MatrixStudio/Codeforces-Python-Submissions

527

A

Playing with Paper

PROGRAMMING

1,100

[ "implementation", "math" ]

null

One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part. After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop. Can you determine how many ships Vasya will make during the lesson?

The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.

Print a single integer — the number of ships that Vasya will make.

[ "2 1\n", "10 7\n", "1000000000000 1\n" ]

[ "2\n", "6\n", "1000000000000\n" ]

Pictures to the first and second sample test.

500

[ { "input": "2 1", "output": "2" }, { "input": "10 7", "output": "6" }, { "input": "1000000000000 1", "output": "1000000000000" }, { "input": "3 1", "output": "3" }, { "input": "4 1", "output": "4" }, { "input": "3 2", "output": "3" }, { "input": "4 2", "output": "2" }, { "input": "1000 700", "output": "6" }, { "input": "959986566087 524054155168", "output": "90" }, { "input": "4 3", "output": "4" }, { "input": "7 6", "output": "7" }, { "input": "1000 999", "output": "1000" }, { "input": "1000 998", "output": "500" }, { "input": "1000 997", "output": "336" }, { "input": "42 1", "output": "42" }, { "input": "1000 1", "output": "1000" }, { "input": "8 5", "output": "5" }, { "input": "13 8", "output": "6" }, { "input": "987 610", "output": "15" }, { "input": "442 42", "output": "22" }, { "input": "754 466", "output": "13" }, { "input": "1000000000000 999999999999", "output": "1000000000000" }, { "input": "1000000000000 999999999998", "output": "500000000000" }, { "input": "941 14", "output": "74" }, { "input": "998 2", "output": "499" }, { "input": "1000 42", "output": "32" }, { "input": "1000 17", "output": "66" }, { "input": "5 1", "output": "5" }, { "input": "5 2", "output": "4" }, { "input": "5 3", "output": "4" }, { "input": "5 4", "output": "5" }, { "input": "293 210", "output": "17" }, { "input": "787878787878 424242424242", "output": "8" }, { "input": "956722026041 591286729879", "output": "58" }, { "input": "956722026041 365435296162", "output": "58" }, { "input": "628625247282 464807889701", "output": "102" }, { "input": "695928431619 424778620208", "output": "167" }, { "input": "1000000000000 42", "output": "23809523821" }, { "input": "987654345678 23", "output": "42941493300" }, { "input": "10000000001 2", "output": "5000000002" }, { "input": "1000000000000 2", "output": "500000000000" }, { "input": "1000000000000 3", "output": "333333333336" }, { "input": "100000000000 3", "output": "33333333336" }, { "input": "100000000000 23", "output": "4347826109" }, { "input": "999999999997 7", "output": "142857142861" }, { "input": "8589934592 4294967296", "output": "2" } ]

1,604,062,479

2,147,483,647

Python 3

OK

TESTS

46

108

0

def solver(a, b): resp = 0 while b != 0: resp += a//b aux = a a = b b = aux % b return resp a, b = list(map(int, input().split())) print(solver(a, b))

Title: Playing with Paper Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part. After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop. Can you determine how many ships Vasya will make during the lesson? Input Specification: The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper. Output Specification: Print a single integer — the number of ships that Vasya will make. Demo Input: ['2 1\n', '10 7\n', '1000000000000 1\n'] Demo Output: ['2\n', '6\n', '1000000000000\n'] Note: Pictures to the first and second sample test.

```python def solver(a, b): resp = 0 while b != 0: resp += a//b aux = a a = b b = aux % b return resp a, b = list(map(int, input().split())) print(solver(a, b)) ```

3

381

A

Sereja and Dima

PROGRAMMING

800

[ "greedy", "implementation", "two pointers" ]

null

Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.

The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.

On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.

[ "4\n4 1 2 10\n", "7\n1 2 3 4 5 6 7\n" ]

[ "12 5\n", "16 12\n" ]

In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

500

[ { "input": "4\n4 1 2 10", "output": "12 5" }, { "input": "7\n1 2 3 4 5 6 7", "output": "16 12" }, { "input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13", "output": "613 418" }, { "input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24", "output": "644 500" }, { "input": "1\n3", "output": "3 0" }, { "input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646", "output": "6848 6568" }, { "input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727", "output": "9562 9561" }, { "input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12", "output": "315 315" }, { "input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304", "output": "3238 2222" }, { "input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325", "output": "5246 4864" }, { "input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71", "output": "8147 7807" }, { "input": "1\n1", "output": "1 0" } ]

1,684,922,467

2,147,483,647

Python 3

OK

TESTS

34

46

0

n = int(input()) arr = list(map(int, input().split())) count = 1 sum_a, sum_b = 0, 0 while len(arr) > 0: if count == 1: x = max(arr[0], arr[len(arr)-1]) sum_a += x arr.pop(arr.index(x)) count = 0 else: y = max(arr[0], arr[len(arr)-1]) sum_b += y arr.pop(arr.index(y)) count = 1 print(sum_a, sum_b)

Title: Sereja and Dima Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. Output Specification: On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. Demo Input: ['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n'] Demo Output: ['12 5\n', '16 12\n'] Note: In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

```python n = int(input()) arr = list(map(int, input().split())) count = 1 sum_a, sum_b = 0, 0 while len(arr) > 0: if count == 1: x = max(arr[0], arr[len(arr)-1]) sum_a += x arr.pop(arr.index(x)) count = 0 else: y = max(arr[0], arr[len(arr)-1]) sum_b += y arr.pop(arr.index(y)) count = 1 print(sum_a, sum_b) ```

3

315

A

Sereja and Bottles

PROGRAMMING

1,400

[ "brute force" ]

null

Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles. Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle. Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.

The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle.

In a single line print a single integer — the answer to the problem.

[ "4\n1 1\n2 2\n3 3\n4 4\n", "4\n1 2\n2 3\n3 4\n4 1\n" ]

[ "4\n", "0\n" ]

none

500

[ { "input": "4\n1 1\n2 2\n3 3\n4 4", "output": "4" }, { "input": "4\n1 2\n2 3\n3 4\n4 1", "output": "0" }, { "input": "3\n2 828\n4 392\n4 903", "output": "3" }, { "input": "4\n2 3\n1 772\n3 870\n3 668", "output": "2" }, { "input": "5\n1 4\n6 6\n4 3\n3 4\n4 758", "output": "2" }, { "input": "6\n4 843\n2 107\n10 943\n9 649\n7 806\n6 730", "output": "6" }, { "input": "7\n351 955\n7 841\n102 377\n394 102\n549 440\n630 324\n624 624", "output": "6" }, { "input": "8\n83 978\n930 674\n542 22\n834 116\n116 271\n640 930\n659 930\n705 987", "output": "6" }, { "input": "9\n162 942\n637 967\n356 108\n768 53\n656 656\n575 32\n32 575\n53 53\n351 222", "output": "6" }, { "input": "10\n423 360\n947 538\n507 484\n31 947\n414 351\n169 901\n901 21\n592 22\n763 200\n656 485", "output": "8" }, { "input": "1\n1000 1000", "output": "1" }, { "input": "1\n500 1000", "output": "1" }, { "input": "11\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11", "output": "11" }, { "input": "49\n1 758\n5 3\n5 3\n4 2\n4 36\n3 843\n5 107\n1 943\n1 649\n2 806\n3 730\n2 351\n2 102\n1 4\n3 4\n3 955\n2 841\n2 377\n5 2\n3 440\n4 324\n3 3\n3 83\n2 2\n2 1\n4 1\n1 931\n3 4\n2 5\n2 5\n4 73\n5 830\n3 4\n3 5\n5 291\n1 2\n5 3\n4 4\n2 3\n3 151\n4 2\n4 431\n5 1\n2 5\n2 4\n4 2\n4 4\n3 1\n5 2", "output": "0" }, { "input": "50\n507 31\n31 250\n414 763\n169 304\n901 9\n592 610\n763 414\n656 789\n411 422\n360 468\n625 504\n538 201\n549 619\n484 797\n596 282\n42 310\n603 656\n351 623\n292 293\n837 180\n375 658\n21 192\n597 729\n22 512\n349 635\n200 56\n669 647\n485 887\n282 939\n735 808\n54 417\n1000 310\n419 652\n939 617\n901 669\n789 390\n128 549\n468 511\n729 837\n894 729\n649 894\n484 22\n808 586\n422 286\n311 427\n618 656\n814 933\n515 901\n310 894\n617 330", "output": "30" }, { "input": "2\n7 7\n5 359", "output": "2" }, { "input": "2\n465 706\n706 706", "output": "1" }, { "input": "2\n1 1\n1 1", "output": "0" }, { "input": "3\n1 1\n1 1\n2 2", "output": "1" }, { "input": "5\n1 1\n1 2\n2 3\n3 4\n4 3", "output": "1" }, { "input": "3\n1 2\n1 2\n1 1", "output": "1" } ]

1,518,702,905

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

122

5,632,000

n=int(input()) a=[] b=[] count=0 for i in range(n): a=a+list(map(int,input().split())) for i in range(0,2*n,2): for j in range(1,2*n,2): if i!=j-1 and a[i]==a[j]: count+=1 print(n-count)

Title: Sereja and Bottles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles. Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle. Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle. Output Specification: In a single line print a single integer — the answer to the problem. Demo Input: ['4\n1 1\n2 2\n3 3\n4 4\n', '4\n1 2\n2 3\n3 4\n4 1\n'] Demo Output: ['4\n', '0\n'] Note: none

```python n=int(input()) a=[] b=[] count=0 for i in range(n): a=a+list(map(int,input().split())) for i in range(0,2*n,2): for j in range(1,2*n,2): if i!=j-1 and a[i]==a[j]: count+=1 print(n-count) ```

0

155

A

I_love_\%username\%

PROGRAMMING

800

[ "brute force" ]

null

Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him. One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously). Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.

The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated. The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.

Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.

[ "5\n100 50 200 150 200\n", "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n" ]

[ "2\n", "4\n" ]

In the first sample the performances number 2 and 3 are amazing. In the second sample the performances number 2, 4, 9 and 10 are amazing.

500

[ { "input": "5\n100 50 200 150 200", "output": "2" }, { "input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242", "output": "4" }, { "input": "1\n6", "output": "0" }, { "input": "2\n2 1", "output": "1" }, { "input": "5\n100 36 53 7 81", "output": "2" }, { "input": "5\n7 36 53 81 100", "output": "4" }, { "input": "5\n100 81 53 36 7", "output": "4" }, { "input": "10\n8 6 3 4 9 10 7 7 1 3", "output": "5" }, { "input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862", "output": "6" }, { "input": "10\n1 3 3 4 6 7 7 8 9 10", "output": "7" }, { "input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137", "output": "9" }, { "input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198", "output": "5" }, { "input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101", "output": "5" }, { "input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498", "output": "7" }, { "input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679", "output": "9" }, { "input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992", "output": "57" }, { "input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83", "output": "4" }, { "input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901", "output": "57" }, { "input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100", "output": "99" }, { "input": "2\n0 10000", "output": "1" }, { "input": "2\n5 5", "output": "0" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n10 10", "output": "0" }, { "input": "1\n0", "output": "0" } ]

1,691,005,037

2,147,483,647

PyPy 3

OK

TESTS

37

156

0

n = int(input()) A = list(map(int,input().strip().split())) ma = mi = A[0] c = 0 for i in A: c += (i > ma or i < mi) ma,mi = max(ma,i),min(mi,i) print(c)

Title: I_love_\%username\% Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him. One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously). Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. Input Specification: The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated. The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. Output Specification: Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. Demo Input: ['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n'] Demo Output: ['2\n', '4\n'] Note: In the first sample the performances number 2 and 3 are amazing. In the second sample the performances number 2, 4, 9 and 10 are amazing.

```python n = int(input()) A = list(map(int,input().strip().split())) ma = mi = A[0] c = 0 for i in A: c += (i > ma or i < mi) ma,mi = max(ma,i),min(mi,i) print(c) ```

3

992

C

Nastya and a Wardrobe

PROGRAMMING

1,600

[ "math" ]

null

Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month). Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year. Nastya owns *x* dresses now, so she became interested in the [expected number](https://en.wikipedia.org/wiki/Expected_value) of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for *k*<=+<=1 months. Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109<=+<=7, because it is easy to see that it is always integer.

The only line contains two integers *x* and *k* (0<=≤<=*x*,<=*k*<=≤<=1018), where *x* is the initial number of dresses and *k*<=+<=1 is the number of months in a year in Byteland.

In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109<=+<=7.

[ "2 0\n", "2 1\n", "3 2\n" ]

[ "4\n", "7\n", "21\n" ]

In the first example a year consists on only one month, so the wardrobe does not eat dresses at all. In the second example after the first month there are 3 dresses with 50% probability and 4 dresses with 50% probability. Thus, in the end of the year there are 6 dresses with 50% probability and 8 dresses with 50% probability. This way the answer for this test is (6 + 8) / 2 = 7.

1,500

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1,532,268,220

2,440

Python 3

WRONG_ANSWER

TESTS

0

93

0

n,k = input().split(" ") j=0 n = int(n) k = int(k)+1 while k >0: n = n*2 j = j+1 k = k-1 n = n-j print(n)

Title: Nastya and a Wardrobe Time Limit: None seconds Memory Limit: None megabytes Problem Description: Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month). Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year. Nastya owns *x* dresses now, so she became interested in the [expected number](https://en.wikipedia.org/wiki/Expected_value) of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for *k*<=+<=1 months. Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109<=+<=7, because it is easy to see that it is always integer. Input Specification: The only line contains two integers *x* and *k* (0<=≤<=*x*,<=*k*<=≤<=1018), where *x* is the initial number of dresses and *k*<=+<=1 is the number of months in a year in Byteland. Output Specification: In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109<=+<=7. Demo Input: ['2 0\n', '2 1\n', '3 2\n'] Demo Output: ['4\n', '7\n', '21\n'] Note: In the first example a year consists on only one month, so the wardrobe does not eat dresses at all. In the second example after the first month there are 3 dresses with 50% probability and 4 dresses with 50% probability. Thus, in the end of the year there are 6 dresses with 50% probability and 8 dresses with 50% probability. This way the answer for this test is (6 + 8) / 2 = 7.

```python n,k = input().split(" ") j=0 n = int(n) k = int(k)+1 while k >0: n = n*2 j = j+1 k = k-1 n = n-j print(n) ```

0

268

A

Games

PROGRAMMING

800

[ "brute force" ]

null

Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different. There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number. You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.

The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.

In a single line print the number of games where the host team is going to play in the guest uniform.

[ "3\n1 2\n2 4\n3 4\n", "4\n100 42\n42 100\n5 42\n100 5\n", "2\n1 2\n1 2\n" ]

[ "1\n", "5\n", "0\n" ]

In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2. In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).

500

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1,680,327,679

2,147,483,647

Python 3

OK

TESTS

31

92

0

n = int(input()) ar = [] c=0 for i in range(n): ar.append(list(map(int, input().split()))) for i in range(n): for j in range(n): if i == j: continue if ar[i][0] == ar[j][1]: c+=1 print(c)

Title: Games Time Limit: None seconds Memory Limit: None megabytes Problem Description: Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different. There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number. You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question. Input Specification: The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively. Output Specification: In a single line print the number of games where the host team is going to play in the guest uniform. Demo Input: ['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n'] Demo Output: ['1\n', '5\n', '0\n'] Note: In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2. In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).

```python n = int(input()) ar = [] c=0 for i in range(n): ar.append(list(map(int, input().split()))) for i in range(n): for j in range(n): if i == j: continue if ar[i][0] == ar[j][1]: c+=1 print(c) ```

3

312

A

Whose sentence is it?

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.

For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.

[ "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n" ]

[ "Freda's\nOMG>...< I don't know!\n" ]

none

500

[ { "input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .", "output": "Freda's\nOMG>...< I don't know!" }, { "input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iapelala.\nCFPlbUgObrXLejPNu.F\nZSUfvisiHyrIMjMlala.\nmiao. lala.\nd,IWSeumytrVlala.", "output": "OMG>......< I don't know!\nFreda's" }, { "input": "10\nmiao.,taUvXPVlala.\nmiao.txEeId.X_lala.\nLZIeAEd JaeBVlala.\ncKPIsWpwIlala.\nfYp.eSvn,g\nKMx,nFEslala.\nmiao.QtMyxYqiajjuM\nDutxNkCqywgcnCYskcd\ngFLKACjeqfD\n,Ss UmY.wJvcX", "output": "OMG>...........< I don't know!" }, { "input": "10\nmiao.grFTpju.jCLRnZ\ng.pVHYA_Usnm\nlloWONolcMFElala.\nAW,n.JJkOTe.Nd\n.bP.HvKlala.\nGziqPGQa,lala.\nmiao.,QkOCH.vFlala.\n.PUtOwImvUsoeh \nmiao.Z,KIds.R\nmiao.,_MDzoaAiJlala.", "output": "Rainbow's\nOMG>.....< I don't know!" }, { "input": "10\nmiao.xWfjV\nHFVrGCDQXyZ,Sbm\nLMDS.xVkTCAY.vm\nmiao.lLBglala.\nnl,jRPyClala.\nFYnHoXlala.\nmiao. oxaHE\n.WTrw_mNpOQCa\nHOk..wHYoyMhl\nQX,XpMuPIROM", "output": "Rainbow's\nOMG>......< I don't know!" }, { "input": "10\nJBQqiXlala.\npUNUWQRiMPCXv\nAiLnfNHWznwkC.lala.\nmiao.Dl_Oy\nxJJJkVkdfOzQBH_SmKh\nfgD_IHvdHiorE,W\nmiao.usBKixglala.\nwCpqPUzEtD\nmiao.rlala.\nmiao.JylcGvWlala.", "output": "Freda's\nOMG>.......< I don't know!" }, { "input": "10\nmiao..FLhPl_Wjslala.\nmiao. tdEGtfdJlala.\nGAzEUlala.\nKCcmOa .aKBlZyYsdu.V\nmiao.lala.\njKylnM,FXK\nmiao.GBWqjGH.v\nmiao.RefxS Cni.\nOxaaEihuHQR_s,\nmiao.a,Axtlala.", "output": "OMG>.......< I don't know!" }, { "input": "10\nNo.I_aTXlala.\nmiao.JKSCoRZS\nnOBMIlala.\nmiao.nlala.\nmiao._xqxoHIIlala.\nmiao.NJPy SWyiUDWc\nmiao.cCnahFaqqj.Xqp\nnreSMDeXPPYAQxI,W\nAktPajWimdd_qRn\nmiao.QHwKCYlala.", "output": "Freda's\nRainbow's\nFreda's\nOMG>.....< I don't know!" }, { "input": "10\n \n,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ \n \nmiao.miao.miao.\nlala.lala.lala.\nlala.miao.\nmiaolala. \nmiao.lala\nmiaolala_\n,.._ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "OMG>.......< I don't know!" }, { "input": "10\nduClyjMIPsEuWmx_Ce.byVoizYlTM,sF\nuZHsNip_,Mwtg,FZjM_LzPC,_pSvEOyTHfAOvoZXvxCZdgYDTCDdCAoSVZWyxXGcLgWlala.\nEGtJFPAvTEcqjkhaGxdduaQ_rmUzF.WaU, EIuX B,aVzFFpFrxpwADXuayRD azDfj \n_tJqYzXyqc.,u.F,mUYukveBPWnPq,f,dJnPHuBazdnbRHfzwNUdRbheAIjcoaPcnLvocrzcioxCapb R\n.YUBeb_zmwUt.QQuUdQIiOXtqshcsycEe,HLytHlala.\ndJndLqGBHt.GfpN.BgvsbXoLh_DIzAJOtFDmLSCYEztvPcS_GHPxivzV,NPMmSAtfk.Mg.w,A UcCt_lCD.csEzyJJBYtSMkzqiA\nmiao.qlala.\nmiao.FmDlY\nmiao.UQI.aJmnanNvRLskuVaMybDMsOlala.\nmiao.lala.", "output": "OMG>.......< I don't know!" }, { "input": "10\nmiao.vyscfysAtWcPkpFHdwZqAQ,UPPcjhKQTlala.\nmiao.KESqus DybUuYFoWVpo..LWZh.UqEdUsTHFlKfzqkThAUPklala.\nUNoE vfZIAdxkiWKhsHPfsqRPTNQoHgAxooVLYxRzugHjo jaEHWQFF\nCCmdIwr.UkoiYWK.Z,,ZesMpISTXNgnpYnJaWquCyL,gO\n.JvOayhXK_bgoYbfAtnXg\nbvdSzRrXoGxVgWvdXnsjEnEfxDzIQo_aZVGDGrzwuAMtzVAHioMBx_DHuTxyieGbGuSRNUojOREqxBBxvCgqAOMzwIWT\nMBuaWduZmRaOGyIPzWOsBVeqtDrblAbXxmM_uRfqMvnVlLEuhVKlhidN_aigiXyq,ZEDqQAx\nmiao.wCHVCuVKNePKmIUFLL_lala.\nmiao.iAqstXHUv\n pMO yvPkNtnNwmUCao W,wW.OvIMVaEeVYHmqaniWq.ivlala.", "output": "OMG>........< I don't know!\nRainbow's\nFreda's" }, { "input": "10\nmiao.\nmiao.jrwLBCpNaDCjyoK.PFzbwWU.h.. wfQquG_P..lala.\nmiao.LGlYdKjw__.Chlala.\nW.wtr qG KDOHj.xWxPbXIXjD_,GJZDaAZ,JBHphsjWJwSKcZAIAi\nmiao.pHsGAZQDWPJQwKC.zHjJituLgp.eUrzObTI.wrpect.FMUJqu,Zuslala.\nmiao.YVlOpXccUA_YU igbsbZbhOVwyYTyOjnWqgiTmxwAuFa.flCHn.,MtVbqxZQl_BGHXWkwijGjuL, ,ezyNlala.\nmiao.xCrVSz.aMv UOSOroDlQxWeBmlWe.FA.ZfUmviMlala.\nxebAlala.\nmiao.qVSxqf vOTlala.\nD.oBUwsLQRgXAoNkQJhQN.w.oMhuvtujnmiwgQYMfjlNTSHh .lSKgI.OEp", "output": "Rainbow's\nOMG>........< I don't know!" }, { "input": "10\nZXXzYlTiQU\nkXE.DdcbOojSaSgjMcFBPubKHefEVAzbi,PDFgSZIz,lala.\nxEfrTCjKhhwBC.UNmJXgTGUdkQeVDlala.\nLfaEw.jvMmuOBWtfoiJNtDIlQAVWNU,xWK_efBBtfkM\nqtBpqKZMWZMX_NKrUAEKYyQcLZWQlqbM\nmiao.PrJEbUtInremuaKRItqXOrfQEjQcAak VQ\nMpGCq awvQaHRvDr uvtVMKsvZI\nmiao.A.RVGu.szCEp.pXQJwL EuTltlN.WradoTvWHJyhcNSoulala.\nmiao.rzlUHzUdxtDRpWRuc,QZwEBfsKKGHMLGtFymPPQdptLFlzZ_ORWqrlfOrlntuDkpXEvz.CxwAsFYUvpnOnFWG\nmiao.VXUoNBwlgBwcna_n.CgAAcKKUuiVA.doOJKHpMdwNwlHAcLpdfN.Awa SthrlEWpUcuOonUTxIQNszYcHDXxnhArrM..A", "output": "OMG>.....< I don't know!\nRainbow's\nRainbow's" }, { "input": "10\nmiao.qbxBFzrjtWv.yOk\nDBgi,loApO AACrGnwssCHN\nmiao.LV.wbQEE_V.BSAtdTIHTQOJVJ_nGOthbL,nJvQ.UeWFpsa.GGsK_Uv,HQxHS,AN_bkrolala.\nmiao.tBEqk rIQuByGKhfq_iP.BW,nySZEfrfySEcqnnIzxC,lrjIiivbxlkoVXJFiegGFRn NO,txGPhVBcv.CVhMmNO zlala.\nmiao.aBZWDWxk.wkR ,NyCzGxJnJDqBZpetdUPAmmBZDXl_Tbflala.\nmiao. XN,uMwWm. VqloYr..jTLszlala.\n.rshcgfZ.eZOdMu_RMh\nmiao.ahiwpECEe.lala.\nLeoUSroTekQAMSO__M L_ZEeRD_tUihYvQETFB,RzJmFtFiKrU\nBtygQG_OoFEFBL.KsVWTYbtqtalXoStFCZ RINHda.NuLmlkRB.vAQJFvelbsfoJ.T,M sJn", "output": "Rainbow's\nOMG>.........< I don't know!" }, { "input": "10\nYoYBCcaqhXLfvKKf.UYMODTHyPZlala.\ncxgWn J.Q\nmiao.nwH.IHntgKYDhdsjU DMTHXEVRyeJP ZaAecCIBJXuv.YjhEmtbjvjKnK.U,oc,x\nmiao.EcQ.FDtRJgmpAzxhq.RwXBLxjyC,IeMqaFoheMPFCGWBcwUAFnbiwlbz_fcsEGPfJaeryCtFocBNEWTlala.\nmiao.W\nmiao. ZQpIeyCXJSnFgAIzu.THfrmyoogYWQzFqblala.\nmiao.ifzdCwnTDcxpvdr OTC.YqPv.MKDp..utICtAsbfYyGlala.\nmiao.\nmiao.tS.U.wH.s,CxORZJsBAHLi,fXeoDJWVBH\nrcUMpeupOVRKrcIRAvU.rP kgUEfoeXcrFPQOBYG.BNvAQPg.XHMWizhLpZNljXc .LQmVXCi", "output": "Freda's\nOMG>.....< I don't know!" }, { "input": "10\nlala.\nmiao.milalala.lmmialamiao.la.o.iao.a.ao.\nmialala.o.\nmiao.millala.allala.amiao..miao.miao.lala.ao.miammiao.iao.o.\nmiao.miaomiao..\nlalmiao.amiao..\nmiao.lala.lamiamiaolala..o.lalala.miao..\nmlala.iao.lalamiao..\nlmlala.iao.alalamiao.lmialala.lala.miao.o.alala..lala..lalmiaomiao..lalmiao.a.lalamiao..miao.alala..\nlalllamiao.la.lala.alamiao.lalalala.lala..miao.lamiao.la.lallalamiao..a..a.", "output": "Freda's\nRainbow's\nOMG>.....< I don't know!" }, { "input": "10\nlalllala.ala.lala.a.mmimiao.aomiao.lllala.ala.amiao.la.mialalala.la.o..imiao.miao.amlala.iao.o.\nmilala.alllala.ala.amiao.lamiao..o.\nlala.lalalala..lalalala..\nlala.miao.\nmimiao.ao.lala.\nlalmiao.amlala.iamialala.o.o..\nlalammlala.iaolammiao.imiao.ao.la..iao..\nmiao.mialala.omiao..mlala.iaolala..\nmiamiao.o.llallala.ala.la.miao.ala.miao.mimialmiao.ala.o.alala.miaomiao..olala..\nmialala.lamiao.la.lala.miao.ollala.allala.ala.lmiaommiao.imiao.ao.lallallala.a.miao.a..a..", "output": "OMG>........< I don't know!" }, { "input": "10\nlamiao.lamiao.mimiao.ao..\nllala.almiaomiao..lala.miao.a.\nlalala.lala.lalala.lala.lalala..la.\nlalala.la.miao.\nmiao.llalallala.miao.a.la.lala.almimiao.ao.a.\nmiao.mlala.ilala.aomilala.ao..\nmilala.ao.\nmmiao.iao.\nmlala.lala.ialalmiao.alala..olala.lala..\nlala.lmlalalalalala...iao.milala.aommiao.iao..alamiaolallala.milala.ao.a..lalalmiao.a..llala.amiao..", "output": "OMG>..................< I don't know!" } ]

1,608,745,072

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

78

0

#again comeback !!! for q in range(int(input())): s = list(map(str, input().split())) print("Freda's"if s[-1] == 'lala.' else "Rainbow's" if s[0] == 'miao.' else "OMG>.< I don't know!")

