
},
{
"input": "4 5 3\n9 9 3 4",
"output": "6\n2 2 2 2 3 3 "
},
{
"input": "4 5 1\n8 8 9 8",
"output": "18\n2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 "
},
{
"input": "4 5 3\n10 10 10 10",
"output": "8\n2 2 2 2 3 3 3 3 "
},
{
"input": "4 5 2\n7 3 8 5",
"output": "7\n2 2 2 2 3 3 3 "
},
{
"input": "4 5 3\n5 10 7 7",
"output": "5\n2 2 3 3 3 "
},
{
"input": "4 3 1\n8 10 9 7",
"output": "17\n2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 "
},
{
"input": "10 9 5\n12 14 11 11 14 14 12 15 14 12",
"output": "10\n2 2 2 4 5 6 7 9 9 9 "
},
{
"input": "10 5 2\n12 10 6 7 11 4 3 5 9 3",
"output": "13\n2 2 2 2 2 2 2 4 5 5 7 9 9 "
},
{
"input": "10 4 1\n5 12 10 5 13 6 5 5 2 10",
"output": "25\n2 2 2 2 2 2 3 4 5 5 5 6 7 8 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 10 1\n10 12 11 4 12 1 15 15 11 12",
"output": "30\n2 2 2 2 2 2 2 2 2 2 2 4 5 5 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 9 1\n6 12 9 3 7 3 3 11 13 10",
"output": "23\n2 2 2 2 2 2 2 3 5 5 5 7 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 9 1\n8 7 9 8 14 1 9 11 8 13",
"output": "28\n2 2 2 2 2 2 2 2 2 4 5 5 7 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 3\n9 11 9 11 4 5 7 13 12 9",
"output": "13\n2 2 2 2 4 4 4 7 7 9 9 9 9 "
},
{
"input": "10 8 2\n11 10 13 12 9 10 9 9 10 12",
"output": "18\n2 2 2 2 2 2 4 4 5 6 7 9 9 9 9 9 9 9 "
},
{
"input": "10 3 1\n9 6 8 7 10 10 9 6 6 7",
"output": "28\n2 2 2 2 2 2 2 2 2 2 4 5 5 5 5 5 6 7 7 7 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 1\n6 5 4 5 5 4 5 4 5 4",
"output": "18\n2 2 2 2 2 2 2 4 5 5 7 7 7 9 9 9 9 9 "
},
{
"input": "10 4 3\n2 1 2 4 2 4 3 2 2 4",
"output": "6\n2 5 5 7 9 9 "
},
{
"input": "10 3 1\n4 4 3 3 3 3 2 1 3 1",
"output": "11\n2 2 2 2 2 4 5 6 7 9 9 "
},
{
"input": "10 7 1\n3 3 2 1 3 1 2 2 3 1",
"output": "9\n2 2 2 2 5 5 7 9 9 "
},
{
"input": "10 10 1\n8 8 8 8 8 8 8 8 8 8",
"output": "22\n2 2 2 2 2 2 2 2 2 4 5 6 7 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 1\n11 9 11 10 10 11 9 10 9 11",
"output": "33\n2 2 2 2 2 2 2 2 2 2 2 2 4 4 5 5 5 6 6 7 7 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 2\n10 9 14 9 13 11 14 10 14 10",
"output": "21\n2 2 2 2 2 2 4 4 5 6 6 6 7 7 8 9 9 9 9 9 9 "
},
{
"input": "10 8 6\n14 12 14 12 10 8 10 13 9 12",
"output": "10\n2 2 2 4 5 7 8 9 9 9 "
},
{
"input": "10 4 1\n7 8 9 8 8 7 8 9 7 7",
"output": "23\n2 2 2 2 2 2 2 2 4 4 5 5 6 7 7 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n9 10 9 9 10 9 9 10 9 10",
"output": "34\n2 2 2 2 2 2 2 2 2 2 4 4 5 5 5 5 5 5 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 9 4\n11 10 10 10 10 12 10 10 10 12",
"output": "11\n2 2 2 4 5 6 7 9 9 9 9 "
},
{
"input": "10 10 4\n1 1 1 1 1 1 1 1 1 1",
"output": "4\n2 5 8 9 "
},
{
"input": "10 2 1\n9 12 12 8 8 5 14 10 7 3",
"output": "28\n2 2 2 2 2 2 2 2 2 2 4 4 4 5 5 5 7 7 7 7 7 7 7 8 9 9 9 9 "
},
{
"input": "10 2 1\n14 15 15 14 15 15 15 14 14 14",
"output": "49\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5 6 6 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 6 3\n9 8 8 8 11 11 9 10 9 11",
"output": "13\n2 2 2 2 5 5 5 7 7 9 9 9 9 "
},
{
"input": "10 6 2\n11 8 10 11 10 8 8 13 9 13",
"output": "19\n2 2 2 2 2 2 4 5 5 5 7 7 9 9 9 9 9 9 9 "
},
{
"input": "10 3 1\n3 7 9 12 11 3 4 3 14 8",
"output": "22\n2 2 2 2 3 4 4 4 5 5 5 7 7 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 1\n6 7 10 7 6 8 9 8 6 9",
"output": "25\n2 2 2 2 2 2 2 3 4 4 5 6 6 7 7 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 7 2\n2 9 2 6 8 7 6 5 6 2",
"output": "8\n2 2 4 5 6 7 9 9 "
},
{
"input": "10 7 3\n2 7 2 7 3 4 3 2 4 3",
"output": "7\n2 3 5 5 7 9 9 "
},
{
"input": "10 6 3\n8 9 8 9 8 9 10 9 8 9",
"output": "12\n2 2 2 4 5 5 7 7 9 9 9 9 "
},
{
"input": "10 2 1\n10 9 10 9 9 9 10 8 8 10",
"output": "35\n2 2 2 2 2 2 2 2 2 2 2 4 4 4 5 5 5 5 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 3\n4 4 5 6 4 6 4 5 5 4",
"output": "8\n2 2 4 5 7 7 9 9 "
},
{
"input": "10 9 2\n5 7 8 8 7 5 7 4 4 5",
"output": "11\n2 2 2 4 4 6 6 7 9 9 9 "
},
{
"input": "10 9 5\n8 7 5 9 8 7 8 11 11 8",
"output": "7\n2 2 5 5 7 9 9 "
},
{
"input": "10 7 4\n5 6 6 6 7 7 6 5 5 5",
"output": "7\n2 2 5 5 7 9 9 "
},
{
"input": "10 9 1\n10 11 11 11 11 11 11 11 11 11",
"output": "29\n2 2 2 2 2 2 2 2 2 2 2 4 4 5 6 7 7 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 5 1\n6 5 6 5 6 6 6 5 6 6",
"output": "19\n2 2 2 2 2 2 2 4 5 6 7 8 9 9 9 9 9 9 9 "
},
{
"input": "10 3 1\n8 7 9 7 9 12 12 6 8 8",
"output": "29\n2 2 2 2 2 2 2 2 2 4 4 5 5 6 6 7 7 7 7 7 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 2\n7 3 5 3 5 5 3 4 2 4",
"output": "10\n2 2 2 2 5 5 7 9 9 9 "
},
{
"input": "10 7 2\n5 2 5 3 2 3 4 3 5 3",
"output": "8\n2 2 2 5 5 7 9 9 "
},
{
"input": "10 2 1\n5 3 6 6 7 4 4 4 3 3",
"output": "18\n2 2 2 2 2 2 4 4 5 5 5 7 7 8 9 9 9 9 "
},
{
"input": "10 6 1\n13 13 13 13 13 13 13 13 13 13",
"output": "36\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 5 5 6 6 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n14 11 11 11 15 15 12 15 12 14",
"output": "47\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 5 5 5 5 5 5 5 5 6 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 7 1\n9 15 15 11 8 10 13 9 15 9",
"output": "28\n2 2 2 2 2 2 2 2 2 2 3 4 4 5 6 7 7 7 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 10 4\n12 12 14 13 14 12 14 14 11 14",
"output": "13\n2 2 2 2 4 5 6 7 8 9 9 9 9 "
},
{
"input": "10 6 4\n5 5 5 5 5 5 5 6 4 4",
"output": "7\n2 2 5 5 7 9 9 "
},
{
"input": "10 8 7\n15 15 15 15 15 15 15 15 15 15",
"output": "11\n2 2 2 5 5 5 7 7 9 9 9 "
},
{
"input": "10 9 5\n11 10 4 4 6 9 11 4 10 8",
"output": "8\n2 2 2 5 7 8 9 9 "
},
{
"input": "10 6 3\n9 12 8 11 7 14 8 5 15 10",
"output": "13\n2 2 2 2 4 5 5 6 7 9 9 9 9 "
},
{
"input": "10 3 1\n4 4 4 4 3 4 3 3 3 3",
"output": "14\n2 2 2 2 2 4 5 5 6 7 9 9 9 9 "
},
{
"input": "10 6 4\n11 10 10 10 12 12 12 10 10 10",
"output": "11\n2 2 2 4 4 6 6 7 9 9 9 "
},
{
"input": "10 6 1\n3 2 4 4 8 12 5 10 12 6",
"output": "17\n2 2 2 2 4 5 6 6 7 8 9 9 9 9 9 9 9 "
},
{
"input": "10 9 5\n13 13 13 13 13 12 12 12 12 12",
"output": "10\n2 2 2 4 5 7 7 9 9 9 "
},
{
"input": "10 4 1\n7 7 6 6 6 8 6 7 6 7",
"output": "23\n2 2 2 2 2 2 2 2 4 5 5 5 6 7 7 9 9 9 9 9 9 9 9 "
},
{
"input": "10 6 3\n13 10 12 10 9 12 11 8 12 12",
"output": "15\n2 2 2 2 2 4 4 6 6 7 9 9 9 9 9 "
},
{
"input": "10 6 2\n1 4 5 4 4 2 3 6 6 4",
"output": "8\n2 4 4 6 8 9 9 9 "
},
{
"input": "10 8 1\n12 6 7 9 3 12 5 9 5 11",
"output": "30\n2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 6 6 7 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 4 2\n13 14 10 6 8 7 8 8 11 5",
"output": "16\n2 2 2 2 2 2 2 4 5 5 7 7 8 9 9 9 "
},
{
"input": "10 8 1\n3 4 5 6 4 6 5 6 5 4",
"output": "15\n2 2 2 2 4 4 5 6 7 9 9 9 9 9 9 "
},
{
"input": "10 7 2\n12 10 9 9 15 15 10 14 15 15",
"output": "21\n2 2 2 2 2 2 2 4 5 5 6 6 7 9 9 9 9 9 9 9 9 "
},
{
"input": "10 3 1\n9 9 8 8 8 9 9 8 8 8",
"output": "29\n2 2 2 2 2 2 2 2 2 2 4 4 5 5 5 6 7 7 7 7 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n5 4 5 4 4 4 4 4 4 5",
"output": "18\n2 2 2 2 2 2 4 4 5 6 7 7 9 9 9 9 9 9 "
},
{
"input": "10 6 5\n11 8 5 13 8 9 11 15 11 12",
"output": "11\n2 2 2 5 5 5 7 7 9 9 9 "
},
{
"input": "10 5 1\n7 10 15 5 15 5 5 5 11 7",
"output": "24\n2 2 2 2 2 2 2 2 3 3 5 5 5 5 7 7 9 9 9 9 9 9 9 9 "
},
{
"input": "10 3 2\n5 5 4 4 4 4 4 4 5 4",
"output": "10\n2 2 2 4 5 7 7 9 9 9 "
},
{
"input": "10 6 2\n5 8 4 5 1 3 6 7 5 3",
"output": "8\n2 2 2 4 7 7 9 9 "
},
{
"input": "10 5 2\n10 12 10 10 11 9 11 11 9 9",
"output": "18\n2 2 2 2 2 2 4 5 5 5 7 7 8 9 9 9 9 9 "
},
{
"input": "10 6 5\n9 8 10 10 11 11 8 9 10 11",
"output": "9\n2 2 4 5 6 7 9 9 9 "
},
{
"input": "10 3 2\n3 5 1 4 5 3 3 1 3 4",
"output": "9\n2 2 4 5 6 8 9 9 9 "
},
{
"input": "10 3 2\n4 9 6 9 6 8 4 5 6 9",
"output": "15\n2 2 2 3 5 5 5 5 7 8 9 9 9 9 9 "
},
{
"input": "10 3 2\n8 9 8 9 8 8 8 8 8 8",
"output": "17\n2 2 2 2 2 4 4 5 5 7 7 7 9 9 9 9 9 "
},
{
"input": "10 2 1\n11 6 9 9 11 10 7 13 11 9",
"output": "35\n2 2 2 2 2 2 2 2 2 2 2 2 4 5 5 5 5 5 5 5 5 7 7 7 7 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 6 1\n4 5 5 3 7 5 6 5 6 8",
"output": "18\n2 2 2 2 2 4 5 6 7 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 9 6\n15 14 14 12 15 10 9 14 13 8",
"output": "9\n2 2 2 4 5 6 8 9 9 "
},
{
"input": "10 7 1\n9 9 9 9 9 9 9 9 9 9",
"output": "26\n2 2 2 2 2 2 2 2 2 2 4 4 5 6 7 7 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 6 5\n4 5 4 1 3 6 3 2 2 2",
"output": "4\n2 5 7 9 "
},
{
"input": "10 10 3\n10 8 11 11 10 11 11 9 7 10",
"output": "12\n2 2 2 2 4 5 6 7 9 9 9 9 "
},
{
"input": "10 3 2\n7 8 11 6 8 7 2 3 8 7",
"output": "13\n2 2 2 2 4 4 5 6 6 9 9 9 9 "
},
{
"input": "10 8 6\n9 9 8 10 7 13 7 11 13 12",
"output": "8\n2 2 5 5 7 9 9 9 "
},
{
"input": "10 3 2\n9 13 9 10 12 10 14 13 11 11",
"output": "21\n2 2 2 2 2 4 5 5 5 5 7 7 7 7 7 9 9 9 9 9 9 "
},
{
"input": "10 3 2\n12 12 14 15 15 12 12 14 12 14",
"output": "26\n2 2 2 2 2 2 2 4 4 4 4 5 5 6 7 7 7 8 9 9 9 9 9 9 9 9 "
},
{
"input": "10 7 4\n6 4 8 4 8 7 10 6 8 6",
"output": "8\n2 2 4 6 6 8 9 9 "
},
{
"input": "10 5 4\n14 14 14 10 13 15 13 13 10 14",
"output": "14\n2 2 2 2 5 5 5 7 7 8 9 9 9 9 "
},
{
"input": "10 10 2\n10 10 13 10 10 12 6 8 11 12",
"output": "17\n2 2 2 2 2 2 4 5 6 7 9 9 9 9 9 9 9 "
},
{
"input": "10 6 4\n5 6 5 6 6 7 5 7 9 9",
"output": "8\n2 2 5 5 7 9 9 9 "
},
{
"input": "10 7 5\n10 10 10 10 10 10 10 10 10 10",
"output": "10\n2 2 2 4 5 7 7 9 9 9 "
},
{
"input": "10 5 1\n11 10 10 10 11 11 11 10 11 11",
"output": "32\n2 2 2 2 2 2 2 2 2 2 2 2 4 4 5 5 6 6 7 7 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 10 4\n13 13 13 13 13 13 13 13 13 13",
"output": "12\n2 2 2 2 4 5 6 7 9 9 9 9 "
},
{
"input": "10 5 2\n4 5 3 6 7 8 4 4 9 6",
"output": "11\n2 2 2 4 5 6 7 9 9 9 9 "
},
{
"input": "10 7 3\n5 9 8 8 8 7 6 7 6 5",
"output": "8\n2 2 4 5 7 7 9 9 "
},
{
"input": "10 2 1\n12 13 15 14 14 11 12 14 11 15",
"output": "46\n2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n13 12 15 12 14 14 14 15 13 15",
"output": "48\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n14 15 15 14 14 15 14 15 14 15",
"output": "50\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n9 7 14 8 14 15 15 9 12 13",
"output": "42\n2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n14 15 15 14 15 14 15 14 15 15",
"output": "50\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n9 7 14 8 14 15 15 9 12 13",
"output": "42\n2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
},
{
"input": "10 2 1\n4 14 13 15 14 5 8 11 12 14",
"output": "38\n2 2 2 2 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 "
}
] | 1,428,173,181
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 20
| 1,528
| 6,144,000
|
'''codeforce task'''
import math
steps=0
def base_gen(b,l):
n=[0]*l
yield n[:]
while not all(ne==b-1 for ne in n):
p=l-1
while n[p]==b-1:
n[p]=0
p-=1
n[p]+=1
yield n[:]
def deplete(pos,val,n,a,b,gs):
gs[pos-1]-=b*val
gs[pos]-=a*val
gs[pos+1]-=b*val
#global steps
#steps+=1
def fill_gaps(s,n,a,b,hs):
gs=hs[:]
for i in range(2,len(s)-1,2):
deplete(i,s[i],n,a,b,gs)
for i in range(1,len(s)-1,2):
val=max([0]+[math.ceil(x) for x in [gs[i-1]/b,gs[i]/a,gs[i+1]/b]])
s[i]=val
deplete(i,s[i],n,a,b,gs)
return (s,gs)
def merged_base(bs,n):
st=[0]*n
for bi,be in enumerate(bs):
st[2+bi*2]=be
return st
#n,a,b=8,3,1
#hs=[2,5,5,6,2,5,5,2]
n,a,b=[int(x) for x in input().split()]
hs=[int(x) for x in input().split()]
hs=[hse+1 for hse in hs]
min_final_state=[]
min_score=float("inf")
for bs in base_gen(max(hs)+1,(n-2)//2):
st=merged_base(bs,n)
final_state,remained_health=fill_gaps(st,n,a,b,hs)
if all(rh<=0 for rh in remained_health):
if sum(final_state)<min_score:
min_score=sum(final_state)
min_final_state=final_state
print(sum(min_final_state))
res=[]
for i,v in enumerate(min_final_state):
res+=[i+1]*v
print(*res)
#print('Total steps:',steps)
|
Title: Lizards and Basements 2
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
This is simplified version of the problem used on the original contest. The original problem seems to have too difiicult solution. The constraints for input data have been reduced.
Polycarp likes to play computer role-playing game «Lizards and Basements». At the moment he is playing it as a magician. At one of the last levels he has to fight the line of archers. The only spell with which he can damage them is a fire ball. If Polycarp hits the *i*-th archer with his fire ball (they are numbered from left to right), the archer loses *a* health points. At the same time the spell damages the archers adjacent to the *i*-th (if any) — they lose *b* (1<=≤<=*b*<=<<=*a*<=≤<=10) health points each.
As the extreme archers (i.e. archers numbered 1 and *n*) are very far, the fire ball cannot reach them. Polycarp can hit any other archer with his fire ball.
The amount of health points for each archer is known. An archer will be killed when this amount is less than 0. What is the minimum amount of spells Polycarp can use to kill all the enemies?
Polycarp can throw his fire ball into an archer if the latter is already killed.
Input Specification:
The first line of the input contains three integers *n*,<=*a*,<=*b* (3<=≤<=*n*<=≤<=10; 1<=≤<=*b*<=<<=*a*<=≤<=10). The second line contains a sequence of *n* integers — *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=15), where *h**i* is the amount of health points the *i*-th archer has.
Output Specification:
In the first line print *t* — the required minimum amount of fire balls.
In the second line print *t* numbers — indexes of the archers that Polycarp should hit to kill all the archers in *t* shots. All these numbers should be between 2 and *n*<=-<=1. Separate numbers with spaces. If there are several solutions, output any of them. Print numbers in any order.
Demo Input:
['3 2 1\n2 2 2\n', '4 3 1\n1 4 1 1\n']
Demo Output:
['3\n2 2 2 ', '4\n2 2 3 3 ']
Note:
none
|
```python
'''codeforce task'''
import math
steps=0
def base_gen(b,l):
n=[0]*l
yield n[:]
while not all(ne==b-1 for ne in n):
p=l-1
while n[p]==b-1:
n[p]=0
p-=1
n[p]+=1
yield n[:]
def deplete(pos,val,n,a,b,gs):
gs[pos-1]-=b*val
gs[pos]-=a*val
gs[pos+1]-=b*val
#global steps
#steps+=1
def fill_gaps(s,n,a,b,hs):
gs=hs[:]
for i in range(2,len(s)-1,2):
deplete(i,s[i],n,a,b,gs)
for i in range(1,len(s)-1,2):
val=max([0]+[math.ceil(x) for x in [gs[i-1]/b,gs[i]/a,gs[i+1]/b]])
s[i]=val
deplete(i,s[i],n,a,b,gs)
return (s,gs)
def merged_base(bs,n):
st=[0]*n
for bi,be in enumerate(bs):
st[2+bi*2]=be
return st
#n,a,b=8,3,1
#hs=[2,5,5,6,2,5,5,2]
n,a,b=[int(x) for x in input().split()]
hs=[int(x) for x in input().split()]
hs=[hse+1 for hse in hs]
min_final_state=[]
min_score=float("inf")
for bs in base_gen(max(hs)+1,(n-2)//2):
st=merged_base(bs,n)
final_state,remained_health=fill_gaps(st,n,a,b,hs)
if all(rh<=0 for rh in remained_health):
if sum(final_state)<min_score:
min_score=sum(final_state)
min_final_state=final_state
print(sum(min_final_state))
res=[]
for i,v in enumerate(min_final_state):
res+=[i+1]*v
print(*res)
#print('Total steps:',steps)
```
| 0
|
339
|
A
|
Helpful Maths
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings",
"strings"
] | null | null |
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
|
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
|
Print the new sum that Xenia can count.
|
[
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] |
[
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,697,603,542
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
str = input()
# str.replace("+","")
lis = str.split("+")
for i in range(len(lis)):
lis[i] = int(lis[i])
lis = sorted(lis)
for i in range(len(lis)-1):
print(f"{lis[i]}+",end="")
print(lis[-1])
|
Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none
|
```python
str = input()
# str.replace("+","")
lis = str.split("+")
for i in range(len(lis)):
lis[i] = int(lis[i])
lis = sorted(lis)
for i in range(len(lis)-1):
print(f"{lis[i]}+",end="")
print(lis[-1])
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,665,319,283
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 124
| 0
|
word=str(input())
Ucounter=0
Lcounter=0
for i in word:
if i.isupper():
Ucounter+=1
else:
Lcounter+=1
if Ucounter>Lcounter:
print(word.upper())
else:
print(word.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
word=str(input())
Ucounter=0
Lcounter=0
for i in word:
if i.isupper():
Ucounter+=1
else:
Lcounter+=1
if Ucounter>Lcounter:
print(word.upper())
else:
print(word.lower())
```
| 3.969
|
854
|
B
|
Maxim Buys an Apartment
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"math"
] | null | null |
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
|
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
|
Print the minimum possible and the maximum possible number of apartments good for Maxim.