Title: Whose sentence is it? Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. Output Specification: For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions. Demo Input: ['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n'] Demo Output: ["Freda's\nOMG>...< I don't know!\n"] Note: none

```python #again comeback !!! for q in range(int(input())): s = list(map(str, input().split())) print("Freda's"if s[-1] == 'lala.' else "Rainbow's" if s[0] == 'miao.' else "OMG>.< I don't know!") ```

0

508

A

Pasha and Pixels

PROGRAMMING

1,100

[ "brute force" ]

null

Pasha loves his phone and also putting his hair up... But the hair is now irrelevant. Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed. Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move. Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.

The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform. The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.

If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed. If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.

[ "2 2 4\n1 1\n1 2\n2 1\n2 2\n", "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n", "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n" ]

[ "4\n", "5\n", "0\n" ]

none

500

[ { "input": "2 2 4\n1 1\n1 2\n2 1\n2 2", "output": "4" }, { "input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1", "output": "5" }, { "input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2", "output": "0" }, { "input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3", "output": "9" }, { "input": "2 2 5\n1 1\n2 1\n2 1\n1 2\n2 2", "output": "5" }, { "input": "518 518 10\n37 97\n47 278\n17 467\n158 66\n483 351\n83 123\n285 219\n513 187\n380 75\n304 352", "output": "0" }, { "input": "1 1 5\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "0" }, { "input": "1 5 5\n1 1\n1 2\n1 3\n1 4\n1 5", "output": "0" }, { "input": "5 1 5\n1 1\n2 1\n3 1\n4 1\n5 1", "output": "0" }, { "input": "1 1 1\n1 1", "output": "0" }, { "input": "10 10 4\n5 9\n6 9\n6 10\n5 10", "output": "4" }, { "input": "1000 1000 4\n999 999\n999 1000\n1000 999\n1000 1000", "output": "4" }, { "input": "2 3 5\n2 3\n1 3\n1 2\n1 1\n2 2", "output": "5" }, { "input": "1000 1000 4\n1000 1000\n999 999\n1000 999\n999 1000", "output": "4" } ]

1,668,090,126

2,147,483,647

Python 3

OK

TESTS

52

327

4,505,600

n, m, k = map(int, input().split()) matrix = [[0]*(m + 2) for _ in range(n + 2)] def check(i,j): if matrix[i-1][j-1] and matrix[i-1][j] and matrix[i][j-1]: return True if matrix[i-1][j] and matrix[i-1][j+1] and matrix[i][j+1]: return True if matrix[i][j-1] and matrix[i+1][j-1] and matrix[i+1][j]: return True if matrix[i][j+1] and matrix[i+1][j] and matrix[i+1][j+1]: return True return False for s in range(k): i, j = map(int, input().split()) matrix[i][j] = 1 if check(i, j): print(s + 1) break else: print(0)

Title: Pasha and Pixels Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pasha loves his phone and also putting his hair up... But the hair is now irrelevant. Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed. Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move. Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed. Input Specification: The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform. The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move. Output Specification: If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed. If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0. Demo Input: ['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n'] Demo Output: ['4\n', '5\n', '0\n'] Note: none

```python n, m, k = map(int, input().split()) matrix = [[0]*(m + 2) for _ in range(n + 2)] def check(i,j): if matrix[i-1][j-1] and matrix[i-1][j] and matrix[i][j-1]: return True if matrix[i-1][j] and matrix[i-1][j+1] and matrix[i][j+1]: return True if matrix[i][j-1] and matrix[i+1][j-1] and matrix[i+1][j]: return True if matrix[i][j+1] and matrix[i+1][j] and matrix[i+1][j+1]: return True return False for s in range(k): i, j = map(int, input().split()) matrix[i][j] = 1 if check(i, j): print(s + 1) break else: print(0) ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,622,015,400

2,147,483,647

PyPy 3

OK

TESTS

81

216

0

n = int(input()) li = [input().split() for j in range(n)] xsum = 0 i = 0 while i<n: xsum += int(li[i][0]) i+=1 ysum = 0 j = 0 while j<n: ysum += int(li[j][1]) j+=1 zsum = 0 k = 0 while k<n: zsum += int(li[k][2]) k+=1 if xsum == 0 and ysum == 0 and zsum == 0: print("YES") else: print("NO")

Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none

```python n = int(input()) li = [input().split() for j in range(n)] xsum = 0 i = 0 while i<n: xsum += int(li[i][0]) i+=1 ysum = 0 j = 0 while j<n: ysum += int(li[j][1]) j+=1 zsum = 0 k = 0 while k<n: zsum += int(li[k][2]) k+=1 if xsum == 0 and ysum == 0 and zsum == 0: print("YES") else: print("NO") ```

3.946

746

B

Decoding

PROGRAMMING

900

[ "implementation", "strings" ]

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" } ]

1,622,232,701

2,147,483,647

Python 3

OK

TESTS

34

62

0

n = int(input()) s = input() q = s[::-1] l = [] for i in range(len(q)): if(l == []): l.append(q[i]) else: l.insert(len(l)//2,q[i]) print("".join(l))

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 n = int(input()) s = input() q = s[::-1] l = [] for i in range(len(q)): if(l == []): l.append(q[i]) else: l.insert(len(l)//2,q[i]) print("".join(l)) ```

3

488

A

Giga Tower

PROGRAMMING

1,100

[ "brute force" ]

null

Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view. In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?). Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number.

The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109).

Print the minimum *b* in a line.

[ "179\n", "-1\n", "18\n" ]

[ "1\n", "9\n", "10\n" ]

For the first sample, he has to arrive at the floor numbered 180. For the second sample, he will arrive at 8. Note that *b* should be positive, so the answer for the third sample is 10, not 0.

500

[ { "input": "179", "output": "1" }, { "input": "-1", "output": "9" }, { "input": "18", "output": "10" }, { "input": "-410058385", "output": "1" }, { "input": "-586825624", "output": "1" }, { "input": "852318890", "output": "1" }, { "input": "919067153", "output": "5" }, { "input": "690422411", "output": "7" }, { "input": "-408490162", "output": "1" }, { "input": "-8", "output": "16" }, { "input": "-6", "output": "14" }, { "input": "-4", "output": "12" }, { "input": "-2", "output": "10" }, { "input": "0", "output": "8" }, { "input": "2", "output": "6" }, { "input": "4", "output": "4" }, { "input": "6", "output": "2" }, { "input": "8", "output": "10" }, { "input": "1000000000", "output": "8" }, { "input": "-1000000000", "output": "2" }, { "input": "88888", "output": "1" }, { "input": "89", "output": "9" }, { "input": "-80000000", "output": "2" }, { "input": "-8888", "output": "1" }, { "input": "-17", "output": "9" }, { "input": "78", "output": "2" }, { "input": "-19", "output": "1" }, { "input": "-999999998", "output": "9" }, { "input": "-999999997", "output": "8" }, { "input": "999999997", "output": "1" }, { "input": "811111111", "output": "1" }, { "input": "-8", "output": "16" }, { "input": "-5", "output": "13" }, { "input": "-7", "output": "15" }, { "input": "1000000000", "output": "8" } ]

1,597,817,730

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

108

0

# import sys # sys.stdin = open('input.txt', 'r') # sys.stdout = open('output.txt', 'w') n = int(input()) if n < 0: print(8-n) elif n == 0: print(8) else: for i in range(1, 11): if (n+i) % 10 == 8: print(i) break # -----------------------------------------------------------------------------------------!

Title: Giga Tower Time Limit: None seconds Memory Limit: None megabytes Problem Description: Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view. In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?). Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number. Input Specification: The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109). Output Specification: Print the minimum *b* in a line. Demo Input: ['179\n', '-1\n', '18\n'] Demo Output: ['1\n', '9\n', '10\n'] Note: For the first sample, he has to arrive at the floor numbered 180. For the second sample, he will arrive at 8. Note that *b* should be positive, so the answer for the third sample is 10, not 0.

```python # import sys # sys.stdin = open('input.txt', 'r') # sys.stdout = open('output.txt', 'w') n = int(input()) if n < 0: print(8-n) elif n == 0: print(8) else: for i in range(1, 11): if (n+i) % 10 == 8: print(i) break # -----------------------------------------------------------------------------------------! ```

0

none

0

[ "none" ]

null

This is an interactive problem. Jury has hidden a permutation *p* of integers from 0 to *n*<=-<=1. You know only the length *n*. Remind that in permutation all integers are distinct. Let *b* be the inverse permutation for *p*, i.e. *p**b**i*<==<=*i* for all *i*. The only thing you can do is to ask xor of elements *p**i* and *b**j*, printing two indices *i* and *j* (not necessarily distinct). As a result of the query with indices *i* and *j* you'll get the value , where denotes the xor operation. You can find the description of xor operation in notes. Note that some permutations can remain indistinguishable from the hidden one, even if you make all possible *n*2 queries. You have to compute the number of permutations indistinguishable from the hidden one, and print one of such permutations, making no more than 2*n* queries. The hidden permutation does not depend on your queries.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=5000) — the length of the hidden permutation. You should read this integer first.

When your program is ready to print the answer, print three lines. In the first line print "!". In the second line print single integer *answers*_*cnt* — the number of permutations indistinguishable from the hidden one, including the hidden one. In the third line print *n* integers *p*0,<=*p*1,<=...,<=*p**n*<=-<=1 (0<=≤<=*p**i*<=<<=*n*, all *p**i* should be distinct) — one of the permutations indistinguishable from the hidden one. Your program should terminate after printing the answer.

[ "3\n0\n0\n3\n2\n3\n2", "4\n2\n3\n2\n0\n2\n3\n2\n0" ]

[ "? 0 0\n? 1 1\n? 1 2\n? 0 2\n? 2 1\n? 2 0\n!\n1\n0 1 2", "? 0 1\n? 1 2\n? 2 3\n? 3 3\n? 3 2\n? 2 1\n? 1 0\n? 0 0\n!\n2\n3 1 2 0" ]

xor operation, or bitwise exclusive OR, is an operation performed over two integers, in which the *i*-th digit in binary representation of the result is equal to 1 if and only if exactly one of the two integers has the *i*-th digit in binary representation equal to 1. For more information, see [here](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). In the first example *p* = [0, 1, 2], thus *b* = [0, 1, 2], the values <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5dae53b8ed0e62c9af6c21130fc022b384e359ff.png" style="max-width: 100.0%;max-height: 100.0%;"/> are correct for the given *i*, *j*. There are no other permutations that give the same answers for the given queries. The answers for the queries are: - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b36cf4ccf629eaa3a225cda06a5697eb030eb4d7.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/88e848d2e69ecfb3b8c8d63fdc8949c7cc1e9f28.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/233e40a031e99efe33bdf68f4a383163a23e1e7b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3ec81fca9911a6bde9c2c7cae6c189514105085f.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/be5d5595e56564765e3db6b865de2219cdadadeb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3355af6093e240986c338bf13071b479e2604ca9.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second example *p* = [3, 1, 2, 0], and *b* = [3, 1, 2, 0], the values <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5dae53b8ed0e62c9af6c21130fc022b384e359ff.png" style="max-width: 100.0%;max-height: 100.0%;"/> match for all pairs *i*, *j*. But there is one more suitable permutation *p* = [0, 2, 1, 3], *b* = [0, 2, 1, 3] that matches all *n*2 possible queries as well. All other permutations do not match even the shown queries.

0

[ { "input": "3\n0 1 2", "output": "1\n0 1 2 " }, { "input": "4\n3 1 2 0", "output": "2\n0 2 1 3 " }, { "input": "4\n3 2 1 0", "output": "4\n0 1 2 3 " }, { "input": "8\n2 3 0 1 4 5 6 7", "output": "4\n0 1 2 3 6 7 4 5 " }, { "input": "1\n0", "output": "1\n0 " }, { "input": "10\n0 1 2 3 4 5 6 7 8 9", "output": "2\n0 1 2 3 4 5 6 7 8 9 " }, { "input": "6\n0 1 2 4 3 5", "output": "1\n0 1 2 4 3 5 " }, { "input": "2\n0 1", "output": "2\n0 1 " }, { "input": "10\n6 7 4 5 0 1 8 9 2 3", "output": "2\n6 7 4 5 0 1 8 9 2 3 " }, { "input": "10\n9 8 1 7 6 4 5 2 0 3", "output": "1\n9 8 1 7 6 4 5 2 0 3 " }, { "input": "64\n0 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 62 63 60 61", "output": "4\n0 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 62 63 60 61 " }, { "input": "128\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 126 123 124 125 122 127", "output": "2\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 126 123 124 125 122 127 " } ]

1,508,061,764

7,064

Python 3

TIME_LIMIT_EXCEEDED

PRETESTS

14

2,000

5,529,600

import sys n = int(input()) def ask(a, b): print('?', a, b) sys.stdout.flush() return int(input()) ret = [] for i in range(n): ret.append(ask(0, i)) y = ret.index(0) ret2 = [ret[y], ] for i in range(1, n): ret2.append(ask(i, y)) cnt = 0 for i in range(n): ps = [x ^ i for x in ret2] bs = [y ^ i for y in ret] if any(x >= n for x in ps) or any(y >= n for y in bs): continue if all(bs[y] == x for x, y in enumerate(ps)): cnt += 1 ans = ps print('!') print(cnt) print(*ans)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: This is an interactive problem. Jury has hidden a permutation *p* of integers from 0 to *n*<=-<=1. You know only the length *n*. Remind that in permutation all integers are distinct. Let *b* be the inverse permutation for *p*, i.e. *p**b**i*<==<=*i* for all *i*. The only thing you can do is to ask xor of elements *p**i* and *b**j*, printing two indices *i* and *j* (not necessarily distinct). As a result of the query with indices *i* and *j* you'll get the value , where denotes the xor operation. You can find the description of xor operation in notes. Note that some permutations can remain indistinguishable from the hidden one, even if you make all possible *n*2 queries. You have to compute the number of permutations indistinguishable from the hidden one, and print one of such permutations, making no more than 2*n* queries. The hidden permutation does not depend on your queries. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=5000) — the length of the hidden permutation. You should read this integer first. Output Specification: When your program is ready to print the answer, print three lines. In the first line print "!". In the second line print single integer *answers*_*cnt* — the number of permutations indistinguishable from the hidden one, including the hidden one. In the third line print *n* integers *p*0,<=*p*1,<=...,<=*p**n*<=-<=1 (0<=≤<=*p**i*<=<<=*n*, all *p**i* should be distinct) — one of the permutations indistinguishable from the hidden one. Your program should terminate after printing the answer. Demo Input: ['3\n0\n0\n3\n2\n3\n2', '4\n2\n3\n2\n0\n2\n3\n2\n0'] Demo Output: ['? 0 0\n? 1 1\n? 1 2\n? 0 2\n? 2 1\n? 2 0\n!\n1\n0 1 2', '? 0 1\n? 1 2\n? 2 3\n? 3 3\n? 3 2\n? 2 1\n? 1 0\n? 0 0\n!\n2\n3 1 2 0'] Note: xor operation, or bitwise exclusive OR, is an operation performed over two integers, in which the *i*-th digit in binary representation of the result is equal to 1 if and only if exactly one of the two integers has the *i*-th digit in binary representation equal to 1. For more information, see [here](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). In the first example *p* = [0, 1, 2], thus *b* = [0, 1, 2], the values <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5dae53b8ed0e62c9af6c21130fc022b384e359ff.png" style="max-width: 100.0%;max-height: 100.0%;"/> are correct for the given *i*, *j*. There are no other permutations that give the same answers for the given queries. The answers for the queries are: - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b36cf4ccf629eaa3a225cda06a5697eb030eb4d7.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/88e848d2e69ecfb3b8c8d63fdc8949c7cc1e9f28.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/233e40a031e99efe33bdf68f4a383163a23e1e7b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3ec81fca9911a6bde9c2c7cae6c189514105085f.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/be5d5595e56564765e3db6b865de2219cdadadeb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3355af6093e240986c338bf13071b479e2604ca9.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second example *p* = [3, 1, 2, 0], and *b* = [3, 1, 2, 0], the values <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5dae53b8ed0e62c9af6c21130fc022b384e359ff.png" style="max-width: 100.0%;max-height: 100.0%;"/> match for all pairs *i*, *j*. But there is one more suitable permutation *p* = [0, 2, 1, 3], *b* = [0, 2, 1, 3] that matches all *n*2 possible queries as well. All other permutations do not match even the shown queries.

```python import sys n = int(input()) def ask(a, b): print('?', a, b) sys.stdout.flush() return int(input()) ret = [] for i in range(n): ret.append(ask(0, i)) y = ret.index(0) ret2 = [ret[y], ] for i in range(1, n): ret2.append(ask(i, y)) cnt = 0 for i in range(n): ps = [x ^ i for x in ret2] bs = [y ^ i for y in ret] if any(x >= n for x in ps) or any(y >= n for y in bs): continue if all(bs[y] == x for x, y in enumerate(ps)): cnt += 1 ans = ps print('!') print(cnt) print(*ans) ```

0

916

A

Jamie and Alarm Snooze

PROGRAMMING

900

[ "brute force", "implementation", "math" ]

null

Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button. A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky. Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*. Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'. Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.

The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60). The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).

Print the minimum number of times he needs to press the button.

[ "3\n11 23\n", "5\n01 07\n" ]

[ "2\n", "0\n" ]

In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20. In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.

500

[ { "input": "3\n11 23", "output": "2" }, { "input": "5\n01 07", "output": "0" }, { "input": "34\n09 24", "output": "3" }, { "input": "2\n14 37", "output": "0" }, { "input": "14\n19 54", "output": "9" }, { "input": "42\n15 44", "output": "12" }, { "input": "46\n02 43", "output": "1" }, { "input": "14\n06 41", "output": "1" }, { "input": "26\n04 58", "output": "26" }, { "input": "54\n16 47", "output": "0" }, { "input": "38\n20 01", "output": "3" }, { "input": "11\n02 05", "output": "8" }, { "input": "55\n22 10", "output": "5" }, { "input": "23\n10 08", "output": "6" }, { "input": "23\n23 14", "output": "9" }, { "input": "51\n03 27", "output": "0" }, { "input": "35\n15 25", "output": "13" }, { "input": "3\n12 15", "output": "6" }, { "input": "47\n00 28", "output": "3" }, { "input": "31\n13 34", "output": "7" }, { "input": "59\n17 32", "output": "0" }, { "input": "25\n11 03", "output": "8" }, { "input": "9\n16 53", "output": "4" }, { "input": "53\n04 06", "output": "3" }, { "input": "37\n00 12", "output": "5" }, { "input": "5\n13 10", "output": "63" }, { "input": "50\n01 59", "output": "10" }, { "input": "34\n06 13", "output": "4" }, { "input": "2\n18 19", "output": "1" }, { "input": "46\n06 16", "output": "17" }, { "input": "14\n03 30", "output": "41" }, { "input": "40\n13 37", "output": "0" }, { "input": "24\n17 51", "output": "0" }, { "input": "8\n14 57", "output": "0" }, { "input": "52\n18 54", "output": "2" }, { "input": "20\n15 52", "output": "24" }, { "input": "20\n03 58", "output": "30" }, { "input": "48\n07 11", "output": "0" }, { "input": "32\n04 01", "output": "2" }, { "input": "60\n08 15", "output": "1" }, { "input": "44\n20 20", "output": "4" }, { "input": "55\n15 35", "output": "9" }, { "input": "55\n03 49", "output": "11" }, { "input": "23\n16 39", "output": "4" }, { "input": "7\n20 36", "output": "7" }, { "input": "35\n16 42", "output": "1" }, { "input": "35\n05 56", "output": "21" }, { "input": "3\n17 45", "output": "0" }, { "input": "47\n05 59", "output": "6" }, { "input": "15\n10 13", "output": "9" }, { "input": "59\n06 18", "output": "9" }, { "input": "34\n17 18", "output": "0" }, { "input": "18\n05 23", "output": "2" }, { "input": "46\n17 21", "output": "0" }, { "input": "30\n06 27", "output": "0" }, { "input": "14\n18 40", "output": "3" }, { "input": "58\n22 54", "output": "6" }, { "input": "26\n19 44", "output": "5" }, { "input": "10\n15 57", "output": "0" }, { "input": "54\n20 47", "output": "0" }, { "input": "22\n08 45", "output": "3" }, { "input": "48\n18 08", "output": "1" }, { "input": "32\n07 06", "output": "0" }, { "input": "60\n19 19", "output": "2" }, { "input": "45\n07 25", "output": "0" }, { "input": "29\n12 39", "output": "8" }, { "input": "13\n08 28", "output": "3" }, { "input": "41\n21 42", "output": "5" }, { "input": "41\n09 32", "output": "3" }, { "input": "9\n21 45", "output": "2" }, { "input": "37\n10 43", "output": "5" }, { "input": "3\n20 50", "output": "1" }, { "input": "47\n00 04", "output": "1" }, { "input": "15\n13 10", "output": "21" }, { "input": "15\n17 23", "output": "0" }, { "input": "43\n22 13", "output": "2" }, { "input": "27\n10 26", "output": "6" }, { "input": "55\n22 24", "output": "5" }, { "input": "55\n03 30", "output": "11" }, { "input": "24\n23 27", "output": "0" }, { "input": "52\n11 33", "output": "3" }, { "input": "18\n22 48", "output": "17" }, { "input": "1\n12 55", "output": "8" }, { "input": "1\n04 27", "output": "0" }, { "input": "1\n12 52", "output": "5" }, { "input": "1\n20 16", "output": "9" }, { "input": "1\n04 41", "output": "4" }, { "input": "1\n20 21", "output": "4" }, { "input": "1\n04 45", "output": "8" }, { "input": "1\n12 18", "output": "1" }, { "input": "1\n04 42", "output": "5" }, { "input": "1\n02 59", "output": "2" }, { "input": "1\n18 24", "output": "7" }, { "input": "1\n02 04", "output": "7" }, { "input": "1\n18 28", "output": "1" }, { "input": "1\n18 01", "output": "2" }, { "input": "1\n10 25", "output": "8" }, { "input": "1\n02 49", "output": "2" }, { "input": "1\n02 30", "output": "3" }, { "input": "1\n18 54", "output": "7" }, { "input": "1\n02 19", "output": "2" }, { "input": "1\n05 25", "output": "8" }, { "input": "60\n23 55", "output": "6" }, { "input": "60\n08 19", "output": "1" }, { "input": "60\n00 00", "output": "7" }, { "input": "60\n08 24", "output": "1" }, { "input": "60\n16 13", "output": "9" }, { "input": "60\n08 21", "output": "1" }, { "input": "60\n16 45", "output": "9" }, { "input": "60\n08 26", "output": "1" }, { "input": "60\n08 50", "output": "1" }, { "input": "60\n05 21", "output": "12" }, { "input": "60\n13 29", "output": "6" }, { "input": "60\n05 18", "output": "12" }, { "input": "60\n13 42", "output": "6" }, { "input": "60\n05 07", "output": "0" }, { "input": "60\n05 47", "output": "0" }, { "input": "60\n21 55", "output": "4" }, { "input": "60\n05 36", "output": "12" }, { "input": "60\n21 08", "output": "4" }, { "input": "60\n21 32", "output": "4" }, { "input": "60\n16 31", "output": "9" }, 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1,516,372,814

314

Python 3

OK

TESTS

254

77

5,632,000

x = int(input()) h, m = [int(x) for x in input().split()] for y in range(3600): t = h * 60 + m - x * y if t < 0: t += 60 * 24 h_new = t // 60 m_new = t % 60 if '7' in str(h_new) + str(m_new): print(y) break

Title: Jamie and Alarm Snooze Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button. A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky. Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*. Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'. Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00. Input Specification: The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60). The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59). Output Specification: Print the minimum number of times he needs to press the button. Demo Input: ['3\n11 23\n', '5\n01 07\n'] Demo Output: ['2\n', '0\n'] Note: In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20. In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.

```python x = int(input()) h, m = [int(x) for x in input().split()] for y in range(3600): t = h * 60 + m - x * y if t < 0: t += 60 * 24 h_new = t // 60 m_new = t % 60 if '7' in str(h_new) + str(m_new): print(y) break ```

3

899

C

Dividing the numbers

PROGRAMMING

1,300

[ "constructive algorithms", "graphs", "math" ]

null

Petya has *n* integers: 1,<=2,<=3,<=...,<=*n*. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible. Help Petya to split the integers. Each of *n* integers should be exactly in one group.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of integers Petya has.

Print the smallest possible absolute difference in the first line. In the second line print the size of the first group, followed by the integers in that group. You can print these integers in arbitrary order. If there are multiple answers, print any of them.

[ "4\n", "2\n" ]

[ "0\n2 1 4 \n", "1\n1 1 \n" ]

In the first example you have to put integers 1 and 4 in the first group, and 2 and 3 in the second. This way the sum in each group is 5, and the absolute difference is 0. In the second example there are only two integers, and since both groups should be non-empty, you have to put one integer in the first group and one in the second. This way the absolute difference of sums of integers in each group is 1.

1,500

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249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 2..." }, { "input": "53520", "output": "0\n26760 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 ..." }, { "input": "37193", "output": "1\n18597\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "12645", "output": "1\n6323\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 15..." }, { "input": "53237", "output": "1\n26619\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "28693", "output": "1\n14347\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "4145", "output": "1\n2073\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 15..." }, { "input": "36042", "output": "1\n18021 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 ..." }, { "input": "16646", "output": "1\n8323 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 2..." }, { "input": "57238", "output": "1\n28619 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 ..." }, { "input": "27542", "output": "1\n13771 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 ..." }, { "input": "8146", "output": "1\n4073 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 2..." }, { "input": "46659", "output": "0\n23329\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "27259", "output": "0\n13629\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "2715", "output": "0\n1357\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 15..." }, { "input": "38159", "output": "0\n19079\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "18759", "output": "0\n9379\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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 15..." } ]

1,516,620,378

2,147,483,647

Python 3

OK

TESTS

60

62

6,758,400

n=int(input()) if n==2: print(1) print('1 1') exit() dif=from_pos=to_pos=0 mod=n%4 if mod==0 or mod==3: from_pos=n//4+1 if mod==0: to_pos=from_pos+n//2 else: to_pos=from_pos+n//2+1 else: dif=1 if mod==1: from_pos=n//4+2 to_pos=from_pos+n//2 else: from_pos=n//4 to_pos=from_pos+n//2+1 print(dif) print(to_pos-from_pos,' '.join(map(str,range(from_pos,to_pos))))

Title: Dividing the numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has *n* integers: 1,<=2,<=3,<=...,<=*n*. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible. Help Petya to split the integers. Each of *n* integers should be exactly in one group. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of integers Petya has. Output Specification: Print the smallest possible absolute difference in the first line. In the second line print the size of the first group, followed by the integers in that group. You can print these integers in arbitrary order. If there are multiple answers, print any of them. Demo Input: ['4\n', '2\n'] Demo Output: ['0\n2 1 4 \n', '1\n1 1 \n'] Note: In the first example you have to put integers 1 and 4 in the first group, and 2 and 3 in the second. This way the sum in each group is 5, and the absolute difference is 0. In the second example there are only two integers, and since both groups should be non-empty, you have to put one integer in the first group and one in the second. This way the absolute difference of sums of integers in each group is 1.

```python n=int(input()) if n==2: print(1) print('1 1') exit() dif=from_pos=to_pos=0 mod=n%4 if mod==0 or mod==3: from_pos=n//4+1 if mod==0: to_pos=from_pos+n//2 else: to_pos=from_pos+n//2+1 else: dif=1 if mod==1: from_pos=n//4+2 to_pos=from_pos+n//2 else: from_pos=n//4 to_pos=from_pos+n//2+1 print(dif) print(to_pos-from_pos,' '.join(map(str,range(from_pos,to_pos)))) ```

3

770

A

New Password

PROGRAMMING

800

[ "*special", "implementation" ]

null

Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help. Innokentiy decides that new password should satisfy the following conditions: - the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct. Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.

The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it. Pay attention that a desired new password always exists.

Print any password which satisfies all conditions given by Innokentiy.

[ "4 3\n", "6 6\n", "5 2\n" ]

[ "java\n", "python\n", "phphp\n" ]

In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it. In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters. In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it. Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.

500

[ { "input": "4 3", "output": "abca" }, { "input": "6 6", "output": "abcdef" }, { "input": "5 2", "output": "ababa" }, { "input": "3 2", "output": "aba" }, { "input": "10 2", "output": "ababababab" }, { "input": "26 13", "output": "abcdefghijklmabcdefghijklm" }, { "input": "100 2", "output": "abababababababababababababababababababababababababababababababababababababababababababababababababab" }, { "input": "100 10", "output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij" }, { "input": "3 3", "output": "abc" }, { "input": "6 3", "output": "abcabc" }, { "input": "10 3", "output": "abcabcabca" }, { "input": "50 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab" }, { "input": "90 2", "output": "ababababababababababababababababababababababababababababababababababababababababababababab" }, { "input": "6 2", "output": "ababab" }, { "input": "99 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc" }, { "input": "4 2", "output": "abab" }, { "input": "100 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca" }, { "input": "40 22", "output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr" }, { "input": "13 8", "output": "abcdefghabcde" }, { "input": "16 15", "output": "abcdefghijklmnoa" }, { "input": "17 17", "output": "abcdefghijklmnopq" }, { "input": "19 4", "output": "abcdabcdabcdabcdabc" }, { "input": "100 26", "output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv" }, { "input": "100 25", "output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy" }, { "input": "26 26", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "27 26", "output": "abcdefghijklmnopqrstuvwxyza" }, { "input": "2 2", "output": "ab" }, { "input": "26 25", "output": "abcdefghijklmnopqrstuvwxya" }, { "input": "99 2", "output": "abababababababababababababababababababababababababababababababababababababababababababababababababa" }, { "input": "99 26", "output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu" }, { "input": "4 4", "output": "abcd" }, { "input": "5 3", "output": "abcab" }, { "input": "5 4", "output": "abcda" }, { "input": "5 5", "output": "abcde" }, { "input": "24 22", "output": "abcdefghijklmnopqrstuvab" }, { "input": "26 14", "output": "abcdefghijklmnabcdefghijkl" }, { "input": "26 15", "output": "abcdefghijklmnoabcdefghijk" }, { "input": "30 12", "output": "abcdefghijklabcdefghijklabcdef" }, { "input": "35 4", "output": "abcdabcdabcdabcdabcdabcdabcdabcdabc" }, { "input": "79 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca" }, { "input": "79 14", "output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi" }, { "input": "85 13", "output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg" }, { "input": "90 25", "output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno" }, { "input": "90 19", "output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn" }, { "input": "26 24", "output": "abcdefghijklmnopqrstuvwxab" }, { "input": "100 17", "output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno" }, { "input": "26 2", "output": "ababababababababababababab" } ]

1,619,768,182

2,147,483,647

Python 3

OK

TESTS

47

62

0

n,k = map(int,input().split()) pw = '' for i in range(k): pw += chr(97+i) pw *= n//k+1 print(pw[0:n])

Title: New Password Time Limit: None seconds Memory Limit: None megabytes Problem Description: Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help. Innokentiy decides that new password should satisfy the following conditions: - the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct. Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. Input Specification: The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it. Pay attention that a desired new password always exists. Output Specification: Print any password which satisfies all conditions given by Innokentiy. Demo Input: ['4 3\n', '6 6\n', '5 2\n'] Demo Output: ['java\n', 'python\n', 'phphp\n'] Note: In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it. In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters. In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it. Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.

```python n,k = map(int,input().split()) pw = '' for i in range(k): pw += chr(97+i) pw *= n//k+1 print(pw[0:n]) ```

3

554

B

Ohana Cleans Up

PROGRAMMING

1,200

[ "brute force", "greedy", "strings" ]

null

Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column. Return the maximum number of rows that she can make completely clean.

The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100). The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty.

The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.

[ "4\n0101\n1000\n1111\n0101\n", "3\n111\n111\n111\n" ]

[ "2\n", "3\n" ]

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean. In the second sample, everything is already clean, so Ohana doesn't need to do anything.