|
[
"6 3\n"
] |
[
"1 3\n"
] |
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
| 1,000
|
[
{
"input": "6 3",
"output": "1 3"
},
{
"input": "10 1",
"output": "1 2"
},
{
"input": "10 9",
"output": "1 1"
},
{
"input": "8 0",
"output": "0 0"
},
{
"input": "8 8",
"output": "0 0"
},
{
"input": "966871928 890926970",
"output": "1 75944958"
},
{
"input": "20 2",
"output": "1 4"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 0",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "9 4",
"output": "1 5"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "10 5",
"output": "1 5"
},
{
"input": "1000 1000",
"output": "0 0"
},
{
"input": "1000 333",
"output": "1 666"
},
{
"input": "1000 334",
"output": "1 666"
},
{
"input": "999 333",
"output": "1 666"
},
{
"input": "999 334",
"output": "1 665"
},
{
"input": "998 332",
"output": "1 664"
},
{
"input": "998 333",
"output": "1 665"
},
{
"input": "89 4",
"output": "1 8"
},
{
"input": "66 50",
"output": "1 16"
},
{
"input": "88 15",
"output": "1 30"
},
{
"input": "95 43",
"output": "1 52"
},
{
"input": "900 344",
"output": "1 556"
},
{
"input": "777 113",
"output": "1 226"
},
{
"input": "964 42",
"output": "1 84"
},
{
"input": "982 867",
"output": "1 115"
},
{
"input": "1000000000 0",
"output": "0 0"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 333333333",
"output": "1 666666666"
},
{
"input": "1000000000 333333334",
"output": "1 666666666"
},
{
"input": "999999999 333333333",
"output": "1 666666666"
},
{
"input": "999999999 333333334",
"output": "1 666666665"
},
{
"input": "999999998 333333332",
"output": "1 666666664"
},
{
"input": "999999998 333333333",
"output": "1 666666665"
},
{
"input": "78602604 42160832",
"output": "1 36441772"
},
{
"input": "35679021 9137902",
"output": "1 18275804"
},
{
"input": "41949373 13173511",
"output": "1 26347022"
},
{
"input": "77855558 49163875",
"output": "1 28691683"
},
{
"input": "87187123 2851901",
"output": "1 5703802"
},
{
"input": "66849627 25004217",
"output": "1 41845410"
},
{
"input": "873046672 517064947",
"output": "1 355981725"
},
{
"input": "639857373 1393427",
"output": "1 2786854"
},
{
"input": "637563683 69636269",
"output": "1 139272538"
},
{
"input": "911669737 141068293",
"output": "1 282136586"
},
{
"input": "547575919 313272818",
"output": "1 234303101"
},
{
"input": "955020006 297895809",
"output": "1 595791618"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "11 3",
"output": "1 6"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "4 1",
"output": "1 2"
},
{
"input": "9 3",
"output": "1 6"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "7 3",
"output": "1 4"
},
{
"input": "12 5",
"output": "1 7"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "1000 8",
"output": "1 16"
}
] | 1,579,602,166
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 108
| 0
|
n ,k = map(int,input().split())
if k == 0 or k == n:
print(0,end = " ")
else:
print(1,end = " ")
if k < (n + 1)//2:
print(k + 1)
else:
print(n - k)
|
Title: Maxim Buys an Apartment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input Specification:
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Demo Input:
['6 3\n']
Demo Output:
['1 3\n']
Note:
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
|
```python
n ,k = map(int,input().split())
if k == 0 or k == n:
print(0,end = " ")
else:
print(1,end = " ")
if k < (n + 1)//2:
print(k + 1)
else:
print(n - k)
```
| 0
|
|
900
|
A
|
Find Extra One
|
PROGRAMMING
| 800
|
[
"geometry",
"implementation"
] | null | null |
You have *n* distinct points on a plane, none of them lie on *OY* axis. Check that there is a point after removal of which the remaining points are located on one side of the *OY* axis.
|
The first line contains a single positive integer *n* (2<=≤<=*n*<=≤<=105).
The following *n* lines contain coordinates of the points. The *i*-th of these lines contains two single integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109, *x**i*<=≠<=0). No two points coincide.
|
Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower).
|
[
"3\n1 1\n-1 -1\n2 -1\n",
"4\n1 1\n2 2\n-1 1\n-2 2\n",
"3\n1 2\n2 1\n4 60\n"
] |
[
"Yes",
"No",
"Yes"
] |
In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed.
| 500
|
[
{
"input": "3\n1 1\n-1 -1\n2 -1",
"output": "Yes"
},
{
"input": "4\n1 1\n2 2\n-1 1\n-2 2",
"output": "No"
},
{
"input": "3\n1 2\n2 1\n4 60",
"output": "Yes"
},
{
"input": "10\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n-1 -1",
"output": "Yes"
},
{
"input": "2\n1000000000 -1000000000\n1000000000 1000000000",
"output": "Yes"
},
{
"input": "23\n-1 1\n-1 2\n-2 4\n-7 -8\n-3 3\n-9 -14\n-5 3\n-6 2\n-7 11\n-4 4\n-8 5\n1 1\n-1 -1\n-1 -2\n-2 -4\n-7 8\n-3 -3\n-9 14\n-5 -3\n-6 -2\n-7 -11\n-4 -4\n-8 -5",
"output": "Yes"
},
{
"input": "4\n-1000000000 -1000000000\n1000000000 1000000000\n-1000000000 1000000000\n1000000000 -1000000000",
"output": "No"
},
{
"input": "2\n-1000000000 1000000000\n-1000000000 -1000000000",
"output": "Yes"
},
{
"input": "5\n-1 -1\n-2 2\n2 2\n2 -2\n3 2",
"output": "No"
},
{
"input": "2\n1 0\n-1 0",
"output": "Yes"
},
{
"input": "4\n-1 1\n-1 2\n-1 3\n-1 4",
"output": "Yes"
},
{
"input": "2\n-1 0\n1 0",
"output": "Yes"
},
{
"input": "2\n1 2\n-1 2",
"output": "Yes"
},
{
"input": "2\n8 0\n7 0",
"output": "Yes"
},
{
"input": "6\n-1 0\n-2 0\n-1 -1\n-1 5\n1 0\n1 1",
"output": "No"
},
{
"input": "4\n1 0\n2 0\n-1 0\n-2 0",
"output": "No"
},
{
"input": "4\n-2 0\n-1 0\n1 0\n2 0",
"output": "No"
},
{
"input": "2\n1 1\n-1 1",
"output": "Yes"
},
{
"input": "4\n-1 0\n-2 0\n1 0\n2 0",
"output": "No"
},
{
"input": "2\n4 3\n-4 -2",
"output": "Yes"
},
{
"input": "4\n1 0\n2 0\n-1 1\n-1 2",
"output": "No"
},
{
"input": "5\n1 1\n2 1\n3 1\n-1 1\n-2 1",
"output": "No"
},
{
"input": "2\n1 1\n-1 -1",
"output": "Yes"
},
{
"input": "4\n1 2\n1 0\n1 -2\n-1 2",
"output": "Yes"
},
{
"input": "5\n-2 3\n-3 3\n4 2\n3 2\n1 2",
"output": "No"
},
{
"input": "3\n2 0\n3 0\n4 0",
"output": "Yes"
},
{
"input": "5\n-3 1\n-2 1\n-1 1\n1 1\n2 1",
"output": "No"
},
{
"input": "4\n-3 0\n1 0\n2 0\n3 0",
"output": "Yes"
},
{
"input": "2\n1 0\n-1 1",
"output": "Yes"
},
{
"input": "3\n-1 0\n1 0\n2 0",
"output": "Yes"
},
{
"input": "5\n1 0\n3 0\n-1 0\n-6 0\n-4 1",
"output": "No"
},
{
"input": "5\n-1 2\n-2 2\n-3 1\n1 2\n2 3",
"output": "No"
},
{
"input": "3\n1 0\n-1 0\n-2 0",
"output": "Yes"
},
{
"input": "4\n1 0\n2 0\n3 1\n4 1",
"output": "Yes"
},
{
"input": "4\n1 0\n1 2\n1 3\n-1 5",
"output": "Yes"
},
{
"input": "4\n2 2\n2 5\n-2 3\n-2 0",
"output": "No"
},
{
"input": "4\n1 1\n-1 1\n-1 0\n-1 -1",
"output": "Yes"
},
{
"input": "4\n2 0\n3 0\n-3 -3\n-3 -4",
"output": "No"
},
{
"input": "4\n-1 0\n-2 0\n-3 0\n-4 0",
"output": "Yes"
},
{
"input": "2\n-1 1\n1 1",
"output": "Yes"
},
{
"input": "5\n1 1\n2 2\n3 3\n-4 -4\n-5 -5",
"output": "No"
},
{
"input": "5\n2 0\n3 0\n4 0\n5 0\n6 0",
"output": "Yes"
},
{
"input": "2\n-1 2\n1 2",
"output": "Yes"
},
{
"input": "4\n1 1\n2 1\n-3 0\n-4 0",
"output": "No"
},
{
"input": "4\n-1 0\n-2 0\n3 0\n4 0",
"output": "No"
},
{
"input": "3\n3 0\n2 0\n1 0",
"output": "Yes"
},
{
"input": "4\n-2 0\n-3 0\n1 -1\n3 1",
"output": "No"
},
{
"input": "3\n-1 -1\n1 1\n2 2",
"output": "Yes"
},
{
"input": "4\n-2 0\n-1 0\n2 0\n1 0",
"output": "No"
},
{
"input": "2\n-3 5\n3 5",
"output": "Yes"
},
{
"input": "2\n-1 5\n1 5",
"output": "Yes"
},
{
"input": "4\n2 0\n3 0\n-2 0\n-3 0",
"output": "No"
},
{
"input": "3\n-1 1\n1 1\n1 -1",
"output": "Yes"
},
{
"input": "2\n1 0\n2 0",
"output": "Yes"
},
{
"input": "4\n-1 1\n-2 1\n2 -1\n3 -1",
"output": "No"
},
{
"input": "5\n1 0\n2 0\n3 0\n-1 0\n-2 0",
"output": "No"
},
{
"input": "4\n-3 0\n-4 0\n-5 0\n-6 0",
"output": "Yes"
},
{
"input": "6\n-3 0\n-2 0\n-1 0\n1 0\n2 0\n3 0",
"output": "No"
},
{
"input": "4\n5 0\n5 1\n6 0\n6 1",
"output": "Yes"
}
] | 1,631,291,641
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 186
| 6,758,400
|
n = int(input())
num = sum(int(input().split()[0]) > 0 for i in range(n))
print("Yes" if num <= 1 or num >= n - 1 else "No")
|
Title: Find Extra One
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* distinct points on a plane, none of them lie on *OY* axis. Check that there is a point after removal of which the remaining points are located on one side of the *OY* axis.
Input Specification:
The first line contains a single positive integer *n* (2<=≤<=*n*<=≤<=105).
The following *n* lines contain coordinates of the points. The *i*-th of these lines contains two single integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109, *x**i*<=≠<=0). No two points coincide.
Output Specification:
Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower).
Demo Input:
['3\n1 1\n-1 -1\n2 -1\n', '4\n1 1\n2 2\n-1 1\n-2 2\n', '3\n1 2\n2 1\n4 60\n']
Demo Output:
['Yes', 'No', 'Yes']
Note:
In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed.
|
```python
n = int(input())
num = sum(int(input().split()[0]) > 0 for i in range(n))
print("Yes" if num <= 1 or num >= n - 1 else "No")
```
| 3
|
|
801
|
B
|
Valued Keys
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
|
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
|
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
|
[
"ab\naa\n",
"nzwzl\nniwel\n",
"ab\nba\n"
] |
[
"ba\n",
"xiyez\n",
"-1\n"
] |
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
| 1,000
|
[
{
"input": "ab\naa",
"output": "ba"
},
{
"input": "nzwzl\nniwel",
"output": "xiyez"
},
{
"input": "ab\nba",
"output": "-1"
},
{
"input": "r\nl",
"output": "l"
},
{
"input": "d\ny",
"output": "-1"
},
{
"input": "yvowz\ncajav",
"output": "cajav"
},
{
"input": "lwzjp\ninjit",
"output": "-1"
},
{
"input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm",
"output": "eodnlemiicdedmkcgavqbnqmm"
},
{
"input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs",
"output": "-1"
},
{
"input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf",
"output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf"
},
{
"input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen",
"output": "-1"
},
{
"input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb",
"output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb"
},
{
"input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym",
"output": "-1"
},
{
"input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib",
"output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib"
},
{
"input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew",
"output": "-1"
},
{
"input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv",
"output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv"
},
{
"input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns",
"output": "-1"
},
{
"input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf",
"output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf"
},
{
"input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp",
"output": "-1"
},
{
"input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp",
"output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp"
},
{
"input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej",
"output": "-1"
},
{
"input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe",
"output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe"
},
{
"input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang",
"output": "-1"
},
{
"input": "ftfr\nftfr",
"output": "ftfr"
},
{
"input": "ftr\nftr",
"output": "ftr"
},
{
"input": "shftr\nshftr",
"output": "shftr"
},
{
"input": "vkvkkv\nvkvkkv",
"output": "vkvkkv"
},
{
"input": "ftrd\nftrd",
"output": "ftrd"
},
{
"input": "fztr\nfztr",
"output": "fztr"
},
{
"input": "frtr\nfrtr",
"output": "frtr"
},
{
"input": "shdftr\nshdftr",
"output": "shdftr"
},
{
"input": "zzz\nzzz",
"output": "zzz"
},
{
"input": "shtr\nshtr",
"output": "shtr"
},
{
"input": "aaaaa\nzzzzz",
"output": "-1"
},
{
"input": "efr\nefr",
"output": "efr"
}
] | 1,675,781,860
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
# LUOGU_RID: 101671665
s, t = input(), input()
for x, y in zip(s, t):
if x < y:
exit(print(-1))
print(y)
|
Title: Valued Keys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
Input Specification:
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
Output Specification:
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
Demo Input:
['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n']
Demo Output:
['ba\n', 'xiyez\n', '-1\n']
Note:
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
|
```python
# LUOGU_RID: 101671665
s, t = input(), input()
for x, y in zip(s, t):
if x < y:
exit(print(-1))
print(y)
```
| 0
|
|
749
|
A
|
Bachgold Problem
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
|
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
|
[
"5\n",
"6\n"
] |
[
"2\n2 3\n",
"3\n2 2 2\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\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 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 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 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": "100000",
"output": "50000\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 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 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 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": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\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 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 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 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": "99996",
"output": "49998\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 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 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 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": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\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 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 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 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": "99998",
"output": "49999\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 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 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 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": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\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 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 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 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": "99994",
"output": "49997\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 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 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 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": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\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 3"
},
{
"input": "57",
"output": "28\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 2 3"
},
{
"input": "61",
"output": "30\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 2 2 2 3"
},
{
"input": "774",
"output": "387\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 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 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 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": "202",
"output": "101\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 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": "530",
"output": "265\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 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 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 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": "7166",
"output": "3583\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 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 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 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": "9294",
"output": "4647\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 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 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 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": "2422",
"output": "1211\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 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 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 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": "15326",
"output": "7663\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 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 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 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": "11454",
"output": "5727\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 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 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 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": "14878",
"output": "7439\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 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 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 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": "90672",
"output": "45336\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 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 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 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": "99544",
"output": "49772\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 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 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 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": "90472",
"output": "45236\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 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 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 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": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\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 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 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 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..."
}
] | 1,697,114,545
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 93
| 512,000
|
def isEven(number):
if number %2 ==0:
return True
else :
return False
result = []
number = int(input())
if isEven(number):
for i in range(number//2):
result.append(2)
else :
for i in range((number//2)-1):
result.append(2)
result.append(3)
print(len(result))
for value in result:
print(value, end=' ')
|
Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none
|
```python
def isEven(number):
if number %2 ==0:
return True
else :
return False
result = []
number = int(input())
if isEven(number):
for i in range(number//2):
result.append(2)
else :
for i in range((number//2)-1):
result.append(2)
result.append(3)
print(len(result))
for value in result:
print(value, end=' ')
```
| 3
|
|
144
|
A
|
Arrival of the General
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
|
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
|
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
|
[
"4\n33 44 11 22\n",
"7\n10 10 58 31 63 40 76\n"
] |
[
"2\n",
"10\n"
] |
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
| 500
|
[
{
"input": "4\n33 44 11 22",
"output": "2"
},
{
"input": "7\n10 10 58 31 63 40 76",
"output": "10"
},
{
"input": "2\n88 89",
"output": "1"
},
{
"input": "5\n100 95 100 100 88",
"output": "0"
},
{
"input": "7\n48 48 48 48 45 45 45",
"output": "0"
},
{
"input": "10\n68 47 67 29 63 71 71 65 54 56",
"output": "10"
},
{
"input": "15\n77 68 96 60 92 75 61 60 66 79 80 65 60 95 92",
"output": "4"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "20\n30 30 30 14 30 14 30 30 30 14 30 14 14 30 14 14 30 14 14 14",
"output": "0"
},
{
"input": "35\n37 41 46 39 47 39 44 47 44 42 44 43 47 39 46 39 38 42 39 37 40 44 41 42 41 42 39 42 36 36 42 36 42 42 42",
"output": "7"
},
{
"input": "40\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 98 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99",
"output": "47"
},
{
"input": "50\n48 52 44 54 53 56 62 49 39 41 53 39 40 64 53 50 62 48 40 52 51 48 40 52 61 62 62 61 48 64 55 57 56 40 48 58 41 60 60 56 64 50 64 45 48 45 46 63 59 57",
"output": "50"
},
{
"input": "57\n7 24 17 19 6 19 10 11 12 22 14 5 5 11 13 10 24 19 24 24 24 11 21 20 4 14 24 24 18 13 24 3 20 3 3 3 3 9 3 9 22 22 16 3 3 3 15 11 3 3 8 17 10 13 3 14 13",
"output": "3"
},
{
"input": "65\n58 50 35 44 35 37 36 58 38 36 58 56 56 49 48 56 58 43 40 44 52 44 58 58 57 50 43 35 55 39 38 49 53 56 50 42 41 56 34 57 49 38 34 51 56 38 58 40 53 46 48 34 38 43 49 49 58 56 41 43 44 34 38 48 36",
"output": "3"
},
{
"input": "69\n70 48 49 48 49 71 48 53 55 69 48 53 54 58 53 63 48 48 69 67 72 75 71 75 74 74 57 63 65 60 48 48 65 48 48 51 50 49 62 53 76 68 76 56 76 76 64 76 76 57 61 76 73 51 59 76 65 50 69 50 76 67 76 63 62 74 74 58 73",
"output": "73"
},
{
"input": "75\n70 65 64 71 71 64 71 64 68 71 65 64 65 68 71 66 66 69 68 63 69 65 71 69 68 68 71 67 71 65 65 65 71 71 65 69 63 66 62 67 64 63 62 64 67 65 62 69 62 64 69 62 67 64 67 70 64 63 64 64 69 62 62 64 70 62 62 68 67 69 62 64 66 70 68",
"output": "7"
},
{
"input": "84\n92 95 84 85 94 80 90 86 80 92 95 84 86 83 86 83 93 91 95 92 84 88 82 84 84 84 80 94 93 80 94 80 95 83 85 80 95 95 80 84 86 92 83 81 90 87 81 89 92 93 80 87 90 85 93 85 93 94 93 89 94 83 93 91 80 83 90 94 95 80 95 92 85 84 93 94 94 82 91 95 95 89 85 94",
"output": "15"
},
{
"input": "90\n86 87 72 77 82 71 75 78 61 67 79 90 64 94 94 74 85 87 73 76 71 71 60 69 77 73 76 80 82 57 62 57 57 83 76 72 75 87 72 94 77 85 59 82 86 69 62 80 95 73 83 94 79 85 91 68 85 74 93 95 68 75 89 93 83 78 95 78 83 77 81 85 66 92 63 65 75 78 67 91 77 74 59 86 77 76 90 67 70 64",
"output": "104"
},
{
"input": "91\n94 98 96 94 95 98 98 95 98 94 94 98 95 95 99 97 97 94 95 98 94 98 96 98 96 98 97 95 94 94 94 97 94 96 98 98 98 94 96 95 94 95 97 97 97 98 94 98 96 95 98 96 96 98 94 97 96 98 97 95 97 98 94 95 94 94 97 94 96 97 97 93 94 95 95 94 96 98 97 96 94 98 98 96 96 96 96 96 94 96 97",
"output": "33"
},
{
"input": "92\n44 28 32 29 41 41 36 39 40 39 41 35 41 28 35 27 41 34 28 38 43 43 41 38 27 26 28 36 30 29 39 32 35 35 32 30 39 30 37 27 41 41 28 30 43 31 35 33 36 28 44 40 41 35 31 42 37 38 37 34 39 40 27 40 33 33 44 43 34 33 34 34 35 38 38 37 30 39 35 41 45 42 41 32 33 33 31 30 43 41 43 43",
"output": "145"
},
{
"input": "93\n46 32 52 36 39 30 57 63 63 30 32 44 27 59 46 38 40 45 44 62 35 36 51 48 39 58 36 51 51 51 48 58 59 36 29 35 31 49 64 60 34 38 42 56 33 42 52 31 63 34 45 51 35 45 33 53 33 62 31 38 66 29 51 54 28 61 32 45 57 41 36 34 47 36 31 28 67 48 52 46 32 40 64 58 27 53 43 57 34 66 43 39 26",
"output": "76"
},
{
"input": "94\n56 55 54 31 32 42 46 29 24 54 40 40 20 45 35 56 32 33 51 39 26 56 21 56 51 27 29 39 56 52 54 43 43 55 48 51 44 49 52 49 23 19 19 28 20 26 45 33 35 51 42 36 25 25 38 23 21 35 54 50 41 20 37 28 42 20 22 43 37 34 55 21 24 38 19 41 45 34 19 33 44 54 38 31 23 53 35 32 47 40 39 31 20 34",
"output": "15"
},
{
"input": "95\n57 71 70 77 64 64 76 81 81 58 63 75 81 77 71 71 71 60 70 70 69 67 62 64 78 64 69 62 76 76 57 70 68 77 70 68 73 77 79 73 60 57 69 60 74 65 58 75 75 74 73 73 65 75 72 57 81 62 62 70 67 58 76 57 79 81 68 64 58 77 70 59 79 64 80 58 71 59 81 71 80 64 78 80 78 65 70 68 78 80 57 63 64 76 81",
"output": "11"
},
{
"input": "96\n96 95 95 95 96 97 95 97 96 95 98 96 97 95 98 96 98 96 98 96 98 95 96 95 95 95 97 97 95 95 98 98 95 96 96 95 97 96 98 96 95 97 97 95 97 97 95 94 96 96 97 96 97 97 96 94 94 97 95 95 95 96 95 96 95 97 97 95 97 96 95 94 97 97 97 96 97 95 96 94 94 95 97 94 94 97 97 97 95 97 97 95 94 96 95 95",
"output": "13"
},
{
"input": "97\n14 15 12 12 13 15 12 15 12 12 12 12 12 14 15 15 13 12 15 15 12 12 12 13 14 15 15 13 14 15 14 14 14 14 12 13 12 13 13 12 15 12 13 13 15 12 15 13 12 13 13 13 14 13 12 15 14 13 14 15 13 14 14 13 14 12 15 12 14 12 13 14 15 14 13 15 13 12 15 15 15 13 15 15 13 14 16 16 16 13 15 13 15 14 15 15 15",
"output": "104"
},
{
"input": "98\n37 69 35 70 58 69 36 47 41 63 60 54 49 35 55 50 35 53 52 43 35 41 40 49 38 35 48 70 42 35 35 65 56 54 44 59 59 48 51 49 59 67 35 60 69 35 58 50 35 44 48 69 41 58 44 45 35 47 70 61 49 47 37 39 35 51 44 70 72 65 36 41 63 63 48 66 45 50 50 71 37 52 72 67 72 39 72 39 36 64 48 72 69 49 45 72 72 67",
"output": "100"
},
{
"input": "99\n31 31 16 15 19 31 19 22 29 27 12 22 28 30 25 33 26 25 19 22 34 21 17 33 31 22 16 26 22 30 31 17 13 33 13 17 28 25 18 33 27 22 31 22 13 27 20 22 23 15 24 32 29 13 16 20 32 33 14 33 19 27 16 28 25 17 17 28 18 26 32 33 19 23 30 13 14 23 24 28 14 28 22 20 30 14 24 23 17 29 18 28 29 21 28 18 16 24 32",
"output": "107"
},
{
"input": "100\n37 54 39 29 32 49 21 13 34 21 16 42 34 27 16 26 7 34 51 9 11 27 16 40 36 7 48 52 30 42 42 52 51 11 32 26 6 7 28 54 48 51 6 54 42 20 51 48 46 4 4 31 47 6 9 16 8 23 36 50 49 30 47 37 45 24 48 26 8 50 18 24 27 13 39 7 37 34 5 43 42 19 34 23 52 38 35 6 29 11 53 49 21 29 36 22 45 33 51 22",
"output": "50"
},
{
"input": "100\n41 5 20 23 5 47 28 28 64 5 12 14 34 56 22 48 6 52 39 55 62 40 24 54 39 28 56 27 35 51 14 24 38 10 55 20 39 62 52 2 30 55 49 48 15 23 59 44 31 28 66 29 62 59 31 13 28 12 50 45 41 47 29 27 64 66 14 39 62 22 44 63 24 5 63 17 42 7 25 41 49 33 31 51 63 3 37 66 19 9 2 40 16 45 56 21 34 29 39 61",
"output": "59"
},
{
"input": "100\n66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95",
"output": "99"
},
{
"input": "100\n81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 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",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "2\n100 99",
"output": "0"
},
{
"input": "100\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 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "197"
},
{
"input": "100\n87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87",
"output": "0"
},
{
"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": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "3\n1 3 2",
"output": "2"
}
] | 1,685,619,480
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 92
| 0
|
a = eval(input())
s = input()
s = s.split()
min=int(s[0])
minidx = 0
maxidx = 0
max = int(s[0])
for i in range(a):
if int(s[i]) > max:
max = int(s[i])
maxidx=i
elif int(s[i])<=min:
min = int(s[i])
minidx = i
if minidx<maxidx:
print(maxidx+(a-minidx)-2)
else:
print(maxidx +a- minidx-1)
|
Title: Arrival of the General
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
Input Specification:
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
Output Specification:
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
Demo Input:
['4\n33 44 11 22\n', '7\n10 10 58 31 63 40 76\n']
Demo Output:
['2\n', '10\n']
Note:
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
|
```python
a = eval(input())
s = input()
s = s.split()
min=int(s[0])
minidx = 0
maxidx = 0
max = int(s[0])
for i in range(a):
if int(s[i]) > max:
max = int(s[i])
maxidx=i
elif int(s[i])<=min:
min = int(s[i])
minidx = i
if minidx<maxidx:
print(maxidx+(a-minidx)-2)
else:
print(maxidx +a- minidx-1)
```
| 3
|
|
99
|
A
|
Help Far Away Kingdom
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Help Far Away Kingdom
|
2
|
256
|
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
|
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
|
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
|
[
"0.0\n",
"1.49\n",
"1.50\n",
"2.71828182845904523536\n",
"3.14159265358979323846\n",
"12345678901234567890.1\n",
"123456789123456789.999\n"
] |
[
"0",
"1",
"2",
"3",
"3",
"12345678901234567890",
"GOTO Vasilisa."