500

[ { "input": "4\n0101\n1000\n1111\n0101", "output": "2" }, { "input": "3\n111\n111\n111", "output": "3" }, { "input": "10\n0100000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000", "output": "9" }, { "input": "1\n1", "output": "1" }, { "input": "10\n0111010011\n0111010011\n1010010001\n0111010011\n0000110000\n0111010011\n0111010011\n0000110000\n1010010001\n0111010011", "output": "6" }, { "input": "20\n10101011101000011010\n11111010001100110101\n01011100010000001111\n10110100000101010011\n11010001110111101101\n00100110011011101010\n01000110101011001100\n01101100111101101101\n10111010010100111100\n00010010110001101110\n10111110010000101010\n10010111110100100100\n11010111001111110100\n11110111101100000001\n00011010100111011000\n11110001011000011010\n10001101010000011011\n01010101110010000111\n11100110111101101111\n11011111110010001111", "output": "1" }, { "input": "10\n1001000000\n0111101111\n1111001011\n1000010100\n0111101111\n0101100110\n1001000000\n1000010100\n0111101111\n1001000000", "output": "3" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "10\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000", "output": "10" }, { "input": "10\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111", "output": "10" }, { "input": "10\n1000000000\n0100000000\n0010000000\n0001000000\n0000100000\n0000010000\n0000001000\n0000000100\n0000000010\n0000000001", "output": "1" }, { "input": "2\n10\n01", "output": "1" }, { "input": "1\n0", "output": "1" }, { "input": "4\n0000\n0000\n1111\n1111", "output": "2" }, { "input": "11\n10000000001\n10000000001\n10000000001\n10000000001\n10001000001\n10001000000\n10001000001\n10001000001\n10001000000\n10001000000\n10001000100", "output": "4" } ]

1,527,710,368

2,147,483,647

Python 3

OK

TESTS

36

78

7,065,600

n=int(input()) l=[] for i in range(n): s=input() l.append(s) result=0 for i in l: if l.count(i)>result: result=l.count(i) print(result)

Title: Ohana Cleans Up Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column. Return the maximum number of rows that she can make completely clean. Input Specification: The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100). The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty. Output Specification: The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean. Demo Input: ['4\n0101\n1000\n1111\n0101\n', '3\n111\n111\n111\n'] Demo Output: ['2\n', '3\n'] Note: In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean. In the second sample, everything is already clean, so Ohana doesn't need to do anything.

```python n=int(input()) l=[] for i in range(n): s=input() l.append(s) result=0 for i in l: if l.count(i)>result: result=l.count(i) print(result) ```

3

34

A

Reconnaissance 2

PROGRAMMING

800

[ "implementation" ]

A. Reconnaissance 2

2

256

*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.

The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.

Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.

[ "5\n10 12 13 15 10\n", "4\n10 20 30 40\n" ]

[ "5 1\n", "1 2\n" ]

none

500

[ { "input": "5\n10 12 13 15 10", "output": "5 1" }, { "input": "4\n10 20 30 40", "output": "1 2" }, { "input": "6\n744 359 230 586 944 442", "output": "2 3" }, { "input": "5\n826 747 849 687 437", "output": "1 2" }, { "input": "5\n999 999 993 969 999", "output": "1 2" }, { "input": "5\n4 24 6 1 15", "output": "3 4" }, { "input": "2\n511 32", "output": "1 2" }, { "input": "3\n907 452 355", "output": "2 3" }, { "input": "4\n303 872 764 401", "output": "4 1" }, { "input": "10\n684 698 429 694 956 812 594 170 937 764", "output": "1 2" }, { "input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695", "output": "7 8" }, { "input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000", "output": "12 13" }, { "input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58", "output": "16 17" }, { "input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000", "output": "1 2" }, { "input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4", "output": "1 2" }, { "input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438", "output": "86 87" }, { "input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25", "output": "7 8" }, { "input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964", "output": "1 2" }, { "input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967", "output": "81 82" }, { "input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17", "output": "36 37" } ]

1,594,095,008

2,147,483,647

PyPy 3

OK

TESTS

20

278

20,172,800

n=int(input()) l=list(map(int,input().split())) inf=1e9 a,b=0,0 for i in range(n): if abs(l[i]-l[(i+1)%n])<inf: inf=abs(l[i]-l[(i+1)%n]) a,b=i+1,(i+1)%n+1 print(a,b)

Title: Reconnaissance 2 Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. Output Specification: Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. Demo Input: ['5\n10 12 13 15 10\n', '4\n10 20 30 40\n'] Demo Output: ['5 1\n', '1 2\n'] Note: none

```python n=int(input()) l=list(map(int,input().split())) inf=1e9 a,b=0,0 for i in range(n): if abs(l[i]-l[(i+1)%n])<inf: inf=abs(l[i]-l[(i+1)%n]) a,b=i+1,(i+1)%n+1 print(a,b) ```

3.892925

334

A

Candy Bags

PROGRAMMING

1,000

[ "implementation" ]

null

Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies. Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.

The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.

Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order. It is guaranteed that the solution exists at the given limits.

[ "2\n" ]

[ "1 4\n2 3\n" ]

The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

500

[ { "input": "2", "output": "1 4\n2 3" }, { "input": "4", "output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9" }, { "input": "6", "output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19" }, { "input": "8", "output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33" }, { "input": "10", "output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51" }, { "input": "100", "output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..." }, { "input": "62", "output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..." }, { "input": "66", "output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..." }, { "input": "18", "output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..." }, { "input": "68", "output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..." }, { "input": "86", "output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..." }, { "input": "96", "output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..." }, { "input": "12", "output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73" }, { "input": "88", "output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..." }, { "input": "28", "output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..." }, { "input": "80", "output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..." }, { "input": "48", "output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..." }, { "input": "54", "output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..." }, { "input": "58", "output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..." }, { "input": "64", "output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..." }, { "input": "44", "output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..." } ]

1,690,565,366

2,147,483,647

Python 3

OK

TESTS

21

92

102,400

n = int(input()) y = (n*n)//2 for i in range(y): print(i+1,n*n-i)

Title: Candy Bags Time Limit: None seconds Memory Limit: None megabytes Problem Description: Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies. Help him give *n* bags of candies to each brother so that all brothers got the same number of candies. Input Specification: The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers. Output Specification: Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order. It is guaranteed that the solution exists at the given limits. Demo Input: ['2\n'] Demo Output: ['1 4\n2 3\n'] Note: The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

```python n = int(input()) y = (n*n)//2 for i in range(y): print(i+1,n*n-i) ```

3

160

A

Twins

PROGRAMMING

900

[ "greedy", "sortings" ]

null

Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like. Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally. As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.

The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.

In the single line print the single number — the minimum needed number of coins.

[ "2\n3 3\n", "3\n2 1 2\n" ]

[ "2\n", "2\n" ]

In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum. In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.

500

[ { "input": "2\n3 3", "output": "2" }, { "input": "3\n2 1 2", "output": "2" }, { "input": "1\n5", "output": "1" }, { "input": "5\n4 2 2 2 2", "output": "3" }, { "input": "7\n1 10 1 2 1 1 1", "output": "1" }, { "input": "5\n3 2 3 3 1", "output": "3" }, { "input": "2\n2 1", "output": "1" }, { "input": "3\n2 1 3", "output": "2" }, { "input": "6\n1 1 1 1 1 1", "output": "4" }, { "input": "7\n10 10 5 5 5 5 1", "output": "3" }, { "input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1", "output": "8" }, { "input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4", "output": "8" }, { "input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40", "output": "8" }, { "input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90", "output": "6" }, { "input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28", "output": "14" }, { "input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42", "output": "22" }, { "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": "51" }, { "input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2", "output": "37" }, { "input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2", "output": "36" }, { "input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3", "output": "33" }, { "input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2", "output": "30" }, { "input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2", "output": "29" }, { "input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1", "output": "30" }, { "input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52", "output": "29" }, { "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": "30" }, { "input": "1\n1", "output": "1" }, { "input": "4\n4 1 2 3", "output": "2" }, { "input": "5\n2 2 2 4 4", "output": "2" }, { "input": "4\n100 99 99 100", "output": "2" } ]

1,687,827,993

2,147,483,647

Python 3

OK

TESTS

29

92

307,200

# !python3.8 """ Heap max Em Python """ from math import floor class MaxHeap: def __init__(self, arr=[], key=lambda item: item): """ Nesta implementação o Heapinicializa com um valor; """ self.heap_list = [None]+arr self.current_size = len(arr) self.key = key # Monta nosso Heap for i in range(floor(len(self.heap_list)/2), 0, -1): self.max_heapify(i) def getMax(self): if(self.current_size == 0): return 'Vazio!' return self.heap_list[1] def parent(self, i): return i//2 def left(self, i): ''' Retorna o filho Esquerdo ''' return 2*i def right(self, i): ''' Retorna o filho direito ''' return (2*i)+1 def heapSort(self): for i in range(self.current_size, 1, -1): # O(nlogn) self.heap_list[1], self.heap_list[i] = self.heap_list[i], self.heap_list[1] self.current_size -= 1 self.max_heapify(1) return self.heap_list[1:] def heap_increse_key(self, i): """ Move o valor para cima para manter a propriedade do heapFy. """ # Rearanjando os elementos. while self.parent(i) > 0: # Se o elemento for maior que seu pai. Então troque. if (self.key(self.heap_list[i]) > self.key(self.heap_list[self.parent(i)])): self.heap_list[i], self.heap_list[self.parent( i)] = self.heap_list[self.parent(i)], self.heap_list[i] # Move o index para o pai para manter a propriedade. i = self.parent(i) def insert(self, k): """ Insere um valor no nosso Heap """ # Inserindo o novo valor. self.heap_list.append(k) # Incrementando o tamanho do heap. self.current_size += 1 # Movendo o elemento de sua posição de baixo para cima. self.heap_increse_key(self.current_size) def max_heapify(self, i): ''' Mantém a propriedade Heap. ''' l = self.left(i) r = self.right(i) if(l <= self.current_size and self.key(self.heap_list[l]) > self.key(self.heap_list[i])): maior = l else: maior = i if(r <= self.current_size and self.key(self.heap_list[r]) > self.key(self.heap_list[maior])): maior = r if maior != i: self.heap_list[i], self.heap_list[maior] = self.heap_list[maior], self.heap_list[i] self.max_heapify(maior) def heap_extract_max(self): ''' Extrai o valor mínimo do Heap ''' if len(self.heap_list) == 1: return 'Heap Vazio!' # Pegando a raiz do heap. (O maior valor) root = self.heap_list[1] # Move o último valor do Heap para raiz. self.heap_list[1] = self.heap_list[self.current_size] # Removendo o último valor, pois fizemos uma cópia para raiz. *self.heap_list, _ = self.heap_list # Diminuindo o tamanho do heap self.current_size -= 1 # Mantendo a propriedade heap self.max_heapify(1) # retornando o maior valor return root def Twins(): ''' ENTRADA: n Representando o número de moedas Saída: Número mínimo de moedas que podem ser pego. ''' n = int(input()) sumCoin = 0 # Representa a soma de todas moedas que irá entrar. sumGetCoin = 0 #Soma total de moedas capturada até o momento. totalCoin =0# Total de moedas que podemos pegar. line = input().split() for i in range(len(line)): number = int(line[i]) sumCoin += number heapMax = MaxHeap(list(map(int, line))) #Define nosso Heap while(sumGetCoin <= sumCoin/2): #Enquanto a soma total de moedas captura não atingir o máximo. sumGetCoin += heapMax.heap_extract_max() #Captura a maior moeda e armazena na soma. totalCoin += 1 #Incrementa mais uma moeda return totalCoin print(Twins())

Title: Twins Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like. Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally. As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces. Output Specification: In the single line print the single number — the minimum needed number of coins. Demo Input: ['2\n3 3\n', '3\n2 1 2\n'] Demo Output: ['2\n', '2\n'] Note: In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum. In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.

```python # !python3.8 """ Heap max Em Python """ from math import floor class MaxHeap: def __init__(self, arr=[], key=lambda item: item): """ Nesta implementação o Heapinicializa com um valor; """ self.heap_list = [None]+arr self.current_size = len(arr) self.key = key # Monta nosso Heap for i in range(floor(len(self.heap_list)/2), 0, -1): self.max_heapify(i) def getMax(self): if(self.current_size == 0): return 'Vazio!' return self.heap_list[1] def parent(self, i): return i//2 def left(self, i): ''' Retorna o filho Esquerdo ''' return 2*i def right(self, i): ''' Retorna o filho direito ''' return (2*i)+1 def heapSort(self): for i in range(self.current_size, 1, -1): # O(nlogn) self.heap_list[1], self.heap_list[i] = self.heap_list[i], self.heap_list[1] self.current_size -= 1 self.max_heapify(1) return self.heap_list[1:] def heap_increse_key(self, i): """ Move o valor para cima para manter a propriedade do heapFy. """ # Rearanjando os elementos. while self.parent(i) > 0: # Se o elemento for maior que seu pai. Então troque. if (self.key(self.heap_list[i]) > self.key(self.heap_list[self.parent(i)])): self.heap_list[i], self.heap_list[self.parent( i)] = self.heap_list[self.parent(i)], self.heap_list[i] # Move o index para o pai para manter a propriedade. i = self.parent(i) def insert(self, k): """ Insere um valor no nosso Heap """ # Inserindo o novo valor. self.heap_list.append(k) # Incrementando o tamanho do heap. self.current_size += 1 # Movendo o elemento de sua posição de baixo para cima. self.heap_increse_key(self.current_size) def max_heapify(self, i): ''' Mantém a propriedade Heap. ''' l = self.left(i) r = self.right(i) if(l <= self.current_size and self.key(self.heap_list[l]) > self.key(self.heap_list[i])): maior = l else: maior = i if(r <= self.current_size and self.key(self.heap_list[r]) > self.key(self.heap_list[maior])): maior = r if maior != i: self.heap_list[i], self.heap_list[maior] = self.heap_list[maior], self.heap_list[i] self.max_heapify(maior) def heap_extract_max(self): ''' Extrai o valor mínimo do Heap ''' if len(self.heap_list) == 1: return 'Heap Vazio!' # Pegando a raiz do heap. (O maior valor) root = self.heap_list[1] # Move o último valor do Heap para raiz. self.heap_list[1] = self.heap_list[self.current_size] # Removendo o último valor, pois fizemos uma cópia para raiz. *self.heap_list, _ = self.heap_list # Diminuindo o tamanho do heap self.current_size -= 1 # Mantendo a propriedade heap self.max_heapify(1) # retornando o maior valor return root def Twins(): ''' ENTRADA: n Representando o número de moedas Saída: Número mínimo de moedas que podem ser pego. ''' n = int(input()) sumCoin = 0 # Representa a soma de todas moedas que irá entrar. sumGetCoin = 0 #Soma total de moedas capturada até o momento. totalCoin =0# Total de moedas que podemos pegar. line = input().split() for i in range(len(line)): number = int(line[i]) sumCoin += number heapMax = MaxHeap(list(map(int, line))) #Define nosso Heap while(sumGetCoin <= sumCoin/2): #Enquanto a soma total de moedas captura não atingir o máximo. sumGetCoin += heapMax.heap_extract_max() #Captura a maior moeda e armazena na soma. totalCoin += 1 #Incrementa mais uma moeda return totalCoin print(Twins()) ```

3

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,664,422,231

2,147,483,647

Python 3

OK

TESTS

34

92

0

test=int(input()) teams=[] ele=[] for i in range(test): team=input() if team not in teams: teams.append(team) ele.append(team) maxc=0 ans="" for i in teams: c=ele.count(i) if(c>maxc): maxc=c ans=i print(ans)

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python test=int(input()) teams=[] ele=[] for i in range(test): team=input() if team not in teams: teams.append(team) ele.append(team) maxc=0 ans="" for i in teams: c=ele.count(i) if(c>maxc): maxc=c ans=i print(ans) ```

3.977

294

A

Shaass and Oskols

PROGRAMMING

800

[ "implementation", "math" ]

null

Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.

The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.

On the *i*-th line of the output print the number of birds on the *i*-th wire.

[ "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n", "3\n2 4 1\n1\n2 2\n" ]

[ "0\n12\n5\n0\n16\n", "3\n0\n3\n" ]

none

500

[ { "input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6", "output": "0\n12\n5\n0\n16" }, { "input": "3\n2 4 1\n1\n2 2", "output": "3\n0\n3" }, { "input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43", "output": "0\n66\n57\n7\n0" }, { "input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51", "output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52" }, { "input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1", "output": "0\n0" }, { "input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5", "output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21" }, { "input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0", "output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95" }, { "input": "1\n100\n1\n1 100", "output": "0" }, { "input": "1\n100\n1\n1 1", "output": "0" }, { "input": "1\n50\n1\n1 25", "output": "0" }, { "input": "2\n50 0\n1\n1 1", "output": "0\n49" }, { "input": "1\n10\n0", "output": "10" } ]

1,644,161,718

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

2

92

0

def main(): n=int(input()) l=input() l=l.split() for i in range(n): l[i]=int(l[i]) m=int(input()) for i in range(m): xy=input() xy=xy.split() wire=int(xy[0])-1 bird=int(xy[1]) if wire!=0: #before l[wire-1]+=bird-1 if wire!=n: l[wire+1]+=l[wire]-bird l[wire]=0 for i in range(n): print(l[i]) if __name__ =="__main__": main()

Title: Shaass and Oskols Time Limit: None seconds Memory Limit: None megabytes Problem Description: Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. Input Specification: The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. Output Specification: On the *i*-th line of the output print the number of birds on the *i*-th wire. Demo Input: ['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n'] Demo Output: ['0\n12\n5\n0\n16\n', '3\n0\n3\n'] Note: none

```python def main(): n=int(input()) l=input() l=l.split() for i in range(n): l[i]=int(l[i]) m=int(input()) for i in range(m): xy=input() xy=xy.split() wire=int(xy[0])-1 bird=int(xy[1]) if wire!=0: #before l[wire-1]+=bird-1 if wire!=n: l[wire+1]+=l[wire]-bird l[wire]=0 for i in range(n): print(l[i]) if __name__ =="__main__": main() ```

-1

887

A

Div. 64

PROGRAMMING

1,000

[ "implementation" ]

null

Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills. Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.

In the only line given a non-empty binary string *s* with length up to 100.

Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.

[ "100010001\n", "100\n" ]

[ "yes", "no" ]

In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system. You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)

500

[ { "input": "100010001", "output": "yes" }, { "input": "100", "output": "no" }, { "input": "0000001000000", "output": "yes" }, { "input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "no" }, { "input": "1111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111111111", "output": "no" }, { "input": "0111111101111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "no" }, { "input": "1111011111111111111111111111110111110111111111111111111111011111111111111110111111111111111111111111", "output": "no" }, { "input": "1111111111101111111111111111111111111011111111111111111111111101111011111101111111111101111111111111", "output": "yes" }, { "input": "0110111111111111111111011111111110110111110111111111111111111111111111111111111110111111111111111111", "output": "yes" }, { "input": "1100110001111011001101101000001110111110011110111110010100011000100101000010010111100000010001001101", "output": "yes" }, { "input": "000000", "output": "no" }, { "input": "0001000", "output": "no" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "1000000", "output": "yes" }, { "input": "0", "output": "no" }, { "input": "1", "output": "no" }, { "input": "10000000000", "output": "yes" }, { "input": "0000000000", "output": "no" }, { "input": "0010000", "output": "no" }, { "input": "000000011", "output": "no" }, { "input": "000000000", "output": "no" }, { "input": "00000000", "output": "no" }, { "input": "000000000011", "output": "no" }, { "input": "0000000", "output": "no" }, { "input": "00000000011", "output": "no" }, { "input": "000000001", "output": "no" }, { "input": "000000000000000000000000000", "output": "no" }, { "input": "0000001", "output": "no" }, { "input": "00000001", "output": "no" }, { "input": "00000000100", "output": "no" }, { "input": "00000000000000000000", "output": "no" }, { "input": "0000000000000000000", "output": "no" }, { "input": "00001000", "output": "no" }, { "input": "0000000000010", "output": "no" }, { "input": "000000000010", "output": "no" }, { "input": "000000000000010", "output": "no" }, { "input": "0100000", "output": "no" }, { "input": "00010000", "output": "no" }, { "input": "00000000000000000", "output": "no" }, { "input": "00000000000", "output": "no" }, { "input": "000001000", "output": "no" }, { "input": "000000000000", "output": "no" }, { "input": "100000000000000", "output": "yes" }, { "input": "000010000", "output": "no" }, { "input": "00000100", "output": "no" }, { "input": "0001100000", "output": "no" }, { "input": "000000000000000000000000001", "output": "no" }, { "input": "000000100", "output": "no" }, { "input": "0000000000001111111111", "output": "no" }, { "input": "00000010", "output": "no" }, { "input": "0001110000", "output": "no" }, { "input": "0000000000000000000000", "output": "no" }, { "input": "000000010010", "output": "no" }, { "input": "0000100", "output": "no" }, { "input": "0000000001", "output": "no" }, { "input": "000000111", "output": "no" }, { "input": "0000000000000", "output": "no" }, { "input": "000000000000000000", "output": "no" }, { "input": "0000000000000000000000000", "output": "no" }, { "input": "000000000000000", "output": "no" }, { "input": "0010000000000100", "output": "yes" }, { "input": "0000001000", "output": "no" }, { "input": "00000000000000000001", "output": "no" }, { "input": "100000000", "output": "yes" }, { "input": "000000000001", "output": "no" }, { "input": "0000011001", "output": "no" }, { "input": "000", "output": "no" }, { "input": "000000000000000000000", "output": "no" }, { "input": "0000000000011", "output": "no" }, { "input": "0000000000000000", "output": "no" }, { "input": "00000000000000001", "output": "no" }, { "input": "00000000000000", "output": "no" }, { "input": "0000000000000000010", "output": "no" }, { "input": "00000000000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "000011000", "output": "no" }, { "input": "00000011", "output": "no" }, { "input": "0000000000001100", "output": "no" }, { "input": "00000", "output": "no" }, { "input": "000000000000000000000000000111111111111111", "output": "no" }, { "input": "000000010", "output": "no" }, { "input": "00000000111", "output": "no" }, { "input": "000000000000001", "output": "no" }, { "input": "0000000000000011111111111111111", "output": "no" }, { "input": "0000000010", "output": "no" }, { "input": "0000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "00000000010", "output": "no" }, { "input": "101000000000", "output": "yes" }, { "input": "00100000", "output": "no" }, { "input": "00000000000001", "output": "no" }, { "input": "0000000000100", "output": "no" }, { "input": "0000", "output": "no" }, { "input": "00000000000111", "output": "no" }, { "input": "0000000000000011", "output": "no" }, { "input": "0000000000000000000000000000000000000000", "output": "no" }, { "input": "0000000000000010", "output": "no" }, { "input": "0010101010", "output": "no" }, { "input": "0000000000000001", "output": "no" }, { "input": "1010101", "output": "no" } ]

1,510,850,980

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

62

0

s = input() f = 0 x = '1' for i in s: if i == '0': x += '0' if len(x) >= 6: print("yes") else: print("no")

Title: Div. 64 Time Limit: None seconds Memory Limit: None megabytes Problem Description: Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills. Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system. Input Specification: In the only line given a non-empty binary string *s* with length up to 100. Output Specification: Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise. Demo Input: ['100010001\n', '100\n'] Demo Output: ['yes', 'no'] Note: In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system. You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)

```python s = input() f = 0 x = '1' for i in s: if i == '0': x += '0' if len(x) >= 6: print("yes") else: print("no") ```

0

765

A

Neverending competitions

PROGRAMMING

900

[ "implementation", "math" ]

null

There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back. Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that: - this list contains all Jinotega's flights in this year (in arbitrary order), - Jinotega has only flown from his hometown to a snooker contest and back, - after each competition Jinotega flies back home (though they may attend a competition in one place several times), - and finally, at the beginning of the year Jinotega was at home. Please help them to determine Jinotega's location!

In the first line of input there is a single integer *n*: the number of Jinotega's flights (1<=≤<=*n*<=≤<=100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next *n* lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport. It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.

If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".

[ "4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO\n", "3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP\n" ]

[ "home\n", "contest\n" ]

In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.

500

[ { "input": "4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO", "output": "home" }, { "input": "3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP", "output": "contest" }, { "input": "1\nESJ\nESJ->TSJ", "output": "contest" }, { "input": "2\nXMR\nFAJ->XMR\nXMR->FAJ", "output": "home" }, { "input": "3\nZIZ\nDWJ->ZIZ\nZIZ->DWJ\nZIZ->DWJ", "output": "contest" }, { "input": "10\nPVO\nDMN->PVO\nDMN->PVO\nPVO->DMN\nDMN->PVO\nPVO->DMN\nPVO->DMN\nPVO->DMN\nDMN->PVO\nPVO->DMN\nDMN->PVO", "output": "home" }, { "input": "11\nIAU\nIAU->RUQ\nIAU->RUQ\nRUQ->IAU\nRUQ->IAU\nIAU->RUQ\nRUQ->IAU\nIAU->RUQ\nRUQ->IAU\nIAU->RUQ\nIAU->RUQ\nRUQ->IAU", "output": "contest" }, { "input": "10\nHPN\nDFI->HPN\nHPN->KAB\nHPN->DFI\nVSO->HPN\nHPN->KZX\nHPN->VSO\nKZX->HPN\nLDW->HPN\nKAB->HPN\nHPN->LDW", "output": "home" }, { "input": "11\nFGH\nFGH->BRZ\nUBK->FGH\nQRE->FGH\nFGH->KQK\nFGH->QRE\nKQK->FGH\nFGH->UBK\nBRZ->FGH\nFGH->ALX\nALX->FGH\nFGH->KQK", "output": "contest" }, { "input": "50\nPFH\nJFV->PFH\nBVP->PFH\nPFH->BVP\nPFH->JFV\nPFH->ETQ\nPFH->LQJ\nZTO->PFH\nPFH->BVP\nPFH->RXO\nPFH->ZTO\nHWL->PFH\nPFH->HIV\nPFH->AFP\nPFH->HWL\nOBB->PFH\nHIV->PFH\nPFH->LSR\nAFP->PFH\nLQJ->PFH\nHWL->PFH\nETQ->PFH\nPFH->HWL\nLSR->PFH\nWBR->PFH\nBNZ->PFH\nHQR->PFH\nZTO->PFH\nPFH->WBR\nPFH->BYJ\nRXO->PFH\nFHZ->PFH\nFHZ->PFH\nPFN->PFH\nPFH->GMB\nPFH->JFV\nJFV->PFH\nGNZ->PFH\nPFH->BNZ\nPFH->GNZ\nPFH->HQR\nBYJ->PFH\nGMB->PFH\nPFH->FHZ\nPFH->FHZ\nPFH->ZTO\nPFH->UGD\nBVP->PFH\nUGD->PFH\nPFH->PFN\nPFH->OBB", "output": "home" }, { "input": "1\nAAK\nAAK->ABA", "output": "contest" }, { "input": "1\nXYZ\nXYZ->XYR", "output": "contest" } ]

1,553,492,754

2,147,483,647

PyPy 3

OK

TESTS

23

124

0

n=int(input()) l=list(input()) z=[] for i in range(n): x,y=map(str,input().split("->")) z.append(x) z.append(y) z1=list(set(z)) z2=[] for i in range(len(z1)): z2.append(z.count(z1[i])) for i in range(len(z2)): if z2[i]%2==1: print("contest") break else: print("home")

Title: Neverending competitions Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back. Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that: - this list contains all Jinotega's flights in this year (in arbitrary order), - Jinotega has only flown from his hometown to a snooker contest and back, - after each competition Jinotega flies back home (though they may attend a competition in one place several times), - and finally, at the beginning of the year Jinotega was at home. Please help them to determine Jinotega's location! Input Specification: In the first line of input there is a single integer *n*: the number of Jinotega's flights (1<=≤<=*n*<=≤<=100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next *n* lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport. It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement. Output Specification: If Jinotega is now at home, print "home" (without quotes), otherwise print "contest". Demo Input: ['4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO\n', '3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP\n'] Demo Output: ['home\n', 'contest\n'] Note: In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.

```python n=int(input()) l=list(input()) z=[] for i in range(n): x,y=map(str,input().split("->")) z.append(x) z.append(y) z1=list(set(z)) z2=[] for i in range(len(z1)): z2.append(z.count(z1[i])) for i in range(len(z2)): if z2[i]%2==1: print("contest") break else: print("home") ```

3

785

A

Anton and Polyhedrons

PROGRAMMING

800

[ "implementation", "strings" ]

null

Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: - Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces. All five kinds of polyhedrons are shown on the picture below: Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection. Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this: - "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.

Output one number — the total number of faces in all the polyhedrons in Anton's collection.

[ "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n", "3\nDodecahedron\nOctahedron\nOctahedron\n" ]

[ "42\n", "28\n" ]

In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.

500

[ { "input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron", "output": "42" }, { "input": "3\nDodecahedron\nOctahedron\nOctahedron", "output": "28" }, { "input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron", "output": "256" }, { "input": "1\nTetrahedron", "output": "4" }, { "input": "1\nCube", "output": "6" }, { "input": "1\nOctahedron", "output": "8" }, { "input": "1\nDodecahedron", "output": "12" }, { "input": "1\nIcosahedron", "output": "20" }, { "input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron", "output": "340" } ]

1,697,563,684

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

31

0

n=int(input()) l=[] k={} k['Tetrahedron']=4 k['Cube']=6 k['Octahedron']=8 k['Dodecahedron']=12 k['Icosahedron']=20 for i in range(0,n): p=input() l.append(p) sum=sum+k[l[i]] print(sum)

Title: Anton and Polyhedrons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: - Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces. All five kinds of polyhedrons are shown on the picture below: Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection. Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this: - "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. Output Specification: Output one number — the total number of faces in all the polyhedrons in Anton's collection. Demo Input: ['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n'] Demo Output: ['42\n', '28\n'] Note: In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.

```python n=int(input()) l=[] k={} k['Tetrahedron']=4 k['Cube']=6 k['Octahedron']=8 k['Dodecahedron']=12 k['Icosahedron']=20 for i in range(0,n): p=input() l.append(p) sum=sum+k[l[i]] print(sum) ```

-1

702

B

Powers of Two

PROGRAMMING

1,500

[ "brute force", "data structures", "implementation", "math" ]

null

You are given *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Find the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2 (i. e. some integer *x* exists so that *a**i*<=+<=*a**j*<==<=2*x*).

The first line contains the single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers. The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).

Print the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2.

[ "4\n7 3 2 1\n", "3\n1 1 1\n" ]

[ "2\n", "3\n" ]

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4). In the second example all pairs of indexes (*i*, *j*) (where *i* < *j*) include in answer.

0

[ { "input": "4\n7 3 2 1", "output": "2" }, { "input": "3\n1 1 1", "output": "3" }, { "input": "1\n1000000000", "output": "0" }, { "input": "10\n2827343 1373647 96204862 723505 796619138 71550121 799843967 5561265 402690754 446173607", "output": "2" }, { "input": "10\n6 6 7 3 9 14 15 7 2 2", "output": "9" }, { "input": "100\n3 6 12 1 16 4 9 5 4 4 5 8 12 4 6 14 5 1 2 2 2 1 7 1 9 10 6 13 7 8 3 11 8 11 7 5 15 6 14 10 4 2 10 9 1 8 14 9 5 11 3 4 1 12 6 8 13 4 8 5 4 13 13 1 3 9 14 7 14 10 7 3 12 8 9 8 6 15 9 10 12 14 15 4 16 8 8 4 8 7 5 10 16 4 10 13 6 16 16 5", "output": "532" }, { "input": "1\n2", "output": "0" }, { "input": "2\n1 1", "output": "1" } ]

1,689,331,284

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

8

3,000

6,246,400

n=int(input()) A=list(map(int,input().split())) k=0 for i in range(n-1): for j in range(i+1,n): r=A[i]+A[j] if r&(r-1)==0: k+=1 print(k)

Title: Powers of Two Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Find the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2 (i. e. some integer *x* exists so that *a**i*<=+<=*a**j*<==<=2*x*). Input Specification: The first line contains the single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers. The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: Print the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2. Demo Input: ['4\n7 3 2 1\n', '3\n1 1 1\n'] Demo Output: ['2\n', '3\n'] Note: In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4). In the second example all pairs of indexes (*i*, *j*) (where *i* < *j*) include in answer.

```python n=int(input()) A=list(map(int,input().split())) k=0 for i in range(n-1): for j in range(i+1,n): r=A[i]+A[j] if r&(r-1)==0: k+=1 print(k) ```

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" } ]

1,676,589,345

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

word = input() Count_Upper = 0 Count_Lower = 0 for char in word: if char.isupper(): Count_Upper += 1 for char in word: if char.iflower(): Count_Lower += 1 if Count_Upper <= Count_Lower: print(word.Lower()) else: print(word.Upper())

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 = input() Count_Upper = 0 Count_Lower = 0 for char in word: if char.isupper(): Count_Upper += 1 for char in word: if char.iflower(): Count_Lower += 1 if Count_Upper <= Count_Lower: print(word.Lower()) else: print(word.Upper()) ```

-1

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,598,383,117

2,147,483,647

Python 3

OK

TESTS

34

216

0

n = int(input()) a=[] for i in range(n): a.append(input()) a.sort() team_1 = a.count(a[0]) team_2 = a.count(a[-1]) if team_1 > team_2: print(a[0]) else: print(a[-1])

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python n = int(input()) a=[] for i in range(n): a.append(input()) a.sort() team_1 = a.count(a[0]) team_2 = a.count(a[-1]) if team_1 > team_2: print(a[0]) else: print(a[-1]) ```

3.946

670

B

Game of Robots

PROGRAMMING

1,000

[ "implementation" ]

null

In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109. At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier. Your task is to determine the *k*-th identifier to be pronounced.

The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2). The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different.

Print the *k*-th pronounced identifier (assume that the numeration starts from 1).

[ "2 2\n1 2\n", "4 5\n10 4 18 3\n" ]

[ "1\n", "4\n" ]

In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1. In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4.

750

[ { "input": "2 2\n1 2", "output": "1" }, { "input": "4 5\n10 4 18 3", "output": "4" }, { "input": "1 1\n4", "output": "4" }, { "input": "2 1\n5 1", "output": "5" }, { "input": "2 2\n1 4", "output": "1" }, { "input": "2 3\n6 7", "output": "7" }, { "input": "3 1\n4 5 6", "output": "4" }, { "input": "3 2\n4 5 6", "output": "4" }, { "input": "3 3\n4 5 6", "output": "5" }, { "input": "3 4\n4 5 6", "output": "4" }, { "input": "3 5\n4 5 6", "output": "5" }, { "input": "3 6\n4 5 6", "output": "6" }, { "input": "4 1\n5 1000000000 999999999 12", "output": "5" }, { "input": "4 2\n5 1000000000 999999999 12", "output": "5" }, { "input": "4 3\n5 1000000000 999999999 12", "output": "1000000000" }, { "input": "4 4\n5 1000000000 999999999 12", "output": "5" }, { "input": "4 5\n5 1000000000 999999999 12", "output": "1000000000" }, { "input": "4 6\n5 1000000000 999999999 12", "output": "999999999" }, { "input": "4 7\n5 1000000000 999999999 12", "output": "5" }, { "input": "4 8\n5 1000000000 999999999 12", "output": "1000000000" }, { "input": "4 9\n5 1000000000 999999999 12", "output": "999999999" }, { "input": "4 10\n5 1000000000 999999999 12", "output": "12" } ]

1,638,781,433

2,147,483,647

Python 3

OK

TESTS

143

92

7,884,800

n,k=[int(x) for x in input().split()] a=[int(x) for x in input().split()] k1=int((2*k)**0.5) if k1*(k1+1)>=2*k: k1-=1 k-=k1*(k1+1)//2 print(a[k-1]) else: k-=k1*(k1+1)//2 print(a[k-1])

Title: Game of Robots Time Limit: None seconds Memory Limit: None megabytes Problem Description: In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109. At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier. Your task is to determine the *k*-th identifier to be pronounced. Input Specification: The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2). The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different. Output Specification: Print the *k*-th pronounced identifier (assume that the numeration starts from 1). Demo Input: ['2 2\n1 2\n', '4 5\n10 4 18 3\n'] Demo Output: ['1\n', '4\n'] Note: In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1. In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4.

```python n,k=[int(x) for x in input().split()] a=[int(x) for x in input().split()] k1=int((2*k)**0.5) if k1*(k1+1)>=2*k: k1-=1 k-=k1*(k1+1)//2 print(a[k-1]) else: k-=k1*(k1+1)//2 print(a[k-1]) ```

3

219

D

Choosing Capital for Treeland

PROGRAMMING

1,700

[ "dfs and similar", "dp", "graphs", "trees" ]

null

The country Treeland consists of *n* cities, some pairs of them are connected with unidirectional roads. Overall there are *n*<=-<=1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any city to any other one. The council of the elders has recently decided to choose the capital of Treeland. Of course it should be a city of this country. The council is supposed to meet in the capital and regularly move from the capital to other cities (at this stage nobody is thinking about getting back to the capital from these cities). For that reason if city *a* is chosen a capital, then all roads must be oriented so that if we move along them, we can get from city *a* to any other city. For that some roads may have to be inversed. Help the elders to choose the capital so that they have to inverse the minimum number of roads in the country.

The first input line contains integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in Treeland. Next *n*<=-<=1 lines contain the descriptions of the roads, one road per line. A road is described by a pair of integers *s**i*,<=*t**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*; *s**i*<=≠<=*t**i*) — the numbers of cities, connected by that road. The *i*-th road is oriented from city *s**i* to city *t**i*. You can consider cities in Treeland indexed from 1 to *n*.

In the first line print the minimum number of roads to be inversed if the capital is chosen optimally. In the second line print all possible ways to choose the capital — a sequence of indexes of cities in the increasing order.

[ "3\n2 1\n2 3\n", "4\n1 4\n2 4\n3 4\n" ]

[ "0\n2 \n", "2\n1 2 3 \n" ]

none

2,000

[ { "input": "3\n2 1\n2 3", "output": "0\n2 " }, { "input": "4\n1 4\n2 4\n3 4", "output": "2\n1 2 3 " }, { "input": "2\n1 2", "output": "0\n1 " }, { "input": "8\n1 2\n3 2\n4 3\n4 5\n6 5\n6 7\n8 7", "output": "3\n4 6 8 " }, { "input": "10\n2 3\n1 8\n9 5\n5 4\n6 10\n4 8\n5 6\n7 6\n5 3", "output": "3\n9 " }, { "input": "10\n9 3\n3 8\n4 3\n3 5\n7 8\n10 3\n2 3\n6 2\n3 1", "output": "4\n6 " }, { "input": "10\n5 4\n6 5\n10 1\n2 5\n9 8\n1 3\n7 10\n2 3\n5 9", "output": "2\n7 " }, { "input": "10\n2 3\n1 8\n9 5\n5 4\n6 10\n4 8\n5 6\n7 6\n5 3", "output": "3\n9 " }, { "input": "11\n3 10\n4 7\n11 7\n8 2\n6 9\n9 3\n5 9\n1 3\n3 8\n7 3", "output": "5\n4 5 6 11 " }, { "input": "12\n10 7\n6 10\n8 10\n4 12\n10 4\n3 12\n1 2\n2 7\n4 5\n11 5\n11 9", "output": "5\n1 6 8 " }, { "input": "13\n7 13\n5 4\n5 1\n11 12\n8 3\n13 4\n10 7\n11 5\n8 13\n2 8\n10 6\n4 9", "output": "4\n2 10 " }, { "input": "14\n13 10\n5 14\n9 4\n8 12\n10 1\n5 2\n3 6\n11 2\n13 2\n13 7\n8 2\n4 5\n1 3", "output": "3\n9 " }, { "input": "15\n15 6\n2 15\n14 15\n15 12\n1 2\n10 15\n7 4\n13 7\n11 2\n5 2\n7 2\n3 2\n7 8\n9 15", "output": "7\n13 " } ]

1,684,964,920

2,147,483,647

PyPy 3-64

RUNTIME_ERROR

TESTS

7

2,682

73,113,600

n = int(input()) from collections import defaultdict g = defaultdict(list) for _ in range(n-1): a,b = [int(x) for x in input().split()] g[a].append([b,1]) # green edge g[b].append([a,0]) # red edge dist = [1e9 for i in range(n+1)] def helper(node, par, cur): dist[node] = cur for child, nofarkpadtamereko in g[node]: if child == par: continue helper(child, node, cur+1) helper(1,-1,0) red = [0 for i in range(n+1)] def helper2(node, par,cur): red[node] = cur for child, check in g[node]: if child == par: continue helper2(child, node, cur + int(check == 0) ) helper2(1,-1,0) def findtotalred(node, par): cur = 0 for child, check in g[node]: if child == par: continue cur += int(check == 0) + findtotalred(child, node) return cur total = findtotalred(1,-1) # print('total red ', total) dp = defaultdict(lambda: 1e9) mn = 1e9 for i in range(1,n+1): dp[i] = total + dist[i] - 2*red[i] mn = min(mn, dp[i]) print(mn) for i in range(1,n+1): if dp[i] == mn: print(i, end = " ") print()

Title: Choosing Capital for Treeland Time Limit: None seconds Memory Limit: None megabytes Problem Description: The country Treeland consists of *n* cities, some pairs of them are connected with unidirectional roads. Overall there are *n*<=-<=1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any city to any other one. The council of the elders has recently decided to choose the capital of Treeland. Of course it should be a city of this country. The council is supposed to meet in the capital and regularly move from the capital to other cities (at this stage nobody is thinking about getting back to the capital from these cities). For that reason if city *a* is chosen a capital, then all roads must be oriented so that if we move along them, we can get from city *a* to any other city. For that some roads may have to be inversed. Help the elders to choose the capital so that they have to inverse the minimum number of roads in the country. Input Specification: The first input line contains integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in Treeland. Next *n*<=-<=1 lines contain the descriptions of the roads, one road per line. A road is described by a pair of integers *s**i*,<=*t**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*; *s**i*<=≠<=*t**i*) — the numbers of cities, connected by that road. The *i*-th road is oriented from city *s**i* to city *t**i*. You can consider cities in Treeland indexed from 1 to *n*. Output Specification: In the first line print the minimum number of roads to be inversed if the capital is chosen optimally. In the second line print all possible ways to choose the capital — a sequence of indexes of cities in the increasing order. Demo Input: ['3\n2 1\n2 3\n', '4\n1 4\n2 4\n3 4\n'] Demo Output: ['0\n2 \n', '2\n1 2 3 \n'] Note: none

```python n = int(input()) from collections import defaultdict g = defaultdict(list) for _ in range(n-1): a,b = [int(x) for x in input().split()] g[a].append([b,1]) # green edge g[b].append([a,0]) # red edge dist = [1e9 for i in range(n+1)] def helper(node, par, cur): dist[node] = cur for child, nofarkpadtamereko in g[node]: if child == par: continue helper(child, node, cur+1) helper(1,-1,0) red = [0 for i in range(n+1)] def helper2(node, par,cur): red[node] = cur for child, check in g[node]: if child == par: continue helper2(child, node, cur + int(check == 0) ) helper2(1,-1,0) def findtotalred(node, par): cur = 0 for child, check in g[node]: if child == par: continue cur += int(check == 0) + findtotalred(child, node) return cur total = findtotalred(1,-1) # print('total red ', total) dp = defaultdict(lambda: 1e9) mn = 1e9 for i in range(1,n+1): dp[i] = total + dist[i] - 2*red[i] mn = min(mn, dp[i]) print(mn) for i in range(1,n+1): if dp[i] == mn: print(i, end = " ") print() ```

-1

262

B

Roma and Changing Signs

PROGRAMMING

1,200

[ "greedy" ]

null

Roma works in a company that sells TVs. Now he has to prepare a report for the last year. Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times. The operation of changing a number's sign is the operation of multiplying this number by -1. Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made. The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104). The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.

In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.

[ "3 2\n-1 -1 1\n", "3 1\n-1 -1 1\n" ]

[ "3\n", "1\n" ]

In the first sample we can get sequence [1, 1, 1], thus the total income equals 3. In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.

1,000

[ { "input": "3 2\n-1 -1 1", "output": "3" }, { "input": "3 1\n-1 -1 1", "output": "1" }, { "input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343", "output": "81852" }, { "input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -8981 -8521 -8313 -8257 -8253 -7591 -7587 -7301 -7161 -7001 -6847 -6441 -6241 -5949 -5896 -5713 -5692 -5644 -5601 -5545 -5525 -5331 -5253 -5041 -5000 -4951 -4855 -4384 -4293 -4251 -4001 -3991 -3762 -3544 -3481 -3261 -2983 -2882 -2857 -2713 -2691 -2681 -2653 -2221 -2043 -2011 -1997 -1601 -1471 -1448 -1363 -1217 -1217 -1129 -961 -926 -801 -376 -327 -305 -174 -91", "output": "102443" }, { "input": "12 28\n-6652 -6621 -6471 -5559 -5326 -4551 -4401 -4326 -3294 -1175 -1069 -43", "output": "49488" }, { "input": "78 13\n-9961 -9922 -9817 -9813 -9521 -9368 -9361 -9207 -9153 -9124 -9008 -8981 -8951 -8911 -8551 -8479 -8245 -8216 -7988 -7841 -7748 -7741 -7734 -7101 -6846 -6804 -6651 -6526 -6519 -6463 -6297 -6148 -6090 -5845 -5209 -5201 -5161 -5061 -4537 -4529 -4433 -4370 -4266 -4189 -4125 -3945 -3843 -3777 -3751 -3476 -3461 -3279 -3205 -3001 -2889 -2761 -2661 -2521 -2481 -2305 -2278 -2269 -2225 -1648 -1524 -1476 -1353 -1097 -867 -785 -741 -711 -692 -440 -401 -225 -65 -41", "output": "-147832" }, { "input": "4 1\n218 3441 4901 7601", "output": "15725" }, { "input": "73 26\n-8497 -8363 -7603 -7388 -6830 -6827 -6685 -6389 -6237 -6099 -6013 -5565 -5465 -4965 -4947 -4201 -3851 -3793 -3421 -3410 -3201 -3169 -3156 -2976 -2701 -2623 -2321 -2169 -1469 -1221 -950 -926 -9 47 236 457 773 1321 1485 1545 1671 1736 2014 2137 2174 2301 2625 3181 3536 3851 4041 4685 4981 4987 5145 5163 5209 5249 6011 6337 6790 7254 7361 7407 7969 7982 8083 8251 8407 8735 9660 9855 9957", "output": "315919" }, { "input": "53 5\n-9821 -9429 -9146 -8973 -8807 -8801 -8321 -7361 -7222 -7161 -6913 -5961 -4877 -4756 -4753 -4661 -3375 -3031 -2950 -2661 -2161 -2041 -1111 -1071 -905 -697 -397 323 772 1617 1752 2736 2737 3201 3465 4029 4121 4463 4561 4637 4814 6119 6610 6641 6961 7217 7523 8045 8610 8915 9004 9265 9576", "output": "92703" }, { "input": "1 1\n0", "output": "0" }, { "input": "1 1\n10000", "output": "-10000" }, { "input": "1 2\n-1", "output": "-1" }, { "input": "2 1\n0 1", "output": "1" }, { "input": "2 2\n-1 0", "output": "1" }, { "input": "5 6\n-3 -2 -1 5 6", "output": "15" }, { "input": "3 3\n-50 -10 30", "output": "70" }, { "input": "4 4\n-100 -90 -80 1", "output": "269" }, { "input": "4 3\n-3 -2 1 2", "output": "6" }, { "input": "6 4\n-6 -3 -2 1 2 3", "output": "15" }, { "input": "5 6\n-10 -9 -8 1 2", "output": "28" }, { "input": "1 2\n1", "output": "1" }, { "input": "2 2\n-1 3", "output": "2" }, { "input": "4 7\n-3 -2 1 6", "output": "10" }, { "input": "4 3\n-7 -6 1 5", "output": "17" }, { "input": "4 4\n-5 -1 1 2", "output": "9" } ]

1,662,106,820

2,147,483,647

Python 3

OK

TESTS

56

154

7,270,400

n, m = map(int, input(). split()) arr = list(map(int, input(). split())) q = 0 for i in range(min(n, m)): if arr[i] < 0: arr[i] *= (-1) q += 1 else: break if m <= q: print(sum(arr)) else: if (m - q) % 2: arr. sort() print(sum(arr) - (arr[0] * 2)) else: print(sum(arr))

Title: Roma and Changing Signs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Roma works in a company that sells TVs. Now he has to prepare a report for the last year. Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times. The operation of changing a number's sign is the operation of multiplying this number by -1. Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made. The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104). The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order. Output Specification: In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes. Demo Input: ['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n'] Demo Output: ['3\n', '1\n'] Note: In the first sample we can get sequence [1, 1, 1], thus the total income equals 3. In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.

```python n, m = map(int, input(). split()) arr = list(map(int, input(). split())) q = 0 for i in range(min(n, m)): if arr[i] < 0: arr[i] *= (-1) q += 1 else: break if m <= q: print(sum(arr)) else: if (m - q) % 2: arr. sort() print(sum(arr) - (arr[0] * 2)) else: print(sum(arr)) ```

3

453

A

Little Pony and Expected Maximum

PROGRAMMING

1,600

[ "probabilities" ]

null

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.

A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).

Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.

[ "6 1\n", "6 3\n", "2 2\n" ]

[ "3.500000000000\n", "4.958333333333\n", "1.750000000000\n" ]

Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value

500

[ { "input": "6 1", "output": "3.500000000000" }, { "input": "6 3", "output": "4.958333333333" }, { "input": "2 2", "output": "1.750000000000" }, { "input": "5 4", "output": "4.433600000000" }, { "input": "5 8", "output": "4.814773760000" }, { "input": "3 10", "output": "2.982641534996" }, { "input": "3 6", "output": "2.910836762689" }, { "input": "1 8", "output": "1.000000000000" }, { "input": "24438 9", "output": "21994.699969310015" }, { "input": "94444 9", "output": "85000.099992058866" }, { "input": "8 66716", "output": "8.000000000000" }, { "input": "4 25132", "output": "4.000000000000" }, { "input": "51520 73331", "output": "51519.682650242677" }, { "input": "54230 31747", "output": "54228.743352775018" }, { "input": "24236 90163", "output": "24235.975171545670" }, { "input": "26946 99523", "output": "26945.974480086279" }, { "input": "50323 7", "output": "44033.124988408454" }, { "input": "53033 3", "output": "39775.249995286234" }, { "input": "55743 5", "output": "46452.999992525307" }, { "input": "59964 79", "output": "59214.949890211828" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "3 1", "output": "2.000000000000" }, { "input": "1 2", "output": "1.000000000000" }, { "input": "53513 34040", "output": "53511.875329020870" }, { "input": "100000 100000", "output": "99999.418033254507" }, { "input": "1 100000", "output": "1.000000000000" }, { "input": "100000 1", "output": "50000.499999999935" }, { "input": "2 100000", "output": "2.000000000000" }, { "input": "100000 2", "output": "66667.166665000332" }, { "input": "50000 100000", "output": "49999.843487110789" }, { "input": "99999 1111", "output": "99909.571915885972" }, { "input": "99999 99999", "output": "99998.418033254609" }, { "input": "1000 1000", "output": "999.419018443269" }, { "input": "50000 50000", "output": "49999.418043215679" }, { "input": "88888 88888", "output": "88887.418034499773" }, { "input": "99999 100000", "output": "99998.418042461126" }, { "input": "100000 555", "output": "99820.643422392372" }, { "input": "10000 10000", "output": "9999.418122897887" }, { "input": "100000 5001", "output": "99980.503829474910" }, { "input": "100000 1000", "output": "99900.599066768002" } ]

1,408,883,571

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

10

607

204,800

def power(a, n): if n == 0: return 1 if not n % 2: return power(a, n // 2) ** 2 else: return power(a, n - 1) * a m, n = tuple(map(int, input().split())) evidence = 1 / power(m, n) s = 0 for i in range(1, m + 1): s += (power(i, n) - power(i - 1, n)) * i print(s * evidence)

Title: Little Pony and Expected Maximum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times. Input Specification: A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105). Output Specification: Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4. Demo Input: ['6 1\n', '6 3\n', '2 2\n'] Demo Output: ['3.500000000000\n', '4.958333333333\n', '1.750000000000\n'] Note: Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value

```python def power(a, n): if n == 0: return 1 if not n % 2: return power(a, n // 2) ** 2 else: return power(a, n - 1) * a m, n = tuple(map(int, input().split())) evidence = 1 / power(m, n) s = 0 for i in range(1, m + 1): s += (power(i, n) - power(i - 1, n)) * i print(s * evidence) ```

-1

822

A

I'm bored with life

PROGRAMMING

800

[ "implementation", "math", "number theory" ]

null

Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom! Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*. Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?

The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).

Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.

[ "4 3\n" ]

[ "6\n" ]

Consider the sample. 4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.

500

[ { "input": "4 3", "output": "6" }, { "input": "10 399603090", "output": "3628800" }, { "input": "6 973151934", "output": "720" }, { "input": "2 841668075", "output": "2" }, { "input": "7 415216919", "output": "5040" }, { "input": "3 283733059", "output": "6" }, { "input": "11 562314608", "output": "39916800" }, { "input": "3 990639260", "output": "6" }, { "input": "11 859155400", "output": "39916800" }, { "input": "1 1", "output": "1" }, { "input": "5 3", "output": "6" }, { "input": "1 4", "output": "1" }, { "input": "5 4", "output": "24" }, { "input": "1 12", "output": "1" }, { "input": "9 7", "output": "5040" }, { "input": "2 3", "output": "2" }, { "input": "6 11", "output": "720" }, { "input": "6 7", "output": "720" }, { "input": "11 11", "output": "39916800" }, { "input": "4 999832660", "output": "24" }, { "input": "7 999228288", "output": "5040" }, { "input": "11 999257105", "output": "39916800" }, { "input": "11 999286606", "output": "39916800" }, { "input": "3 999279109", "output": "6" }, { "input": "999632727 11", "output": "39916800" }, { "input": "999625230 7", "output": "5040" }, { "input": "999617047 3", "output": "6" }, { "input": "999646548 7", "output": "5040" }, { "input": "999639051 3", "output": "6" }, { "input": "12 12", "output": "479001600" }, { "input": "12 1", "output": "1" }, { "input": "1213 5", "output": "120" }, { "input": "8 9", "output": "40320" }, { "input": "12 9", "output": "362880" }, { "input": "12 1000000000", "output": "479001600" }, { "input": "1000000000 1", "output": "1" }, { "input": "12 13", "output": "479001600" }, { "input": "2 29845", "output": "2" }, { "input": "10 21", "output": "3628800" }, { "input": "12 20", "output": "479001600" }, { "input": "15 12", "output": "479001600" }, { "input": "1 1", "output": "1" }, { "input": "1000000000 12", "output": "479001600" }, { "input": "11 30", "output": "39916800" }, { "input": "17 12", "output": "479001600" }, { "input": "4 19", "output": "24" }, { "input": "12 15", "output": "479001600" }, { "input": "20 6", "output": "720" }, { "input": "10 20", "output": "3628800" }, { "input": "10 10", "output": "3628800" }, { "input": "22 12", "output": "479001600" }, { "input": "20 12", "output": "479001600" }, { "input": "12 23", "output": "479001600" }, { "input": "12 22", "output": "479001600" }, { "input": "18 3", "output": "6" }, { "input": "14 10", "output": "3628800" }, { "input": "14 12", "output": "479001600" }, { "input": "8 3", "output": "6" }, { "input": "5 5", "output": "120" } ]

1,663,421,072

2,147,483,647

PyPy 3-64

OK

TESTS

59

62

0

import math arr = list(map(int,input().split())) ans = min(arr) print(math.factorial(ans))

Title: I'm bored with life Time Limit: None seconds Memory Limit: None megabytes Problem Description: Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom! Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*. Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? Input Specification: The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). Output Specification: Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. Demo Input: ['4 3\n'] Demo Output: ['6\n'] Note: Consider the sample. 4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.

```python import math arr = list(map(int,input().split())) ans = min(arr) print(math.factorial(ans)) ```

3

863

B

Kayaking

PROGRAMMING

1,500

[ "brute force", "greedy", "sortings" ]

null

Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers. Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash. Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks. Help the party to determine minimum possible total instability!

The first line contains one number *n* (2<=≤<=*n*<=≤<=50). The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).

Print minimum possible total instability.

[ "2\n1 2 3 4\n", "4\n1 3 4 6 3 4 100 200\n" ]

[ "1\n", "5\n" ]

none

0

[ { "input": "2\n1 2 3 4", "output": "1" }, { "input": "4\n1 3 4 6 3 4 100 200", "output": "5" }, { "input": "3\n305 139 205 406 530 206", "output": "102" }, { "input": "3\n610 750 778 6 361 407", "output": "74" }, { "input": "5\n97 166 126 164 154 98 221 7 51 47", "output": "35" }, { "input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2", "output": "0" }, { "input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4", "output": "1" }, { "input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476", "output": "368" }, { "input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597", "output": "393" }, { "input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21", "output": "387" }, { "input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637", "output": "376" }, { "input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61", "output": "351" }, { "input": "3\n1 2 10 11 100 100", "output": "1" }, { "input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858", "output": "318" }, { "input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636", "output": "355" }, { "input": "2\n55 5 25 51", "output": "4" }, { "input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973", "output": "348" }, { "input": "4\n35 48 71 44 78 79 57 48", "output": "10" }, { "input": "3\n58 89 73 15 5 47", "output": "21" }, { "input": "2\n1 20 99 100", "output": "1" } ]

1,646,620,293

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

#from code force # https://codeforces.com/contest/863/problem/B import sys input = sys.stdin.readline ############ ---- Input Functions ---- ############ def inp(): return(int(input())) def inlt(): return(list(map(int,input().split()))) def insr(): s = input() return(list(s[:len(s) - 1])) def invr(): return(map(int,input().split())) n = inp() N2 = n*2 #total length of the array arrPeople = inlt() arrPeople.sort() totalStability = 0 currentDistanceBar = 0 while(len(arrPeople)>2): for i in range(len(arrPeople)-1, 0, -1): distance = abs(arrPeople[i]-arrPeople[i-1]) if distance == currentDistanceBar: totalStability += distance arrPeople.pop(i) arrPeople.pop(i-1) i = i-1 #skip next item currentDistanceBar += 1 print(totalStability)