] |
none
| 500
|
[
{
"input": "0.0",
"output": "0"
},
{
"input": "1.49",
"output": "1"
},
{
"input": "1.50",
"output": "2"
},
{
"input": "2.71828182845904523536",
"output": "3"
},
{
"input": "3.14159265358979323846",
"output": "3"
},
{
"input": "12345678901234567890.1",
"output": "12345678901234567890"
},
{
"input": "123456789123456789.999",
"output": "GOTO Vasilisa."
},
{
"input": "12345678901234567890.9",
"output": "12345678901234567891"
},
{
"input": "123456789123456788.999",
"output": "123456789123456789"
},
{
"input": "9.000",
"output": "GOTO Vasilisa."
},
{
"input": "0.1",
"output": "0"
},
{
"input": "0.2",
"output": "0"
},
{
"input": "0.3",
"output": "0"
},
{
"input": "0.4",
"output": "0"
},
{
"input": "0.5",
"output": "1"
},
{
"input": "0.6",
"output": "1"
},
{
"input": "0.7",
"output": "1"
},
{
"input": "0.8",
"output": "1"
},
{
"input": "0.9",
"output": "1"
},
{
"input": "1.0",
"output": "1"
},
{
"input": "1.1",
"output": "1"
},
{
"input": "1.2",
"output": "1"
},
{
"input": "1.3",
"output": "1"
},
{
"input": "1.4",
"output": "1"
},
{
"input": "1.5",
"output": "2"
},
{
"input": "1.6",
"output": "2"
},
{
"input": "1.7",
"output": "2"
},
{
"input": "1.8",
"output": "2"
},
{
"input": "1.9",
"output": "2"
},
{
"input": "2.0",
"output": "2"
},
{
"input": "2.1",
"output": "2"
},
{
"input": "2.2",
"output": "2"
},
{
"input": "2.3",
"output": "2"
},
{
"input": "2.4",
"output": "2"
},
{
"input": "2.5",
"output": "3"
},
{
"input": "2.6",
"output": "3"
},
{
"input": "2.7",
"output": "3"
},
{
"input": "2.8",
"output": "3"
},
{
"input": "2.9",
"output": "3"
},
{
"input": "3.0",
"output": "3"
},
{
"input": "3.1",
"output": "3"
},
{
"input": "3.2",
"output": "3"
},
{
"input": "3.3",
"output": "3"
},
{
"input": "3.4",
"output": "3"
},
{
"input": "3.5",
"output": "4"
},
{
"input": "3.6",
"output": "4"
},
{
"input": "3.7",
"output": "4"
},
{
"input": "3.8",
"output": "4"
},
{
"input": "3.9",
"output": "4"
},
{
"input": "4.0",
"output": "4"
},
{
"input": "4.1",
"output": "4"
},
{
"input": "4.2",
"output": "4"
},
{
"input": "4.3",
"output": "4"
},
{
"input": "4.4",
"output": "4"
},
{
"input": "4.5",
"output": "5"
},
{
"input": "4.6",
"output": "5"
},
{
"input": "4.7",
"output": "5"
},
{
"input": "4.8",
"output": "5"
},
{
"input": "4.9",
"output": "5"
},
{
"input": "5.0",
"output": "5"
},
{
"input": "5.1",
"output": "5"
},
{
"input": "5.2",
"output": "5"
},
{
"input": "5.3",
"output": "5"
},
{
"input": "5.4",
"output": "5"
},
{
"input": "5.5",
"output": "6"
},
{
"input": "5.6",
"output": "6"
},
{
"input": "5.7",
"output": "6"
},
{
"input": "5.8",
"output": "6"
},
{
"input": "5.9",
"output": "6"
},
{
"input": "6.0",
"output": "6"
},
{
"input": "6.1",
"output": "6"
},
{
"input": "6.2",
"output": "6"
},
{
"input": "6.3",
"output": "6"
},
{
"input": "6.4",
"output": "6"
},
{
"input": "6.5",
"output": "7"
},
{
"input": "6.6",
"output": "7"
},
{
"input": "6.7",
"output": "7"
},
{
"input": "6.8",
"output": "7"
},
{
"input": "6.9",
"output": "7"
},
{
"input": "7.0",
"output": "7"
},
{
"input": "7.1",
"output": "7"
},
{
"input": "7.2",
"output": "7"
},
{
"input": "7.3",
"output": "7"
},
{
"input": "7.4",
"output": "7"
},
{
"input": "7.5",
"output": "8"
},
{
"input": "7.6",
"output": "8"
},
{
"input": "7.7",
"output": "8"
},
{
"input": "7.8",
"output": "8"
},
{
"input": "7.9",
"output": "8"
},
{
"input": "8.0",
"output": "8"
},
{
"input": "8.1",
"output": "8"
},
{
"input": "8.2",
"output": "8"
},
{
"input": "8.3",
"output": "8"
},
{
"input": "8.4",
"output": "8"
},
{
"input": "8.5",
"output": "9"
},
{
"input": "8.6",
"output": "9"
},
{
"input": "8.7",
"output": "9"
},
{
"input": "8.8",
"output": "9"
},
{
"input": "8.9",
"output": "9"
},
{
"input": "9.0",
"output": "GOTO Vasilisa."
},
{
"input": "9.1",
"output": "GOTO Vasilisa."
},
{
"input": "9.2",
"output": "GOTO Vasilisa."
},
{
"input": "9.3",
"output": "GOTO Vasilisa."
},
{
"input": "9.4",
"output": "GOTO Vasilisa."
},
{
"input": "9.5",
"output": "GOTO Vasilisa."
},
{
"input": "9.6",
"output": "GOTO Vasilisa."
},
{
"input": "9.7",
"output": "GOTO Vasilisa."
},
{
"input": "9.8",
"output": "GOTO Vasilisa."
},
{
"input": "9.9",
"output": "GOTO Vasilisa."
},
{
"input": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098.1664915772983166314490532653577560222779830866949001942720729759794777105570672781798092416748052690224813237139640723361527601154465287615917169132637313918577673651098507390501962",
"output": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098"
},
{
"input": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394.20211856263503281388748282682120712214711232598021393495443628276945042110862480888110959179019986486690931930108026302665438087068150666835901617457150158918705186964935221768346957536540345814875615118637945520917367155931078965",
"output": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394"
},
{
"input": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616067.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616068"
},
{
"input": "718130341896330596635811874410345440628950330.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "718130341896330596635811874410345440628950331"
},
{
"input": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280722.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280723"
},
{
"input": "68289614863244584294178637364598054554769889.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "7536521504744364134984603189602839063535643888645969434165019366202558753840519.4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
},
{
"input": "646188694587964249318078225173.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "646188694587964249318078225174"
},
{
"input": "7017472758334494297677068672222822550374695787171163207025095950296957261530.50",
"output": "7017472758334494297677068672222822550374695787171163207025095950296957261531"
},
{
"input": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774368.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774369"
},
{
"input": "682500858233333594535201113441004740771119672961581796618069185960107115823662126812159957094407454522028503739299.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "41203422675619090661099806687619.49999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
}
] | 1,634,939,167
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 126
| 186
| 17,715,200
|
i, f = [int(i) for i in input().split(".")]
if int(str(i)[-1]) == 9:
print('GOTO Vasilisa.')
elif int(str(f)[0]) < 5:
print(i)
else:
print(int(i) + 1)
|
Title: Help Far Away Kingdom
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
Input Specification:
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
Output Specification:
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
Demo Input:
['0.0\n', '1.49\n', '1.50\n', '2.71828182845904523536\n', '3.14159265358979323846\n', '12345678901234567890.1\n', '123456789123456789.999\n']
Demo Output:
['0', '1', '2', '3', '3', '12345678901234567890', 'GOTO Vasilisa.']
Note:
none
|
```python
i, f = [int(i) for i in input().split(".")]
if int(str(i)[-1]) == 9:
print('GOTO Vasilisa.')
elif int(str(f)[0]) < 5:
print(i)
else:
print(int(i) + 1)
```
| 0
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,559,871,580
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 0
|
if __name__ == '__main__':
n = int(input().strip())
matrix = []
for i in range(n):
k = [1]
matrix.append(k)
for j in range(1, n):
k.append(0)
matrix[0] = [1 for i in range(n)]
for i in range(1, n):
for j in range(1, n):
matrix[i][j] = matrix[i-1][j] + matrix[i][j-1]
print(matrix[n-1][n-1])
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
if __name__ == '__main__':
n = int(input().strip())
matrix = []
for i in range(n):
k = [1]
matrix.append(k)
for j in range(1, n):
k.append(0)
matrix[0] = [1 for i in range(n)]
for i in range(1, n):
for j in range(1, n):
matrix[i][j] = matrix[i-1][j] + matrix[i][j-1]
print(matrix[n-1][n-1])
```
| 0
|
|
556
|
A
|
Case of the Zeros and Ones
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
|
Output the minimum length of the string that may remain after applying the described operations several times.
|
[
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] |
[
"0\n",
"1\n",
"6\n"
] |
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 250
|
[
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,664,376,877
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 62
| 512,000
|
n=int(input())
s=input()
one=0
for i in s:
if(i=='1'):
one+=1
print((n - (min(one, n - one) * 2)))
|
Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n=int(input())
s=input()
one=0
for i in s:
if(i=='1'):
one+=1
print((n - (min(one, n - one) * 2)))
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,674,949,706
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
s1 = [*input()]
s2 = [*input()]
res = ""
for c1, c2 in zip(s1, s2):
if c1 == c2:
if c1 == "1": res += c1
else: res += c2
else: res += "1"
print(res)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
s1 = [*input()]
s2 = [*input()]
res = ""
for c1, c2 in zip(s1, s2):
if c1 == c2:
if c1 == "1": res += c1
else: res += c2
else: res += "1"
print(res)
```
| 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"
}
] | 1,678,714,972
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 31
| 0
|
b=int(input())
s=list(map(int,input().split()))
s=sorted(list(set(s)))
for i in range(b-2):
if s[i]+1==s[i+1] and s[i+1]+1==s[i+2] and s[i]+2==s[i+2]:
print('YES')
exit()
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
b=int(input())
s=list(map(int,input().split()))
s=sorted(list(set(s)))
for i in range(b-2):
if s[i]+1==s[i+1] and s[i+1]+1==s[i+2] and s[i]+2==s[i+2]:
print('YES')
exit()
print('NO')
```
| -1
|
|
230
|
B
|
T-primes
|
PROGRAMMING
| 1,300
|
[
"binary search",
"implementation",
"math",
"number theory"
] | null | null |
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
|
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
|
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
|
[
"3\n4 5 6\n"
] |
[
"YES\nNO\nNO\n"
] |
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
| 500
|
[
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,697,294,361
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
import math
# 化院 荆屹然 2300011884
def find_max():
global max_num
for k in range(0, n):
if array[k] > max_num:
max_num = array[k]
def euler():
for k in range(2, max_num + 1):
if if_primes[k]:
primes.append(k)
for prime in primes:
if prime * k <= max_num:
if_primes[prime * k] = False
else:
break
else:
for prime in primes:
if prime * k <= max_num:
if_primes[prime * k] = False
else:
break
if k % prime == 0:
break
def if_t_prime(num: int):
root = math.floor(math.sqrt(num))
if root ** 2 == num and if_primes[root]:
return True
else:
return False
n = int(input())
array = list(map(int, input().split()))
max_num = 0
find_max()
if_primes: list[bool] = [False] * 2 + [True] * (max_num - 1)
primes = []
euler()
for i in range(0, n):
if array[i] != 1 and if_t_prime(array[i]):
print("YES")
else:
print("NO")
|
Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
|
```python
import math
# 化院 荆屹然 2300011884
def find_max():
global max_num
for k in range(0, n):
if array[k] > max_num:
max_num = array[k]
def euler():
for k in range(2, max_num + 1):
if if_primes[k]:
primes.append(k)
for prime in primes:
if prime * k <= max_num:
if_primes[prime * k] = False
else:
break
else:
for prime in primes:
if prime * k <= max_num:
if_primes[prime * k] = False
else:
break
if k % prime == 0:
break
def if_t_prime(num: int):
root = math.floor(math.sqrt(num))
if root ** 2 == num and if_primes[root]:
return True
else:
return False
n = int(input())
array = list(map(int, input().split()))
max_num = 0
find_max()
if_primes: list[bool] = [False] * 2 + [True] * (max_num - 1)
primes = []
euler()
for i in range(0, n):
if array[i] != 1 and if_t_prime(array[i]):
print("YES")
else:
print("NO")
```
| -1
|
|
961
|
B
|
Lecture Sleep
|
PROGRAMMING
| 1,200
|
[
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null |
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
|
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
|
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
|
[
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] |
[
"16\n"
] |
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
| 0
|
[
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "20"
},
{
"input": "5 2\n1 2 3 4 20\n0 0 0 1 0",
"output": "24"
},
{
"input": "3 1\n1 2 3\n0 0 1",
"output": "5"
},
{
"input": "4 2\n4 5 6 8\n1 0 1 0",
"output": "18"
},
{
"input": "6 3\n1 3 5 2 1 15\n1 1 0 1 0 0",
"output": "22"
},
{
"input": "5 5\n1 2 3 4 5\n1 1 1 0 1",
"output": "15"
},
{
"input": "3 3\n3 3 3\n1 0 1",
"output": "9"
},
{
"input": "5 5\n500 44 3 4 50\n1 0 0 0 0",
"output": "601"
},
{
"input": "2 2\n3 2\n1 0",
"output": "5"
},
{
"input": "7 6\n4 9 1 7 1 8 4\n0 0 0 1 0 1 0",
"output": "30"
},
{
"input": "4 3\n6 5 9 6\n1 1 0 1",
"output": "26"
},
{
"input": "2 1\n3 2\n0 0",
"output": "3"
},
{
"input": "1 1\n10\n0",
"output": "10"
},
{
"input": "2 1\n3 2\n1 0",
"output": "5"
},
{
"input": "4 2\n3 6 7 2\n0 0 1 1",
"output": "18"
},
{
"input": "10 5\n3 5 9 2 5 9 3 8 8 1\n0 1 1 1 0 1 0 0 0 0",
"output": "49"
},
{
"input": "10 4\n9 5 6 4 3 9 5 1 10 7\n0 0 0 0 0 0 1 0 0 1",
"output": "36"
},
{
"input": "9 8\n3 3 7 7 1 9 10 7 1\n1 1 1 1 1 1 1 1 1",
"output": "48"
},
{
"input": "2 1\n3 4\n0 0",
"output": "4"
},
{
"input": "2 1\n3 2\n0 1",
"output": "5"
},
{
"input": "10 1\n6 6 8 7 6 6 3 2 5 6\n0 0 1 0 0 1 0 1 1 1",
"output": "34"
},
{
"input": "3 2\n10 10 6\n0 0 0",
"output": "20"
},
{
"input": "6 3\n1 3 5 2 5 4\n1 1 1 1 1 1",
"output": "20"
},
{
"input": "10 5\n1 1 1 1 1 1 1 1 10000 1\n1 1 1 1 1 1 1 1 0 1",
"output": "10009"
}
] | 1,620,148,840
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 233
| 6,348,800
|
l1 = [int(x) for x in input().split()]
n = l1[0]
k = l1[1]
l2 = [int(x) for x in input().split()]
l3 = [int(x) for x in input().split()]
res = 0
for i in range(0,n):
if l3[i] == 1:
res += l2[i]
l2[i] = 0
if k==0:
print(res)
else:
prefix_sum = [0 for i in range(0,n+1)]
for i in range(0,n):
prefix_sum[i] = prefix_sum[i-1] + l2[i-1]
val = 0
for i in range(0,n-k+1):
val = max(val,prefix_sum[i+k] - prefix_sum[i-1])
print(res + val)
|
Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
|
```python
l1 = [int(x) for x in input().split()]
n = l1[0]
k = l1[1]
l2 = [int(x) for x in input().split()]
l3 = [int(x) for x in input().split()]
res = 0
for i in range(0,n):
if l3[i] == 1:
res += l2[i]
l2[i] = 0
if k==0:
print(res)
else:
prefix_sum = [0 for i in range(0,n+1)]
for i in range(0,n):
prefix_sum[i] = prefix_sum[i-1] + l2[i-1]
val = 0
for i in range(0,n-k+1):
val = max(val,prefix_sum[i+k] - prefix_sum[i-1])
print(res + val)
```
| 0
|
|
979
|
A
|
Pizza, Pizza, Pizza!!!
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
|
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
|
A single integer — the number of straight cuts Shiro needs.
|
[
"3\n",
"4\n"
] |
[
"2",
"5"
] |
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
| 500
|
[
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "3754606979"
},
{
"input": "99999999999999999",
"output": "50000000000000000"
},
{
"input": "21",
"output": "11"
},
{
"input": "712394453192",
"output": "712394453193"
},
{
"input": "172212168",
"output": "172212169"
},
{
"input": "822981260158260519",
"output": "411490630079130260"
},
{
"input": "28316250877914571",
"output": "14158125438957286"
},
{
"input": "779547116602436424",
"output": "779547116602436425"
},
{
"input": "578223540024979436",
"output": "578223540024979437"
},
{
"input": "335408917861648766",
"output": "335408917861648767"
},
{
"input": "74859962623690078",
"output": "74859962623690079"
},
{
"input": "252509054433933439",
"output": "126254527216966720"
},
{
"input": "760713016476190622",
"output": "760713016476190623"
},
{
"input": "919845426262703496",
"output": "919845426262703497"
},
{
"input": "585335723211047194",
"output": "585335723211047195"
},
{
"input": "522842184971407769",
"output": "261421092485703885"
},
{
"input": "148049062628894320",
"output": "148049062628894321"
},
{
"input": "84324828731963974",
"output": "84324828731963975"
},
{
"input": "354979173822804781",
"output": "177489586911402391"
},
{
"input": "1312150450968413",
"output": "656075225484207"
},
{
"input": "269587449430302150",
"output": "269587449430302151"
},
{
"input": "645762258982631926",
"output": "645762258982631927"
},
{
"input": "615812229161735895",
"output": "307906114580867948"
},
{
"input": "0",
"output": "0"
},
{
"input": "349993004923078531",
"output": "174996502461539266"
},
{
"input": "891351282707723851",
"output": "445675641353861926"
},
{
"input": "563324731189330734",
"output": "563324731189330735"
},
{
"input": "520974001910286909",
"output": "260487000955143455"
},
{
"input": "666729339802329204",
"output": "666729339802329205"
},
{
"input": "856674611404539671",
"output": "428337305702269836"
},
{
"input": "791809296303238499",
"output": "395904648151619250"
},
{
"input": "711066337317063338",
"output": "711066337317063339"
},
{
"input": "931356503492686566",
"output": "931356503492686567"
},
{
"input": "234122432773361866",
"output": "234122432773361867"
},
{
"input": "1000000000000000000",
"output": "1000000000000000001"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "7",
"output": "4"
},
{
"input": "63",
"output": "32"
},
{
"input": "24",
"output": "25"
},
{
"input": "8",
"output": "9"
},
{
"input": "15",
"output": "8"
}
] | 1,570,214,341
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 28
| 124
| 0
|
n=int(input())
if (n+1)%2==0:
print((n+1)//2)
elif (n+1)%2==1:
print(n+1)
|
Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
|
```python
n=int(input())
if (n+1)%2==0:
print((n+1)//2)
elif (n+1)%2==1:
print(n+1)
```
| 0
|
|
486
|
A
|
Calculating Function
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
|
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
|
Print *f*(*n*) in a single line.
|
[
"4\n",
"5\n"
] |
[
"2\n",
"-3\n"
] |
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
| 500
|
[
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,691,246,070
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 46
| 0
|
n: int = int(input())
ans = (n // 2) - [0, n][n % 2]
print(ans)
|
Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3
|
```python
n: int = int(input())
ans = (n // 2) - [0, n][n % 2]
print(ans)
```
| 3
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,615,463,027
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 1,808
| 409,600
|
def getSum(n):
sm = 0
for digit in str(n):
sm += int(digit)
return sm
n = input()
num = int(n)
sm = getSum(num)
ans = 0
while len(n) > 1:
num = sm
n = str(num)
sm = getSum(num)
ans += 1
print(ans)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
def getSum(n):
sm = 0
for digit in str(n):
sm += int(digit)
return sm
n = input()
num = int(n)
sm = getSum(num)
ans = 0
while len(n) > 1:
num = sm
n = str(num)
sm = getSum(num)
ans += 1
print(ans)
```
| 3.547263
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,696,362,690
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
c=0
for x in range(n):
if (n>=1) and (n<=1000):
a=(input())
if (a[0]==1) or (a[2]==1) or (a[4]==1):
c+=1
elif (a[0]==1) or (a[2]==1) or (a[4]==0):
c+=1
elif (a[0]==1) or (a[2]==0) or (a[4]==1):
c+=1
elif (a[0]==0) or (a[2]==1) or (a[4]==1):
c+=1
elif (a[0]==0) and (a[2]==0) or (a[4]==1):
pass
elif (a[0]==0) or (a[2]==1) or (a[4]==0):
pass
elif (a[0]==1) or (a[2]==0) or (a[4]==0):
pass
print(c)
else:
print("unexpected input.")