Title: Kayaking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers. Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash. Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks. Help the party to determine minimum possible total instability! Input Specification: The first line contains one number *n* (2<=≤<=*n*<=≤<=50). The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000). Output Specification: Print minimum possible total instability. Demo Input: ['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n'] Demo Output: ['1\n', '5\n'] Note: none

```python #from code force # https://codeforces.com/contest/863/problem/B import sys input = sys.stdin.readline ############ ---- Input Functions ---- ############ def inp(): return(int(input())) def inlt(): return(list(map(int,input().split()))) def insr(): s = input() return(list(s[:len(s) - 1])) def invr(): return(map(int,input().split())) n = inp() N2 = n*2 #total length of the array arrPeople = inlt() arrPeople.sort() totalStability = 0 currentDistanceBar = 0 while(len(arrPeople)>2): for i in range(len(arrPeople)-1, 0, -1): distance = abs(arrPeople[i]-arrPeople[i-1]) if distance == currentDistanceBar: totalStability += distance arrPeople.pop(i) arrPeople.pop(i-1) i = i-1 #skip next item currentDistanceBar += 1 print(totalStability) ```

-1

378

A

Playing with Dice

PROGRAMMING

800

[ "brute force" ]

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" } ]

1,690,148,921

2,147,483,647

Python 3

OK

TESTS

38

46

0

a, b = map(int, input().split()) count_a = 0 count_b = 0 count_c = 0 for i in range(1, 7): if abs(a-i) < abs(b-i): count_a += 1 elif abs(b-i) < abs(a-i): count_b += 1 else: count_c += 1 print(count_a, count_c, count_b)

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()) count_a = 0 count_b = 0 count_c = 0 for i in range(1, 7): if abs(a-i) < abs(b-i): count_a += 1 elif abs(b-i) < abs(a-i): count_b += 1 else: count_c += 1 print(count_a, count_c, count_b) ```

3

523

B

Mean Requests

PROGRAMMING

1,500

[ "*special", "implementation" ]

null

In this problem you will have to deal with a real algorithm that is used in the VK social network. As in any other company that creates high-loaded websites, the VK developers have to deal with request statistics regularly. An important indicator reflecting the load of the site is the mean number of requests for a certain period of time of *T* seconds (for example, *T*<==<=60 *seconds*<==<=1 *min* and *T*<==<=86400 *seconds*<==<=1 *day*). For example, if this value drops dramatically, that shows that the site has access problem. If this value grows, that may be a reason to analyze the cause for the growth and add more servers to the website if it is really needed. However, even such a natural problem as counting the mean number of queries for some period of time can be a challenge when you process the amount of data of a huge social network. That's why the developers have to use original techniques to solve problems approximately, but more effectively at the same time. Let's consider the following formal model. We have a service that works for *n* seconds. We know the number of queries to this resource *a**t* at each moment of time *t* (1<=≤<=*t*<=≤<=*n*). Let's formulate the following algorithm of calculating the mean with exponential decay. Let *c* be some real number, strictly larger than one. Thus, the mean variable is recalculated each second using the number of queries that came at that second. We can make some mathematical calculations and prove that choosing the value of constant *c* correctly will make the value of mean not very different from the real mean value *a**x* at *t*<=-<=*T*<=+<=1<=≤<=*x*<=≤<=*t*. The advantage of such approach is that it only uses the number of requests at the current moment of time and doesn't require storing the history of requests for a large time range. Also, it considers the recent values with the weight larger than the weight of the old ones, which helps to react to dramatic change in values quicker. However before using the new theoretical approach in industrial programming, there is an obligatory step to make, that is, to test its credibility practically on given test data sets. Your task is to compare the data obtained as a result of the work of an approximate algorithm to the real data. You are given *n* values *a**t*, integer *T* and real number *c*. Also, you are given *m* moments *p**j* (1<=≤<=*j*<=≤<=*m*), where we are interested in the mean value of the number of queries for the last *T* seconds. Implement two algorithms. The first one should calculate the required value by definition, i.e. by the formula . The second algorithm should calculate the mean value as is described above. Print both values and calculate the relative error of the second algorithm by the formula , where *approx* is the approximate value, obtained by the second algorithm, and *real* is the exact value obtained by the first algorithm.

The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105), integer *T* (1<=≤<=*T*<=≤<=*n*) and real number *c* (1<=<<=*c*<=≤<=100) — the time range when the resource should work, the length of the time range during which we need the mean number of requests and the coefficient *c* of the work of approximate algorithm. Number *c* is given with exactly six digits after the decimal point. The next line contains *n* integers *a**t* (1<=≤<=*a**t*<=≤<=106) — the number of queries to the service at each moment of time. The next line contains integer *m* (1<=≤<=*m*<=≤<=*n*) — the number of moments of time when we are interested in the mean number of queries for the last *T* seconds. The next line contains *m* integers *p**j* (*T*<=≤<=*p**j*<=≤<=*n*), representing another moment of time for which we need statistics. Moments *p**j* are strictly increasing.

Print *m* lines. The *j*-th line must contain three numbers *real*, *approx* and *error*, where: - is the real mean number of queries for the last *T* seconds; - *approx* is calculated by the given algorithm and equals *mean* at the moment of time *t*<==<=*p**j* (that is, after implementing the *p**j*-th iteration of the cycle); - is the relative error of the approximate algorithm. The numbers you printed will be compared to the correct numbers with the relative or absolute error 10<=-<=4. It is recommended to print the numbers with at least five digits after the decimal point.

[ "1 1 2.000000\n1\n1\n1\n", "11 4 1.250000\n9 11 7 5 15 6 6 6 6 6 6\n8\n4 5 6 7 8 9 10 11\n", "13 4 1.250000\n3 3 3 3 3 20 3 3 3 3 3 3 3\n10\n4 5 6 7 8 9 10 11 12 13\n" ]

[ "1.000000 0.500000 0.500000\n", "8.000000 4.449600 0.443800\n9.500000 6.559680 0.309507\n8.250000 6.447744 0.218455\n8.000000 6.358195 0.205226\n8.250000 6.286556 0.237993\n6.000000 6.229245 0.038207\n6.000000 6.183396 0.030566\n6.000000 6.146717 0.024453\n", "3.000000 1.771200 0.409600\n3.000000 2.016960 0.327680\n7.250000 5.613568 0.225715\n7.250000 5.090854 0.297813\n7.250000 4.672684 0.355492\n7.250000 4.338147 0.401635\n3.000000 4.070517 0.356839\n3.000000 3.856414 0.285471\n3.000000 3.685131 0.228377\n3.000000 3.548105 0.182702\n" ]

none

1,000

[ { "input": "1 1 2.000000\n1\n1\n1", "output": "1.000000 0.500000 0.500000" }, { "input": "11 4 1.250000\n9 11 7 5 15 6 6 6 6 6 6\n8\n4 5 6 7 8 9 10 11", "output": "8.000000 4.449600 0.443800\n9.500000 6.559680 0.309507\n8.250000 6.447744 0.218455\n8.000000 6.358195 0.205226\n8.250000 6.286556 0.237993\n6.000000 6.229245 0.038207\n6.000000 6.183396 0.030566\n6.000000 6.146717 0.024453" }, { "input": "13 4 1.250000\n3 3 3 3 3 20 3 3 3 3 3 3 3\n10\n4 5 6 7 8 9 10 11 12 13", "output": "3.000000 1.771200 0.409600\n3.000000 2.016960 0.327680\n7.250000 5.613568 0.225715\n7.250000 5.090854 0.297813\n7.250000 4.672684 0.355492\n7.250000 4.338147 0.401635\n3.000000 4.070517 0.356839\n3.000000 3.856414 0.285471\n3.000000 3.685131 0.228377\n3.000000 3.548105 0.182702" }, { "input": "1 1 2.000000\n4\n1\n1", "output": "4.000000 2.000000 0.500000" }, { "input": "1 1 2.000000\n1121\n1\n1", "output": "1121.000000 560.500000 0.500000" }, { "input": "1 1 2.000000\n758432\n1\n1", "output": "758432.000000 379216.000000 0.500000" }, { "input": "3 1 2.000000\n8 25 21\n3\n1 2 3", "output": "8.000000 4.000000 0.500000\n25.000000 14.500000 0.420000\n21.000000 17.750000 0.154762" }, { "input": "19 3 1.333333\n12 15 11 10 16 4 9 2 24 3 6 3 21 21 2 16 13 12 2\n17\n3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19", "output": "12.666667 7.250003 0.427631\n12.000000 7.937505 0.338541\n12.333333 9.953131 0.192989\n10.000000 8.464850 0.153515\n9.666667 8.598640 0.110486\n5.000000 6.948982 0.389796\n11.666667 11.211739 0.038994\n9.666667 9.158807 0.052537\n11.000000 8.369107 0.239172\n4.000000 7.026832 0.756708\n10.000000 10.520127 0.052013\n15.000000 13.140098 0.123993\n14.666667 10.355076 0.293972\n13.000000 11.766310 0.094899\n10.333333 12.074736 0.168523\n13.666667 12.056055 0.117850\n9.000000 9.542043 0.060227" }, { "input": "64 3 1.333333\n1337 1913 135 885 1567 1049 1116 368 350 725 517 1874 588 918 1923 998 1237 1098 121 1304 1459 942 538 1480 293 178 958 728 1240 1721 1549 825 928 1189 194 626 1872 670 1145 200 333 1772 1136 614 174 1448 249 1783 798 1375 1574 870 360 398 1387 1092 314 294 1056 1890 1170 697 668 1570\n62\n3 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", "output": "1128.333333 580.453454 0.485566\n977.666667 656.590254 0.328411\n862.333333 884.192912 0.025349\n1167.000000 925.394915 0.207031\n1244.000000 973.046430 0.217808\n844.333333 821.785028 0.026705\n611.333333 703.838947 0.151318\n481.000000 709.129387 0.474281\n530.666667 661.097206 0.245786\n1038.666667 964.323145 0.071576\n993.000000 870.242577 0.123623\n1126.666667 882.182153 0.216998\n1143.000000 1142.386900 0.000536\n1279.666667 1106.290452 0.135485\n1386.000000 1138.968124 0.178234\n1111.000000 1128.726375 0.015955\n81..." } ]

1,426,355,079

9,879

Python 3

TIME_LIMIT_EXCEEDED

PRETESTS

9

4,000

17,817,600

n, T, c = map(float, input().split()) n = int(n) T = int(T) a = [eval(i) for i in input().split()] m = int(input()) q = [eval(i) for i in input().split()] res = [] for i in range(n): res.append([0] * 2) for i in range(m): d = q[i] s = 0 for j in range(d - T, d): s += a[j] real = s / T res[d-1][0] = real mean = 0 for i in range(n): mean = (mean + a[i] / T) / c res[i][1] = mean for i in range(len(q)): print(res[q[i] - 1][0], res[q[i] - 1][1], abs((res[q[i] - 1][1] - res[q[i] - 1][0]) / res[q[i] - 1][0]))

Title: Mean Requests Time Limit: None seconds Memory Limit: None megabytes Problem Description: In this problem you will have to deal with a real algorithm that is used in the VK social network. As in any other company that creates high-loaded websites, the VK developers have to deal with request statistics regularly. An important indicator reflecting the load of the site is the mean number of requests for a certain period of time of *T* seconds (for example, *T*<==<=60 *seconds*<==<=1 *min* and *T*<==<=86400 *seconds*<==<=1 *day*). For example, if this value drops dramatically, that shows that the site has access problem. If this value grows, that may be a reason to analyze the cause for the growth and add more servers to the website if it is really needed. However, even such a natural problem as counting the mean number of queries for some period of time can be a challenge when you process the amount of data of a huge social network. That's why the developers have to use original techniques to solve problems approximately, but more effectively at the same time. Let's consider the following formal model. We have a service that works for *n* seconds. We know the number of queries to this resource *a**t* at each moment of time *t* (1<=≤<=*t*<=≤<=*n*). Let's formulate the following algorithm of calculating the mean with exponential decay. Let *c* be some real number, strictly larger than one. Thus, the mean variable is recalculated each second using the number of queries that came at that second. We can make some mathematical calculations and prove that choosing the value of constant *c* correctly will make the value of mean not very different from the real mean value *a**x* at *t*<=-<=*T*<=+<=1<=≤<=*x*<=≤<=*t*. The advantage of such approach is that it only uses the number of requests at the current moment of time and doesn't require storing the history of requests for a large time range. Also, it considers the recent values with the weight larger than the weight of the old ones, which helps to react to dramatic change in values quicker. However before using the new theoretical approach in industrial programming, there is an obligatory step to make, that is, to test its credibility practically on given test data sets. Your task is to compare the data obtained as a result of the work of an approximate algorithm to the real data. You are given *n* values *a**t*, integer *T* and real number *c*. Also, you are given *m* moments *p**j* (1<=≤<=*j*<=≤<=*m*), where we are interested in the mean value of the number of queries for the last *T* seconds. Implement two algorithms. The first one should calculate the required value by definition, i.e. by the formula . The second algorithm should calculate the mean value as is described above. Print both values and calculate the relative error of the second algorithm by the formula , where *approx* is the approximate value, obtained by the second algorithm, and *real* is the exact value obtained by the first algorithm. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105), integer *T* (1<=≤<=*T*<=≤<=*n*) and real number *c* (1<=<<=*c*<=≤<=100) — the time range when the resource should work, the length of the time range during which we need the mean number of requests and the coefficient *c* of the work of approximate algorithm. Number *c* is given with exactly six digits after the decimal point. The next line contains *n* integers *a**t* (1<=≤<=*a**t*<=≤<=106) — the number of queries to the service at each moment of time. The next line contains integer *m* (1<=≤<=*m*<=≤<=*n*) — the number of moments of time when we are interested in the mean number of queries for the last *T* seconds. The next line contains *m* integers *p**j* (*T*<=≤<=*p**j*<=≤<=*n*), representing another moment of time for which we need statistics. Moments *p**j* are strictly increasing. Output Specification: Print *m* lines. The *j*-th line must contain three numbers *real*, *approx* and *error*, where: - is the real mean number of queries for the last *T* seconds; - *approx* is calculated by the given algorithm and equals *mean* at the moment of time *t*<==<=*p**j* (that is, after implementing the *p**j*-th iteration of the cycle); - is the relative error of the approximate algorithm. The numbers you printed will be compared to the correct numbers with the relative or absolute error 10<=-<=4. It is recommended to print the numbers with at least five digits after the decimal point. Demo Input: ['1 1 2.000000\n1\n1\n1\n', '11 4 1.250000\n9 11 7 5 15 6 6 6 6 6 6\n8\n4 5 6 7 8 9 10 11\n', '13 4 1.250000\n3 3 3 3 3 20 3 3 3 3 3 3 3\n10\n4 5 6 7 8 9 10 11 12 13\n'] Demo Output: ['1.000000 0.500000 0.500000\n', '8.000000 4.449600 0.443800\n9.500000 6.559680 0.309507\n8.250000 6.447744 0.218455\n8.000000 6.358195 0.205226\n8.250000 6.286556 0.237993\n6.000000 6.229245 0.038207\n6.000000 6.183396 0.030566\n6.000000 6.146717 0.024453\n', '3.000000 1.771200 0.409600\n3.000000 2.016960 0.327680\n7.250000 5.613568 0.225715\n7.250000 5.090854 0.297813\n7.250000 4.672684 0.355492\n7.250000 4.338147 0.401635\n3.000000 4.070517 0.356839\n3.000000 3.856414 0.285471\n3.000000 3.685131 0.228377\n3.000000 3.548105 0.182702\n'] Note: none

```python n, T, c = map(float, input().split()) n = int(n) T = int(T) a = [eval(i) for i in input().split()] m = int(input()) q = [eval(i) for i in input().split()] res = [] for i in range(n): res.append([0] * 2) for i in range(m): d = q[i] s = 0 for j in range(d - T, d): s += a[j] real = s / T res[d-1][0] = real mean = 0 for i in range(n): mean = (mean + a[i] / T) / c res[i][1] = mean for i in range(len(q)): print(res[q[i] - 1][0], res[q[i] - 1][1], abs((res[q[i] - 1][1] - res[q[i] - 1][0]) / res[q[i] - 1][0])) ```

0

136

A

Presents

PROGRAMMING

800

[ "implementation" ]

null

Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.

Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.

[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]

[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]

none

500

[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 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26 56 29 3 90 10 91 28 85 47 71 50 43 18 2" }, { "input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53", "output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5" }, { "input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39", "output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12" }, { "input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87", "output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82" }, { "input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25", "output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97" }, { "input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61", "output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96" }, { "input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92", "output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96" }, { "input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6", "output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36" }, { "input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1" } ]

1,697,817,427

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

62

0

n = int(input()); inp = input().split() dict = {} for i in range(1,n+1): dict[i]=inp[i-1] list =[] for i in sorted(dict.values()): for k in dict.keys(): if dict[k] == i: list.append(str(k)) print(" ".join(list))

Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none

```python n = int(input()); inp = input().split() dict = {} for i in range(1,n+1): dict[i]=inp[i-1] list =[] for i in sorted(dict.values()): for k in dict.keys(): if dict[k] == i: list.append(str(k)) print(" ".join(list)) ```

0

914

B

Conan and Agasa play a Card Game

PROGRAMMING

1,200

[ "games", "greedy", "implementation" ]

null

Edogawa Conan got tired of solving cases, and invited his friend, Professor Agasa, over. They decided to play a game of cards. Conan has *n* cards, and the *i*-th card has a number *a**i* written on it. They take turns playing, starting with Conan. In each turn, the player chooses a card and removes it. Also, he removes all cards having a number strictly lesser than the number on the chosen card. Formally, if the player chooses the *i*-th card, he removes that card and removes the *j*-th card for all *j* such that *a**j*<=<<=*a**i*. A player loses if he cannot make a move on his turn, that is, he loses if there are no cards left. Predict the outcome of the game, assuming both players play optimally.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of cards Conan has. The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105), where *a**i* is the number on the *i*-th card.

If Conan wins, print "Conan" (without quotes), otherwise print "Agasa" (without quotes).

[ "3\n4 5 7\n", "2\n1 1\n" ]

[ "Conan\n", "Agasa\n" ]

In the first example, Conan can just choose the card having number 7 on it and hence remove all the cards. After that, there are no cards left on Agasa's turn. In the second example, no matter which card Conan chooses, there will be one one card left, which Agasa can choose. After that, there are no cards left when it becomes Conan's turn again.

1,000

[ { "input": "3\n4 5 7", "output": "Conan" }, { "input": "2\n1 1", "output": "Agasa" }, { "input": "10\n38282 53699 38282 38282 38282 38282 38282 38282 38282 38282", "output": "Conan" }, { "input": "10\n50165 50165 50165 50165 50165 50165 50165 50165 50165 50165", "output": "Agasa" }, { "input": "10\n83176 83176 83176 23495 83176 8196 83176 23495 83176 83176", "output": "Conan" }, { "input": "10\n32093 36846 32093 32093 36846 36846 36846 36846 36846 36846", "output": "Conan" }, { "input": "3\n1 2 3", "output": "Conan" }, { "input": "4\n2 3 4 5", "output": "Conan" }, { "input": "10\n30757 30757 33046 41744 39918 39914 41744 39914 33046 33046", "output": "Conan" }, { "input": "10\n50096 50096 50096 50096 50096 50096 28505 50096 50096 50096", "output": "Conan" }, { "input": "10\n54842 54842 54842 54842 57983 54842 54842 57983 57983 54842", "output": "Conan" }, { "input": "10\n87900 87900 5761 87900 87900 87900 5761 87900 87900 87900", "output": "Agasa" }, { "input": "10\n53335 35239 26741 35239 35239 26741 35239 35239 53335 35239", "output": "Agasa" }, { "input": "10\n75994 64716 75994 64716 75994 75994 56304 64716 56304 64716", "output": "Agasa" }, { "input": "1\n1", "output": "Conan" }, { "input": "5\n2 2 1 1 1", "output": "Conan" }, { "input": "5\n1 4 4 5 5", "output": "Conan" }, { "input": "3\n1 3 3", "output": "Conan" }, { "input": "3\n2 2 2", "output": "Conan" }, { "input": "5\n1 1 1 2 2", "output": "Conan" }, { "input": "4\n1 2 1 2", "output": "Agasa" }, { "input": "7\n7 7 7 7 6 6 6", "output": "Conan" }, { "input": "3\n2 3 3", "output": "Conan" }, { "input": "3\n1 1 100000", "output": "Conan" }, { "input": "1\n100000", "output": "Conan" }, { "input": "5\n3 3 3 4 4", "output": "Conan" }, { "input": "3\n1 2 2", "output": "Conan" }, { "input": "3\n4 4 5", "output": "Conan" }, { "input": "1\n2", "output": "Conan" }, { "input": "3\n97 97 100", "output": "Conan" }, { "input": "5\n100000 100000 100000 1 1", "output": "Conan" }, { "input": "7\n7 7 6 6 5 5 4", "output": "Conan" }, { "input": "5\n100000 100000 100000 2 2", "output": "Conan" }, { "input": "4\n3 3 2 1", "output": "Conan" }, { "input": "1\n485", "output": "Conan" }, { "input": "3\n4 4 100000", "output": "Conan" }, { "input": "3\n1 1 2", "output": "Conan" }, { "input": "3\n1 1 1", "output": "Conan" }, { "input": "5\n1 1 2 2 2", "output": "Conan" } ]

1,572,452,747

2,147,483,647

Python 3

OK

TESTS

91

187

10,035,200

input() d={} c=0 for i in input().split(): d[i] = d.get(i,0)+1;c+=d[i]%2*2-1 print(["Conan","Agasa"][not c])

Title: Conan and Agasa play a Card Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Edogawa Conan got tired of solving cases, and invited his friend, Professor Agasa, over. They decided to play a game of cards. Conan has *n* cards, and the *i*-th card has a number *a**i* written on it. They take turns playing, starting with Conan. In each turn, the player chooses a card and removes it. Also, he removes all cards having a number strictly lesser than the number on the chosen card. Formally, if the player chooses the *i*-th card, he removes that card and removes the *j*-th card for all *j* such that *a**j*<=<<=*a**i*. A player loses if he cannot make a move on his turn, that is, he loses if there are no cards left. Predict the outcome of the game, assuming both players play optimally. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of cards Conan has. The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105), where *a**i* is the number on the *i*-th card. Output Specification: If Conan wins, print "Conan" (without quotes), otherwise print "Agasa" (without quotes). Demo Input: ['3\n4 5 7\n', '2\n1 1\n'] Demo Output: ['Conan\n', 'Agasa\n'] Note: In the first example, Conan can just choose the card having number 7 on it and hence remove all the cards. After that, there are no cards left on Agasa's turn. In the second example, no matter which card Conan chooses, there will be one one card left, which Agasa can choose. After that, there are no cards left when it becomes Conan's turn again.

```python input() d={} c=0 for i in input().split(): d[i] = d.get(i,0)+1;c+=d[i]%2*2-1 print(["Conan","Agasa"][not c]) ```

3

820

A

Mister B and Book Reading

PROGRAMMING

900

[ "implementation" ]

null

Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages. At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day. Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time. Help Mister B to calculate how many days he needed to finish the book.

First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.

Print one integer — the number of days Mister B needed to finish the book.

[ "5 5 10 5 4\n", "12 4 12 4 1\n", "15 1 100 0 0\n" ]

[ "1\n", "3\n", "15\n" ]

In the first sample test the book contains 5 pages, so Mister B read it right at the first day. In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book. In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.

500

[ { "input": "5 5 10 5 4", "output": "1" }, { "input": "12 4 12 4 1", "output": "3" }, { "input": "15 1 100 0 0", "output": "15" }, { "input": "1 1 1 0 0", "output": "1" }, { "input": "1000 999 1000 1000 998", "output": "2" }, { "input": "1000 2 2 5 1", "output": "999" }, { "input": "1000 1 1 1000 0", "output": "1000" }, { "input": "737 41 74 12 11", "output": "13" }, { "input": "1000 1000 1000 0 999", "output": "1" }, { "input": "765 12 105 5 7", "output": "17" }, { "input": "15 2 2 1000 0", "output": "8" }, { "input": "1000 1 1000 1000 0", "output": "2" }, { "input": "20 3 7 1 2", "output": "6" }, { "input": "1000 500 500 1000 499", "output": "501" }, { "input": "1 1000 1000 1000 0", "output": "1" }, { "input": "1000 2 1000 56 0", "output": "7" }, { "input": "1000 2 1000 802 0", "output": "3" }, { "input": "16 1 8 2 0", "output": "4" }, { "input": "20 6 10 2 2", "output": "3" }, { "input": "8 2 12 4 1", "output": "3" }, { "input": "8 6 13 2 5", "output": "2" }, { "input": "70 4 20 87 0", "output": "5" }, { "input": "97 8 13 234 5", "output": "13" }, { "input": "16 4 23 8 3", "output": "3" }, { "input": "65 7 22 7 4", "output": "5" }, { "input": "93 10 18 11 7", "output": "9" }, { "input": "86 13 19 15 9", "output": "9" }, { "input": "333 17 50 10 16", "output": "12" }, { "input": "881 16 55 10 12", "output": "23" }, { "input": "528 11 84 3 9", "output": "19" }, { "input": "896 2 184 8 1", "output": "16" }, { "input": "236 10 930 9 8", "output": "8" }, { "input": "784 1 550 14 0", "output": "12" }, { "input": "506 1 10 4 0", "output": "53" }, { "input": "460 1 3 2 0", "output": "154" }, { "input": "701 1 3 1 0", "output": "235" }, { "input": "100 49 50 1000 2", "output": "3" }, { "input": "100 1 100 100 0", "output": "2" }, { "input": "12 1 4 2 0", "output": "4" }, { "input": "22 10 12 0 0", "output": "3" }, { "input": "20 10 15 1 4", "output": "3" }, { "input": "1000 5 10 1 4", "output": "169" }, { "input": "1000 1 1000 1 0", "output": "45" }, { "input": "4 1 2 2 0", "output": "3" }, { "input": "1 5 5 1 1", "output": "1" }, { "input": "19 10 11 0 2", "output": "3" }, { "input": "1 2 3 0 0", "output": "1" }, { "input": "10 1 4 10 0", "output": "4" }, { "input": "20 3 100 1 1", "output": "5" }, { "input": "1000 5 9 5 0", "output": "112" }, { "input": "1 11 12 0 10", "output": "1" }, { "input": "1 1 1 1 0", "output": "1" }, { "input": "1000 1 20 1 0", "output": "60" }, { "input": "9 1 4 2 0", "output": "4" }, { "input": "129 2 3 4 0", "output": "44" }, { "input": "4 2 2 0 1", "output": "3" }, { "input": "1000 1 10 100 0", "output": "101" }, { "input": "100 1 100 1 0", "output": "14" }, { "input": "8 3 4 2 0", "output": "3" }, { "input": "20 1 6 4 0", "output": "5" }, { "input": "8 2 4 2 0", "output": "3" }, { "input": "11 5 6 7 2", "output": "3" }, { "input": "100 120 130 120 0", "output": "1" }, { "input": "7 1 4 1 0", "output": "4" }, { "input": "5 3 10 0 2", "output": "3" }, { "input": "5 2 2 0 0", "output": "3" }, { "input": "1000 10 1000 10 0", "output": "14" }, { "input": "25 3 50 4 2", "output": "4" }, { "input": "9 10 10 10 9", "output": "1" }, { "input": "17 10 12 6 5", "output": "2" }, { "input": "15 5 10 3 0", "output": "3" }, { "input": "8 3 5 1 0", "output": "3" }, { "input": "19 1 12 5 0", "output": "4" }, { "input": "1000 10 1000 1 0", "output": "37" }, { "input": "100 1 2 1000 0", "output": "51" }, { "input": "20 10 11 1000 9", "output": "6" }, { "input": "16 2 100 1 1", "output": "5" }, { "input": "18 10 13 2 5", "output": "3" }, { "input": "12 3 5 3 1", "output": "4" }, { "input": "17 3 11 2 0", "output": "4" }, { "input": "4 2 100 1 1", "output": "2" }, { "input": "7 4 5 2 3", "output": "3" }, { "input": "100 1 2 2 0", "output": "51" }, { "input": "50 4 5 5 0", "output": "11" }, { "input": "1 2 2 0 1", "output": "1" }, { "input": "1000 2 3 10 1", "output": "500" }, { "input": "500 10 500 1000 0", "output": "2" }, { "input": "1000 4 12 1 0", "output": "87" }, { "input": "18 10 13 1 5", "output": "3" }, { "input": "7 3 6 2 2", "output": "3" }, { "input": "15 5 100 1 2", "output": "4" }, { "input": "100 1 10 1 0", "output": "15" }, { "input": "8 2 7 5 1", "output": "2" }, { "input": "11 2 4 1 1", "output": "5" }, { "input": "1000 500 900 100 300", "output": "3" }, { "input": "7 1 2 5 0", "output": "4" }, { "input": "7 3 5 3 2", "output": "3" }, { "input": "7 3 10 2 1", "output": "2" }, { "input": "1000 501 510 1 499", "output": "50" }, { "input": "1000 1 1000 2 0", "output": "32" }, { "input": "1 5 5 0 0", "output": "1" }, { "input": "18 10 15 1 5", "output": "3" }, { "input": "100 4 1000 1 2", "output": "13" }, { "input": "20 2 40 1 1", "output": "6" }, { "input": "1 11 1000 100 1", "output": "1" }, { "input": "6 4 4 1 2", "output": "2" }, { "input": "8 3 5 3 1", "output": "3" }, { "input": "10 5 7 1 2", "output": "3" }, { "input": "400 100 198 1 99", "output": "25" }, { "input": "3 1 2 5 0", "output": "2" } ]

1,584,892,972

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

5

108

0

c,v0,v1,a,l=map(int,input().split()) # 12 4 12 4 1 i,speed=0,v0 # speed=4 pages=c-v0 # pages=12-4=8 i=1 if c<=v0: print(1) else: while speed<=v1 and pages>0: # 4<12,8<12,12==12 i+=1 # i=2,3 speed=speed+a # speed=4+4=8,8+4=12,12+4=16 pages=pages-speed+l # pages=8-8+1=1,9-8=1,1-12=-11 print(i)

Title: Mister B and Book Reading Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages. At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day. Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time. Help Mister B to calculate how many days he needed to finish the book. Input Specification: First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading. Output Specification: Print one integer — the number of days Mister B needed to finish the book. Demo Input: ['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n'] Demo Output: ['1\n', '3\n', '15\n'] Note: In the first sample test the book contains 5 pages, so Mister B read it right at the first day. In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book. In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.

```python c,v0,v1,a,l=map(int,input().split()) # 12 4 12 4 1 i,speed=0,v0 # speed=4 pages=c-v0 # pages=12-4=8 i=1 if c<=v0: print(1) else: while speed<=v1 and pages>0: # 4<12,8<12,12==12 i+=1 # i=2,3 speed=speed+a # speed=4+4=8,8+4=12,12+4=16 pages=pages-speed+l # pages=8-8+1=1,9-8=1,1-12=-11 print(i) ```

0

723

A

The New Year: Meeting Friends

PROGRAMMING

800

[ "implementation", "math", "sortings" ]

null

There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer.

The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.

Print one integer — the minimum total distance the friends need to travel in order to meet together.

[ "7 1 4\n", "30 20 10\n" ]

[ "6\n", "20\n" ]

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

500

[ { "input": "7 1 4", "output": "6" }, { "input": "30 20 10", "output": "20" }, { "input": "1 4 100", "output": "99" }, { "input": "100 1 91", "output": "99" }, { "input": "1 45 100", "output": "99" }, { "input": "1 2 3", "output": "2" }, { "input": "71 85 88", "output": "17" }, { "input": "30 38 99", "output": "69" }, { "input": "23 82 95", "output": "72" }, { "input": "22 41 47", "output": "25" }, { "input": "9 94 77", "output": "85" }, { "input": "1 53 51", "output": "52" }, { "input": "25 97 93", "output": "72" }, { "input": "42 53 51", "output": "11" }, { "input": "81 96 94", "output": "15" }, { "input": "21 5 93", "output": "88" }, { "input": "50 13 75", "output": "62" }, { "input": "41 28 98", "output": "70" }, { "input": "69 46 82", "output": "36" }, { "input": "87 28 89", "output": "61" }, { "input": "44 45 40", "output": "5" }, { "input": "86 97 68", "output": "29" }, { "input": "43 92 30", "output": "62" }, { "input": "16 70 1", "output": "69" }, { "input": "40 46 19", "output": "27" }, { "input": "71 38 56", "output": "33" }, { "input": "82 21 80", "output": "61" }, { "input": "75 8 35", "output": "67" }, { "input": "75 24 28", "output": "51" }, { "input": "78 23 56", "output": "55" }, { "input": "85 31 10", "output": "75" }, { "input": "76 50 9", "output": "67" }, { "input": "95 37 34", "output": "61" }, { "input": "84 61 35", "output": "49" }, { "input": "87 85 37", "output": "50" }, { "input": "1 3 2", "output": "2" }, { "input": "4 2 6", "output": "4" }, { "input": "6 9 3", "output": "6" }, { "input": "12 4 8", "output": "8" }, { "input": "15 10 5", "output": "10" }, { "input": "1 50 17", "output": "49" }, { "input": "10 5 15", "output": "10" }, { "input": "8 1 9", "output": "8" }, { "input": "3 5 4", "output": "2" }, { "input": "2 1 3", "output": "2" }, { "input": "1 8 2", "output": "7" }, { "input": "1 100 2", "output": "99" }, { "input": "1 4 6", "output": "5" } ]

1,693,131,530

2,147,483,647

Python 3

OK

TESTS

48

46

0

x,y,z = map(int,input().split()) min_dist = float('inf') xy = abs(x-y) xz = abs(x-z) yz = abs(y-z) if xy+xz < min_dist: min_dist = xy+xz if xy+yz < min_dist: min_dist = xy+yz if xz+yz < min_dist: min_dist = xz + yz print(min_dist)

Title: The New Year: Meeting Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer. Input Specification: The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. Output Specification: Print one integer — the minimum total distance the friends need to travel in order to meet together. Demo Input: ['7 1 4\n', '30 20 10\n'] Demo Output: ['6\n', '20\n'] Note: In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

```python x,y,z = map(int,input().split()) min_dist = float('inf') xy = abs(x-y) xz = abs(x-z) yz = abs(y-z) if xy+xz < min_dist: min_dist = xy+xz if xy+yz < min_dist: min_dist = xy+yz if xz+yz < min_dist: min_dist = xz + yz print(min_dist) ```

3

753

A

Santa Claus and Candies

PROGRAMMING

1,000

[ "dp", "greedy", "math" ]

null

Santa Claus has *n* candies, he dreams to give them as gifts to children. What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.