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n=int(input())
c=0
for x in range(n):
if (n>=1) and (n<=1000):
a=(input())
if (a[0]==1) or (a[2]==1) or (a[4]==1):
c+=1
elif (a[0]==1) or (a[2]==1) or (a[4]==0):
c+=1
elif (a[0]==1) or (a[2]==0) or (a[4]==1):
c+=1
elif (a[0]==0) or (a[2]==1) or (a[4]==1):
c+=1
elif (a[0]==0) and (a[2]==0) or (a[4]==1):
pass
elif (a[0]==0) or (a[2]==1) or (a[4]==0):
pass
elif (a[0]==1) or (a[2]==0) or (a[4]==0):
pass
print(c)
else:
print("unexpected input.")
```
| -1
|
|
965
|
C
|
Greedy Arkady
|
PROGRAMMING
| 2,000
|
[
"math"
] | null | null |
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
|
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
|
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
|
[
"20 4 5 2\n",
"30 9 4 1\n"
] |
[
"8\n",
"4\n"
] |
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
| 1,500
|
[
{
"input": "20 4 5 2",
"output": "8"
},
{
"input": "30 9 4 1",
"output": "4"
},
{
"input": "2 2 1 1",
"output": "1"
},
{
"input": "42 20 5 29",
"output": "5"
},
{
"input": "1000000000000000000 135 1000000000000000 1000",
"output": "8325624421831635"
},
{
"input": "100 33 100 100",
"output": "100"
},
{
"input": "1000000000 1000000000 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 32428 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 324934 1000 1000",
"output": "4000"
},
{
"input": "1000000000000000000 32400093004 10000000 1000",
"output": "40000000"
},
{
"input": "885 2 160 842",
"output": "504"
},
{
"input": "216 137 202 208",
"output": "202"
},
{
"input": "72 66 28 9",
"output": "28"
},
{
"input": "294 4 13 8",
"output": "80"
},
{
"input": "9 2 2 3",
"output": "4"
},
{
"input": "31 3 2 8",
"output": "10"
},
{
"input": "104 2 5 11",
"output": "50"
},
{
"input": "1000000000000000000 1000000000000000000 1000 1000",
"output": "1000"
},
{
"input": "1000000000000000000 100000000000000000 1 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 23925738098196565 1000",
"output": "23925738098196565"
},
{
"input": "576460752303423488 576460752303423488 351082447248993993 1000",
"output": "351082447248993993"
},
{
"input": "962768465676381898 72057594037927936 586039918340257175 256",
"output": "586039918340257175"
},
{
"input": "1000000000000000000 1000000000000000000 10 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 1 1000",
"output": "1"
},
{
"input": "1000000000000000000 1000000000000000000 1 1000",
"output": "1"
}
] | 1,524,726,972
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 78
| 7,065,600
|
n, k, M, D = map(int, input().split())
res = 0
for day in range(1, D + 1):
ma = n // ((day-1) * k + 1)
# print(day, ' -- ', mi, ma)
if ma <= M :
res = max(res, ma * day)
if (n//M - 1)//k + 1 <= D :
res = max(res, M * ((n//M-1)//k + 1))
print(res)
|
Title: Greedy Arkady
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
Input Specification:
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
Output Specification:
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
Demo Input:
['20 4 5 2\n', '30 9 4 1\n']
Demo Output:
['8\n', '4\n']
Note:
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
|
```python
n, k, M, D = map(int, input().split())
res = 0
for day in range(1, D + 1):
ma = n // ((day-1) * k + 1)
# print(day, ' -- ', mi, ma)
if ma <= M :
res = max(res, ma * day)
if (n//M - 1)//k + 1 <= D :
res = max(res, M * ((n//M-1)//k + 1))
print(res)
```
| 3
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,609,928,248
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 218
| 0
|
import sys
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
arr = get_ints()
n = arr[0]
m = arr[1]
arr = get_ints()
arr = sorted(arr)
ans = 0
for i in range(m):
ans += ( arr[i] * -1)
print(ans)
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
import sys
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
arr = get_ints()
n = arr[0]
m = arr[1]
arr = get_ints()
arr = sorted(arr)
ans = 0
for i in range(m):
ans += ( arr[i] * -1)
print(ans)
```
| 0
|
980
|
E
|
The Number Games
|
PROGRAMMING
| 2,200
|
[
"data structures",
"greedy",
"trees"
] | null | null |
The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant.
The nation has $n$ districts numbered from $1$ to $n$, each district has exactly one path connecting it to every other district. The number of fans of a contestant from district $i$ is equal to $2^i$.
This year, the president decided to reduce the costs. He wants to remove $k$ contestants from the games. However, the districts of the removed contestants will be furious and will not allow anyone to cross through their districts.
The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants.
Which contestants should the president remove?
|
The first line of input contains two integers $n$ and $k$ ($1 \leq k < n \leq 10^6$) — the number of districts in Panel, and the number of contestants the president wishes to remove, respectively.
The next $n-1$ lines each contains two integers $a$ and $b$ ($1 \leq a, b \leq n$, $a \ne b$), that describe a road that connects two different districts $a$ and $b$ in the nation. It is guaranteed that there is exactly one path between every two districts.
|
Print $k$ space-separated integers: the numbers of the districts of which the contestants should be removed, in increasing order of district number.
|
[
"6 3\n2 1\n2 6\n4 2\n5 6\n2 3\n",
"8 4\n2 6\n2 7\n7 8\n1 2\n3 1\n2 4\n7 5\n"
] |
[
"1 3 4\n",
"1 3 4 5\n"
] |
In the first sample, the maximum possible total number of fans is $2^2 + 2^5 + 2^6 = 100$. We can achieve it by removing the contestants of the districts 1, 3, and 4.
| 2,500
|
[
{
"input": "6 3\n2 1\n2 6\n4 2\n5 6\n2 3",
"output": "1 3 4"
},
{
"input": "8 4\n2 6\n2 7\n7 8\n1 2\n3 1\n2 4\n7 5",
"output": "1 3 4 5"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "3 1\n2 1\n2 3",
"output": "1"
},
{
"input": "3 2\n1 3\n1 2",
"output": "1 2"
},
{
"input": "4 2\n4 2\n1 4\n3 2",
"output": "1 3"
},
{
"input": "15 3\n9 11\n11 8\n7 9\n9 14\n12 8\n10 7\n1 14\n1 5\n12 15\n10 3\n5 2\n13 15\n4 13\n6 4",
"output": "1 2 5"
},
{
"input": "15 12\n2 3\n2 14\n4 3\n4 10\n3 5\n1 4\n1 12\n4 15\n3 9\n10 7\n11 2\n12 8\n15 13\n1 6",
"output": "1 2 3 5 6 7 8 9 10 11 12 14"
},
{
"input": "32 16\n32 8\n11 32\n22 8\n22 17\n22 3\n16 22\n8 12\n22 7\n8 27\n11 6\n32 4\n9 8\n10 22\n22 31\n1 22\n21 11\n22 15\n14 32\n32 30\n22 29\n24 11\n18 11\n25 32\n13 8\n2 32\n28 8\n32 5\n11 20\n11 19\n22 23\n26 32",
"output": "1 2 3 4 5 6 7 9 10 12 13 14 15 16 17 18"
},
{
"input": "32 1\n30 25\n30 8\n8 22\n22 20\n21 20\n6 21\n29 6\n4 29\n2 4\n13 2\n1 13\n1 11\n11 24\n31 24\n31 15\n15 14\n27 14\n16 27\n5 16\n12 5\n9 12\n9 18\n3 18\n3 17\n17 19\n19 32\n32 10\n10 26\n7 26\n7 23\n23 28",
"output": "25"
},
{
"input": "32 2\n7 20\n15 20\n7 18\n31 20\n28 15\n20 25\n27 31\n27 6\n27 12\n6 16\n22 6\n21 22\n13 6\n16 5\n23 5\n23 26\n23 24\n23 17\n24 14\n17 4\n29 4\n2 24\n30 29\n1 29\n8 4\n30 32\n11 2\n32 3\n1 9\n11 10\n19 32",
"output": "3 8"
},
{
"input": "64 46\n52 22\n38 52\n28 38\n46 38\n30 38\n30 37\n7 37\n37 48\n48 27\n2 7\n27 11\n32 2\n32 35\n8 11\n59 32\n58 59\n59 24\n58 40\n40 4\n40 49\n40 41\n49 16\n9 16\n5 9\n12 9\n9 62\n3 62\n12 63\n63 25\n64 63\n63 42\n36 42\n45 36\n14 45\n53 45\n39 53\n60 39\n50 39\n50 23\n23 10\n19 50\n20 10\n19 56\n21 56\n31 56\n44 31\n26 31\n1 31\n43 44\n15 26\n15 6\n6 29\n47 6\n18 6\n29 55\n18 54\n33 54\n61 55\n54 17\n61 34\n13 17\n17 51\n57 34",
"output": "1 2 3 4 5 6 7 8 10 11 13 14 15 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 37 38 43 44 46 47 48 51 52 54 55 56 57 61"
},
{
"input": "64 63\n11 51\n64 11\n64 39\n11 6\n24 64\n51 63\n43 51\n64 29\n6 22\n47 6\n1 47\n41 22\n41 38\n4 47\n35 47\n41 23\n45 1\n52 35\n26 45\n15 35\n21 35\n23 32\n12 21\n21 62\n25 21\n28 25\n15 54\n57 12\n20 28\n48 57\n8 57\n20 44\n46 28\n18 57\n8 9\n27 9\n48 34\n50 46\n9 56\n9 61\n7 56\n19 61\n61 17\n42 17\n19 58\n58 30\n7 14\n7 59\n31 19\n17 5\n55 14\n13 31\n2 31\n59 40\n55 49\n2 60\n2 53\n36 2\n37 13\n2 16\n53 33\n37 10\n60 3",
"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"
},
{
"input": "64 21\n33 48\n33 64\n55 48\n11 55\n18 11\n62 55\n48 57\n17 62\n64 52\n30 48\n49 48\n23 18\n13 11\n9 64\n11 5\n11 45\n2 55\n62 56\n64 36\n33 10\n34 18\n55 43\n3 48\n60 62\n48 32\n33 8\n62 41\n44 11\n33 46\n15 33\n14 62\n11 6\n64 31\n62 20\n64 25\n62 51\n53 33\n62 40\n18 4\n16 62\n21 11\n33 22\n37 33\n59 62\n55 7\n27 64\n24 64\n39 64\n19 62\n54 55\n33 61\n47 48\n58 55\n42 55\n38 11\n50 18\n28 48\n11 26\n11 1\n62 35\n64 29\n12 33\n18 63",
"output": "1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 19 20 21 22 23"
}
] | 1,525,839,486
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 77
| 7,065,600
|
class PathNode:
def __init__(self, node,length):
self.node = node
self.length = length
self.pathway = []
class Node:
def __init__(self, id):
self.number = id
self.links = []
def link_count(self):
return len(self.links)
def add_link(self, link):
self.links.append(link)
def node_for_id(id):
for n in nodes:
if n.number == id:
return n
def node_index_for_id(id):
for i,n in enumerate(nodes):
if n.number == id:
return i
def remove_node(id):
node = node_index_for_id(id)
for l in nodes[node].links:
node_for_id(l).links.remove(id)
nodes.pop(node)
n, k = list(map(int,input().split()))
nodes = [Node(i) for i in range(1,n+1)]
for _ in range(n-1):
link = list(map(lambda x : int(x),input().split()))
node_for_id(link[0]).add_link(link[1])
node_for_id(link[1]).add_link(link[0])
steps = []
for _ in range(k):
center_node = None
max_link = 0
for i,node in enumerate(nodes):
if node.link_count()>max_link:
max_link = node.link_count()
center_node = node
states = [PathNode(center_node,0)]
goods = []
while True:
if len(states) ==0:break
fuck = states.pop(0)
if fuck.node.link_count() == 1:
goods.append(fuck)
continue
for l in fuck.node.links:
if l in fuck.pathway: continue
acc = node_for_id(l)
pn = PathNode(acc,fuck.length + acc.number)
pn.pathway.append(fuck.node.number)
states.append(pn)
low = 999999999
good = None
for g in goods:
if g.length < low:
low = g.length
good = g.node
steps.append(good.number)
remove_node(good.number)
steps.sort()
print(' '.join(list(map(str,steps))))
|
Title: The Number Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant.
The nation has $n$ districts numbered from $1$ to $n$, each district has exactly one path connecting it to every other district. The number of fans of a contestant from district $i$ is equal to $2^i$.
This year, the president decided to reduce the costs. He wants to remove $k$ contestants from the games. However, the districts of the removed contestants will be furious and will not allow anyone to cross through their districts.
The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants.
Which contestants should the president remove?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \leq k < n \leq 10^6$) — the number of districts in Panel, and the number of contestants the president wishes to remove, respectively.
The next $n-1$ lines each contains two integers $a$ and $b$ ($1 \leq a, b \leq n$, $a \ne b$), that describe a road that connects two different districts $a$ and $b$ in the nation. It is guaranteed that there is exactly one path between every two districts.
Output Specification:
Print $k$ space-separated integers: the numbers of the districts of which the contestants should be removed, in increasing order of district number.
Demo Input:
['6 3\n2 1\n2 6\n4 2\n5 6\n2 3\n', '8 4\n2 6\n2 7\n7 8\n1 2\n3 1\n2 4\n7 5\n']
Demo Output:
['1 3 4\n', '1 3 4 5\n']
Note:
In the first sample, the maximum possible total number of fans is $2^2 + 2^5 + 2^6 = 100$. We can achieve it by removing the contestants of the districts 1, 3, and 4.
|
```python
class PathNode:
def __init__(self, node,length):
self.node = node
self.length = length
self.pathway = []
class Node:
def __init__(self, id):
self.number = id
self.links = []
def link_count(self):
return len(self.links)
def add_link(self, link):
self.links.append(link)
def node_for_id(id):
for n in nodes:
if n.number == id:
return n
def node_index_for_id(id):
for i,n in enumerate(nodes):
if n.number == id:
return i
def remove_node(id):
node = node_index_for_id(id)
for l in nodes[node].links:
node_for_id(l).links.remove(id)
nodes.pop(node)
n, k = list(map(int,input().split()))
nodes = [Node(i) for i in range(1,n+1)]
for _ in range(n-1):
link = list(map(lambda x : int(x),input().split()))
node_for_id(link[0]).add_link(link[1])
node_for_id(link[1]).add_link(link[0])
steps = []
for _ in range(k):
center_node = None
max_link = 0
for i,node in enumerate(nodes):
if node.link_count()>max_link:
max_link = node.link_count()
center_node = node
states = [PathNode(center_node,0)]
goods = []
while True:
if len(states) ==0:break
fuck = states.pop(0)
if fuck.node.link_count() == 1:
goods.append(fuck)
continue
for l in fuck.node.links:
if l in fuck.pathway: continue
acc = node_for_id(l)
pn = PathNode(acc,fuck.length + acc.number)
pn.pathway.append(fuck.node.number)
states.append(pn)
low = 999999999
good = None
for g in goods:
if g.length < low:
low = g.length
good = g.node
steps.append(good.number)
remove_node(good.number)
steps.sort()
print(' '.join(list(map(str,steps))))
```
| 0
|
|
27
|
A
|
Next Test
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] |
A. Next Test
|
2
|
256
|
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
|
Output the required default value for the next test index.
|
[
"3\n1 7 2\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
{
"input": "7\n3 2 1 6 5 7 4",
"output": "8"
},
{
"input": "8\n2 8 3 7 6 9 1 5",
"output": "4"
},
{
"input": "9\n10 5 9 3 8 7 1 2 4",
"output": "6"
},
{
"input": "10\n7 2 3 8 9 6 5 4 1 10",
"output": "11"
},
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 4 1",
"output": "3"
},
{
"input": "4\n4 2 3 1",
"output": "5"
},
{
"input": "5\n3 1 4 2 5",
"output": "6"
},
{
"input": "6\n1 3 6 7 2 4",
"output": "5"
},
{
"input": "7\n1 5 4 7 2 3 6",
"output": "8"
},
{
"input": "8\n12 1 6 5 2 8 3 4",
"output": "7"
},
{
"input": "9\n3 2 7 5 6 4 1 9 10",
"output": "8"
},
{
"input": "10\n1 7 13 6 5 10 3 8 2 4",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "1\n3000",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3000 1",
"output": "2"
}
] | 1,586,199,578
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 307,200
|
if __name__ == "__main__":
n = int(input())
v = list(map(int, input().split()))
for i in range(n):
if v[i]>0 and v[i]<=n:
v[v[i]-1] = v[i]
flag = 0
for i in range(n):
if v[i] != i+1:
flag = 1
print(i+1)
break
if flag == 0:
print(n)
|
Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
if __name__ == "__main__":
n = int(input())
v = list(map(int, input().split()))
for i in range(n):
if v[i]>0 and v[i]<=n:
v[v[i]-1] = v[i]
flag = 0
for i in range(n):
if v[i] != i+1:
flag = 1
print(i+1)
break
if flag == 0:
print(n)
```
| 0
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
You are given a sequence *a* consisting of *n* integers. Find the maximum possible value of (integer remainder of *a**i* divided by *a**j*), where 1<=≤<=*i*,<=*j*<=≤<=*n* and *a**i*<=≥<=*a**j*.
|
The first line contains integer *n* — the length of the sequence (1<=≤<=*n*<=≤<=2·105).
The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=106).
|
Print the answer to the problem.
|
[
"3\n3 4 5\n"
] |
[
"2\n"
] |
none
| 0
|
[
{
"input": "3\n3 4 5",
"output": "2"
},
{
"input": "3\n1 2 4",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"input": "2\n1000000 999999",
"output": "1"
},
{
"input": "12\n4 4 10 13 28 30 41 43 58 61 70 88",
"output": "30"
},
{
"input": "7\n2 13 22 32 72 91 96",
"output": "27"
},
{
"input": "5\n5 11 12 109 110",
"output": "10"
}
] | 1,691,564,784
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
n = int(input())
s = list(map(int, input().split()))
s.sort(reverse=True)
max_c = 1
for i in range(n):
for j in range(i, n):
c = s[i] % s[j]
if c > max_c:
max_c = c
print(max_c)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a* consisting of *n* integers. Find the maximum possible value of (integer remainder of *a**i* divided by *a**j*), where 1<=≤<=*i*,<=*j*<=≤<=*n* and *a**i*<=≥<=*a**j*.
Input Specification:
The first line contains integer *n* — the length of the sequence (1<=≤<=*n*<=≤<=2·105).
The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=106).
Output Specification:
Print the answer to the problem.
Demo Input:
['3\n3 4 5\n']
Demo Output:
['2\n']
Note:
none
|
```python
n = int(input())
s = list(map(int, input().split()))
s.sort(reverse=True)
max_c = 1
for i in range(n):
for j in range(i, n):
c = s[i] % s[j]
if c > max_c:
max_c = c
print(max_c)
```
| 0
|
|
701
|
A
|
Cards
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
|
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
|
[
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] |
[
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] |
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
| 500
|
[
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\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 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": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n82 46 8 44",
"output": "3 1\n4 2"
},
{
"input": "2\n35 50",
"output": "1 2"
},
{
"input": "8\n24 39 49 38 44 64 44 50",
"output": "1 6\n4 8\n2 3\n5 7"
},
{
"input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14",
"output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15"
},
{
"input": "12\n22 83 2 67 55 12 40 93 83 73 12 28",
"output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5"
},
{
"input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43",
"output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7"
},
{
"input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58",
"output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16"
},
{
"input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62",
"output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12"
},
{
"input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76",
"output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3"
},
{
"input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8",
"output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5"
},
{
"input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41",
"output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36"
},
{
"input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55",
"output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12"
},
{
"input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25",
"output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27"
},
{
"input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30",
"output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10"
},
{
"input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39",
"output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32"
},
{
"input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26",
"output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54"
},
{
"input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30",
"output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10"
},
{
"input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18",
"output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34"
},
{
"input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65",
"output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53"
},
{
"input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50",
"output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14"
},
{
"input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57",
"output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7"
},
{
"input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49",
"output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24"
},
{
"input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69",
"output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9"
},
{
"input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57",
"output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20"
},
{
"input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34",
"output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11"
},
{
"input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43",
"output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3"
},
{
"input": "4\n100 100 1 1",
"output": "3 2\n4 1"
},
{
"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": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"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": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n3 4 4 5",
"output": "1 4\n2 3"
},
{
"input": "4\n1 1 2 2",
"output": "1 4\n2 3"
},
{
"input": "4\n1 2 3 4",
"output": "1 4\n2 3"
}
] | 1,699,245,033
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
x = int(input())
y = list(map(int,input().split()))
n = int(x/2)
p = sum(y)/3
for i in range(0,n):
for j in range(0,x):
if p == y[j]+y[i]:
print(i+1 , j+16)
break
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
|
```python
x = int(input())
y = list(map(int,input().split()))
n = int(x/2)
p = sum(y)/3
for i in range(0,n):
for j in range(0,x):
if p == y[j]+y[i]:
print(i+1 , j+16)
break
```
| 0
|
|
61
|
B
|
Hard Work
|
PROGRAMMING
| 1,300
|
[
"strings"
] |
B. Hard Work
|
2
|
256
|
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
|
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
|
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
|
[
"Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n",
"Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n"
] |
[
"ACC\nACC\nACC\nWA\nACC\nACC\nWA\n",
"WA\nACC\nACC\n"
] |
none
| 1,000
|
[
{
"input": "Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful",
"output": "ACC\nACC\nACC\nWA\nACC\nACC\nWA"
},
{
"input": "Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;",
"output": "WA\nACC\nACC"
},
{
"input": "rr\nrrx\nab\n1\nrabrrrx",
"output": "WA"
},
{
"input": "AB\nBC\nCD\n1\nABCDZZ",
"output": "WA"
},
{
"input": "aa\naaa\nz\n1\naazaaa",
"output": "ACC"
},
{
"input": "aa\naaa\nz\n1\naaazaa",
"output": "ACC"
},
{
"input": "as\nav\nax\n1\n-------",
"output": "WA"
},
{
"input": "a\nab\nb\n1\nabcd",
"output": "WA"
},
{
"input": "c\naba\ncc\n2\nccabac\nabcacc",
"output": "ACC\nWA"
},
{
"input": "ab\na\nb\n1\nabcd",
"output": "WA"
},
{
"input": "ACB\nTB\nAC\n1\nATBACBC",
"output": "WA"
},
{
"input": "cc\naba\ncc\n1\nccabaxx",
"output": "WA"
}
] | 1,643,278,025
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
strings = []
length = 0
for i in range(3):
s = input().replace(';',"")
s = s.replace('_',"")
s = s.lower()
strings.append(s)
length += len(s)
n = int(input())
#print(n)
#print(strings)
concat = []
concat.append(strings[0]+strings[1]+strings[2])
concat.append(strings[0]+strings[2]+strings[1])
concat.append(strings[1]+strings[0]+strings[2])
concat.append(strings[1]+strings[2]+strings[0])
concat.append(strings[2]+strings[0]+strings[1])
concat.append(strings[2]+strings[1]+strings[0])
for i in range(n):
s = input().replace(';',"")
s = s.replace('_',"")
s = s.lower()
#print(s)
if(s in concat):
print('ACC')
else:
print('WA')
|
Title: Hard Work
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input Specification:
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output Specification:
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Demo Input:
['Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n', 'Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n']
Demo Output:
['ACC\nACC\nACC\nWA\nACC\nACC\nWA\n', 'WA\nACC\nACC\n']
Note:
none
|
```python
strings = []
length = 0
for i in range(3):
s = input().replace(';',"")
s = s.replace('_',"")
s = s.lower()
strings.append(s)
length += len(s)
n = int(input())
#print(n)
#print(strings)
concat = []
concat.append(strings[0]+strings[1]+strings[2])
concat.append(strings[0]+strings[2]+strings[1])
concat.append(strings[1]+strings[0]+strings[2])
concat.append(strings[1]+strings[2]+strings[0])
concat.append(strings[2]+strings[0]+strings[1])
concat.append(strings[2]+strings[1]+strings[0])
for i in range(n):
s = input().replace(';',"")
s = s.replace('_',"")
s = s.lower()
#print(s)
if(s in concat):
print('ACC')
else:
print('WA')
```
| 0
|
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"
}
] | 1,694,492,644
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
def square(n, m, a):
x = a
z = a
count = 1
count2 = 1
while a < n:
a += x
count += 1
while z < m:
z += x
count2 += 1
res = count2 * count
return res
|
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
def square(n, m, a):
x = a
z = a
count = 1
count2 = 1
while a < n:
a += x
count += 1
while z < m:
z += x
count2 += 1
res = count2 * count
return res
```
| 0
|
625
|
A
|
Guest From the Past
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math"
] | null | null |
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
|
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
|
[
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] |
[
"2\n",
"2\n"
] |
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
| 750
|
[
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"output": "5"
},
{
"input": "9\n2\n4\n1",
"output": "4"
},
{
"input": "9\n2\n2\n1",
"output": "8"
},
{
"input": "9\n10\n10\n1",
"output": "0"
},
{
"input": "10\n2\n2\n1",
"output": "9"
},
{
"input": "1000000000000000000\n2\n10\n9",
"output": "999999999999999995"
},
{
"input": "501000000000000000\n300000000000000000\n301000000000000000\n100000000000000000",
"output": "2"
},
{
"input": "10\n1\n9\n8",
"output": "10"
},
{
"input": "10\n8\n8\n7",
"output": "3"
},
{
"input": "10\n5\n5\n1",
"output": "2"
},
{
"input": "29\n3\n3\n1",
"output": "14"
},
{
"input": "45\n9\n9\n8",
"output": "37"
},
{
"input": "45\n9\n9\n1",
"output": "5"
},
{
"input": "100\n10\n10\n9",
"output": "91"
},
{
"input": "179\n10\n9\n1",
"output": "22"
},
{
"input": "179\n2\n2\n1",
"output": "178"
},
{
"input": "179\n179\n179\n1",
"output": "1"
},
{
"input": "179\n59\n59\n58",
"output": "121"
},
{
"input": "500\n250\n250\n1",
"output": "2"
},
{
"input": "500\n1\n250\n1",
"output": "500"
},
{
"input": "501\n500\n500\n499",
"output": "2"
},
{
"input": "501\n450\n52\n1",
"output": "9"
},
{
"input": "501\n300\n301\n100",
"output": "2"
},
{
"input": "500\n179\n10\n1",
"output": "55"
},
{
"input": "1000\n500\n10\n9",
"output": "991"
},
{
"input": "1000\n2\n10\n9",
"output": "995"
},
{
"input": "1001\n1000\n1000\n999",
"output": "2"
},
{
"input": "10000\n10000\n10000\n1",
"output": "1"
},
{
"input": "10000\n10\n5000\n4999",
"output": "5500"
},
{
"input": "1000000000\n999999998\n999999999\n999999998",
"output": "3"
},
{
"input": "1000000000\n50\n50\n49",
"output": "999999951"
},
{
"input": "1000000000\n500\n5000\n4999",
"output": "999995010"
},
{
"input": "1000000000\n51\n100\n98",
"output": "499999952"
},
{
"input": "1000000000\n100\n51\n50",
"output": "999999950"
},
{
"input": "1000000000\n2\n5\n4",
"output": "999999998"
},
{
"input": "1000000000000000000\n999999998000000000\n999999999000000000\n999999998000000000",
"output": "3"
},
{
"input": "1000000000\n2\n2\n1",
"output": "999999999"
},
{
"input": "999999999\n2\n999999998\n1",
"output": "499999999"
},
{
"input": "999999999999999999\n2\n2\n1",
"output": "999999999999999998"
},
{
"input": "999999999999999999\n10\n10\n9",
"output": "999999999999999990"
},
{
"input": "999999999999999999\n999999999999999998\n999999999999999998\n999999999999999997",
"output": "2"
},
{
"input": "999999999999999999\n501\n501\n1",
"output": "1999999999999999"
},
{
"input": "999999999999999999\n2\n50000000000000000\n49999999999999999",
"output": "974999999999999999"
},
{
"input": "999999999999999999\n180\n180\n1",
"output": "5586592178770949"
},
{
"input": "1000000000000000000\n42\n41\n1",
"output": "24999999999999999"
},
{
"input": "1000000000000000000\n41\n40\n1",
"output": "25641025641025641"
},
{
"input": "100000000000000000\n79\n100\n25",
"output": "1333333333333333"
},
{
"input": "1\n100\n5\n4",
"output": "0"
},
{
"input": "1000000000000000000\n1000000000000000000\n10000000\n9999999",
"output": "999999999990000001"
},
{
"input": "999999999999999999\n999999999000000000\n900000000000000000\n899999999999999999",
"output": "100000000000000000"
},
{
"input": "13\n10\n15\n11",
"output": "1"
},
{
"input": "1\n1000\n5\n4",
"output": "0"
},
{
"input": "10\n100\n10\n1",
"output": "1"
},
{
"input": "3\n2\n100000\n99999",
"output": "1"
},
{
"input": "4\n2\n4\n2",
"output": "2"
},
{
"input": "5\n3\n6\n4",
"output": "1"
},
{
"input": "1\n7\n65\n49",
"output": "0"
},
{
"input": "10\n20\n100\n99",
"output": "0"
},
{
"input": "10000000000\n10000000000\n9000000000\n8999999999",
"output": "1000000001"
},
{
"input": "90\n30\n101\n100",
"output": "3"
},
{
"input": "999999999999999\n5\n500000000000000\n499999999999999",
"output": "599999999999999"
},
{
"input": "1000000000000000000\n1000000000000000000\n1000000000\n999999999",
"output": "999999999000000001"
},
{
"input": "1\n1000000000000000000\n1000000000\n999999999",
"output": "0"
},
{
"input": "100000000000000000\n100000000000000000\n1000000000\n999999999",
"output": "99999999000000001"
},
{
"input": "100000000000000009\n100\n1000000000000000\n999999999999999",
"output": "99010000000000009"
},
{
"input": "10\n20\n10\n9",
"output": "1"
},
{
"input": "10\n4\n14\n13",
"output": "2"
},
{
"input": "11\n3\n9\n7",
"output": "4"
},
{
"input": "1000000000\n5\n7\n4",
"output": "333333332"
},
{
"input": "12155\n1943\n28717\n24074",
"output": "6"
},
{
"input": "1000000000000000000\n10\n20\n5",
"output": "100000000000000000"
},
{
"input": "98\n33\n440\n314",
"output": "2"
},
{
"input": "1070252292\n57449678\n237309920\n221182550",
"output": "56"
},
{
"input": "100\n3\n102\n101",
"output": "33"
},
{
"input": "100000000000000000\n100000000000000001\n1000000000000000\n999999999999999",
"output": "99000000000000001"
},
{
"input": "66249876257975628\n302307316\n406102416\n182373516",
"output": "296116756"
},
{
"input": "10\n5\n10\n1",
"output": "2"
},
{
"input": "1000000000000000000\n10\n1000000000\n999999998",
"output": "499999999600000000"
}
] | 1,454,841,606
| 5,706
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 1
| 46
| 0
|
n = int(input())
a = int(input())
b = int(input())
c = int(input())
if (a <= b - c):
print(n // a)
exit
otv = 10000000000000000000
razn = b - c
razn1 = n - b + 1
if ((razn1 // razn) >= 0):
otv = razn1 // razn
else:
otv = 0
otv += (n - otv * (b - c)) // a
print(otv)
|
Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
|
```python
n = int(input())
a = int(input())
b = int(input())
c = int(input())
if (a <= b - c):
print(n // a)
exit
otv = 10000000000000000000
razn = b - c
razn1 = n - b + 1
if ((razn1 // razn) >= 0):
otv = razn1 // razn
else:
otv = 0
otv += (n - otv * (b - c)) // a
print(otv)
```
| 0
|
|
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"
}
] | 1,677,660,338
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
def theater(n,m,a):
if n%a==0:
x=n/a
else:
x=n//a+1
if m%a==0:
y=m/a
else:
y=m//a+1
print (int(x*y))
str=input()
j=0
for i in range(len(str)):
if str[i]==" ":
if j==0:
n1=int(str[:i])
j=i+1
else:
m1=int(str[j:i])
a1=int(str[i+1:])
theater(n1,m1,a1)
|
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
def theater(n,m,a):
if n%a==0:
x=n/a
else:
x=n//a+1
if m%a==0:
y=m/a
else:
y=m//a+1
print (int(x*y))
str=input()
j=0
for i in range(len(str)):
if str[i]==" ":
if j==0:
n1=int(str[:i])
j=i+1
else:
m1=int(str[j:i])
a1=int(str[i+1:])
theater(n1,m1,a1)
```
| 3.977
|
431
|
A
|
Black Square
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
|
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104).
The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
|
Print a single integer — the total number of calories that Jury wastes.
|
[
"1 2 3 4\n123214\n",
"1 5 3 2\n11221\n"
] |
[
"13\n",
"13\n"
] |
none
| 500
|
[
{
"input": "1 2 3 4\n123214",
"output": "13"
},
{
"input": "1 5 3 2\n11221",
"output": "13"
},
{
"input": "5 5 5 1\n3422",
"output": "16"
},
{
"input": "4 3 2 1\n2",
"output": "3"
},
{
"input": "5651 6882 6954 4733\n2442313421",
"output": "60055"
},
{
"input": "0 0 0 0\n4132",
"output": "0"
},
{
"input": "3163 5778 83 7640\n11141442444",
"output": "64270"
},
{
"input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442",
"output": "420780"
},
{
"input": "0 0 0 0\n1",
"output": "0"
},
{
"input": "1 2 3 4\n4",
"output": "4"
},
{
"input": "2343 7653 1242 5432\n1",
"output": "2343"
},
{
"input": "2343 7653 1242 5432\n2",
"output": "7653"
},
{
"input": "2343 7653 1242 5432\n3",
"output": "1242"
},
{
"input": "2343 7653 1242 5432\n4",
"output": "5432"
},
{
"input": "1 2 3 4\n123412",
"output": "13"
},
{
"input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111",
"output": "2650"
},
{
"input": "1 2 3 4\n11111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "1 2 3 4\n23123231321231231231231231221232123121312321",
"output": "87"
},
{
"input": "1 2 3 4\n1111111111111222222222233333333333444444444444444",
"output": "126"
},
{
"input": "2 3 1 4\n121321232412342112312313213123123412131231231232",
"output": "105"
}
] | 1,680,056,271
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 78
| 0
|
N=list(map(int,input().split()))
s=input()
sum=0
for c in s:
idx=ord(c)-ord('1')
sum+=N[idx]
print(sum)
|
Title: Black Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
Input Specification:
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104).
The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Output Specification:
Print a single integer — the total number of calories that Jury wastes.
Demo Input:
['1 2 3 4\n123214\n', '1 5 3 2\n11221\n']
Demo Output:
['13\n', '13\n']
Note:
none
|
```python
N=list(map(int,input().split()))
s=input()
sum=0
for c in s:
idx=ord(c)-ord('1')
sum+=N[idx]
print(sum)
```
| 3
|
|
792
|
B
|
Counting-out Rhyme
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
|
The first line contains two integer numbers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=109).
|
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
|
[
"7 5\n10 4 11 4 1\n",
"3 2\n2 5\n"
] |
[
"4 2 5 6 1 \n",
"3 2 \n"
] |
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
| 0
|
[
{
"input": "7 5\n10 4 11 4 1",
"output": "4 2 5 6 1 "
},
{
"input": "3 2\n2 5",
"output": "3 2 "
},
{
"input": "2 1\n1",
"output": "2 "
},
{
"input": "2 1\n2",
"output": "1 "
},
{
"input": "2 1\n3",
"output": "2 "
},
{
"input": "10 7\n5 10 4 3 8 10 6",
"output": "6 8 3 9 2 4 10 "
},
{
"input": "10 8\n12 6 12 15 20 8 17 12",
"output": "3 10 6 8 2 9 4 5 "
},
{
"input": "12 10\n76 58 82 54 97 46 17 40 36 15",
"output": "5 9 12 1 3 10 8 11 2 4 "
},
{
"input": "12 6\n76 61 94 15 66 26",
"output": "5 12 6 2 7 3 "
},
{
"input": "90 10\n1045 8705 6077 3282 1459 9809 383 6206 2674 7274",
"output": "56 39 45 20 17 55 14 85 51 33 "
},
{
"input": "100 30\n601771 913885 829106 91674 465657 367068 142461 873149 294276 916519 720701 370006 551782 321506 68525 570684 81178 724855 564907 661130 10112 983124 799801 100639 766045 862312 513021 232094 979480 408554",
"output": "72 89 16 26 85 73 29 99 63 30 8 46 70 19 100 93 36 54 65 77 17 79 62 64 21 69 42 82 68 1 "
},
{
"input": "3 2\n20148340 81473314",
"output": "2 3 "
},
{
"input": "3 2\n301633543 643389490",
"output": "2 3 "
},
{
"input": "6 5\n532623340 628883728 583960589 690950241 488468353",
"output": "5 3 6 1 4 "
},
{
"input": "6 2\n458995521 294343587",
"output": "4 1 "
},
{
"input": "68 1\n5",
"output": "6 "
}
] | 1,571,034,851
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 22
| 109
| 0
|
import math
n,k= map(int,input().split())
a= list(map(int,input().split()))
l1=[x for x in range(1,n+1)]
j=0
for i in a:
k-=1
j=(j+i)%n
if k!=0:
print(l1[j],end=' ')
else:
print (l1[j],end='')
n-=1
l1.pop(j)
|
Title: Counting-out Rhyme
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
Input Specification:
The first line contains two integer numbers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
Demo Input:
['7 5\n10 4 11 4 1\n', '3 2\n2 5\n']
Demo Output:
['4 2 5 6 1 \n', '3 2 \n']
Note:
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
|
```python
import math
n,k= map(int,input().split())
a= list(map(int,input().split()))
l1=[x for x in range(1,n+1)]
j=0
for i in a:
k-=1
j=(j+i)%n
if k!=0:
print(l1[j],end=' ')
else:
print (l1[j],end='')
n-=1
l1.pop(j)
```
| 3
|
|
204
|
A
|
Little Elephant and Interval
|
PROGRAMMING
| 1,500
|
[
"binary search",
"combinatorics",
"dp"
] | null | null |
The Little Elephant very much loves sums on intervals.
This time he has a pair of integers *l* and *r* (*l*<=≤<=*r*). The Little Elephant has to find the number of such integers *x* (*l*<=≤<=*x*<=≤<=*r*), that the first digit of integer *x* equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not.
Help him and count the number of described numbers *x* for a given pair *l* and *r*.
|
The single line contains a pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018) — the boundaries of the interval.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
On a single line print a single integer — the answer to the problem.
|
[
"2 47\n",
"47 1024\n"
] |
[
"12\n",
"98\n"
] |
In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.
| 500
|
[
{
"input": "2 47",
"output": "12"
},
{
"input": "47 1024",
"output": "98"
},
{
"input": "1 1000",
"output": "108"
},
{
"input": "1 10000",
"output": "1008"
},
{
"input": "47 8545",
"output": "849"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "47547 4587554587754542",
"output": "458755458770699"
},
{
"input": "1 1000000",
"output": "100008"
},
{
"input": "47 74",
"output": "2"
},
{
"input": "10001 10000002",
"output": "999001"
},
{
"input": "10000 100000",
"output": "9000"
},
{
"input": "458754 4588754",
"output": "413001"
},
{
"input": "111 111",
"output": "1"
},
{
"input": "110 147",
"output": "4"
},
{
"input": "1 1000000000",
"output": "100000008"
},
{
"input": "12 10000000000",
"output": "999999998"
},
{
"input": "1000000000 1000000000",
"output": "0"
},
{
"input": "1 1000000000000000000",
"output": "100000000000000008"
},
{
"input": "11 111111111111111100",
"output": "11111111111111109"
},
{
"input": "100000000000000000 1000000000000000000",
"output": "90000000000000000"
},
{
"input": "45481484484 848469844684844",
"output": "84842436320036"
},
{
"input": "975400104587000 48754000000000001",
"output": "4777859989541300"
},
{
"input": "11220451511 51511665251233335",
"output": "5151165403078183"
},
{
"input": "77 77",
"output": "1"
},
{
"input": "99 102",
"output": "2"
},
{
"input": "9997 87878000008",
"output": "8787799002"
},
{
"input": "10000000001 111111111111100001",
"output": "11111110111110001"
},
{
"input": "7777 88888",
"output": "8112"
},
{
"input": "999999999 10000000000",
"output": "900000001"
},
{
"input": "235 236",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "4 7",
"output": "4"
},
{
"input": "7 10",
"output": "3"
},
{
"input": "1 11",
"output": "10"
},
{
"input": "1 10",
"output": "9"
},
{
"input": "7 8",
"output": "2"
},
{
"input": "88 990",
"output": "91"
},
{
"input": "458985985498001244 985458425544874008",
"output": "52647244004687276"
},
{
"input": "115998725487587451 245744899758754501",
"output": "12974617427116705"
},
{
"input": "595754249475458004 615044544745124547",
"output": "1929029526966655"
},
{
"input": "9754875457700 1000000000000000000",
"output": "99999024512454230"
},
{
"input": "8758754570000 999999999999999999",
"output": "99999124124543000"
},
{
"input": "111111111111111111 333333333444444445",
"output": "22222222233333334"
},
{
"input": "822981258385599125 841978899930248528",
"output": "1899764154464941"
},
{
"input": "779547115376367013 980561039207670775",
"output": "20101392383130376"
},
{
"input": "335408916782916802 416495628489807285",
"output": "8108671170689049"
},
{
"input": "252509053898415172 285803555062529649",
"output": "3329450116411448"
},
{
"input": "919845424847912645 970651082117950285",
"output": "5080565727003764"
},
{
"input": "522842183413115088 853628713003942530",
"output": "33078652959082744"
},
{
"input": "84324827171274023 607953653548585226",
"output": "52362882637731121"
},
{
"input": "1312148742261681 277460340506883334",
"output": "27614819176462166"
},
{
"input": "645762257531682046 885295120956158518",
"output": "23953286342447648"
},
{
"input": "819875140559301752 946247219812473271",
"output": "12637207925317152"
},
{
"input": "4 19",
"output": "7"
},
{
"input": "5 45",
"output": "9"
},
{
"input": "9999999999999987 99999999999999711",
"output": "8999999999999973"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1827171 232817181719384635",
"output": "23281718171755747"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "1"
},
{
"input": "73 678",
"output": "61"
}
] | 1,645,718,615
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 17
| 154
| 0
|
a , b =map(int,input().split())
d = str(a)
e = str(b)
c = 0
if a<=10 and b<=10:
print(abs(a-b)+1)
else:
if a<10 and b>10:
c +=9-a
if d[0]>=d[-1]:
if e[0]<=e[-1]:
b -= int(e[-1])
c +=2
a += 10 - int(d[-1])
print(int((b-a)/10+c))
else:
a += 10 - int(d[-1])
b -= int(e[-1])
c+=1
print(int((b-a)/10+c))
elif e[0]<=e[-1]:
a += 10 - int(d[-1])
b -= int(e[-1])
c+=1
print(int((b-a)/10+c))
else:
a += 10 - int(d[-1])
b -= int(e[-1])
print(int((b-a)/10+c))
|
Title: Little Elephant and Interval
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant very much loves sums on intervals.
This time he has a pair of integers *l* and *r* (*l*<=≤<=*r*). The Little Elephant has to find the number of such integers *x* (*l*<=≤<=*x*<=≤<=*r*), that the first digit of integer *x* equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not.
Help him and count the number of described numbers *x* for a given pair *l* and *r*.
Input Specification:
The single line contains a pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018) — the boundaries of the interval.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Output Specification:
On a single line print a single integer — the answer to the problem.