The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.

Print to the first line integer number *k* — maximal number of kids which can get candies. Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*. If there are many solutions, print any of them.

[ "5\n", "9\n", "2\n" ]

[ "2\n2 3\n", "3\n3 5 1\n", "1\n2 \n" ]

none

500

[ { "input": "5", "output": "2\n1 4 " }, { "input": "9", "output": "3\n1 2 6 " }, { "input": "2", "output": "1\n2 " }, { "input": "1", "output": "1\n1 " }, { "input": "3", "output": "2\n1 2 " }, { "input": "1000", "output": "44\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 54 " }, { "input": "4", "output": "2\n1 3 " }, { "input": "6", "output": "3\n1 2 3 " }, { "input": "7", "output": "3\n1 2 4 " }, { "input": "8", "output": "3\n1 2 5 " }, { "input": "10", "output": "4\n1 2 3 4 " }, { "input": "11", "output": "4\n1 2 3 5 " }, { "input": "12", "output": "4\n1 2 3 6 " }, { "input": "13", "output": "4\n1 2 3 7 " }, { "input": "14", "output": "4\n1 2 3 8 " }, { "input": "15", "output": "5\n1 2 3 4 5 " }, { "input": "16", "output": "5\n1 2 3 4 6 " }, { "input": "20", "output": "5\n1 2 3 4 10 " }, { "input": "21", "output": "6\n1 2 3 4 5 6 " }, { "input": "22", "output": "6\n1 2 3 4 5 7 " }, { "input": "27", "output": "6\n1 2 3 4 5 12 " }, { "input": "28", "output": "7\n1 2 3 4 5 6 7 " }, { "input": "29", "output": "7\n1 2 3 4 5 6 8 " }, { "input": "35", "output": "7\n1 2 3 4 5 6 14 " }, { "input": "36", "output": "8\n1 2 3 4 5 6 7 8 " }, { "input": "37", "output": "8\n1 2 3 4 5 6 7 9 " }, { "input": "44", "output": "8\n1 2 3 4 5 6 7 16 " }, { "input": "45", "output": "9\n1 2 3 4 5 6 7 8 9 " }, { "input": "46", "output": "9\n1 2 3 4 5 6 7 8 10 " }, { "input": "230", "output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 " }, { "input": "231", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 " }, { "input": "232", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 " }, { "input": "239", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 " }, { "input": "629", "output": "34\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 68 " }, { "input": "630", "output": "35\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 " }, { "input": "631", "output": "35\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 36 " }, { "input": "945", "output": "42\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 84 " }, { "input": "946", "output": "43\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 " }, { "input": "947", "output": "43\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 44 " }, { "input": "989", "output": "43\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 86 " }, { "input": "990", "output": "44\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 " }, { "input": "991", "output": "44\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 45 " }, { "input": "956", "output": "43\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 53 " }, { "input": "981", "output": "43\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 78 " }, { "input": "867", "output": "41\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 47 " }, { "input": "906", "output": "42\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 45 " }, { "input": "999", "output": "44\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 53 " }, { "input": "100", "output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 " }, { "input": "126", "output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 " } ]

1,483,195,942

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

62

4,608,000

k = int(input()) def bagi(n): i = 1 k = 1 while n > i: k += 1 i += k return k-1 n = bagi(k) print(n) hasil = [i+1 for i in range(n-1)] hasil.append(k-sum(hasil)) hasil = [str(i) for i in hasil] print(" ".join(hasil))

Title: Santa Claus and Candies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus has *n* candies, he dreams to give them as gifts to children. What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has. Input Specification: The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has. Output Specification: Print to the first line integer number *k* — maximal number of kids which can get candies. Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*. If there are many solutions, print any of them. Demo Input: ['5\n', '9\n', '2\n'] Demo Output: ['2\n2 3\n', '3\n3 5 1\n', '1\n2 \n'] Note: none

```python k = int(input()) def bagi(n): i = 1 k = 1 while n > i: k += 1 i += k return k-1 n = bagi(k) print(n) hasil = [i+1 for i in range(n-1)] hasil.append(k-sum(hasil)) hasil = [str(i) for i in hasil] print(" ".join(hasil)) ```

0

5

C

Longest Regular Bracket Sequence

PROGRAMMING

1,900

[ "constructive algorithms", "data structures", "dp", "greedy", "sortings", "strings" ]

C. Longest Regular Bracket Sequence

2

256

This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

[ ")((())))(()())\n", "))(\n" ]

[ "6 2\n", "0 1\n" ]

none

0

[ { "input": ")((())))(()())", "output": "6 2" }, { "input": "))(", "output": "0 1" }, { "input": "()(())()", "output": "8 1" }, { "input": "((((()(((", "output": "2 1" }, { "input": "))))()())))", "output": "4 1" }, { "input": "(()())()(())()()())())()((()(()(())()()())((()(())()(()()()()))()(())()(((()())()(()((())()(())(()))", "output": "28 1" }, { "input": ")(()(((((())))))(((()))))))((()(((((((())())()))))((((((((())))((((((())))))(()((((()))))(()))))))))", "output": "68 1" }, { "input": "())(((((())())((((()))(())))())())(((()(()()()())(())()))((()(())())()()))()(()())))))(()))((())((((", "output": "80 1" }, { "input": "))(((((((()((((())(((((((((()()))))(()())()((())())))((((()))(()()()))))(()()))())))))((())())()))((", "output": "92 1" } ]

1,510,791,551

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

14

2,000

8,601,600

import sys from collections import defaultdict sys.setrecursionlimit(100000) OPEN, CLOSE = '(', ')' def mistery(data): terminated_counter = 0 counter = 0 open_index = 0 for char in data: if char == OPEN: terminated_counter += 1 open_index = counter elif char == CLOSE and terminated_counter > 0: terminated_counter -= 1 data[open_index] = 1 data[counter] = 1 return mistery(data) counter += 1 return data def solve(string): groups = defaultdict(int) group = 0 for c in [x for x in mistery(list(string.strip())) if x]: if c not in [OPEN, CLOSE]: group += c elif group: groups[group] += 1 group = 0 if group: groups[group] += 1 if groups: print("%s %s" % (max(groups), groups[max(groups)])) else: print("0 1") solve(sys.stdin.readline())

Title: Longest Regular Bracket Sequence Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well. Input Specification: The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106. Output Specification: Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1". Demo Input: [')((())))(()())\n', '))(\n'] Demo Output: ['6 2\n', '0 1\n'] Note: none

```python import sys from collections import defaultdict sys.setrecursionlimit(100000) OPEN, CLOSE = '(', ')' def mistery(data): terminated_counter = 0 counter = 0 open_index = 0 for char in data: if char == OPEN: terminated_counter += 1 open_index = counter elif char == CLOSE and terminated_counter > 0: terminated_counter -= 1 data[open_index] = 1 data[counter] = 1 return mistery(data) counter += 1 return data def solve(string): groups = defaultdict(int) group = 0 for c in [x for x in mistery(list(string.strip())) if x]: if c not in [OPEN, CLOSE]: group += c elif group: groups[group] += 1 group = 0 if group: groups[group] += 1 if groups: print("%s %s" % (max(groups), groups[max(groups)])) else: print("0 1") solve(sys.stdin.readline()) ```

0

25

A

IQ test

PROGRAMMING

1,300

[ "brute force" ]

A. IQ test

2

256

Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.

The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.

Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.

[ "5\n2 4 7 8 10\n", "4\n1 2 1 1\n" ]

[ "3\n", "2\n" ]

none

0

[ { "input": "5\n2 4 7 8 10", "output": "3" }, { "input": "4\n1 2 1 1", "output": "2" }, { "input": "3\n1 2 2", "output": "1" }, { "input": "3\n100 99 100", "output": "2" }, { "input": "3\n5 3 2", "output": "3" }, { "input": "4\n43 28 1 91", "output": "2" }, { "input": "4\n75 13 94 77", "output": "3" }, { "input": "4\n97 8 27 3", "output": "2" }, { "input": "10\n95 51 12 91 85 3 1 31 25 7", "output": "3" }, { "input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46", "output": "4" }, { "input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36", "output": "26" }, { "input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83", "output": "48" }, { "input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25", "output": "1" }, { "input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46", "output": "25" }, { "input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73", "output": "13" }, { "input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76", "output": "99" }, { "input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14", "output": "1" }, { "input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22", "output": "93" }, { "input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23", "output": "1" }, { "input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98", "output": "1" }, { "input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94", "output": "93" }, { "input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29", "output": "32" }, { "input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47", "output": "26" }, { "input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72", "output": "56" }, { "input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86", "output": "97" }, { "input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42", "output": "79" }, { "input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52", "output": "25" }, { "input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86", "output": "78" }, { "input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68", "output": "70" }, { "input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81", "output": "39" }, { "input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83", "output": "88" }, { "input": "3\n100 100 1", "output": "3" } ]

1,544,188,931

2,147,483,647

Python 3

OK

TESTS

32

218

0

x=eval(input()) l=list(map(int,input().split())) i = 0 ev=0 odd=0 while(i<x): if (l[i]%2==0): ev+=1 else: odd+=1 i+=1 i=0 if ev==1: while(i<x): if (l[i]%2==0): print(i+1) break i+=1 else: while(i<x): if (l[i]%2!=0): print(i+1) break i+=1

Title: IQ test Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. Output Specification: Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. Demo Input: ['5\n2 4 7 8 10\n', '4\n1 2 1 1\n'] Demo Output: ['3\n', '2\n'] Note: none

```python x=eval(input()) l=list(map(int,input().split())) i = 0 ev=0 odd=0 while(i<x): if (l[i]%2==0): ev+=1 else: odd+=1 i+=1 i=0 if ev==1: while(i<x): if (l[i]%2==0): print(i+1) break i+=1 else: while(i<x): if (l[i]%2!=0): print(i+1) break i+=1 ```

3.9455

385

A

Bear and Raspberry

PROGRAMMING

1,000

[ "brute force", "greedy", "implementation" ]

null

The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry. Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel. The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.

The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel. The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*.

Print a single integer — the answer to the problem.

[ "5 1\n5 10 7 3 20\n", "6 2\n100 1 10 40 10 40\n", "3 0\n1 2 3\n" ]

[ "3\n", "97\n", "0\n" ]

In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3. In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.

500

[ { "input": "5 1\n5 10 7 3 20", "output": "3" }, { "input": "6 2\n100 1 10 40 10 40", "output": "97" }, { "input": "3 0\n1 2 3", "output": "0" }, { "input": "2 0\n2 1", "output": "1" }, { "input": "10 5\n10 1 11 2 12 3 13 4 14 5", "output": "4" }, { "input": "100 4\n2 57 70 8 44 10 88 67 50 44 93 79 72 50 69 19 21 9 71 47 95 13 46 10 68 72 54 40 15 83 57 92 58 25 4 22 84 9 8 55 87 0 16 46 86 58 5 21 32 28 10 46 11 29 13 33 37 34 78 33 33 21 46 70 77 51 45 97 6 21 68 61 87 54 8 91 37 12 76 61 57 9 100 45 44 88 5 71 98 98 26 45 37 87 34 50 33 60 64 77", "output": "87" }, { "input": "100 5\n15 91 86 53 18 52 26 89 8 4 5 100 11 64 88 91 35 57 67 72 71 71 69 73 97 23 11 1 59 86 37 82 6 67 71 11 7 31 11 68 21 43 89 54 27 10 3 33 8 57 79 26 90 81 6 28 24 7 33 50 24 13 27 85 4 93 14 62 37 67 33 40 7 48 41 4 14 9 95 10 64 62 7 93 23 6 28 27 97 64 26 83 70 0 97 74 11 82 70 93", "output": "84" }, { "input": "6 100\n10 9 8 7 6 5", "output": "0" }, { "input": "100 9\n66 71 37 41 23 38 77 11 74 13 51 26 93 56 81 17 12 70 85 37 54 100 14 99 12 83 44 16 99 65 13 48 92 32 69 33 100 57 58 88 25 45 44 85 5 41 82 15 37 18 21 45 3 68 33 9 52 64 8 73 32 41 87 99 26 26 47 24 79 93 9 44 11 34 85 26 14 61 49 38 25 65 49 81 29 82 28 23 2 64 38 13 77 68 67 23 58 57 83 46", "output": "78" }, { "input": "100 100\n9 72 46 37 26 94 80 1 43 85 26 53 58 18 24 19 67 2 100 52 61 81 48 15 73 41 97 93 45 1 73 54 75 51 28 79 0 14 41 42 24 50 70 18 96 100 67 1 68 48 44 39 63 77 78 18 10 51 32 53 26 60 1 13 66 39 55 27 23 71 75 0 27 88 73 31 16 95 87 84 86 71 37 40 66 70 65 83 19 4 81 99 26 51 67 63 80 54 23 44", "output": "0" }, { "input": "43 65\n32 58 59 75 85 18 57 100 69 0 36 38 79 95 82 47 7 55 28 88 27 88 63 71 80 86 67 53 69 37 99 54 81 19 55 12 2 17 84 77 25 26 62", "output": "4" }, { "input": "12 64\n14 87 40 24 32 36 4 41 38 77 68 71", "output": "0" }, { "input": "75 94\n80 92 25 48 78 17 69 52 79 73 12 15 59 55 25 61 96 27 98 43 30 43 36 94 67 54 86 99 100 61 65 8 65 19 18 21 75 31 2 98 55 87 14 1 17 97 94 11 57 29 34 71 76 67 45 0 78 29 86 82 29 23 77 100 48 43 65 62 88 34 7 28 13 1 1", "output": "0" }, { "input": "59 27\n76 61 24 66 48 18 69 84 21 8 64 90 19 71 36 90 9 36 30 37 99 37 100 56 9 79 55 37 54 63 11 11 49 71 91 70 14 100 10 44 52 23 21 19 96 13 93 66 52 79 76 5 62 6 90 35 94 7 27", "output": "63" }, { "input": "86 54\n41 84 16 5 20 79 73 13 23 24 42 73 70 80 69 71 33 44 62 29 86 88 40 64 61 55 58 19 16 23 84 100 38 91 89 98 47 50 55 87 12 94 2 12 0 1 4 26 50 96 68 34 94 80 8 22 60 3 72 84 65 89 44 52 50 9 24 34 81 28 56 17 38 85 78 90 62 60 1 40 91 2 7 41 84 22", "output": "38" }, { "input": "37 2\n65 36 92 92 92 76 63 56 15 95 75 26 15 4 73 50 41 92 26 20 19 100 63 55 25 75 61 96 35 0 14 6 96 3 28 41 83", "output": "91" }, { "input": "19 4\n85 2 56 70 33 75 89 60 100 81 42 28 18 92 29 96 49 23 14", "output": "79" }, { "input": "89 1\n50 53 97 41 68 27 53 66 93 19 11 78 46 49 38 69 96 9 43 16 1 63 95 64 96 6 34 34 45 40 19 4 53 8 11 18 95 25 50 16 64 33 97 49 23 81 63 10 30 73 76 55 7 70 9 98 6 36 75 78 3 92 85 75 40 75 55 71 9 91 15 17 47 55 44 35 55 88 53 87 61 22 100 56 14 87 36 84 24", "output": "91" }, { "input": "67 0\n40 48 15 46 90 7 65 52 24 15 42 81 2 6 71 94 32 18 97 67 83 98 48 51 10 47 8 68 36 46 65 75 90 30 62 9 5 35 80 60 69 58 62 68 58 73 80 9 22 46 56 64 44 11 93 73 62 54 15 20 17 69 16 33 85 62 49", "output": "83" }, { "input": "96 0\n38 97 82 43 80 40 1 99 50 94 81 63 92 13 57 24 4 10 25 32 79 56 96 19 25 14 69 56 66 22 23 78 87 76 37 30 75 77 61 64 35 64 62 32 44 62 6 84 91 44 99 5 71 19 17 12 35 52 1 14 35 18 8 36 54 42 4 67 80 11 88 44 34 35 12 38 66 42 4 90 45 10 1 44 37 96 23 28 100 90 75 17 27 67 51 70", "output": "94" }, { "input": "14 14\n87 63 62 31 59 47 40 89 92 43 80 30 99 42", "output": "43" }, { "input": "12 0\n100 1 100 2 100 3 100 4 100 5 100 0", "output": "100" }, { "input": "3 1\n1 2 3", "output": "0" }, { "input": "3 2\n3 3 3", "output": "0" }, { "input": "3 3\n3 2 1", "output": "0" }, { "input": "3 100\n1 2 3", "output": "0" }, { "input": "2 100\n0 0", "output": "0" }, { "input": "2 90\n10 5", "output": "0" }, { "input": "2 5\n5 4", "output": "0" }, { "input": "3 1\n19 20 1", "output": "18" }, { "input": "5 1\n5 10 7 4 20", "output": "2" }, { "input": "5 1\n1 2 3 4 5", "output": "0" } ]

1,620,718,612

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

46

6,963,200

y,x=list(map(int,input().split())) nums=list(map(int,input().split())) def func(x,y,nums): if x>y: return 0 ans=0 for i in range((y-x)): s=nums[i:x+i+1] for j in range(1,x+1): if s[j]<s[j-1]: ans=max(ans,s[j-1]-s[j]) return ans print(func(x,y,nums)-x)

Title: Bear and Raspberry Time Limit: None seconds Memory Limit: None megabytes Problem Description: The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry. Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel. The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan. Input Specification: The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel. The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*. Output Specification: Print a single integer — the answer to the problem. Demo Input: ['5 1\n5 10 7 3 20\n', '6 2\n100 1 10 40 10 40\n', '3 0\n1 2 3\n'] Demo Output: ['3\n', '97\n', '0\n'] Note: In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3. In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.

```python y,x=list(map(int,input().split())) nums=list(map(int,input().split())) def func(x,y,nums): if x>y: return 0 ans=0 for i in range((y-x)): s=nums[i:x+i+1] for j in range(1,x+1): if s[j]<s[j-1]: ans=max(ans,s[j-1]-s[j]) return ans print(func(x,y,nums)-x) ```

0

166

A

Rank List

PROGRAMMING

1,100

[ "binary search", "implementation", "sortings" ]

null

Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place. You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*. It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place. Your task is to count what number of teams from the given list shared the *k*-th place.

The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.

In the only line print the sought number of teams that got the *k*-th place in the final results' table.

[ "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n", "5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n" ]

[ "3\n", "4\n" ]

The final results' table for the first sample is: - 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10 The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams. The final table for the second sample is: - 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1 The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.

500

[ { "input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10", "output": "3" }, { "input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1", "output": "4" }, { "input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2", "output": "2" }, { "input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5", "output": "1" }, { "input": "5 5\n3 1\n10 2\n2 2\n1 10\n10 2", "output": "1" }, { "input": "3 2\n3 3\n3 3\n3 3", "output": "3" }, { "input": "4 3\n10 3\n6 10\n5 2\n5 2", "output": "2" }, { "input": "5 3\n10 10\n10 10\n1 1\n10 10\n4 3", "output": "3" }, { "input": "3 1\n2 1\n1 1\n1 2", "output": "1" }, { "input": "1 1\n28 28", "output": "1" }, { "input": "2 2\n1 2\n1 2", "output": "2" }, { "input": "5 3\n2 3\n4 2\n5 3\n2 4\n3 5", "output": "1" }, { "input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8", "output": "1" }, { "input": "50 6\n11 20\n18 13\n1 13\n3 11\n4 17\n15 10\n15 8\n9 16\n11 17\n16 3\n3 20\n14 13\n12 15\n9 10\n14 2\n12 12\n13 17\n6 10\n20 9\n2 8\n13 7\n7 20\n15 3\n1 20\n2 13\n2 5\n14 7\n10 13\n15 12\n15 5\n17 6\n9 11\n18 5\n10 1\n15 14\n3 16\n6 12\n4 1\n14 9\n7 14\n8 17\n17 13\n4 6\n19 16\n5 6\n3 15\n4 19\n15 20\n2 10\n20 10", "output": "1" }, { "input": "50 12\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "50" }, { "input": "50 28\n2 2\n1 1\n2 1\n1 2\n1 1\n1 1\n1 1\n2 2\n2 2\n2 2\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n2 2\n2 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n2 2\n2 1\n2 1\n2 2\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n2 2\n2 2\n2 2\n2 2", "output": "13" }, { "input": "50 40\n2 3\n3 1\n2 1\n2 1\n2 1\n3 1\n1 1\n1 2\n2 3\n1 3\n1 3\n2 1\n3 1\n1 1\n3 1\n3 1\n2 2\n1 1\n3 3\n3 1\n3 2\n2 3\n3 3\n3 1\n1 3\n2 3\n2 1\n3 2\n3 3\n3 1\n2 1\n2 2\n1 3\n3 3\n1 1\n3 2\n1 2\n2 3\n2 1\n2 2\n3 2\n1 3\n3 1\n1 1\n3 3\n2 3\n2 1\n2 3\n2 3\n1 2", "output": "5" }, { "input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3", "output": "1" }, { "input": "50 32\n6 6\n4 2\n5 5\n1 1\n2 4\n6 5\n2 3\n6 5\n2 3\n6 3\n1 4\n1 6\n3 3\n2 4\n3 2\n6 2\n4 1\n3 3\n3 1\n5 5\n1 2\n2 1\n5 4\n3 1\n4 4\n5 6\n4 1\n2 5\n3 1\n4 6\n2 3\n1 1\n6 5\n2 6\n3 3\n2 6\n2 3\n2 6\n3 4\n2 6\n4 5\n5 4\n1 6\n3 2\n5 1\n4 1\n4 6\n4 2\n1 2\n5 2", "output": "1" }, { "input": "50 48\n5 1\n6 4\n3 2\n2 1\n4 7\n3 6\n7 1\n7 5\n6 5\n5 6\n4 7\n5 7\n5 7\n5 5\n7 3\n3 5\n4 3\n5 4\n6 2\n1 6\n6 3\n6 5\n5 2\n4 2\n3 1\n1 1\n5 6\n1 3\n6 5\n3 7\n1 5\n7 5\n6 5\n3 6\n2 7\n5 3\n5 3\n4 7\n5 2\n6 5\n5 7\n7 1\n2 3\n5 5\n2 6\n4 1\n6 2\n6 5\n3 3\n1 6", "output": "1" }, { "input": "50 8\n5 3\n7 3\n4 3\n7 4\n2 2\n4 4\n5 4\n1 1\n7 7\n4 8\n1 1\n6 3\n1 5\n7 3\n6 5\n4 5\n8 6\n3 6\n2 1\n3 2\n2 5\n7 6\n5 8\n1 3\n5 5\n8 4\n4 5\n4 4\n8 8\n7 2\n7 2\n3 6\n2 8\n8 3\n3 2\n4 5\n8 1\n3 2\n8 7\n6 3\n2 3\n5 1\n3 4\n7 2\n6 3\n7 3\n3 3\n6 4\n2 2\n5 1", "output": "3" }, { "input": "20 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "20" }, { "input": "20 20\n1 2\n2 2\n1 1\n2 1\n2 2\n1 1\n1 1\n2 1\n1 1\n1 2\n2 2\n1 2\n1 2\n2 2\n2 2\n1 2\n2 1\n2 1\n1 2\n2 2", "output": "6" }, { "input": "30 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "30" }, { "input": "30 22\n2 1\n1 2\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n2 2\n1 2\n2 2\n1 2\n1 2\n2 1\n1 2\n2 2\n2 2\n1 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n2 2\n1 2\n2 2\n2 1\n1 1", "output": "13" }, { "input": "30 22\n1 1\n1 3\n2 3\n3 1\n2 3\n3 1\n1 2\n3 3\n2 1\n2 1\n2 2\n3 1\n3 2\n2 3\n3 1\n1 3\n2 3\n3 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 2\n1 3\n3 3\n3 3\n3 3\n3 3\n3 1", "output": "5" }, { "input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3", "output": "1" }, { "input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8", "output": "1" }, { "input": "50 22\n29 15\n18 10\n6 23\n38 28\n34 40\n40 1\n16 26\n22 33\n14 30\n26 7\n15 16\n22 40\n14 15\n6 28\n32 27\n33 3\n38 22\n40 17\n16 27\n21 27\n34 26\n5 15\n34 9\n38 23\n7 36\n17 6\n19 37\n40 1\n10 28\n9 14\n8 31\n40 8\n14 2\n24 16\n38 33\n3 37\n2 9\n21 21\n40 26\n28 33\n24 31\n10 12\n27 27\n17 4\n38 5\n21 31\n5 12\n29 7\n39 12\n26 14", "output": "1" }, { "input": "50 14\n4 20\n37 50\n46 19\n20 25\n47 10\n6 34\n12 41\n47 9\n22 28\n41 34\n47 40\n12 42\n9 4\n15 15\n27 8\n38 9\n4 17\n8 13\n47 7\n9 38\n30 48\n50 7\n41 34\n23 11\n16 37\n2 32\n18 46\n37 48\n47 41\n13 9\n24 50\n46 14\n33 49\n9 50\n35 30\n49 44\n42 49\n39 15\n33 42\n3 18\n44 15\n44 28\n9 17\n16 4\n10 36\n4 22\n47 17\n24 12\n2 31\n6 30", "output": "2" }, { "input": "2 1\n50 50\n50 50", "output": "2" }, { "input": "2 2\n50 50\n50 50", "output": "2" }, { "input": "2 1\n50 50\n50 49", "output": "1" }, { "input": "2 2\n50 50\n50 49", "output": "1" }, { "input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50", "output": "50" }, { "input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n49 50", "output": "1" }, { "input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 49", "output": "49" }, { "input": "10 1\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4", "output": "1" }, { "input": "4 3\n1 1\n1 2\n2 1\n2 2", "output": "1" }, { "input": "4 2\n1 2\n1 3\n1 5\n1 2", "output": "2" }, { "input": "6 1\n4 3\n4 2\n5 1\n5 2\n5 3\n5 3", "output": "1" }, { "input": "5 3\n2 1\n1 3\n1 2\n1 1\n1 1", "output": "2" }, { "input": "3 1\n1 1\n1 2\n1 3", "output": "1" }, { "input": "10 10\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4", "output": "4" }, { "input": "5 1\n5 5\n5 6\n5 7\n5 8\n5 5", "output": "2" }, { "input": "4 1\n1 1\n1 1\n1 2\n1 3", "output": "2" } ]

1,609,508,295

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

12

216

307,200

n,k = [int(x) for x in input().split()] a = [] for _ in range(n): a.append([int(x) for x in input().split()]) a.sort(reverse=True) same = 1 flag = False for i in range(0,n): if i>0 and a[i]==a[i-1]: same+=1 elif i>0 and a[i]!=a[i-1] and flag: print(same) break elif not flag: same=1 if i==k-1: flag = True else: print(same)