Demo Input:
['2 47\n', '47 1024\n']
Demo Output:
['12\n', '98\n']
Note:
In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.
|
```python
a , b =map(int,input().split())
d = str(a)
e = str(b)
c = 0
if a<=10 and b<=10:
print(abs(a-b)+1)
else:
if a<10 and b>10:
c +=9-a
if d[0]>=d[-1]:
if e[0]<=e[-1]:
b -= int(e[-1])
c +=2
a += 10 - int(d[-1])
print(int((b-a)/10+c))
else:
a += 10 - int(d[-1])
b -= int(e[-1])
c+=1
print(int((b-a)/10+c))
elif e[0]<=e[-1]:
a += 10 - int(d[-1])
b -= int(e[-1])
c+=1
print(int((b-a)/10+c))
else:
a += 10 - int(d[-1])
b -= int(e[-1])
print(int((b-a)/10+c))
```
| 0
|
|
868
|
A
|
Bark to Unlock
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
|
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
|
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
|
[
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring.
| 250
|
[
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
"input": "ba\n4\ncd\nad\ncc\ncb",
"output": "YES"
},
{
"input": "pg\n4\nzl\nxs\ndi\nxn",
"output": "NO"
},
{
"input": "bn\n100\ndf\nyb\nze\nml\nyr\nof\nnw\nfm\ndw\nlv\nzr\nhu\nzt\nlw\nld\nmo\nxz\ntp\nmr\nou\nme\npx\nvp\nes\nxi\nnr\nbx\nqc\ngm\njs\nkn\ntw\nrq\nkz\nuc\nvc\nqr\nab\nna\nro\nya\nqy\ngu\nvk\nqk\ngs\nyq\nop\nhw\nrj\neo\nlz\nbh\nkr\nkb\nma\nrd\nza\nuf\nhq\nmc\nmn\nti\nwn\nsh\nax\nsi\nnd\ntz\ndu\nfj\nkl\nws\now\nnf\nvr\nye\nzc\niw\nfv\nkv\noo\nsm\nbc\nrs\nau\nuz\nuv\ngh\nsu\njn\ndz\nrl\nwj\nbk\nzl\nas\nms\nit\nwu",
"output": "YES"
},
{
"input": "bb\n1\naa",
"output": "NO"
},
{
"input": "qm\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "NO"
},
{
"input": "mq\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "bb\n1\nbb",
"output": "YES"
},
{
"input": "ba\n1\ncc",
"output": "NO"
},
{
"input": "ha\n1\nha",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "ez\n1\njl",
"output": "NO"
},
{
"input": "aa\n2\nab\nba",
"output": "YES"
},
{
"input": "aa\n2\nca\ncc",
"output": "NO"
},
{
"input": "dd\n2\nac\ndc",
"output": "NO"
},
{
"input": "qc\n2\nyc\nkr",
"output": "NO"
},
{
"input": "aa\n3\nba\nbb\nab",
"output": "YES"
},
{
"input": "ca\n3\naa\nbb\nab",
"output": "NO"
},
{
"input": "ca\n3\nbc\nbd\nca",
"output": "YES"
},
{
"input": "dd\n3\nmt\nrg\nxl",
"output": "NO"
},
{
"input": "be\n20\nad\ncd\ncb\ndb\ndd\naa\nab\nca\nae\ned\ndc\nbb\nba\nda\nee\nea\ncc\nac\nec\neb",
"output": "YES"
},
{
"input": "fc\n20\nca\nbb\nce\nfd\nde\nfa\ncc\nec\nfb\nfc\nff\nbe\ncf\nba\ndb\ned\naf\nae\nda\nef",
"output": "YES"
},
{
"input": "ca\n20\ndc\naf\ndf\neg\naa\nbc\nea\nbd\nab\ndb\ngc\nfb\nba\nbe\nee\ngf\ncf\nag\nga\nca",
"output": "YES"
},
{
"input": "ke\n20\nzk\nra\nbq\nqz\nwt\nzg\nmz\nuk\nge\nuv\nud\nfd\neh\ndm\nsk\nki\nfv\ntp\nat\nfb",
"output": "YES"
},
{
"input": "hh\n50\nag\nhg\ndg\nfh\neg\ngh\ngd\nda\nbh\nab\nhf\ndc\nhb\nfe\nad\nec\nac\nfd\nca\naf\ncg\nhd\neb\nce\nhe\nha\ngb\nea\nae\nfb\nff\nbe\nch\nhh\nee\nde\nge\ngf\naa\ngg\neh\ned\nbf\nfc\nah\nga\nbd\ncb\nbg\nbc",
"output": "YES"
},
{
"input": "id\n50\nhi\ndc\nfg\nee\ngi\nhc\nac\nih\ndg\nfc\nde\ned\nie\neb\nic\ncf\nib\nfa\ngc\nba\nbe\nga\nha\nhg\nia\ndf\nab\nei\neh\nad\nii\nci\ndh\nec\nif\ndi\nbg\nag\nhe\neg\nca\nae\ndb\naa\nid\nfh\nhh\ncc\nfb\ngb",
"output": "YES"
},
{
"input": "fe\n50\nje\nbi\nbg\ngc\nfb\nig\ndf\nji\ndg\nfe\nfc\ncf\ngf\nai\nhe\nac\nch\nja\ngh\njf\nge\ncb\nij\ngb\ncg\naf\neh\nee\nhd\njd\njb\nii\nca\nci\nga\nab\nhi\nag\nfj\nej\nfi\nie\ndj\nfg\nef\njc\njg\njh\nhf\nha",
"output": "YES"
},
{
"input": "rn\n50\nba\nec\nwg\nao\nlk\nmz\njj\ncf\nfa\njk\ndy\nsz\njs\nzr\nqv\ntx\nwv\nrd\nqw\nls\nrr\nvt\nrx\nkc\neh\nnj\niq\nyi\nkh\nue\nnv\nkz\nrn\nes\nua\nzf\nvu\nll\neg\nmj\ncz\nzj\nxz\net\neb\nci\nih\nig\nam\nvd",
"output": "YES"
},
{
"input": "ee\n100\nah\nfb\ncd\nbi\nii\nai\nid\nag\nie\nha\ndi\nec\nae\nce\njb\ndg\njg\ngd\ngf\nda\nih\nbd\nhj\ngg\nhb\ndf\ned\nfh\naf\nja\nci\nfc\nic\nji\nac\nhi\nfj\nch\nbc\njd\naa\nff\nad\ngj\nej\nde\nee\nhe\ncf\nga\nia\ncg\nbb\nhc\nbe\ngi\njf\nbg\naj\njj\nbh\nfe\ndj\nef\ngb\nge\ndb\nig\ncj\ndc\nij\njh\nei\ndd\nib\nhf\neg\nbf\nfg\nab\ngc\nfd\nhd\ngh\neh\njc\neb\nhh\nca\nje\nbj\nif\nea\nhg\nfa\ncc\nba\ndh\ncb\nfi",
"output": "YES"
},
{
"input": "if\n100\njd\nbc\nje\nhi\nga\nde\nkb\nfc\ncd\ngd\naj\ncb\nei\nbf\ncf\ndk\ndb\ncg\nki\ngg\nkg\nfa\nkj\nii\njf\njg\ngb\nbh\nbg\neh\nhj\nhb\ndg\ndj\njc\njb\nce\ndi\nig\nci\ndf\nji\nhc\nfk\naf\nac\ngk\nhd\nae\nkd\nec\nkc\neb\nfh\nij\nie\nca\nhh\nkf\nha\ndd\nif\nef\nih\nhg\nej\nfe\njk\nea\nib\nck\nhf\nak\ngi\nch\ndc\nba\nke\nad\nka\neg\njh\nja\ngc\nfd\ncc\nab\ngj\nik\nfg\nbj\nhe\nfj\nge\ngh\nhk\nbk\ned\nid\nfi",
"output": "YES"
},
{
"input": "kd\n100\nek\nea\nha\nkf\nkj\ngh\ndl\nfj\nal\nga\nlj\nik\ngd\nid\ncb\nfh\ndk\nif\nbh\nkb\nhc\nej\nhk\ngc\ngb\nef\nkk\nll\nlf\nkh\ncl\nlh\njj\nil\nhh\nci\ndb\ndf\ngk\njg\nch\nbd\ncg\nfg\nda\neb\nlg\ndg\nbk\nje\nbg\nbl\njl\ncj\nhb\nei\naa\ngl\nka\nfa\nfi\naf\nkc\nla\ngi\nij\nib\nle\ndi\nck\nag\nlc\nca\nge\nie\nlb\nke\nii\nae\nig\nic\nhe\ncf\nhd\nak\nfb\nhi\ngf\nad\nba\nhg\nbi\nkl\nac\ngg\ngj\nbe\nlk\nld\naj",
"output": "YES"
},
{
"input": "ab\n1\nab",
"output": "YES"
},
{
"input": "ya\n1\nya",
"output": "YES"
},
{
"input": "ay\n1\nyb",
"output": "NO"
},
{
"input": "ax\n2\nii\nxa",
"output": "YES"
},
{
"input": "hi\n1\nhi",
"output": "YES"
},
{
"input": "ag\n1\nag",
"output": "YES"
},
{
"input": "th\n1\nth",
"output": "YES"
},
{
"input": "sb\n1\nsb",
"output": "YES"
},
{
"input": "hp\n1\nhp",
"output": "YES"
},
{
"input": "ah\n1\nah",
"output": "YES"
},
{
"input": "ta\n1\nta",
"output": "YES"
},
{
"input": "tb\n1\ntb",
"output": "YES"
},
{
"input": "ab\n5\nca\nda\nea\nfa\nka",
"output": "NO"
},
{
"input": "ac\n1\nac",
"output": "YES"
},
{
"input": "ha\n2\nha\nzz",
"output": "YES"
},
{
"input": "ok\n1\nok",
"output": "YES"
},
{
"input": "bc\n1\nbc",
"output": "YES"
},
{
"input": "az\n1\nzz",
"output": "NO"
},
{
"input": "ab\n2\nba\ntt",
"output": "YES"
},
{
"input": "ah\n2\nap\nhp",
"output": "NO"
},
{
"input": "sh\n1\nsh",
"output": "YES"
},
{
"input": "az\n1\nby",
"output": "NO"
},
{
"input": "as\n1\nas",
"output": "YES"
},
{
"input": "ab\n2\nab\ncd",
"output": "YES"
},
{
"input": "ab\n2\nxa\nza",
"output": "NO"
},
{
"input": "ab\n2\net\nab",
"output": "YES"
},
{
"input": "ab\n1\naa",
"output": "NO"
},
{
"input": "ab\n2\nab\nde",
"output": "YES"
},
{
"input": "ah\n2\nba\nha",
"output": "YES"
},
{
"input": "ha\n3\ndd\ncc\nha",
"output": "YES"
},
{
"input": "oo\n1\nox",
"output": "NO"
},
{
"input": "ab\n2\nax\nbx",
"output": "NO"
},
{
"input": "ww\n4\nuw\now\npo\nko",
"output": "NO"
},
{
"input": "ay\n1\nay",
"output": "YES"
},
{
"input": "yo\n1\nyo",
"output": "YES"
},
{
"input": "ba\n1\nba",
"output": "YES"
},
{
"input": "qw\n1\nqw",
"output": "YES"
},
{
"input": "la\n1\nla",
"output": "YES"
},
{
"input": "ab\n2\nbb\nbc",
"output": "NO"
},
{
"input": "aa\n2\nab\nac",
"output": "NO"
},
{
"input": "ah\n2\nbb\nha",
"output": "YES"
},
{
"input": "ya\n42\nab\nac\nad\nae\naf\nag\nah\nai\nak\naj\nba\nbc\nbd\nbe\nbf\nbg\nbh\nbi\nbk\nbj\ncb\nca\ncd\nce\ncf\ncg\nch\nci\nck\ncj\ndb\ndc\nda\nde\ndf\ndg\ndh\ndi\ndk\ndj\nef\nek",
"output": "NO"
},
{
"input": "ab\n3\nab\nxx\nyy",
"output": "YES"
},
{
"input": "ab\n2\nab\ncc",
"output": "YES"
},
{
"input": "sa\n2\nxx\nas",
"output": "YES"
},
{
"input": "ma\n1\nma",
"output": "YES"
},
{
"input": "ba\n1\nbb",
"output": "NO"
},
{
"input": "bc\n1\nab",
"output": "NO"
},
{
"input": "fa\n1\nfa",
"output": "YES"
},
{
"input": "ap\n1\nap",
"output": "YES"
},
{
"input": "ab\n1\nbb",
"output": "NO"
},
{
"input": "bk\n1\nbk",
"output": "YES"
},
{
"input": "xy\n2\nxy\naa",
"output": "YES"
},
{
"input": "ab\n2\nza\nbz",
"output": "YES"
}
] | 1,608,922,905
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 90
| 155
| 0
|
def bark():
p = input()
lp = list(p)
dict = {}
for i in range(int(input())):
b = input()
bL = len(b)
rB = b[bL::-1]
listB = list(b)
if(b == p):
return "YES"
elif(rB == p):
return "YES"
elif(lp[0] in listB or lp[1] in listB):
dict.update({listB[0] : listB[1]})
if(lp[0] in dict.values() and lp[1] in dict.keys()):
return 'YES'
else:
return 'NO'
ans = bark()
print(ans)
|
Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring.
|
```python
def bark():
p = input()
lp = list(p)
dict = {}
for i in range(int(input())):
b = input()
bL = len(b)
rB = b[bL::-1]
listB = list(b)
if(b == p):
return "YES"
elif(rB == p):
return "YES"
elif(lp[0] in listB or lp[1] in listB):
dict.update({listB[0] : listB[1]})
if(lp[0] in dict.values() and lp[1] in dict.keys()):
return 'YES'
else:
return 'NO'
ans = bark()
print(ans)
```
| 3
|
|
146
|
A
|
Lucky Ticket
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
|
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
|
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
|
[
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] |
[
"NO\n",
"NO\n",
"YES\n"
] |
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number.
| 500
|
[
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
{
"input": "20\n44444444444444444444",
"output": "YES"
},
{
"input": "2\n44",
"output": "YES"
},
{
"input": "10\n4745474547",
"output": "NO"
},
{
"input": "14\n77770004444444",
"output": "NO"
},
{
"input": "10\n4747777744",
"output": "YES"
},
{
"input": "10\n1234567890",
"output": "NO"
},
{
"input": "50\n44444444444444444444444444444444444444444444444444",
"output": "YES"
},
{
"input": "50\n44444444444444444444444444444444444444444444444447",
"output": "NO"
},
{
"input": "50\n74444444444444444444444444444444444444444444444444",
"output": "NO"
},
{
"input": "50\n07777777777777777777777777777777777777777777777770",
"output": "NO"
},
{
"input": "50\n77777777777777777777777777777777777777777777777777",
"output": "YES"
},
{
"input": "50\n44747747774474747747747447777447774747447477444474",
"output": "YES"
},
{
"input": "48\n447474444777444474747747744774447444747474774474",
"output": "YES"
},
{
"input": "32\n74474474777444474444747774474774",
"output": "YES"
},
{
"input": "40\n4747777444447747777447447747447474774777",
"output": "YES"
},
{
"input": "10\n4477477444",
"output": "YES"
},
{
"input": "18\n447747474447744747",
"output": "YES"
},
{
"input": "26\n44747744444774744774474447",
"output": "YES"
},
{
"input": "50\n44707747774474747747747447777447774747447477444474",
"output": "NO"
},
{
"input": "40\n4747777444447737777447447747447474774777",
"output": "NO"
},
{
"input": "36\n764477744747444444447747747474744444",
"output": "NO"
},
{
"input": "22\n4477407474777477744447",
"output": "NO"
},
{
"input": "32\n74274474777444474444747774474774",
"output": "NO"
},
{
"input": "16\n4744447974444747",
"output": "NO"
},
{
"input": "2\n11",
"output": "NO"
},
{
"input": "2\n22",
"output": "NO"
},
{
"input": "2\n33",
"output": "NO"
},
{
"input": "2\n74",
"output": "NO"
},
{
"input": "2\n55",
"output": "NO"
},
{
"input": "2\n66",
"output": "NO"
},
{
"input": "2\n77",
"output": "YES"
},
{
"input": "2\n88",
"output": "NO"
},
{
"input": "2\n99",
"output": "NO"
},
{
"input": "4\n4004",
"output": "NO"
},
{
"input": "14\n00077774444444",
"output": "NO"
},
{
"input": "6\n004444",
"output": "NO"
},
{
"input": "4\n0044",
"output": "NO"
},
{
"input": "14\n77771114444444",
"output": "NO"
},
{
"input": "14\n44444447777000",
"output": "NO"
},
{
"input": "6\n004774",
"output": "NO"
}
] | 1,586,590,097
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 46
| 280
| 0
|
l = int(input()); a = str(input())
if l % 2 != 0:
print("NO")
else:
f1 = 0
f2 = 0
s1 = 0
s2 = 0
y = 0
while y < int(l / 2):
if a[y] == '4':
f1 += 1
elif a[y] == '7':
s1 += 1
else:
print("NO")
exit()
y += 1
y = int(l / 2)
while y < l:
if a[y] == '4':
f2 += 1
elif a[y] == '7':
s2 += 1
else:
print("NO")
exit()
y += 1
if f1 == f2 and s1 == s2:
print("YES")
else: print("NO")
|
Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number.
|
```python
l = int(input()); a = str(input())
if l % 2 != 0:
print("NO")
else:
f1 = 0
f2 = 0
s1 = 0
s2 = 0
y = 0
while y < int(l / 2):
if a[y] == '4':
f1 += 1
elif a[y] == '7':
s1 += 1
else:
print("NO")
exit()
y += 1
y = int(l / 2)
while y < l:
if a[y] == '4':
f2 += 1
elif a[y] == '7':
s2 += 1
else:
print("NO")
exit()
y += 1
if f1 == f2 and s1 == s2:
print("YES")
else: print("NO")
```
| 3
|
|
602
|
B
|
Approximating a Constant Range
|
PROGRAMMING
| 1,400
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
|
Print a single number — the maximum length of an almost constant range of the given sequence.
|
[
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
| 1,000
|
[
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 2 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "18\n10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "4"
},
{
"input": "4\n4 3 2 3",
"output": "3"
},
{
"input": "4\n8 8 7 7",
"output": "4"
},
{
"input": "3\n99998 99999 100000",
"output": "2"
},
{
"input": "3\n100000 99999 99998",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "2\n99999 100000",
"output": "2"
},
{
"input": "2\n100000 100000",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "15\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "15"
}
] | 1,691,772,830
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
def max_almost_constant_range(n, data):
left = 0
max_len = 0
freq = {} # To keep track of frequency of values
for right in range(n):
if data[right] not in freq:
freq[data[right]] = 0
freq[data[right]] += 1
while max(freq.values()) - min(freq.values()) > 1:
freq[data[left]] -= 1
if freq[data[left]] == 0:
del freq[data[left]]
left += 1
max_len = max(max_len, right - left + 1)
return max_len
n = int(input())
data = list(map(int, input().split()))
result = max_almost_constant_range(n, data)
print(result)
|
Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
|
```python
def max_almost_constant_range(n, data):
left = 0
max_len = 0
freq = {} # To keep track of frequency of values
for right in range(n):
if data[right] not in freq:
freq[data[right]] = 0
freq[data[right]] += 1
while max(freq.values()) - min(freq.values()) > 1:
freq[data[left]] -= 1
if freq[data[left]] == 0:
del freq[data[left]]
left += 1
max_len = max(max_len, right - left + 1)
return max_len
n = int(input())
data = list(map(int, input().split()))
result = max_almost_constant_range(n, data)
print(result)
```
| 0
|
|
712
|
B
|
Memory and Trident
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
|
The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.
|
If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
|
[
"RRU\n",
"UDUR\n",
"RUUR\n"
] |
[
"-1\n",
"1\n",
"2\n"
] |
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
| 1,000
|
[
{
"input": "RRU",
"output": "-1"
},
{
"input": "UDUR",
"output": "1"
},
{
"input": "RUUR",
"output": "2"
},
{
"input": "DDDD",
"output": "2"
},
{
"input": "RRRR",
"output": "2"
},
{
"input": "RRRUUD",
"output": "2"
},
{
"input": "UDURLRDURLRD",
"output": "1"
},
{
"input": "RLRU",
"output": "1"
},
{
"input": "RDDLLDLUUUDDRDRURLUUURLLDDLRLUURRLLRRLDRLLUDRLRULLDLRRLRLRLRUDUUDLULURLLDUURULURLLRRRURRRDRUUDLDRLRDRLRRDDLDLDLLUDRUDRLLLLDRDUULRUURRDLULLULDUDULRURRDDDLLUDRLUDDLDDDRRDDDULLLLDLDRLRRLRRDDRLULURRUDRDUUUULDURUDRDLDDUDUDRRURDULRRUDRLRRDLUURURDLDRLRDUDDDLDDDURURLUULRDUUULRURUDUDRRUDULLLUUUDRLLDRRDDLRUDRDRDLLRURURRRULURURRRLUUULRRRUURUUDURUDDLLDLDRLRDLDRLLDLDRDRRLRRRURUUUDRDLRLRUDRLULUUULUDDLULDLRLLRDUULLRLRURLRURULLLUDUDDLRULRDUURURLDLUURRRDURRLLDRUUDRDLLDUUDLURUDDUUUULRLLURLUDDRLRRDRURLRUDRLDDRLLL",
"output": "-1"
},
{
"input": "LDLDLDLDLDRULD",
"output": "5"
},
{
"input": "LULULURULLLU",
"output": "5"
},
{
"input": "DRDRDDRR",
"output": "4"
},
{
"input": "LR",
"output": "0"
},
{
"input": "UL",
"output": "1"
},
{
"input": "UD",
"output": "0"
},
{
"input": "R",
"output": "-1"
},
{
"input": "LU",
"output": "1"
},
{
"input": "RR",
"output": "1"
},
{
"input": "UDLR",
"output": "0"
},
{
"input": "RRRRRLLUUUUDD",
"output": "-1"
},
{
"input": "UUUUDLLLLR",
"output": "3"
},
{
"input": "LLRUD",
"output": "-1"
},
{
"input": "LLRDDU",
"output": "1"
}
] | 1,474,561,029
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
s = input()
if len(s) % 4 != 0:
print("-1")
else:
ave = len(s) // 4
res = abs(ave-s.count("L")) + abs(ave-s.count("R")) + abs(ave-s.count("U")) + abs(ave-s.count("D"))
print(res//2)
|
Title: Memory and Trident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
Input Specification:
The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.
Output Specification:
If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
Demo Input:
['RRU\n', 'UDUR\n', 'RUUR\n']
Demo Output:
['-1\n', '1\n', '2\n']
Note:
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
|
```python
s = input()
if len(s) % 4 != 0:
print("-1")
else:
ave = len(s) // 4
res = abs(ave-s.count("L")) + abs(ave-s.count("R")) + abs(ave-s.count("U")) + abs(ave-s.count("D"))
print(res//2)
```
| 0
|
|
479
|
A
|
Expression
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
|
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
|
Print the maximum value of the expression that you can obtain.
|
[
"1\n2\n3\n",
"2\n10\n3\n"
] |
[
"9\n",
"60\n"
] |
none
| 500
|
[
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,691,742,172
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 46
| 0
|
(lambda a, b, c: print(max(a + b * c, a * (b + c), a * b * c, (a + b) * c, a + b + c)))(*[int(input()) for _ in range(3)])
|
Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none
|
```python
(lambda a, b, c: print(max(a + b * c, a * (b + c), a * b * c, (a + b) * c, a + b + c)))(*[int(input()) for _ in range(3)])
```
| 3
|
|
825
|
C
|
Multi-judge Solving
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
|
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
|
[
"3 3\n2 1 9\n",
"4 20\n10 3 6 3\n"
] |
[
"1\n",
"0\n"
] |
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
| 0
|
[
{
"input": "3 3\n2 1 9",
"output": "1"
},
{
"input": "4 20\n10 3 6 3",
"output": "0"
},
{
"input": "1 1000000000\n1",
"output": "0"
},
{
"input": "1 1\n3",
"output": "1"
},
{
"input": "50 100\n74 55 33 5 83 24 75 59 30 36 13 4 62 28 96 17 6 35 45 53 33 11 37 93 34 79 61 72 13 31 44 75 7 3 63 46 18 16 44 89 62 25 32 12 38 55 75 56 61 82",
"output": "0"
},
{
"input": "100 10\n246 286 693 607 87 612 909 312 621 37 801 558 504 914 416 762 187 974 976 123 635 488 416 659 988 998 93 662 92 749 889 78 214 786 735 625 921 372 713 617 975 119 402 411 878 138 548 905 802 762 940 336 529 373 745 835 805 880 816 94 166 114 475 699 974 462 61 337 555 805 968 815 392 746 591 558 740 380 668 29 881 151 387 986 174 923 541 520 998 947 535 651 103 584 664 854 180 852 726 93",
"output": "1"
},
{
"input": "2 1\n1 1000000000",
"output": "29"
},
{
"input": "29 2\n1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911",
"output": "27"
},
{
"input": "1 1\n1000000000",
"output": "29"
},
{
"input": "7 6\n4 20 16 14 3 17 4",
"output": "1"
},
{
"input": "2 1\n3 6",
"output": "1"
},
{
"input": "1 1\n20",
"output": "4"
},
{
"input": "5 2\n86 81 53 25 18",
"output": "4"
},
{
"input": "4 1\n88 55 14 39",
"output": "4"
},
{
"input": "3 1\n2 3 6",
"output": "0"
},
{
"input": "3 2\n4 9 18",
"output": "1"
},
{
"input": "5 3\n6 6 6 13 27",
"output": "2"
},
{
"input": "5 1\n23 8 83 26 18",
"output": "4"
},
{
"input": "3 1\n4 5 6",
"output": "1"
},
{
"input": "3 1\n1 3 6",
"output": "1"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "3 2\n4 5 6",
"output": "0"
},
{
"input": "5 1\n100 200 400 1000 2000",
"output": "7"
},
{
"input": "2 1\n1 4",
"output": "1"
},
{
"input": "4 1\n2 4 8 32",
"output": "1"
},
{
"input": "2 10\n21 42",
"output": "1"
},
{
"input": "3 3\n1 7 13",
"output": "1"
},
{
"input": "3 1\n1 4 6",
"output": "1"
},
{
"input": "2 2\n2 8",
"output": "1"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "2 2\n8 16",
"output": "1"
},
{
"input": "3 1\n4 8 16",
"output": "1"
},
{
"input": "3 1\n3 6 9",
"output": "1"
},
{
"input": "2 1\n4 8",
"output": "1"
},
{
"input": "2 2\n7 14",
"output": "1"
},
{
"input": "1 4\n9",
"output": "1"
},
{
"input": "5 3\n1024 4096 16384 65536 536870913",
"output": "24"
},
{
"input": "2 5\n10 11",
"output": "0"
},
{
"input": "2 2\n3 6",
"output": "0"
},
{
"input": "2 2\n8 11",
"output": "1"
},
{
"input": "3 19905705\n263637263 417905394 108361057",
"output": "3"
},
{
"input": "4 25\n100 11 1 13",
"output": "1"
},
{
"input": "10 295206008\n67980321 440051990 883040288 135744260 96431758 242465794 576630162 972797687 356406646 547451696",
"output": "0"
},
{
"input": "4 2\n45 44 35 38",
"output": "4"
},
{
"input": "1 2\n9",
"output": "2"
},
{
"input": "3 6\n13 26 52",
"output": "1"
},
{
"input": "9 30111088\n824713578 11195876 458715185 731769293 680826358 189542586 550198537 860586039 101083021",
"output": "2"
},
{
"input": "3 72014068\n430005292 807436976 828082746",
"output": "2"
},
{
"input": "3 165219745\n737649884 652879952 506420386",
"output": "1"
},
{
"input": "2 60669400\n95037700 337255240",
"output": "1"
},
{
"input": "4 28\n34 1 86 90",
"output": "1"
},
{
"input": "2 1\n5 10",
"output": "2"
},
{
"input": "2 1\n4 1000000000",
"output": "28"
},
{
"input": "2 1\n2 3",
"output": "0"
},
{
"input": "2 1\n3 5",
"output": "1"
},
{
"input": "3 3\n1 5 20",
"output": "1"
},
{
"input": "9 1\n1 2 4 9 15 32 64 128 1024",
"output": "4"
}
] | 1,500,218,939
| 1,439
|
Python 3
|
OK
|
TESTS
| 61
| 62
| 4,915,200
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
curr = 0
ans = 0
while curr < len(a):
if k >= a[curr] / 2:
k = max(k, a[curr])
curr += 1
else:
ans += 1
k *= 2
print(ans)
|
Title: Multi-judge Solving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input Specification:
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Demo Input:
['3 3\n2 1 9\n', '4 20\n10 3 6 3\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
|
```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
curr = 0
ans = 0
while curr < len(a):
if k >= a[curr] / 2:
k = max(k, a[curr])
curr += 1
else:
ans += 1
k *= 2
print(ans)
```
| 3
|
|
843
|
A
|
Sorting by Subsequences
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"dsu",
"implementation",
"math",
"sortings"
] | null | null |
You are given a sequence *a*1,<=*a*2,<=...,<=*a**n* consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
|
The first line of input data contains integer *n* (1<=≤<=*n*<=≤<=105) — the length of the sequence.
The second line of input data contains *n* different integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
|
In the first line print the maximum number of subsequences *k*, which the original sequence can be split into while fulfilling the requirements.
In the next *k* lines print the description of subsequences in the following format: the number of elements in subsequence *c**i* (0<=<<=*c**i*<=≤<=*n*), then *c**i* integers *l*1,<=*l*2,<=...,<=*l**c**i* (1<=≤<=*l**j*<=≤<=*n*) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to *n* must appear in output exactly once.
If there are several possible answers, print any of them.
|
[
"6\n3 2 1 6 5 4\n",
"6\n83 -75 -49 11 37 62\n"
] |
[
"4\n2 1 3\n1 2\n2 4 6\n1 5\n",
"1\n6 1 2 3 4 5 6\n"
] |
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
| 500
|
[
{
"input": "6\n3 2 1 6 5 4",
"output": "4\n2 1 3\n1 2\n2 4 6\n1 5"
},
{
"input": "6\n83 -75 -49 11 37 62",
"output": "1\n6 1 2 3 4 5 6"
},
{
"input": "1\n1",
"output": "1\n1 1"
},
{
"input": "2\n1 2",
"output": "2\n1 1\n1 2"
},
{
"input": "2\n2 1",
"output": "1\n2 1 2"
},
{
"input": "3\n1 2 3",
"output": "3\n1 1\n1 2\n1 3"
},
{
"input": "3\n3 2 1",
"output": "2\n2 1 3\n1 2"
},
{
"input": "3\n3 1 2",
"output": "1\n3 1 2 3"
},
{
"input": "10\n3 7 10 1 9 5 4 8 6 2",
"output": "3\n6 1 4 7 2 10 3\n3 5 6 9\n1 8"
},
{
"input": "20\n363756450 -204491568 95834122 -840249197 -49687658 470958158 -445130206 189801569 802780784 -790013317 -192321079 586260100 -751917965 -354684803 418379342 -253230108 193944314 712662868 853829789 735867677",
"output": "3\n7 1 4 7 2 10 3 13\n11 5 14 15 6 16 12 17 18 20 19 9\n2 8 11"
},
{
"input": "50\n39 7 45 25 31 26 50 11 19 37 8 16 22 33 14 6 12 46 49 48 29 27 41 15 34 24 3 13 20 47 9 36 5 43 40 21 2 38 35 42 23 28 1 32 10 17 30 18 44 4",
"output": "6\n20 1 43 34 25 4 50 7 2 37 10 45 3 27 22 13 28 42 40 35 39\n23 5 33 14 15 24 26 6 16 12 17 46 18 48 20 29 21 36 32 44 49 19 9 31\n2 8 11\n2 23 41\n2 30 47\n1 38"
},
{
"input": "100\n39 77 67 25 81 26 50 11 73 95 86 16 90 33 14 79 12 100 68 64 60 27 41 15 34 24 3 61 83 47 57 65 99 43 40 21 94 72 82 85 23 71 76 32 10 17 30 18 44 59 35 89 6 63 7 69 62 70 4 29 92 87 31 48 36 28 45 97 93 98 56 38 58 80 8 1 74 91 53 55 54 51 96 5 42 52 9 22 78 88 75 13 66 2 37 20 49 19 84 46",
"output": "6\n41 1 76 43 34 25 4 59 50 7 55 80 74 77 2 94 37 95 10 45 67 3 27 22 88 90 13 92 61 28 66 93 69 56 71 42 85 40 35 51 82 39\n45 5 84 99 33 14 15 24 26 6 53 79 16 12 17 46 100 18 48 64 20 96 83 29 60 21 36 65 32 44 49 97 68 19 98 70 58 73 9 87 62 57 31 63 54 81\n8 8 75 91 78 89 52 86 11\n2 23 41\n2 30 47\n2 38 72"
}
] | 1,677,427,116
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 483
| 13,209,600
|
import bisect
n = int(input())
a = list(map(int, input().split()))
sorted_a = sorted(a)
num_list = [bisect.bisect_left(sorted_a, i) for i in a]
val_list = [False] * n
ans = []
for i in range(n):
if not val_list[i]:
val_list[i] = True
que = [i + 1]
ind = num_list[i]
while ind != i:
que.append(ind + 1)
val_list[ind] = True
ind = num_list[ind]
ans.append(que)
print(len(ans))
for i in ans:
print(len(i), *i)
|
Title: Sorting by Subsequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a*1,<=*a*2,<=...,<=*a**n* consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
Input Specification:
The first line of input data contains integer *n* (1<=≤<=*n*<=≤<=105) — the length of the sequence.
The second line of input data contains *n* different integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
Output Specification:
In the first line print the maximum number of subsequences *k*, which the original sequence can be split into while fulfilling the requirements.
In the next *k* lines print the description of subsequences in the following format: the number of elements in subsequence *c**i* (0<=<<=*c**i*<=≤<=*n*), then *c**i* integers *l*1,<=*l*2,<=...,<=*l**c**i* (1<=≤<=*l**j*<=≤<=*n*) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to *n* must appear in output exactly once.
If there are several possible answers, print any of them.
Demo Input:
['6\n3 2 1 6 5 4\n', '6\n83 -75 -49 11 37 62\n']
Demo Output:
['4\n2 1 3\n1 2\n2 4 6\n1 5\n', '1\n6 1 2 3 4 5 6\n']
Note:
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
|
```python
import bisect
n = int(input())
a = list(map(int, input().split()))
sorted_a = sorted(a)
num_list = [bisect.bisect_left(sorted_a, i) for i in a]
val_list = [False] * n
ans = []
for i in range(n):
if not val_list[i]:
val_list[i] = True
que = [i + 1]
ind = num_list[i]
while ind != i:
que.append(ind + 1)
val_list[ind] = True
ind = num_list[ind]
ans.append(que)
print(len(ans))
for i in ans:
print(len(i), *i)
```
| 3
|
|
992
|
A
|
Nastya and an Array
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
|
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
|
[
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
| 500
|
[
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,640,952,998
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 79
| 170
| 18,022,400
|
n= int(input())
l = list(map(int, input().split()))
s = set(l)
if 0 in s:
print(len(s)-1)
else:
print(len(s))
|
Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
|
```python
n= int(input())
l = list(map(int, input().split()))
s = set(l)
if 0 in s:
print(len(s)-1)
else:
print(len(s))
```
| 3
|
|
538
|
B
|
Quasi Binary
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null |
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
|
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
|
[
"9\n",
"32\n"
] |
[
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] |
none
| 1,000
|
[
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201",
"output": "2\n100 10101 "
},
{
"input": "314159",
"output": "9\n1 1 1 1 11 1011 101011 101011 111111 "
},
{
"input": "999999",
"output": "9\n111111 111111 111111 111111 111111 111111 111111 111111 111111 "
},
{
"input": "2",
"output": "2\n1 1 "
},
{
"input": "10",
"output": "1\n10 "
},
{
"input": "21",
"output": "2\n10 11 "
},
{
"input": "98",
"output": "9\n10 11 11 11 11 11 11 11 11 "
},
{
"input": "102030",
"output": "3\n10 1010 101010 "
},
{
"input": "909090",
"output": "9\n101010 101010 101010 101010 101010 101010 101010 101010 101010 "
},
{
"input": "909823",
"output": "9\n101000 101100 101100 101100 101100 101100 101101 101111 101111 "
},
{
"input": "1000000",
"output": "1\n1000000 "
},
{
"input": "111111",
"output": "1\n111111 "
},
{
"input": "123456",
"output": "6\n1 11 111 1111 11111 111111 "
},
{
"input": "987654",
"output": "9\n100000 110000 111000 111100 111110 111111 111111 111111 111111 "
},
{
"input": "908172",
"output": "9\n100000 101000 101010 101010 101010 101010 101010 101011 101111 "
},
{
"input": "8",
"output": "8\n1 1 1 1 1 1 1 1 "
},
{
"input": "100009",
"output": "9\n1 1 1 1 1 1 1 1 100001 "
},
{
"input": "900000",
"output": "9\n100000 100000 100000 100000 100000 100000 100000 100000 100000 "
},
{
"input": "1435",
"output": "5\n1 101 111 111 1111 "
},
{
"input": "1453",
"output": "5\n10 110 111 111 1111 "
}
] | 1,552,824,730
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 93
| 0
|
#maximum you never know
x = input()
x1= x
x= int(x)
list1 =[]
while(x!=0):
l = len(str(x))
re = '1'*l
re2= 10**(l-1)
# print('l')
i=1
while(i<l+1):
if(x>= int(re)):
list1.append(int(re))
x-=int(re)
# print('#2')
break
else:
re = re[0:i]+'0'+re[i+1:]
i = i+1
if(l==1):
# print("i m n ")
for i in range(x):
list1.append(1)
x=0
print(len(list1))
print(" ".join(str(x) for x in list1))
|
Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none
|
```python
#maximum you never know
x = input()
x1= x
x= int(x)
list1 =[]
while(x!=0):
l = len(str(x))
re = '1'*l
re2= 10**(l-1)
# print('l')
i=1
while(i<l+1):
if(x>= int(re)):
list1.append(int(re))
x-=int(re)
# print('#2')
break
else:
re = re[0:i]+'0'+re[i+1:]
i = i+1
if(l==1):
# print("i m n ")
for i in range(x):
list1.append(1)
x=0
print(len(list1))
print(" ".join(str(x) for x in list1))
```
| 0
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,674,916,879
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
song = input()
stripped = song.replace("WUB"," ").strip()
print(stripped.replace(" ", " "))
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
song = input()
stripped = song.replace("WUB"," ").strip()
print(stripped.replace(" ", " "))
```
| 3
|
|
922
|
B
|
Magic Forest
|
PROGRAMMING
| 1,300
|
[
"brute force"
] | null | null |
Imp is in a magic forest, where xorangles grow (wut?)
A xorangle of order *n* is such a non-degenerate triangle, that lengths of its sides are integers not exceeding *n*, and the xor-sum of the lengths is equal to zero. Imp has to count the number of distinct xorangles of order *n* to get out of the forest.
Formally, for a given integer *n* you have to find the number of such triples (*a*,<=*b*,<=*c*), that:
- 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*; - , where denotes the [bitwise xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of integers *x* and *y*. - (*a*,<=*b*,<=*c*) form a non-degenerate (with strictly positive area) triangle.
|
The only line contains a single integer *n* (1<=≤<=*n*<=≤<=2500).
|
Print the number of xorangles of order *n*.
|
[
"6\n",
"10\n"
] |
[
"1\n",
"2\n"
] |
The only xorangle in the first sample is (3, 5, 6).
| 1,000
|
[
{
"input": "6",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "2500",
"output": "700393"
},
{
"input": "952",
"output": "118547"
},
{
"input": "88",
"output": "536"
},
{
"input": "1216",
"output": "160822"
},
{
"input": "2140",
"output": "614785"
},
{
"input": "564",
"output": "35087"
},
{
"input": "1488",
"output": "239580"
},
{
"input": "116",
"output": "1332"
},
{
"input": "1040",
"output": "145820"
},
{
"input": "1965",
"output": "545494"
},
{
"input": "593",
"output": "36605"
},
{
"input": "779",
"output": "63500"
},
{
"input": "1703",
"output": "352045"
},
{
"input": "331",
"output": "9877"
},
{
"input": "1051",
"output": "145985"
},
{
"input": "2179",
"output": "618074"
},
{
"input": "603",
"output": "37312"
},
{
"input": "1731",
"output": "369691"
},
{
"input": "2451",
"output": "681980"
},
{
"input": "1079",
"output": "146833"
},
{
"input": "2207",
"output": "621708"
},
{
"input": "2394",
"output": "663240"
},
{
"input": "818",
"output": "73972"
},
{
"input": "1946",
"output": "529383"
},
{
"input": "166",
"output": "2200"
},
{
"input": "1294",
"output": "175915"
},
{
"input": "2218",
"output": "623386"
},
{
"input": "846",
"output": "82106"
},
{
"input": "1566",
"output": "273341"
},
{
"input": "194",
"output": "3240"
},
{
"input": "1322",
"output": "183405"
},
{
"input": "1508",
"output": "247634"
},
{
"input": "2433",
"output": "675245"
},
{
"input": "857",
"output": "85529"
},
{
"input": "1781",
"output": "402718"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "2444",
"output": "679373"
},
{
"input": "2498",
"output": "699536"
}
] | 1,596,517,551
| 1,911
|
PyPy 3
|
OK
|
TESTS
| 44
| 343
| 49,049,600
|
n = int(input())
ans=[[0]*(n+1) for i in range(n+1)]
aa=0
for i in range(1,n+1):
for j in range(i,n+1):
if i^j<=n:
a=sorted([i,j,i^j])
if a[0]+a[1]>a[2] and a[0]!=0:
if ans[a[0]][a[1]]==0:
aa+=1
ans[a[0]][a[1]]=1
print(aa)
|
Title: Magic Forest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imp is in a magic forest, where xorangles grow (wut?)
A xorangle of order *n* is such a non-degenerate triangle, that lengths of its sides are integers not exceeding *n*, and the xor-sum of the lengths is equal to zero. Imp has to count the number of distinct xorangles of order *n* to get out of the forest.
Formally, for a given integer *n* you have to find the number of such triples (*a*,<=*b*,<=*c*), that:
- 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*; - , where denotes the [bitwise xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of integers *x* and *y*. - (*a*,<=*b*,<=*c*) form a non-degenerate (with strictly positive area) triangle.
Input Specification:
The only line contains a single integer *n* (1<=≤<=*n*<=≤<=2500).
Output Specification:
Print the number of xorangles of order *n*.
Demo Input:
['6\n', '10\n']
Demo Output:
['1\n', '2\n']
Note:
The only xorangle in the first sample is (3, 5, 6).
|
```python
n = int(input())
ans=[[0]*(n+1) for i in range(n+1)]
aa=0
for i in range(1,n+1):
for j in range(i,n+1):
if i^j<=n:
a=sorted([i,j,i^j])
if a[0]+a[1]>a[2] and a[0]!=0:
if ans[a[0]][a[1]]==0:
aa+=1
ans[a[0]][a[1]]=1
print(aa)
```
| 3
|
|
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"
}
] | 1,689,027,327
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 889
| 819,200
|
string1 = []
string2 = []
_ , num = map(int,input().split())
for i in range(num):
x , y = input().split()
string1.append(x)
string2.append(y)
string = input().split()
for i in range(len(string)):
for j in string1:
if j == string[i]:
if len(j) <= len(string2[string1.index(j)]):
print(j,end=" ")
else:
print(string2[string1.index(j)],end=" ")
|
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
string1 = []
string2 = []
_ , num = map(int,input().split())
for i in range(num):
x , y = input().split()
string1.append(x)
string2.append(y)
string = input().split()
for i in range(len(string)):
for j in string1:
if j == string[i]:
if len(j) <= len(string2[string1.index(j)]):
print(j,end=" ")
else:
print(string2[string1.index(j)],end=" ")
```
| 3
|
|
929
|
A
|
Прокат велосипедов
|
PROGRAMMING
| 1,400
|
[
"*special",
"greedy",
"implementation"
] | null | null |
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
|
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
|
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
|
[
"4 4\n3 6 8 10\n",
"2 9\n10 20\n",
"12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n"
] |
[
"2\n",
"-1\n",
"6\n"
] |
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
| 500
|
[
{
"input": "4 4\n3 6 8 10",
"output": "2"
},
{
"input": "2 9\n10 20",
"output": "-1"
},
{
"input": "12 3\n4 6 7 9 10 11 13 15 17 18 20 21",
"output": "6"
},
{
"input": "2 1\n11164 11165",
"output": "1"
},
{
"input": "3 7\n45823 45825 45829",
"output": "1"
},
{
"input": "2 100000\n0 100000",
"output": "1"
},
{
"input": "50 15001\n1269 1580 5431 5916 6642 14145 15760 19922 20589 22062 24138 33454 33505 35916 37012 42577 43140 49457 54720 55188 56205 56639 56870 57997 58379 59088 59297 61805 61861 63005 64013 68848 71182 71497 72715 73008 75616 76042 76614 80690 83089 85033 86408 92392 92763 93833 95085 95815 97152 99379",
"output": "8"
},
{
"input": "5 7\n6219 6222 6229 6231 6236",
"output": "3"
},
{
"input": "10 448\n78449 78573 78599 78742 78748 78759 78853 79091 79298 79324",
"output": "3"
},
{
"input": "20 19191\n11272 12386 14108 24663 24932 26547 29656 44677 45436 45654 48562 62367 71049 71238 78182 84042 88736 92026 96836 99343",
"output": "6"
},
{
"input": "2 100000\n1 100000",
"output": "1"
},
{
"input": "2 99999\n0 100000",
"output": "-1"
},
{
"input": "2 2\n1 3",
"output": "1"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "2 2\n1 4",
"output": "-1"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
}
] | 1,520,010,578
| 5,678
|
Python 3
|
OK
|
TESTS
| 52
| 62
| 5,632,000
|
n,k=map(int, input().split())
x=input().split()
p=0
i=-1
while i+1!=n-1:
q=0
i+=1
b=p
for j in range(n-i-1):
if int(x[i+j+1])-int(x[i])<=k:
q+=1
else:
i+=q-1
if q!=0:
p+=1
break
if b==p and q==0:
p=0
break
elif b==p:
i+=q-1
p+=1
if p!=0:
p+=1
print(p-1)
|
Title: Прокат велосипедов
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
Output Specification:
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
Demo Input:
['4 4\n3 6 8 10\n', '2 9\n10 20\n', '12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
|
```python
n,k=map(int, input().split())
x=input().split()
p=0
i=-1
while i+1!=n-1:
q=0
i+=1
b=p
for j in range(n-i-1):
if int(x[i+j+1])-int(x[i])<=k:
q+=1
else:
i+=q-1
if q!=0:
p+=1
break
if b==p and q==0:
p=0
break
elif b==p:
i+=q-1
p+=1
if p!=0:
p+=1
print(p-1)
```
| 3
|
|
746
|
B
|
Decoding
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding.