Title: Rank List Time Limit: None seconds Memory Limit: None megabytes Problem Description: Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place. You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*. It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place. Your task is to count what number of teams from the given list shared the *k*-th place. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces. Output Specification: In the only line print the sought number of teams that got the *k*-th place in the final results' table. Demo Input: ['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n'] Demo Output: ['3\n', '4\n'] Note: The final results' table for the first sample is: - 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10 The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams. The final table for the second sample is: - 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1 The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.

```python n,k = [int(x) for x in input().split()] a = [] for _ in range(n): a.append([int(x) for x in input().split()]) a.sort(reverse=True) same = 1 flag = False for i in range(0,n): if i>0 and a[i]==a[i-1]: same+=1 elif i>0 and a[i]!=a[i-1] and flag: print(same) break elif not flag: same=1 if i==k-1: flag = True else: print(same) ```

0

152

A

Marks

PROGRAMMING

900

[ "implementation" ]

null

Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group.

The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.

Print the single number — the number of successful students in the given group.

[ "3 3\n223\n232\n112\n", "3 5\n91728\n11828\n11111\n" ]

[ "2\n", "3\n" ]

In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.

500

[ { "input": "3 3\n223\n232\n112", "output": "2" }, { "input": "3 5\n91728\n11828\n11111", "output": "3" }, { "input": "2 2\n48\n27", "output": "1" }, { "input": "2 1\n4\n6", "output": "1" }, { "input": "1 2\n57", "output": "1" }, { "input": "1 1\n5", "output": "1" }, { "input": "3 4\n2553\n6856\n5133", "output": "2" }, { "input": "8 7\n6264676\n7854895\n3244128\n2465944\n8958761\n1378945\n3859353\n6615285", "output": "6" }, { "input": "9 8\n61531121\n43529859\n18841327\n88683622\n98995641\n62741632\n57441743\n49396792\n63381994", "output": "4" }, { "input": "10 20\n26855662887514171367\n48525577498621511535\n47683778377545341138\n47331616748732562762\n44876938191354974293\n24577238399664382695\n42724955594463126746\n79187344479926159359\n48349683283914388185\n82157191115518781898", "output": "9" }, { "input": "20 15\n471187383859588\n652657222494199\n245695867594992\n726154672861295\n614617827782772\n862889444974692\n373977167653235\n645434268565473\n785993468314573\n722176861496755\n518276853323939\n723712762593348\n728935312568886\n373898548522463\n769777587165681\n247592995114377\n182375946483965\n497496542536127\n988239919677856\n859844339819143", "output": "18" }, { "input": "13 9\n514562255\n322655246\n135162979\n733845982\n473117129\n513967187\n965649829\n799122777\n661249521\n298618978\n659352422\n747778378\n723261619", "output": "11" }, { "input": "75 1\n2\n3\n8\n3\n2\n1\n3\n1\n5\n1\n5\n4\n8\n8\n4\n2\n5\n1\n7\n6\n3\n2\n2\n3\n5\n5\n2\n4\n7\n7\n9\n2\n9\n5\n1\n4\n9\n5\n2\n4\n6\n6\n3\n3\n9\n3\n3\n2\n3\n4\n2\n6\n9\n1\n1\n1\n1\n7\n2\n3\n2\n9\n7\n4\n9\n1\n7\n5\n6\n8\n3\n4\n3\n4\n6", "output": "7" }, { "input": "92 3\n418\n665\n861\n766\n529\n416\n476\n676\n561\n995\n415\n185\n291\n176\n776\n631\n556\n488\n118\n188\n437\n496\n466\n131\n914\n118\n766\n365\n113\n897\n386\n639\n276\n946\n759\n169\n494\n837\n338\n351\n783\n311\n261\n862\n598\n132\n246\n982\n575\n364\n615\n347\n374\n368\n523\n132\n774\n161\n552\n492\n598\n474\n639\n681\n635\n342\n516\n483\n141\n197\n571\n336\n175\n596\n481\n327\n841\n133\n142\n146\n246\n396\n287\n582\n556\n996\n479\n814\n497\n363\n963\n162", "output": "23" }, { "input": "100 1\n1\n6\n9\n1\n1\n5\n5\n4\n6\n9\n6\n1\n7\n8\n7\n3\n8\n8\n7\n6\n2\n1\n5\n8\n7\n3\n5\n4\n9\n7\n1\n2\n4\n1\n6\n5\n1\n3\n9\n4\n5\n8\n1\n2\n1\n9\n7\n3\n7\n1\n2\n2\n2\n2\n3\n9\n7\n2\n4\n7\n1\n6\n8\n1\n5\n6\n1\n1\n2\n9\n7\n4\n9\n1\n9\n4\n1\n3\n5\n2\n4\n4\n6\n5\n1\n4\n5\n8\n4\n7\n6\n5\n6\n9\n5\n8\n1\n5\n1\n6", "output": "10" }, { "input": "100 2\n71\n87\n99\n47\n22\n87\n49\n73\n21\n12\n77\n43\n18\n41\n78\n62\n61\n16\n64\n89\n81\n54\n53\n92\n93\n94\n68\n93\n15\n68\n42\n93\n28\n19\n86\n16\n97\n17\n11\n43\n72\n76\n54\n95\n58\n53\n48\n45\n85\n85\n74\n21\n44\n51\n89\n75\n76\n17\n38\n62\n81\n22\n66\n59\n89\n85\n91\n87\n12\n97\n52\n87\n43\n89\n51\n58\n57\n98\n78\n68\n82\n41\n87\n29\n75\n72\n48\n14\n35\n71\n74\n91\n66\n67\n42\n98\n52\n54\n22\n41", "output": "21" }, { "input": "5 20\n11111111111111111111\n11111111111111111111\n11111111111111111111\n99999999999999999999\n11111111111111111119", "output": "2" }, { "input": "3 3\n111\n111\n999", "output": "1" }, { "input": "3 3\n119\n181\n711", "output": "3" }, { "input": "15 5\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111", "output": "15" }, { "input": "2 20\n22222222222222222222\n11111111111111111111", "output": "1" }, { "input": "3 3\n233\n222\n111", "output": "2" }, { "input": "4 15\n222222222222222\n111111111111119\n111111111111119\n111111111111111", "output": "3" }, { "input": "4 1\n1\n9\n9\n9", "output": "3" }, { "input": "3 3\n123\n321\n132", "output": "3" }, { "input": "3 3\n113\n332\n322", "output": "3" }, { "input": "2 100\n2222222222222222222222222222222222222222222222222222222222222222222222221222222222222222222222222222\n1111111111111111111111111111111111111111111111111111111111111111111111119111111111111111111111111111", "output": "2" }, { "input": "3 3\n321\n231\n123", "output": "3" }, { "input": "2 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222", "output": "1" }, { "input": "3 3\n221\n111\n111", "output": "3" } ]

1,547,227,241

2,147,483,647

Python 3

OK

TESTS

44

124

0

n,m=list(map(int,input().split())) l=[] count=0 for i in range(n): l.append(input()) final=[] for i in range(m): f=[] for j in l: f.append(int(j[i])) r=max(f) for k in range(n): if r==f[k] and k+1 not in final: count+=1 final.append(k+1) print(count)

Title: Marks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group. Input Specification: The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces. Output Specification: Print the single number — the number of successful students in the given group. Demo Input: ['3 3\n223\n232\n112\n', '3 5\n91728\n11828\n11111\n'] Demo Output: ['2\n', '3\n'] Note: In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.

```python n,m=list(map(int,input().split())) l=[] count=0 for i in range(n): l.append(input()) final=[] for i in range(m): f=[] for j in l: f.append(int(j[i])) r=max(f) for k in range(n): if r==f[k] and k+1 not in final: count+=1 final.append(k+1) print(count) ```

3

399

A

Pages

PROGRAMMING

0

[ "implementation" ]

null

User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are *n* pages numbered by integers from 1 to *n*. Assume that somebody is on the *p*-th page now. The navigation will look like this: When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page *n*. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: - If page 1 is in the navigation, the button "<<" must not be printed. - If page *n* is in the navigation, the button ">>" must not be printed. - If the page number is smaller than 1 or greater than *n*, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation.

The first and the only line contains three integers *n*, *p*, *k* (3<=≤<=*n*<=≤<=100; 1<=≤<=*p*<=≤<=*n*; 1<=≤<=*k*<=≤<=*n*)

Print the proper navigation. Follow the format of the output from the test samples.

[ "17 5 2\n", "6 5 2\n", "6 1 2\n", "6 2 2\n", "9 6 3\n", "10 6 3\n", "8 5 4\n" ]

[ "<< 3 4 (5) 6 7 >> ", "<< 3 4 (5) 6 ", "(1) 2 3 >> ", "1 (2) 3 4 >>", "<< 3 4 5 (6) 7 8 9", "<< 3 4 5 (6) 7 8 9 >>", "1 2 3 4 (5) 6 7 8 " ]

none

500

[ { "input": "17 5 2", "output": "<< 3 4 (5) 6 7 >> " }, { "input": "6 5 2", "output": "<< 3 4 (5) 6 " }, { "input": "6 1 2", "output": "(1) 2 3 >> " }, { "input": "6 2 2", "output": "1 (2) 3 4 >> " }, { "input": "9 6 3", "output": "<< 3 4 5 (6) 7 8 9 " }, { "input": "10 6 3", "output": "<< 3 4 5 (6) 7 8 9 >> " }, { "input": "8 5 4", "output": "1 2 3 4 (5) 6 7 8 " }, { "input": "100 10 20", "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 >> " }, { "input": "100 25 11", "output": "<< 14 15 16 17 18 19 20 21 22 23 24 (25) 26 27 28 29 30 31 32 33 34 35 36 >> " }, { "input": "5 2 1", "output": "1 (2) 3 >> " }, { "input": "5 3 1", "output": "<< 2 (3) 4 >> " }, { "input": "79 35 12", "output": "<< 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 >> " }, { "input": "100 99 5", "output": "<< 94 95 96 97 98 (99) 100 " }, { "input": "100 99 15", "output": "<< 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 (99) 100 " }, { "input": "100 100 17", "output": "<< 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 (100) " }, { "input": "100 35 28", "output": "<< 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": "100 46 38", "output": "<< 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 >> " }, { "input": "100 46 48", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 (46) 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 >> " }, { "input": "100 10 100", "output": "1 2 3 4 5 6 7 8 9 (10) 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 " }, { "input": "3 1 1", "output": "(1) 2 >> " }, { "input": "3 2 1", "output": "1 (2) 3 " }, { "input": "17 5 3", "output": "<< 2 3 4 (5) 6 7 8 >> " }, { "input": "3 1 3", "output": "(1) 2 3 " }, { "input": "7 5 1", "output": "<< 4 (5) 6 >> " }, { "input": "5 5 5", "output": "1 2 3 4 (5) " }, { "input": "5 3 5", "output": "1 2 (3) 4 5 " } ]

1,579,897,650

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

93

204,800

n, p, k = map(int, input().split()) if (p - k > 1) and (p + k < n): print("<< ", end="") for c in range(p - k, p + k + 1): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="") print(">>") elif p - k <= 1: for c in range(1, p + k + 1): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="") print(">>") else: print("<< ", end="") for c in range(p - k, n): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="")

Title: Pages Time Limit: None seconds Memory Limit: None megabytes Problem Description: User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are *n* pages numbered by integers from 1 to *n*. Assume that somebody is on the *p*-th page now. The navigation will look like this: When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page *n*. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: - If page 1 is in the navigation, the button "<<" must not be printed. - If page *n* is in the navigation, the button ">>" must not be printed. - If the page number is smaller than 1 or greater than *n*, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input Specification: The first and the only line contains three integers *n*, *p*, *k* (3<=≤<=*n*<=≤<=100; 1<=≤<=*p*<=≤<=*n*; 1<=≤<=*k*<=≤<=*n*) Output Specification: Print the proper navigation. Follow the format of the output from the test samples. Demo Input: ['17 5 2\n', '6 5 2\n', '6 1 2\n', '6 2 2\n', '9 6 3\n', '10 6 3\n', '8 5 4\n'] Demo Output: ['<< 3 4 (5) 6 7 >> ', '<< 3 4 (5) 6 ', '(1) 2 3 >> ', '1 (2) 3 4 >>', '<< 3 4 5 (6) 7 8 9', '<< 3 4 5 (6) 7 8 9 >>', '1 2 3 4 (5) 6 7 8 '] Note: none

```python n, p, k = map(int, input().split()) if (p - k > 1) and (p + k < n): print("<< ", end="") for c in range(p - k, p + k + 1): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="") print(">>") elif p - k <= 1: for c in range(1, p + k + 1): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="") print(">>") else: print("<< ", end="") for c in range(p - k, n): if c == p: print('(' + str(p) + ')', end=" ") else: print(str(c) + " ", end="") ```

0

443

A

Anton and Letters

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line. Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.

The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.

Print a single number — the number of distinct letters in Anton's set.

[ "{a, b, c}\n", "{b, a, b, a}\n", "{}\n" ]

[ "3\n", "2\n", "0\n" ]

none

500

[ { "input": "{a, b, c}", "output": "3" }, { "input": "{b, a, b, a}", "output": "2" }, { "input": "{}", "output": "0" }, { "input": "{a, a, c, b, b, b, c, c, c, c}", "output": "3" }, { "input": "{a, c, b, b}", "output": "3" }, { "input": "{a, b}", "output": "2" }, { "input": "{a}", "output": "1" }, { "input": "{b, a, b, a, b, c, c, b, c, b}", "output": "3" }, { "input": "{e, g, c, e}", "output": "3" }, { "input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}", "output": "1" }, { "input": "{a, a, b}", "output": "2" }, { "input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}", "output": "2" }, { "input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}", "output": "16" }, { "input": "{x, i, w, c, p, e, h, z, k, i}", "output": "9" }, { "input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}", "output": "15" }, { "input": "{y}", "output": "1" }, { "input": "{x}", "output": "1" }, { "input": "{b, z, a, z}", "output": "3" }, { "input": "{z}", "output": "1" }, { "input": "{a, z}", "output": "2" }, { "input": "{a, b, z}", "output": "3" }, { "input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}", "output": "11" } ]

1,686,659,293

2,147,483,647

Python 3

OK

TESTS

27

62

0

input_str = input() input_str = input_str[1:-1] if input_str: letters = input_str.split(", ") distinct_letters = set(letters) count = len(distinct_letters) else: count = 0 print(count)

Title: Anton and Letters Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line. Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set. Input Specification: The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space. Output Specification: Print a single number — the number of distinct letters in Anton's set. Demo Input: ['{a, b, c}\n', '{b, a, b, a}\n', '{}\n'] Demo Output: ['3\n', '2\n', '0\n'] Note: none

```python input_str = input() input_str = input_str[1:-1] if input_str: letters = input_str.split(", ") distinct_letters = set(letters) count = len(distinct_letters) else: count = 0 print(count) ```

3

37

A

Towers

PROGRAMMING

1,000

[ "sortings" ]

A. Towers

2

256

Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same. Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.

The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.

In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.

[ "3\n1 2 3\n", "4\n6 5 6 7\n" ]

[ "1 3\n", "2 3\n" ]

none

500

[ { "input": "3\n1 2 3", "output": "1 3" }, { "input": "4\n6 5 6 7", "output": "2 3" }, { "input": "4\n3 2 1 1", "output": "2 3" }, { "input": "4\n1 2 3 3", "output": "2 3" }, { "input": "3\n20 22 36", "output": "1 3" }, { "input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162", "output": "2 23" }, { "input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193", "output": "2 43" }, { "input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111", "output": "2 57" }, { "input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642", "output": "2 80" }, { "input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188", "output": "2 92" }, { "input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233", "output": "3 101" }, { "input": "5\n5 5 5 5 5", "output": "5 1" }, { "input": "3\n1000 1000 1000", "output": "3 1" }, { "input": "1\n1000", "output": "1 1" }, { "input": "1\n1", "output": "1 1" }, { "input": "5\n1 1000 1000 1000 1000", "output": "4 2" }, { "input": "5\n1000 1000 1000 8 7", "output": "3 3" } ]

1,491,108,469

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

92

4,608,000

input() l=map(int,input().split()) res={0:0} for i in l: if i in res.keys(): res[i]+=1 else: res[i]=1 print(max(res.values()),len(set(l)))

Title: Towers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same. Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible. Input Specification: The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000. Output Specification: In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars. Demo Input: ['3\n1 2 3\n', '4\n6 5 6 7\n'] Demo Output: ['1 3\n', '2 3\n'] Note: none

```python input() l=map(int,input().split()) res={0:0} for i in l: if i in res.keys(): res[i]+=1 else: res[i]=1 print(max(res.values()),len(set(l))) ```

0

699

B

One Bomb

PROGRAMMING

1,400

[ "implementation" ]

null

You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*"). You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*. You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field. The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes). Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

[ "3 4\n.*..\n....\n.*..\n", "3 3\n..*\n.*.\n*..\n", "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n" ]

[ "YES\n1 2\n", "NO\n", "YES\n3 3\n" ]

none

1,000

[ { "input": "3 4\n.*..\n....\n.*..", "output": "YES\n1 2" }, { "input": "3 3\n..*\n.*.\n*..", "output": "NO" }, { "input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..", "output": "YES\n3 3" }, { "input": "1 10\n**********", "output": "YES\n1 1" }, { "input": "10 1\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*", "output": "YES\n1 1" }, { "input": "10 10\n.........*\n.........*\n........**\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*", "output": "YES\n3 10" }, { "input": "10 10\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*", "output": "YES\n1 10" }, { "input": "2 2\n.*\n*.", "output": "YES\n2 2" }, { "input": "4 4\n....\n...*\n....\n*..*", "output": "YES\n4 4" }, { "input": "4 4\n*...\n*...\n....\n****", "output": "YES\n4 1" }, { "input": "1 1\n*", "output": "YES\n1 1" }, { "input": "1 1\n.", "output": "YES\n1 1" }, { "input": "1 2\n.*", "output": "YES\n1 2" }, { "input": "2 1\n.\n*", "output": "YES\n1 1" }, { "input": "2 2\n**\n**", "output": "NO" }, { "input": "3 1\n*\n*\n*", "output": "YES\n1 1" }, { "input": "3 2\n*.\n.*\n.*", "output": "YES\n1 2" }, { "input": "3 3\n***\n***\n***", "output": "NO" }, { "input": "2 2\n..\n.*", "output": "YES\n1 2" }, { "input": "6 5\n..*..\n..*..\n**.**\n..*..\n..*..\n..*..", "output": "YES\n3 3" }, { "input": "3 3\n.*.\n*.*\n.*.", "output": "YES\n2 2" }, { "input": "4 4\n*...\n....\n....\n...*", "output": "YES\n4 1" }, { "input": "2 4\n...*\n...*", "output": "YES\n1 4" }, { "input": "2 2\n..\n..", "output": "YES\n1 1" }, { "input": "3 3\n..*\n.*.\n..*", "output": "YES\n2 3" }, { "input": "2 2\n*.\n.*", "output": "YES\n2 1" }, { "input": "3 2\n.*\n*.\n.*", "output": "YES\n2 2" }, { "input": "3 3\n***\n.*.\n.*.", "output": "YES\n1 2" }, { "input": "4 4\n*.*.\n..*.\n.***\n..*.", "output": "NO" }, { "input": "2 3\n..*\n**.", "output": "YES\n2 3" }, { "input": "3 2\n*.\n.*\n*.", "output": "YES\n2 1" }, { "input": "4 4\n..*.\n**.*\n..*.\n..*.", "output": "YES\n2 3" }, { "input": "3 3\n*..\n*..\n***", "output": "YES\n3 1" }, { "input": "3 3\n...\n*.*\n.*.", "output": "YES\n2 2" }, { "input": "3 2\n..\n..\n**", "output": "YES\n3 1" }, { "input": "3 4\n...*\n...*\n...*", "output": "YES\n1 4" }, { "input": "5 5\n..*..\n..*..\n**.**\n..*..\n..*..", "output": "YES\n3 3" }, { "input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*.*", "output": "NO" }, { "input": "3 3\n...\n.*.\n..*", "output": "YES\n3 2" }, { "input": "3 5\n....*\n....*\n....*", "output": "YES\n1 5" }, { "input": "3 3\n...\n...\n.*.", "output": "YES\n1 2" }, { "input": "3 3\n*..\n...\n..*", "output": "YES\n3 1" }, { "input": "2 3\n..*\n..*", "output": "YES\n1 3" }, { "input": "2 2\n**\n.*", "output": "YES\n1 2" }, { "input": "3 3\n..*\n*..\n*..", "output": "YES\n1 1" }, { "input": "5 4\n.*..\n*.**\n.*..\n.*..\n.*..", "output": "YES\n2 2" }, { "input": "6 5\n*.*..\n..*..\n*****\n..*..\n..*..\n..*..", "output": "NO" }, { "input": "4 4\n.*..\n*.**\n....\n.*..", "output": "YES\n2 2" }, { "input": "3 5\n....*\n....*\n*****", "output": "YES\n3 5" }, { "input": "3 3\n..*\n*..\n..*", "output": "YES\n2 3" }, { "input": "6 6\n..*...\n......\n......\n......\n......\n*....*", "output": "YES\n6 3" }, { "input": "4 4\n.*..\n*...\n.*..\n.*..", "output": "YES\n2 2" }, { "input": "3 3\n...\n..*\n.*.", "output": "YES\n3 3" }, { "input": "3 2\n.*\n*.\n*.", "output": "YES\n1 1" }, { "input": "4 2\n**\n.*\n.*\n.*", "output": "YES\n1 2" }, { "input": "5 5\n*...*\n.....\n.....\n.....\n..*..", "output": "YES\n1 3" }, { "input": "3 3\n**.\n...\n..*", "output": "YES\n1 3" }, { "input": "3 3\n*.*\n*..\n*.*", "output": "NO" }, { "input": "5 4\n....\n....\n*..*\n....\n.*..", "output": "YES\n3 2" }, { "input": "5 5\n...*.\n...*.\n...*.\n...*.\n***.*", "output": "YES\n5 4" }, { "input": "5 5\n*****\n*****\n*****\n*****\n*****", "output": "NO" }, { "input": "3 3\n.*.\n..*\n.*.", "output": "YES\n2 2" }, { "input": "3 3\n*.*\n...\n*.*", "output": "NO" }, { "input": "2 3\n.*.\n*.*", "output": "YES\n2 2" }, { "input": "3 10\n.......*..\n........*.\n.........*", "output": "NO" }, { "input": "3 3\n.*.\n.*.\n.**", "output": "YES\n3 2" }, { "input": "4 4\n*...\n....\n....\n..**", "output": "YES\n4 1" }, { "input": "4 4\n****\n****\n****\n****", "output": "NO" }, { "input": "3 2\n.*\n.*\n*.", "output": "YES\n3 2" }, { "input": "3 3\n..*\n..*\n**.", "output": "YES\n3 3" }, { "input": "6 3\n...\n...\n...\n...\n**.\n.*.", "output": "YES\n5 2" }, { "input": "3 4\n****\n..*.\n..*.", "output": "YES\n1 3" }, { "input": "5 5\n*..*.\n.....\n.....\n.....\n...*.", "output": "YES\n1 4" }, { "input": "6 5\n..*..\n..*..\n.*...\n..*..\n..*..\n..*..", "output": "YES\n3 3" } ]

1,644,976,737

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

n,m=input().split(' ') n,m=int(n),int(m) wl=[] for i in range(n): wl.append(input('').split('')) def bombared(r,i,j): for ie in range(len(r)): r[ie][j]='.' for je in range(len(r[i])): r[i][je]='.' return r def alldest(r): for i in r: for j in i: if j=='*': return False return True def choosepos(r): for i in range(len(r)): for j in range(len(i)): nw=bombared(r,i,j) if alldest(nw): return (i,j) return (2147483647,2147483647) if choosepos(wl)==(2147483647,2147483647): print("NO") else: print("YES\n{} {}".format(choosepos(wl)[0],choosepos(wl)[1]))

Title: One Bomb Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*"). You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*. You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall. Input Specification: The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field. The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall. Output Specification: If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes). Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them. Demo Input: ['3 4\n.*..\n....\n.*..\n', '3 3\n..*\n.*.\n*..\n', '6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n'] Demo Output: ['YES\n1 2\n', 'NO\n', 'YES\n3 3\n'] Note: none

```python n,m=input().split(' ') n,m=int(n),int(m) wl=[] for i in range(n): wl.append(input('').split('')) def bombared(r,i,j): for ie in range(len(r)): r[ie][j]='.' for je in range(len(r[i])): r[i][je]='.' return r def alldest(r): for i in r: for j in i: if j=='*': return False return True def choosepos(r): for i in range(len(r)): for j in range(len(i)): nw=bombared(r,i,j) if alldest(nw): return (i,j) return (2147483647,2147483647) if choosepos(wl)==(2147483647,2147483647): print("NO") else: print("YES\n{} {}".format(choosepos(wl)[0],choosepos(wl)[1])) ```

-1

442

B

Andrey and Problem

PROGRAMMING

1,800

[ "greedy", "math", "probabilities" ]

null

Andrey needs one more problem to conduct a programming contest. He has *n* friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him. Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of Andrey's friends. The second line contains *n* real numbers *p**i* (0.0<=≤<=*p**i*<=≤<=1.0) — the probability that the *i*-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.

Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10<=-<=9.

[ "4\n0.1 0.2 0.3 0.8\n", "2\n0.1 0.2\n" ]

[ "0.800000000000\n", "0.260000000000\n" ]

In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one. In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.

1,500

[ { "input": "4\n0.1 0.2 0.3 0.8", "output": "0.800000000000" }, { "input": "2\n0.1 0.2", "output": "0.260000000000" }, { "input": "1\n0.217266", "output": "0.217266000000" }, { "input": "2\n0.608183 0.375030", "output": "0.608183000000" }, { "input": "3\n0.388818 0.399762 0.393874", "output": "0.478724284024" }, { "input": "4\n0.801024 0.610878 0.808545 0.732504", "output": "0.808545000000" }, { "input": "5\n0.239482 0.686259 0.543226 0.764939 0.401318", "output": "0.764939000000" }, { "input": "6\n0.462434 0.775020 0.479749 0.373861 0.492031 0.746333", "output": "0.775020000000" }, { "input": "7\n0.745337 0.892271 0.792853 0.892917 0.768246 0.901623 0.815793", "output": "0.901623000000" }, { "input": "1\n0.057695", "output": "0.057695000000" }, { "input": "2\n0.057750 0.013591", "output": "0.069771239500" }, { "input": "3\n0.087234 0.075148 0.033833", "output": "0.172781711023" }, { "input": "4\n0.016717 0.061051 0.036222 0.096258", "output": "0.181832937456" }, { "input": "5\n0.057095 0.046954 0.054676 0.025927 0.080810", "output": "0.214634688963" }, { "input": "6\n0.010924 0.032857 0.021824 0.020356 0.007107 0.082489", "output": "0.154629381329" }, { "input": "7\n0.016061 0.043107 0.088973 0.014785 0.044298 0.028315 0.086014", "output": "0.246482855791" }, { "input": "100\n0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01", "output": "0.369729637650" }, { "input": "1\n1.0", "output": "1.000000000000" }, { "input": "3\n0.1 0.1 0.1", "output": "0.243000000000" }, { "input": "3\n0.2 0.2 0.2", "output": "0.384000000000" }, { "input": "5\n0.01 0.01 0.01 0.01 0.01", "output": "0.048029800500" }, { "input": "3\n1.0 1.0 0", "output": "1.000000000000" }, { "input": "3\n0.1 0.2 0.3", "output": "0.398000000000" }, { "input": "7\n0.1 0.1 0.1 0.1 0.1 0.1 0.1", "output": "0.372008700000" }, { "input": "5\n0.5 0.5 0.5 1 0.5", "output": "1.000000000000" }, { "input": "3\n0.4 0.2 0.4", "output": "0.480000000000" }, { "input": "10\n0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1", "output": "0.387420489000" }, { "input": "2\n1.0 1.0", "output": "1.000000000000" }, { "input": "10\n0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01", "output": "0.091351724748" }, { "input": "5\n1.0 1.0 1.0 0.1 0", "output": "1.000000000000" }, { "input": "5\n0.0001 0.0001 0.0001 0.0001 0.0001", "output": "0.000499800030" }, { "input": "20\n0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1", "output": "0.387420489000" }, { "input": "2\n0.0 1.0", "output": "1.000000000000" }, { "input": "5\n0.00001 0.00001 0.00001 0.00001 0.00001", "output": "0.000049998000" }, { "input": "3\n0.2 0.8 1", "output": "1.000000000000" }, { "input": "4\n0.1 0.1 0.1 0.1", "output": "0.291600000000" }, { "input": "5\n0.31 0.21 0.05 0.37 0.18", "output": "0.450600000000" }, { "input": "5\n1 1 1 1 1", "output": "1.000000000000" }, { "input": "4\n1 1 1 1", "output": "1.000000000000" }, { "input": "7\n0.14 0.28 0.13 0.31 0.15 0.17 0.27", "output": "0.438108000000" }, { "input": "20\n0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001", "output": "0.019623400697" }, { "input": "100\n0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1", "output": "0.387420489000" } ]

1,629,585,826

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

4

108

20,172,800

n=int(input()) w=[float(k) for k in input().split()] w.sort() z=w.pop() if z>=0.5: print(z) else: while w and z<0.5: z=z-z*w[-1]+(1-z)*w[-1] w.pop() print(z)

Title: Andrey and Problem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Andrey needs one more problem to conduct a programming contest. He has *n* friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him. Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of Andrey's friends. The second line contains *n* real numbers *p**i* (0.0<=≤<=*p**i*<=≤<=1.0) — the probability that the *i*-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point. Output Specification: Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10<=-<=9. Demo Input: ['4\n0.1 0.2 0.3 0.8\n', '2\n0.1 0.2\n'] Demo Output: ['0.800000000000\n', '0.260000000000\n'] Note: In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one. In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.

```python n=int(input()) w=[float(k) for k in input().split()] w.sort() z=w.pop() if z>=0.5: print(z) else: while w and z<0.5: z=z-z*w[-1]+(1-z)*w[-1] w.pop() print(z) ```

0

999

A

Mishka and Contest

PROGRAMMING

800

[ "brute force", "implementation" ]

null

Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve?

The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.

Print one integer — the maximum number of problems Mishka can solve.

[ "8 4\n4 2 3 1 5 1 6 4\n", "5 2\n3 1 2 1 3\n", "5 100\n12 34 55 43 21\n" ]

[ "5\n", "0\n", "5\n" ]

In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.

0

[ { "input": "8 4\n4 2 3 1 5 1 6 4", "output": "5" }, { "input": "5 2\n3 1 2 1 3", "output": "0" }, { "input": "5 100\n12 34 55 43 21", "output": "5" }, { "input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62", "output": "100" }, { "input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98", "output": "98" }, { "input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2", "output": "39" }, { "input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16", "output": "99" }, { "input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1", "output": "1" }, { "input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "1 5\n4", "output": "1" }, { "input": "3 2\n1 4 1", "output": "2" }, { "input": "1 2\n100", "output": "0" }, { "input": "7 4\n4 2 3 4 4 2 3", "output": "7" }, { "input": "1 2\n1", "output": "1" }, { "input": "1 2\n15", "output": "0" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "5 3\n3 4 3 2 1", "output": "4" }, { "input": "1 1\n2", "output": "0" }, { "input": "1 5\n1", "output": "1" }, { "input": "6 6\n7 1 1 1 1 1", "output": "5" }, { "input": "5 5\n6 5 5 5 5", "output": "4" }, { "input": "1 4\n2", "output": "1" }, { "input": "9 4\n1 2 1 2 4 2 1 2 1", "output": "9" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 10\n5", "output": "1" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4", "output": "71" }, { "input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90", "output": "63" }, { "input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8", "output": "61" }, { "input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22", "output": "72" }, { "input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2", "output": "52" }, { "input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90", "output": "60" }, { "input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10", "output": "56" }, { "input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90", "output": "61" }, { "input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24", "output": "99" }, { "input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50", "output": "99" }, { "input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48", "output": "99" }, { "input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57", "output": "99" }, { "input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93", "output": "99" }, { "input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33", "output": "3" }, { "input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51", "output": "3" }, { "input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49", "output": "6" }, { "input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8", "output": "66" }, { "input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95", "output": "0" }, { "input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66", "output": "0" }, { "input": "2 8\n8 8", "output": "2" }, { "input": "1 6\n3", "output": "1" }, { "input": "1 5\n5", "output": "1" } ]

1,618,145,586

2,147,483,647

PyPy 3

OK

TESTS

48

108

0

k,n=[int(i) for i in input().split(" ")] tasks=[int(i) for i in input().split(" ")] stop=False count=0 while((not stop) and (len(tasks)>0)): if(tasks[0]<=n): count+=1 del(tasks[0]) elif(tasks[-1]<=n): count+=1 del(tasks[-1]) else: stop=True print(count)

Title: Mishka and Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve? Input Specification: The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list. Output Specification: Print one integer — the maximum number of problems Mishka can solve. Demo Input: ['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n'] Demo Output: ['5\n', '0\n', '5\n'] Note: In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.