|
Print the word that Polycarp encoded.
|
[
"5\nlogva\n",
"2\nno\n",
"4\nabba\n"
] |
[
"volga\n",
"no\n",
"baba\n"
] |
In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba.
| 1,000
|
[
{
"input": "5\nlogva",
"output": "volga"
},
{
"input": "2\nno",
"output": "no"
},
{
"input": "4\nabba",
"output": "baba"
},
{
"input": "51\nkfsmpaeviowvkdbuhdagquxxqniselafnfbrgbhmsugcbbnlrvv",
"output": "vlbcumbrfflsnxugdudvovamfkspeiwkbhaqxqieanbghsgbnrv"
},
{
"input": "1\nw",
"output": "w"
},
{
"input": "2\ncb",
"output": "cb"
},
{
"input": "3\nqok",
"output": "oqk"
},
{
"input": "4\naegi",
"output": "gaei"
},
{
"input": "5\noqquy",
"output": "uqoqy"
},
{
"input": "6\nulhpnm",
"output": "nhulpm"
},
{
"input": "7\nijvxljt",
"output": "jxjivlt"
},
{
"input": "8\nwwmiwkeo",
"output": "ewmwwiko"
},
{
"input": "9\ngmwqmpfow",
"output": "opqmgwmfw"
},
{
"input": "10\nhncmexsslh",
"output": "lsechnmxsh"
},
{
"input": "20\nrtcjbjlbtjfmvzdqutuw",
"output": "uudvftlbcrtjjbjmzqtw"
},
{
"input": "21\ngjyiqoebcnpsdegxnsauh",
"output": "usxesnboijgyqecpdgnah"
},
{
"input": "30\nudotcwvcwxajkadxqvxvwgmwmnqrby",
"output": "bqmmwxqdkawvcoudtwcxjaxvvgwnry"
},
{
"input": "31\nipgfrxxcgckksfgexlicjvtnhvrfbmb",
"output": "mfvnvclefkccxfpigrxgksgxijthrbb"
},
{
"input": "50\nwobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "vsolrruoeqehviaqtycivhrbwoevvkhujmhagaholrmsatdjjy"
},
{
"input": "200\nhvayscqiwpcfykibwyudkzuzdkgqqvbnrfeupjefevlvojngmlcjwzijrkzbsaovabkvvwmjgoonyhuiphwmqdoiuueuyqtychbsklflnvghipdgaxhuhiiqlqocpvhldgvnsrtcwxpidrjffwvwcirluyyxzxrglheczeuouklzkvnyubsvgvmdbrylimztotdbmjph",
"output": "pmdoziybmgsunkluuzelrzyurcvfjdpwtsvdhpolihhadignfkbctyeuoqwpuyogmvkaoszriwcmnoleeperbqgdukuwiycwqsahvycipfkbydzzkqvnfujfvvjgljzjkbavbvwjonhihmdiuuqyhsllvhpgxuiqqcvlgnrcxirfwwilyxxghceokzvybvvdrlmttbjh"
},
{
"input": "201\nrpkghhfibtmlkpdiklegblbuyshfirheatjkfoqkfayfbxeeqijwqdwkkrkbdxlhzkhyiifemsghwovorlqedngldskfbhmwrnzmtjuckxoqdszmsdnbuqnlqzswdfhagasmfswanifrjjcuwdsplytvmnfarchgqteedgfpumkssindxndliozojzlpznwedodzwrrus",
"output": "urzoenpzoolndismpgetgcanvypdujriasmaafwzlqbdmsqxcjmnwhfslneloohseiykhxbrkdwiexfakokterfsulglipltihgprkhfbmkdkebbyhihajfqfybeqjqwkkdlzhifmgwvrqdgdkbmrztukodzsnunqsdhgsfwnfjcwsltmfrhqedfuksnxdizjlzwddwrs"
},
{
"input": "500\naopxumqciwxewxvlxzebsztskjvjzwyewjztqrsuvamtvklhqrbodtncqdchjrlpywvmtgnkkwtvpggktewdgvnhydkexwoxkgltaesrtifbwpciqsvrgjtqrdnyqkgqwrryacluaqmgdwxinqieiblolyekcbzahlhxdwqcgieyfgmicvgbbitbzhejkshjunzjteyyfngigjwyqqndtjrdykzrnrpinkwtrlchhxvycrhstpecadszilicrqdeyyidohqvzfnsqfyuemigacysxvtrgxyjcvejkjstsnatfqlkeytxgsksgpcooypsmqgcluzwofaupegxppbupvtumjerohdteuenwcmqaoazohkilgpkjavcrjcslhzkyjcgfzxxzjfufichxcodcawonkxhbqgfimmlycswdzwbnmjwhbwihfoftpcqplncavmbxuwnsabiyvpcrhfgtqyaguoaigknushbqjwqmmyvsxwabrub",
"output": "ubwsymwqhukiogytfrpybswxmanpctohwhjnwdsymigbxnwcoxcffzxfcyzlcrvjplkoaamweedoemtpbpgpaozlgmpocgkgtelfasskecygtxyaieyqnzqoiydriisaethcvhcrwnpnzyrtnqwggfytzuhkeztbgcmfegqdhhzcelliinxdmalarwgqnrtgvqcwftsalkoxkyngwtgptkntvyljcqndbqlvmvsqzwyzvktsexvwxiqupaoxmcwexlzbzsjjwejtruatkhrotcdhrpwmgkwvgkedvhdewxgteribpisrjqdykqrycuqgwiqeboykbalxwciygivbibhjsjnjeynijyqdjdkrriktlhxyrspcdzlcqeydhvfsfumgcsvrxjvjjtntqkyxsspoysqcuwfuexpuvujrhtuncqozhigkacjshkjgzxjuihcdaokhqfmlcwzbmwbiffpqlcvbunaivchgqauagnsbjqmvxarb"
},
{
"input": "501\noilesjbgowlnayckhpoaitijewsyhgavnthycaecwnvzpxgjqfjyxnjcjknvvsmjbjwtcoyfbegmnnheeamvtfjkigqoanhvgdfrjchdqgowrstlmrjmcsuuwvvoeucfyhnxivosrxblfoqwikfxjnnyejdiihpenfcahtjwcnzwvxxseicvdfgqhtvefswznuyohmmljlnxubhevywpmnitnkhecsgccpstxkmdzabsnwxkokdfsogzbpnfvgudvqlstxojzfzugxbfrozveaiofrzksegdelxsdhcjlqwqlgjcqiujptoenxozhkqhcpkarretqzfkwuvbmfdcdoqliyzmlfrfhzrnkbhofuctkpeacqejwvdrlyvepudrlzncbhwrgmxrczphsoymvtuzqjscvbtqpymogupgzctepccravjcrfsadueyraqvwasravkubebojqspdntewnjohvccamvoxdauyakvehjhabpdyzyme",
"output": "mzdbhhvaudomcvonenpqoeuvravayuafcvrcecguoyqbcjztmohzrmrhczrueyrvjqaptuobnzffmylocfbukztraphkzxetjicgqqjhsldekroavofxufjxsqdgfpzofkkwsadktpcsektnpyebxllmouzsetqfvisxwnwtafehijynxkwolxsvxhfuovusmrlswgdcrdvnogkfvaenmefotjjsvkcnyfjxzncaytvgywjtapkynwgjeiolsbolachoiieshanhcewvpgqjxjjnvmbwcybgnhemtjiqahgfjhqortmjcuwvecyniorbfqifjnedipnchjczvxecdghvfwnyhmjnuhvwminhcgcsxmzbnxodsgbnvuvltozzgbrzeifzsgexdclwljquponohqckreqfwvmddqizlrhrkhfckecewdlvpdlnbwgxcpsyvuqsvtpmgpztpcajrsderqwsakbbjsdtwjhcavxaykejapyye"
}
] | 1,626,374,366
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 6,963,200
|
def stringBuilder(char_list):
string = ''
for char in char_list:
string += char
return string
def getPlainText(length, ciphertext):
plainText = []
plainText.append(ciphertext[0])
for i in range(1, length):
if i % 2 == 0:
plainText.insert(i, ciphertext[i])
else:
plainText.insert(0, ciphertext[i])
return stringBuilder(plainText)
if __name__ == '__main__':
cipher_length = int(input())
cipherText = input()
plainText = getPlainText(cipher_length, cipherText)
print(f'{plainText}')
|
Title: Decoding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding.
Output Specification:
Print the word that Polycarp encoded.
Demo Input:
['5\nlogva\n', '2\nno\n', '4\nabba\n']
Demo Output:
['volga\n', 'no\n', 'baba\n']
Note:
In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba.
|
```python
def stringBuilder(char_list):
string = ''
for char in char_list:
string += char
return string
def getPlainText(length, ciphertext):
plainText = []
plainText.append(ciphertext[0])
for i in range(1, length):
if i % 2 == 0:
plainText.insert(i, ciphertext[i])
else:
plainText.insert(0, ciphertext[i])
return stringBuilder(plainText)
if __name__ == '__main__':
cipher_length = int(input())
cipherText = input()
plainText = getPlainText(cipher_length, cipherText)
print(f'{plainText}')
```
| 0
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,696,737,899
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 124
| 1,638,400
|
n = list(input())
b = [int(i) for i in n]
count = 0
while len(n) > 1:
result = sum(b)
n = list(str(result))
b = [int(i) for i in n]
count += 1
print(count)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
n = list(input())
b = [int(i) for i in n]
count = 0
while len(n) > 1:
result = sum(b)
n = list(str(result))
b = [int(i) for i in n]
count += 1
print(count)
```
| 3.966052
|
165
|
A
|
Supercentral Point
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
|
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
|
Print the only number — the number of supercentral points of the given set.
|
[
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
| 500
|
[
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,620,149,449
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 154
| 0
|
N = int(input())
X = []
Y = []
for i in range(N):
x,y =list(map(int,input().split()))
X.append(x)
Y.append(y)
#print(X,Y)
count = 0
for i in range(N):
left = right = up = down = 0
temp_x = X[i] ## assumed X
temp_y = Y[i] ## assumed y
for j in range(0,N):
if (X[j] == temp_x):
if(Y[j]>temp_y):
up+=1
if(Y[j]<temp_y):
down+=1
if (Y[j]== temp_y):
if(X[j]>temp_x):
right+=1
if(X[j]<temp_x):
left+=1
if(up and down and left and right ):
count+=1
print(count)
|
Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
|
```python
N = int(input())
X = []
Y = []
for i in range(N):
x,y =list(map(int,input().split()))
X.append(x)
Y.append(y)
#print(X,Y)
count = 0
for i in range(N):
left = right = up = down = 0
temp_x = X[i] ## assumed X
temp_y = Y[i] ## assumed y
for j in range(0,N):
if (X[j] == temp_x):
if(Y[j]>temp_y):
up+=1
if(Y[j]<temp_y):
down+=1
if (Y[j]== temp_y):
if(X[j]>temp_x):
right+=1
if(X[j]<temp_x):
left+=1
if(up and down and left and right ):
count+=1
print(count)
```
| 3
|
|
975
|
A
|
Aramic script
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
|
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
|
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
|
[
"5\na aa aaa ab abb\n",
"3\namer arem mrea\n"
] |
[
"2",
"1"
] |
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".
| 500
|
[
{
"input": "5\na aa aaa ab abb",
"output": "2"
},
{
"input": "3\namer arem mrea",
"output": "1"
},
{
"input": "10\nbda bbb cda dca dda dcb bcd dcb ada ddd",
"output": "6"
},
{
"input": "2\nfhjlqs aceginpr",
"output": "2"
},
{
"input": "2\nbcdfghimn efghijlmo",
"output": "2"
}
] | 1,534,357,227
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 140
| 3,276,800
|
n=input()
op=[]
po=list(map(str,input().split()))
for item in po:
tam=list(set(item))
tam.sort()
op.append("".join(tam))
print(len(set(op)))
|
Title: Aramic script
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
Output Specification:
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
Demo Input:
['5\na aa aaa ab abb\n', '3\namer arem mrea\n']
Demo Output:
['2', '1']
Note:
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".
|
```python
n=input()
op=[]
po=list(map(str,input().split()))
for item in po:
tam=list(set(item))
tam.sort()
op.append("".join(tam))
print(len(set(op)))
```
| 3
|
|
900
|
B
|
Position in Fraction
|
PROGRAMMING
| 1,300
|
[
"math",
"number theory"
] | null | null |
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
|
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
|
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
|
[
"1 2 0\n",
"2 3 7\n"
] |
[
"2",
"-1"
] |
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
| 1,000
|
[
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
},
{
"input": "7 31 3",
"output": "15"
},
{
"input": "8880 81608 9",
"output": "161"
},
{
"input": "4942 62768 5",
"output": "122"
},
{
"input": "69168 84860 4",
"output": "107"
},
{
"input": "971 1883 3",
"output": "130"
},
{
"input": "1636 3269 6",
"output": "150"
},
{
"input": "6873 7769 3",
"output": "163"
},
{
"input": "13805 15538 3",
"output": "164"
},
{
"input": "10958 21926 3",
"output": "117"
},
{
"input": "8 51 0",
"output": "10"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "1 9 0",
"output": "-1"
},
{
"input": "53 101 6",
"output": "-1"
},
{
"input": "1 10001 9",
"output": "5"
},
{
"input": "25102 31579 2",
"output": "174"
},
{
"input": "38790 39359 0",
"output": "212"
},
{
"input": "47117 78718 0",
"output": "213"
},
{
"input": "1 57 0",
"output": "1"
},
{
"input": "1 3 0",
"output": "-1"
},
{
"input": "1 100 0",
"output": "1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "99971 99989 0",
"output": "9"
},
{
"input": "567 1580 0",
"output": "6"
},
{
"input": "45 97 0",
"output": "39"
},
{
"input": "35 111 4",
"output": "-1"
},
{
"input": "1 2 5",
"output": "1"
},
{
"input": "1 7 0",
"output": "-1"
}
] | 1,596,681,821
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 343
| 30,310,400
|
from decimal import *
getcontext().prec=10**5+500
a,b,c=map(int,input().split())
r = Decimal(a)/Decimal(b)
r=(str(r)+'0'*100).split('.')[1][:10**5]
if str(c) not in r:print(-1)
else:print(r.index(str(c))+1)
|
Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
|
```python
from decimal import *
getcontext().prec=10**5+500
a,b,c=map(int,input().split())
r = Decimal(a)/Decimal(b)
r=(str(r)+'0'*100).split('.')[1][:10**5]
if str(c) not in r:print(-1)
else:print(r.index(str(c))+1)
```
| 3
|
|
394
|
A
|
Counting Sticks
|
PROGRAMMING
| 0
|
[
"brute force",
"implementation"
] | null | null |
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
|
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
|
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
|
[
"||+|=|||||\n",
"|||||+||=||\n",
"|+|=||||||\n",
"||||+||=||||||\n"
] |
[
"|||+|=||||\n",
"Impossible\n",
"Impossible\n",
"||||+||=||||||\n"
] |
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
| 500
|
[
{
"input": "||+|=|||||",
"output": "|||+|=||||"
},
{
"input": "|||||+||=||",
"output": "Impossible"
},
{
"input": "|+|=||||||",
"output": "Impossible"
},
{
"input": "||||+||=||||||",
"output": "||||+||=||||||"
},
{
"input": "||||||||||||+|||||||||||=||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||+||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||=|||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|+|=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+|=||",
"output": "|+|=||"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|+|=|||",
"output": "Impossible"
},
{
"input": "|+|=||||",
"output": "||+|=|||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+||||||=|||||",
"output": "|+|||||=||||||"
},
{
"input": "|+|||=||",
"output": "|+||=|||"
},
{
"input": "|+||||=|||",
"output": "|+|||=||||"
},
{
"input": "|+|||||=||||",
"output": "|+||||=|||||"
},
{
"input": "||+||=||",
"output": "|+||=|||"
},
{
"input": "||+|||=|||",
"output": "|+|||=||||"
},
{
"input": "|||||+||=||||||||||",
"output": "Impossible"
},
{
"input": "|||+|||=||||",
"output": "||+|||=|||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|||||+|||||=||||||||||||",
"output": "||||||+|||||=|||||||||||"
},
{
"input": "||+|||||||=|||||||",
"output": "|+|||||||=||||||||"
},
{
"input": "||||+||||=||||||",
"output": "|||+||||=|||||||"
},
{
"input": "||||+|||=|||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|||=|||||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
}
] | 1,584,857,404
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 0
|
s1=list(map(str,input().split()))
a=s1.index('+')
b=s1.index('=')
c=len(s1)
block1=a
block2=b-a-1
block3=c-b-1
# print(block1)
# print(block2)
# print(block3)
if block1+block2 == block3:
print(*s1)
elif block3 == block1+block2+1:
s2=[]
s2.append('|')
for i in s1:
s2.append(i)
print(*s2)
else:
print("impossible")
|
Title: Counting Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
Input Specification:
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
Output Specification:
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
Demo Input:
['||+|=|||||\n', '|||||+||=||\n', '|+|=||||||\n', '||||+||=||||||\n']
Demo Output:
['|||+|=||||\n', 'Impossible\n', 'Impossible\n', '||||+||=||||||\n']
Note:
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
|
```python
s1=list(map(str,input().split()))
a=s1.index('+')
b=s1.index('=')
c=len(s1)
block1=a
block2=b-a-1
block3=c-b-1
# print(block1)
# print(block2)
# print(block3)
if block1+block2 == block3:
print(*s1)
elif block3 == block1+block2+1:
s2=[]
s2.append('|')
for i in s1:
s2.append(i)
print(*s2)
else:
print("impossible")
```
| -1
|
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,665,650,843
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
o=[]
n=3
while 1:
a=180*(n-2)/n
if a>179:
break
if a-a//1==0:
o.append(int(a))
n+=1
p=[]
for i in range(int(input())):
if int(input()) in o:
p.append('YES')
print('\n'.join(p))
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
```python
o=[]
n=3
while 1:
a=180*(n-2)/n
if a>179:
break
if a-a//1==0:
o.append(int(a))
n+=1
p=[]
for i in range(int(input())):
if int(input()) in o:
p.append('YES')
print('\n'.join(p))
```
| 0
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,640,944,942
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 216
| 0
|
s = input()
big = 0
for c in s:
if c.isupper():
big += 1
if big > len(s) // 2:
print(s.upper())
else:
print(s.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
s = input()
big = 0
for c in s:
if c.isupper():
big += 1
if big > len(s) // 2:
print(s.upper())
else:
print(s.lower())
```
| 3.946
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,604,337,329
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 93
| 0
|
def GCD(a ,b):
if b==0:
return a
return GCD(b,a%b)
if __name__ == '__main__':
i, y = map(int,input().split())
print(GCD(i, y))
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
def GCD(a ,b):
if b==0:
return a
return GCD(b,a%b)
if __name__ == '__main__':
i, y = map(int,input().split())
print(GCD(i, y))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
Help the dwarfs find out how many triangle plants that point "upwards" will be in *n* years.
|
The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1018) — the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
Print a single integer — the remainder of dividing the number of plants that will point "upwards" in *n* years by 1000000007 (109<=+<=7).
|
[
"1\n",
"2\n"
] |
[
"3\n",
"10\n"
] |
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
| 0
|
[
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "10"
},
{
"input": "385599124",
"output": "493875375"
},
{
"input": "989464295",
"output": "31966163"
},
{
"input": "376367012",
"output": "523204186"
},
{
"input": "529357306",
"output": "142578489"
},
{
"input": "782916801",
"output": "51174574"
},
{
"input": "74859961358140080",
"output": "478768275"
},
{
"input": "0",
"output": "1"
},
{
"input": "252509053898415171",
"output": "886314547"
},
{
"input": "760713016078377938",
"output": "79611270"
},
{
"input": "919845424847912644",
"output": "388845650"
},
{
"input": "585335721566249104",
"output": "301383716"
},
{
"input": "522842183413115087",
"output": "556012763"
},
{
"input": "148049062285906746",
"output": "913927498"
},
{
"input": "84324827171274022",
"output": "462535280"
},
{
"input": "354979172034763159",
"output": "239287993"
},
{
"input": "1312148742261680",
"output": "799725655"
},
{
"input": "269587448053313253",
"output": "536645997"
},
{
"input": "645762257531682045",
"output": "543988614"
},
{
"input": "615812227854199662",
"output": "357939938"
},
{
"input": "819875140559301751",
"output": "968653685"
},
{
"input": "349993003033420740",
"output": "709392758"
},
{
"input": "891351282398722856",
"output": "70758467"
},
{
"input": "563324730406715801",
"output": "353494903"
},
{
"input": "520974001002628386",
"output": "164118419"
},
{
"input": "666729339260489789",
"output": "784700006"
},
{
"input": "856674609788912527",
"output": "720540265"
},
{
"input": "791809296233191092",
"output": "369199735"
},
{
"input": "711066335916901717",
"output": "15590358"
},
{
"input": "931356501703211379",
"output": "239824013"
},
{
"input": "234122431978145893",
"output": "905163056"
},
{
"input": "1000000000000000000",
"output": "899770636"
},
{
"input": "3",
"output": "36"
},
{
"input": "4",
"output": "136"
},
{
"input": "5",
"output": "528"
},
{
"input": "6",
"output": "2080"
},
{
"input": "7",
"output": "8256"
},
{
"input": "8",
"output": "32896"
},
{
"input": "9",
"output": "131328"
},
{
"input": "10",
"output": "524800"
},
{
"input": "11",
"output": "2098176"
},
{
"input": "12",
"output": "8390656"
},
{
"input": "13",
"output": "33558528"
},
{
"input": "14",
"output": "134225920"
},
{
"input": "15",
"output": "536887296"
},
{
"input": "16",
"output": "147516402"
},
{
"input": "0",
"output": "1"
},
{
"input": "6265",
"output": "980996097"
}
] | 1,687,708,226
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 124
| 0
|
from math import floor
n = int(input())
k = 0
if n%2 == 0:
k = 1
else:
k = 0
if n > 0:
c = ((pow(2, n, 10**9+7))//2 - 1 - k)%(10**9+7)
a = int(pow(3, n, 10**9+7)+c*pow(3, n-2, 10**9+7)+k%(10**9+7))
print(a%(10**9+7))
else:
print("1")
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
Help the dwarfs find out how many triangle plants that point "upwards" will be in *n* years.
Input Specification:
The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1018) — the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Output Specification:
Print a single integer — the remainder of dividing the number of plants that will point "upwards" in *n* years by 1000000007 (109<=+<=7).
Demo Input:
['1\n', '2\n']
Demo Output:
['3\n', '10\n']
Note:
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
```python
from math import floor
n = int(input())
k = 0
if n%2 == 0:
k = 1
else:
k = 0
if n > 0:
c = ((pow(2, n, 10**9+7))//2 - 1 - k)%(10**9+7)
a = int(pow(3, n, 10**9+7)+c*pow(3, n-2, 10**9+7)+k%(10**9+7))
print(a%(10**9+7))
else:
print("1")
```
| 0
|
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