```python k,n=[int(i) for i in input().split(" ")] tasks=[int(i) for i in input().split(" ")] stop=False count=0 while((not stop) and (len(tasks)>0)): if(tasks[0]<=n): count+=1 del(tasks[0]) elif(tasks[-1]<=n): count+=1 del(tasks[-1]) else: stop=True print(count) ```

3

681

A

A Good Contest

PROGRAMMING

800

[ "implementation" ]

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" } ]

1,580,735,872

2,147,483,647

Python 3

OK

TESTS

60

109

0

n = int(input()) for i in range(n): s , x , y = input().split() x = int(x) y = int(y) if x < y and x >= 2400 : print('YES') break 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()) for i in range(n): s , x , y = input().split() x = int(x) y = int(y) if x < y and x >= 2400 : print('YES') break else: print('NO') ```

3

122

A

Lucky Division

PROGRAMMING

1,000

[ "brute force", "number theory" ]

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" } ]

1,698,214,739

2,147,483,647

PyPy 3-64

OK

TESTS

36

124

1,536,000

n = int(input()) def isLucky(num): return bool(str(num).replace('4', '').replace('7', '') == '') yes = False for i in range(n): if isLucky(n) or (isLucky(i) and n%i == 0): print("YES") yes = True break if not yes: 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 n = int(input()) def isLucky(num): return bool(str(num).replace('4', '').replace('7', '') == '') yes = False for i in range(n): if isLucky(n) or (isLucky(i) and n%i == 0): print("YES") yes = True break if not yes: print("NO") ```

3

961

B

Lecture Sleep

PROGRAMMING

1,200

[ "data structures", "dp", "implementation", "two pointers" ]

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" } ]

1,607,331,345

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

3

186

10,444,800

n,k = list(map(int, input().split())) a = list(map(int, input().split())) t = list(map(int, input().split())) x = 0 summ = 0 maxx = 0 for i in range(n): summ += a[i]*t[i] for i in range(k): if not t[i]: x+=a[i] for i in range(n-k): x+=a[i+k-1]*(1-t[i+k-1]) x-=a[i]*(1-t[i]) maxx = max(x, maxx) print(summ+maxx)

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 n,k = list(map(int, input().split())) a = list(map(int, input().split())) t = list(map(int, input().split())) x = 0 summ = 0 maxx = 0 for i in range(n): summ += a[i]*t[i] for i in range(k): if not t[i]: x+=a[i] for i in range(n-k): x+=a[i+k-1]*(1-t[i+k-1]) x-=a[i]*(1-t[i]) maxx = max(x, maxx) print(summ+maxx) ```

0

689

B

Mike and Shortcuts

PROGRAMMING

1,600

[ "dfs and similar", "graphs", "greedy", "shortest paths" ]

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*1 = 0; 2: 1, 2; *m*2 = 1; 3: 1, 3; *m*3 = |3 - 1| = 2. In the second sample case the sequence for any intersection 1 < *i* is always 1, *i* and *m**i* = |1 - *i*|. In the third sample case — consider the following intersection sequences: 1: 1; *m*1 = 0; 2: 1, 2; *m*2 = |2 - 1| = 1; 3: 1, 4, 3; *m*3 = 1 + |4 - 3| = 2; 4: 1, 4; *m*4 = 1; 5: 1, 4, 5; *m*5 = 1 + |4 - 5| = 2; 6: 1, 4, 6; *m*6 = 1 + |4 - 6| = 3; 7: 1, 4, 5, 7; *m*7 = 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 " } ]

1,467,827,185

4,285

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): 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 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*1 = 0; 2: 1, 2; *m*2 = 1; 3: 1, 3; *m*3 = |3 - 1| = 2. In the second sample case the sequence for any intersection 1 < *i* is always 1, *i* and *m**i* = |1 - *i*|. In the third sample case — consider the following intersection sequences: 1: 1; *m*1 = 0; 2: 1, 2; *m*2 = |2 - 1| = 1; 3: 1, 4, 3; *m*3 = 1 + |4 - 3| = 2; 4: 1, 4; *m*4 = 1; 5: 1, 4, 5; *m*5 = 1 + |4 - 5| = 2; 6: 1, 4, 6; *m*6 = 1 + |4 - 6| = 3; 7: 1, 4, 5, 7; *m*7 = 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): 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 i in range(1,n+1): print(ai2[i],end=" ") ```

0

794

A

Bank Robbery

PROGRAMMING

800

[ "brute force", "implementation" ]

null

A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes. Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg. The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.

The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively. The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes. The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct.

Output a single integer: the maximum number of banknotes Oleg can take.

[ "5 3 7\n8\n4 7 5 5 3 6 2 8\n", "6 5 7\n5\n1 5 7 92 3\n" ]

[ "4\n", "0\n" ]

In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes. For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.

500

[ { "input": "5 3 7\n8\n4 7 5 5 3 6 2 8", "output": "4" }, { "input": "6 5 7\n5\n1 5 7 92 3", "output": "0" }, { "input": "3 2 4\n1\n3", "output": "1" }, { "input": "5 3 8\n12\n8 3 4 5 7 6 8 3 5 4 7 6", "output": "8" }, { "input": "7 3 10\n5\n3 3 3 3 3", "output": "0" }, { "input": "3 2 5\n4\n1 3 4 5", "output": "2" }, { "input": "3 2 4\n1\n1", "output": "0" }, { "input": "6 4 8\n1\n4", "output": "0" }, { "input": "2 1 3\n1\n3", "output": "0" } ]

1,646,980,429

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

4

2,000

9,011,200

from collections import Counter a,b,c=map(int,input().split());input() u=[*map(int,input().split())] r=Counter(u);bank=0 for i in range(b+1,c): try:bank+=r[i] except:bank+=0 print(bank)

Title: Bank Robbery Time Limit: None seconds Memory Limit: None megabytes Problem Description: A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes. Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg. The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather. Input Specification: The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively. The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes. The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct. Output Specification: Output a single integer: the maximum number of banknotes Oleg can take. Demo Input: ['5 3 7\n8\n4 7 5 5 3 6 2 8\n', '6 5 7\n5\n1 5 7 92 3\n'] Demo Output: ['4\n', '0\n'] Note: In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes. For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.

```python from collections import Counter a,b,c=map(int,input().split());input() u=[*map(int,input().split())] r=Counter(u);bank=0 for i in range(b+1,c): try:bank+=r[i] except:bank+=0 print(bank) ```

0

431

C

k-Tree

PROGRAMMING

1,600

[ "dp", "implementation", "trees" ]

null

Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree. A *k*-tree is an infinite rooted tree where: - each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*. The picture below shows a part of a 3-tree. Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7).

A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*).

Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).

[ "3 3 2\n", "3 3 3\n", "4 3 2\n", "4 5 2\n" ]

[ "3\n", "1\n", "6\n", "7\n" ]

none

1,500

[ { "input": "3 3 2", "output": "3" }, { "input": "3 3 3", "output": "1" }, { "input": "4 3 2", "output": "6" }, { "input": "4 5 2", "output": "7" }, { "input": "28 6 3", "output": "110682188" }, { "input": "5 100 1", "output": "16" }, { "input": "50 6 3", "output": "295630102" }, { "input": "10 13 6", "output": "48" }, { "input": "20 16 14", "output": "236" }, { "input": "1 10 1", "output": "1" }, { "input": "8 11 4", "output": "47" }, { "input": "16 5 4", "output": "16175" }, { "input": "5 26 17", "output": "0" }, { "input": "35 47 42", "output": "0" }, { "input": "11 6 2", "output": "975" }, { "input": "54 60 16", "output": "931055544" }, { "input": "47 5 1", "output": "164058640" }, { "input": "70 6 1", "output": "592826579" }, { "input": "40 77 77", "output": "0" }, { "input": "96 9 6", "output": "362487247" }, { "input": "52 46 4", "output": "27907693" }, { "input": "74 41 28", "output": "806604424" }, { "input": "100 100 100", "output": "1" }, { "input": "99 99 5", "output": "700732369" }, { "input": "100 100 1", "output": "988185646" }, { "input": "98 98 64", "output": "237643149" }, { "input": "28 74 2", "output": "134217727" }, { "input": "86 69 62", "output": "217513984" }, { "input": "9 17 14", "output": "0" }, { "input": "73 72 12", "output": "426374014" }, { "input": "41 98 76", "output": "0" }, { "input": "1 1 1", "output": "1" }, { "input": "1 100 100", "output": "0" }, { "input": "1 100 1", "output": "1" }, { "input": "1 100 2", "output": "0" }, { "input": "2 100 2", "output": "1" }, { "input": "2 100 1", "output": "2" }, { "input": "50 50 1", "output": "949480669" }, { "input": "100 50 50", "output": "661237556" }, { "input": "3 2 2", "output": "2" }, { "input": "100 50 3", "output": "494224664" }, { "input": "90 97 24", "output": "413496885" }, { "input": "31 8 8", "output": "52532592" }, { "input": "78 90 38", "output": "744021655" }, { "input": "100 13 11", "output": "883875774" }, { "input": "100 45 8", "output": "367847193" }, { "input": "31 8 4", "output": "924947104" }, { "input": "35 95 9", "output": "927164672" }, { "input": "45 5 3", "output": "252804490" }, { "input": "1 5 5", "output": "0" }, { "input": "89 75 59", "output": "179807625" }, { "input": "90 100 30", "output": "697322870" }, { "input": "89 73 32", "output": "152673180" }, { "input": "100 90 80", "output": "11531520" } ]

1,696,098,975

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

4

1,000

0

def dfs(left, state, k, d): if state and left == 0: return 1 sam = 0 i = 1 while i <= k and i <= left: if i >= d: state = True sam += dfs(left - i, state, k, d) i += 1 return sam def main(): inp = input() inp = inp.split() print(dfs(int(inp[0]),False,int(inp[1]),int(inp[2]))) if __name__ == "__main__": main()

Title: k-Tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree. A *k*-tree is an infinite rooted tree where: - each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*. The picture below shows a part of a 3-tree. Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7). Input Specification: A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*). Output Specification: Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7). Demo Input: ['3 3 2\n', '3 3 3\n', '4 3 2\n', '4 5 2\n'] Demo Output: ['3\n', '1\n', '6\n', '7\n'] Note: none

```python def dfs(left, state, k, d): if state and left == 0: return 1 sam = 0 i = 1 while i <= k and i <= left: if i >= d: state = True sam += dfs(left - i, state, k, d) i += 1 return sam def main(): inp = input() inp = inp.split() print(dfs(int(inp[0]),False,int(inp[1]),int(inp[2]))) if __name__ == "__main__": main() ```

0

287

A

IQ Test

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

In the city of Ultima Thule job applicants are often offered an IQ test. The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed. Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.

Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.

Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.

[ "####\n.#..\n####\n....\n", "####\n....\n####\n....\n" ]

[ "YES\n", "NO\n" ]

In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.

500

[ { "input": "###.\n...#\n###.\n...#", "output": "NO" }, { "input": ".##.\n#..#\n.##.\n#..#", "output": "NO" }, { "input": ".#.#\n#.#.\n.#.#\n#.#.", "output": "NO" }, { "input": "##..\n..##\n##..\n..##", "output": "NO" }, { "input": "#.#.\n#.#.\n.#.#\n.#.#", "output": "NO" }, { "input": ".#.#\n#.#.\n#.#.\n#.#.", "output": "NO" }, { "input": ".#.#\n#.#.\n#.#.\n.#.#", "output": "NO" }, { "input": "#.#.\n#.#.\n#.#.\n#.#.", "output": "NO" }, { "input": ".#.#\n.#.#\n.#.#\n.#.#", "output": "NO" }, { "input": "#..#\n.##.\n####\n####", "output": "YES" }, { "input": "#.#.\n.###\n#.#.\n.###", "output": "YES" }, { "input": "#..#\n.##.\n.##.\n#..#", "output": "YES" }, { "input": ".##.\n.#..\n##.#\n#..#", "output": "YES" }, { "input": ".##.\n##..\n#..#\n..##", "output": "YES" }, { "input": "##..\n##..\n..##\n..##", "output": "YES" }, { "input": ".#..\n###.\n.#.#\n..#.", "output": "YES" }, { "input": "###.\n###.\n...#\n...#", "output": "YES" }, { "input": "#.##\n##.#\n#.##\n##.#", "output": "YES" }, { "input": ".#.#\n#.#.\n.#.#\n#.##", "output": "YES" }, { "input": "##..\n..##\n##..\n...#", "output": "YES" }, { "input": ".#..\n..##\n##..\n..##", "output": "YES" }, { "input": "##..\n...#\n##..\n...#", "output": "YES" }, { "input": ".#..\n..#.\n.#..\n..#.", "output": "YES" }, { "input": "....\n....\n....\n.#.#", "output": "YES" }, { "input": "....\n....\n....\n...#", "output": "YES" } ]

1,664,233,858

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

7

77

0

# zdjfhdaskjfhjsdkahfjkdskfjsdkjfhsdfdsfsdfdfd rows = [] for _ in range(4): rows.append(input()) dir_list = [(-1, 0), (0, -1), (1, 0), (0, 1), (-1, 0)] res = False for i in range(4): prevGood = False for j in range(4): key_char = rows[i][j] for dir in dir_list: new_i = i + dir[0] new_j = j + dir[1] if new_i < 0 or new_i > 3 or new_j < 0 or new_j > 3: prevGood = False continue if rows[new_i][new_j] == key_char: if prevGood: res = True else: prevGood = True else: prevGood = False if res: print("YES") else: print("NO")

Title: IQ Test Time Limit: None seconds Memory Limit: None megabytes Problem Description: In the city of Ultima Thule job applicants are often offered an IQ test. The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed. Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. Input Specification: Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. Output Specification: Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. Demo Input: ['####\n.#..\n####\n....\n', '####\n....\n####\n....\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.

```python # zdjfhdaskjfhjsdkahfjkdskfjsdkjfhsdfdsfsdfdfd rows = [] for _ in range(4): rows.append(input()) dir_list = [(-1, 0), (0, -1), (1, 0), (0, 1), (-1, 0)] res = False for i in range(4): prevGood = False for j in range(4): key_char = rows[i][j] for dir in dir_list: new_i = i + dir[0] new_j = j + dir[1] if new_i < 0 or new_i > 3 or new_j < 0 or new_j > 3: prevGood = False continue if rows[new_i][new_j] == key_char: if prevGood: res = True else: prevGood = True else: prevGood = False if res: print("YES") else: print("NO") ```

0

544

B

Sea and Islands

PROGRAMMING

1,400

[ "constructive algorithms", "implementation" ]

null

A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island). Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists.

The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form.

If the answer doesn't exist, print "NO" (without the quotes) in a single line. Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*. If there are multiple answers, you may print any of them. You should not maximize the sizes of islands.

[ "5 2\n", "5 25\n" ]

[ "YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS\n", "NO\n" ]

none

1,000

[ { "input": "5 2", "output": "YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS" }, { "input": "5 25", "output": "NO" }, { "input": "82 6047", "output": "NO" }, { "input": "6 5", "output": "YES\nLSLSLS\nSLSLSS\nSSSSSS\nSSSSSS\nSSSSSS\nSSSSSS" }, { "input": "10 80", "output": "NO" }, { "input": "48 1279", "output": "NO" }, { "input": "40 1092", "output": "NO" }, { "input": "9 12", "output": "YES\nLSLSLSLSL\nSLSLSLSLS\nLSLSLSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS" }, { "input": "43 146", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSS..." }, { "input": "100 5000", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "100 4999", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "100 5001", "output": "NO" }, { "input": "99 4901", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nS..." }, { "input": "99 4900", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nS..." }, { "input": "99 4902", "output": "NO" }, { "input": "99 9801", "output": "NO" }, { "input": "99 10", "output": "YES\nLSLSLSLSLSLSLSLSLSLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "99 1", "output": "YES\nLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "100 10000", "output": "NO" }, { "input": "100 10", "output": "YES\nLSLSLSLSLSLSLSLSLSLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS..." }, { "input": "50 1200", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL..." }, { "input": "50 1438", "output": "NO" }, { "input": "50 2447", "output": "NO" }, { "input": "49 1719", "output": "NO" }, { "input": "51 1996", "output": "NO" }, { "input": "51 1981", "output": "NO" }, { "input": "34 1060", "output": "NO" }, { "input": "74 3901", "output": "NO" }, { "input": "65 617", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "89 497", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "34 621", "output": "NO" }, { "input": "1 0", "output": "YES\nS" }, { "input": "10 0", "output": "YES\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS" }, { "input": "11 0", "output": "YES\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS" }, { "input": "99 0", "output": "YES\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "100 0", "output": "YES\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS..." }, { "input": "1 1", "output": "YES\nL" }, { "input": "2 1", "output": "YES\nLS\nSS" }, { "input": "2 0", "output": "YES\nSS\nSS" }, { "input": "2 2", "output": "YES\nLS\nSL" }, { "input": "2 3", "output": "NO" }, { "input": "2 4", "output": "NO" } ]

1,610,548,599

2,147,483,647

Python 3

OK

TESTS

42

77

409,600

n,k=input().split() n,k=int(n),int(k) if k>(n*n+1)//2: print("NO") exit() print("YES") for i in range(0,n): for j in range(0,n): if((i+j)%2==0and k>0): print('L',end='') k-=1 else: print('S',end='') print()

Title: Sea and Islands Time Limit: None seconds Memory Limit: None megabytes Problem Description: A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island). Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists. Input Specification: The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form. Output Specification: If the answer doesn't exist, print "NO" (without the quotes) in a single line. Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*. If there are multiple answers, you may print any of them. You should not maximize the sizes of islands. Demo Input: ['5 2\n', '5 25\n'] Demo Output: ['YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS\n', 'NO\n'] Note: none

```python n,k=input().split() n,k=int(n),int(k) if k>(n*n+1)//2: print("NO") exit() print("YES") for i in range(0,n): for j in range(0,n): if((i+j)%2==0and k>0): print('L',end='') k-=1 else: print('S',end='') print() ```

3

799

A

Carrot Cakes

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven. Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.

The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.

If it is reasonable to build the second oven, print "YES". Otherwise print "NO".

[ "8 6 4 5\n", "8 6 4 6\n", "10 3 11 4\n", "4 2 1 4\n" ]

[ "YES\n", "NO\n", "NO\n", "YES\n" ]

In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven. In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven. In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.

500

[ { "input": "8 6 4 5", "output": "YES" }, { "input": "8 6 4 6", "output": "NO" }, { "input": "10 3 11 4", "output": "NO" }, { "input": "4 2 1 4", "output": "YES" }, { "input": "28 17 16 26", "output": "NO" }, { "input": "60 69 9 438", "output": "NO" }, { "input": "599 97 54 992", "output": "YES" }, { "input": "11 22 18 17", "output": "NO" }, { "input": "1 13 22 11", "output": "NO" }, { "input": "1 1 1 1", "output": "NO" }, { "input": "3 1 1 1", "output": "YES" }, { "input": "1000 1000 1000 1000", "output": "NO" }, { "input": "1000 1000 1 1", "output": "YES" }, { "input": "1000 1000 1 400", "output": "YES" }, { "input": "1000 1000 1 1000", "output": "YES" }, { "input": "1000 1000 1 999", "output": "YES" }, { "input": "53 11 3 166", "output": "YES" }, { "input": "313 2 3 385", "output": "NO" }, { "input": "214 9 9 412", "output": "NO" }, { "input": "349 9 5 268", "output": "YES" }, { "input": "611 16 8 153", "output": "YES" }, { "input": "877 13 3 191", "output": "YES" }, { "input": "340 9 9 10", "output": "YES" }, { "input": "31 8 2 205", "output": "NO" }, { "input": "519 3 2 148", "output": "YES" }, { "input": "882 2 21 219", "output": "NO" }, { "input": "982 13 5 198", "output": "YES" }, { "input": "428 13 6 272", "output": "YES" }, { "input": "436 16 14 26", "output": "YES" }, { "input": "628 10 9 386", "output": "YES" }, { "input": "77 33 18 31", "output": "YES" }, { "input": "527 36 4 8", "output": "YES" }, { "input": "128 18 2 169", "output": "YES" }, { "input": "904 4 2 288", "output": "YES" }, { "input": "986 4 3 25", "output": "YES" }, { "input": "134 8 22 162", "output": "NO" }, { "input": "942 42 3 69", "output": "YES" }, { "input": "894 4 9 4", "output": "YES" }, { "input": "953 8 10 312", "output": "YES" }, { "input": "43 8 1 121", "output": "YES" }, { "input": "12 13 19 273", "output": "NO" }, { "input": "204 45 10 871", "output": "YES" }, { "input": "342 69 50 425", "output": "NO" }, { "input": "982 93 99 875", "output": "NO" }, { "input": "283 21 39 132", "output": "YES" }, { "input": "1000 45 83 686", "output": "NO" }, { "input": "246 69 36 432", "output": "NO" }, { "input": "607 93 76 689", "output": "NO" }, { "input": "503 21 24 435", "output": "NO" }, { "input": "1000 45 65 989", "output": "NO" }, { "input": "30 21 2 250", "output": "YES" }, { "input": "1000 49 50 995", "output": "NO" }, { "input": "383 69 95 253", "output": "YES" }, { "input": "393 98 35 999", "output": "YES" }, { "input": "1000 22 79 552", "output": "NO" }, { "input": "268 294 268 154", "output": "NO" }, { "input": "963 465 706 146", "output": "YES" }, { "input": "304 635 304 257", "output": "NO" }, { "input": "4 2 1 6", "output": "NO" }, { "input": "1 51 10 50", "output": "NO" }, { "input": "5 5 4 4", "output": "YES" }, { "input": "3 2 1 1", "output": "YES" }, { "input": "3 4 3 3", "output": "NO" }, { "input": "7 3 4 1", "output": "YES" }, { "input": "101 10 1 1000", "output": "NO" }, { "input": "5 1 1 1", "output": "YES" }, { "input": "5 10 5 5", "output": "NO" }, { "input": "19 1 7 1", "output": "YES" }, { "input": "763 572 745 262", "output": "YES" }, { "input": "1 2 1 1", "output": "NO" }, { "input": "5 1 1 3", "output": "YES" }, { "input": "170 725 479 359", "output": "NO" }, { "input": "6 2 1 7", "output": "YES" }, { "input": "6 2 5 1", "output": "YES" }, { "input": "1 2 2 1", "output": "NO" }, { "input": "24 2 8 3", "output": "YES" }, { "input": "7 3 3 3", "output": "YES" }, { "input": "5 2 2 2", "output": "YES" }, { "input": "3 2 1 2", "output": "YES" }, { "input": "1000 2 200 8", "output": "NO" }, { "input": "3 100 2 100", "output": "NO" }, { "input": "2 999 1 1000", "output": "NO" }, { "input": "2 1 1 1", "output": "NO" }, { "input": "2 3 5 1", "output": "NO" }, { "input": "100 1 5 1", "output": "YES" }, { "input": "7 2 3 3", "output": "YES" }, { "input": "4 1 1 3", "output": "NO" }, { "input": "3 2 2 1", "output": "YES" }, { "input": "1 1 1 2", "output": "NO" }, { "input": "91 8 7 13", "output": "YES" }, { "input": "3 1 2 1", "output": "NO" }, { "input": "5 3 2 3", "output": "YES" }, { "input": "9 6 6 3", "output": "YES" } ]

1,681,506,838

2,147,483,647

Python 3

OK

TESTS

93

61

0

def should_build_second_oven(n, t, k, d): single_oven_time = (n + k - 1) // k * t double_oven_time = d + (n - k * (d // t) + 2 * k - 1) // (2 * k) * t return double_oven_time < single_oven_time # Read input n, t, k, d = map(int, input().split()) # Determine if it's reasonable to build the second oven if should_build_second_oven(n, t, k, d): print("YES") else: print("NO")

Title: Carrot Cakes Time Limit: None seconds Memory Limit: None megabytes Problem Description: In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven. Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable. Input Specification: The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven. Output Specification: If it is reasonable to build the second oven, print "YES". Otherwise print "NO". Demo Input: ['8 6 4 5\n', '8 6 4 6\n', '10 3 11 4\n', '4 2 1 4\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n', 'YES\n'] Note: In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven. In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven. In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.

```python def should_build_second_oven(n, t, k, d): single_oven_time = (n + k - 1) // k * t double_oven_time = d + (n - k * (d // t) + 2 * k - 1) // (2 * k) * t return double_oven_time < single_oven_time # Read input n, t, k, d = map(int, input().split()) # Determine if it's reasonable to build the second oven if should_build_second_oven(n, t, k, d): print("YES") else: print("NO") ```

3

796

B

Find The Bone

PROGRAMMING

1,300

[ "implementation" ]

null

Zane the wizard is going to perform a magic show shuffling the cups. There are *n* cups, numbered from 1 to *n*, placed along the *x*-axis on a table that has *m* holes on it. More precisely, cup *i* is on the table at the position *x*<==<=*i*. The problematic bone is initially at the position *x*<==<=1. Zane will confuse the audience by swapping the cups *k* times, the *i*-th time of which involves the cups at the positions *x*<==<=*u**i* and *x*<==<=*v**i*. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations. Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at *x*<==<=4 and the one at *x*<==<=6, they will not be at the position *x*<==<=5 at any moment during the operation. Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.

The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=106, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively. The second line contains *m* distinct integers *h*1,<=*h*2,<=...,<=*h**m* (1<=≤<=*h**i*<=≤<=*n*) — the positions along the *x*-axis where there is a hole on the table. Each of the next *k* lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the positions of the cups to be swapped.

Print one integer — the final position along the *x*-axis of the bone.

[ "7 3 4\n3 4 6\n1 2\n2 5\n5 7\n7 1\n", "5 1 2\n2\n1 2\n2 4\n" ]

[ "1", "2" ]

In the first sample, after the operations, the bone becomes at *x* = 2, *x* = 5, *x* = 7, and *x* = 1, respectively. In the second sample, after the first operation, the bone becomes at *x* = 2, and falls into the hole onto the ground.

750

[ { "input": "7 3 4\n3 4 6\n1 2\n2 5\n5 7\n7 1", "output": "1" }, { "input": "5 1 2\n2\n1 2\n2 4", "output": "2" }, { "input": "10000 1 9\n55\n44 1\n2929 9292\n9999 9998\n44 55\n49 94\n55 53\n100 199\n55 50\n53 11", "output": "55" }, { "input": "100000 3 7\n2 3 4\n1 5\n5 1\n1 5\n5 1\n1 4\n4 3\n3 2", "output": "4" }, { "input": "1000000 9 11\n38 59 999999 199 283 4849 1000000 2 554\n39 94\n3 9\n1 39\n39 40\n40 292\n5399 5858\n292 49949\n49949 222\n222 38\n202 9494\n38 59", "output": "38" }, { "input": "1000000 11 9\n19 28 39 82 99 929384 8298 892849 202020 777777 123123\n19 28\n28 39\n1 123124\n39 28\n28 99\n99 8298\n123124 123122\n2300 3200\n8298 1000000", "output": "123122" }, { "input": "2 1 1\n1\n1 2", "output": "1" }, { "input": "7 3 6\n1 4 5\n1 2\n2 3\n3 5\n4 5\n4 5\n4 5", "output": "1" }, { "input": "10 3 8\n1 5 10\n1 2\n2 3\n3 4\n3 4\n3 4\n4 5\n5 6\n6 5", "output": "1" }, { "input": "5 2 9\n2 4\n1 3\n3 5\n3 5\n3 4\n4 2\n2 4\n1 4\n1 2\n1 4", "output": "4" }, { "input": "10 10 13\n1 2 3 4 5 6 7 8 9 10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n6 7\n6 10\n10 9\n9 1\n1 10\n1 10\n1 10", "output": "1" }, { "input": "3 3 3\n1 2 3\n1 2\n2 3\n3 2", "output": "1" }, { "input": "100 7 7\n17 27 37 47 57 67 77\n49 39\n55 1\n50 3\n89 1\n1 99\n100 55\n98 55", "output": "100" }, { "input": "9 1 9\n9\n1 2\n3 2\n4 3\n8 9\n4 5\n7 4\n8 5\n1 3\n3 2", "output": "8" }, { "input": "300000 1 1\n200000\n300000 1", "output": "300000" }, { "input": "203948 2 14\n203948 203946\n39 38\n4959 3030\n1 203947\n2929 9292\n203944 203948\n203947 203944\n203944 203922\n203922 203948\n2495 20495\n29419 5959\n12949 12\n49 29292\n1 94\n1 203", "output": "203948" }, { "input": "203948 2 14\n203948 203947\n39 38\n4959 3030\n1 203947\n2929 9292\n203944 203948\n203947 203944\n203944 203922\n203922 203948\n2495 20495\n29419 5959\n12949 12\n49 29292\n1 94\n1 203", "output": "203947" }, { "input": "100 2 5\n1 2\n2 39\n39 29\n99 100\n1 2\n2 39", "output": "1" }, { "input": "3 1 1\n1\n1 2", "output": "1" }, { "input": "5 2 2\n1 2\n1 2\n2 3", "output": "1" }, { "input": "2 2 1\n1 2\n2 1", "output": "1" }, { "input": "5 2 1\n1 2\n2 1", "output": "1" }, { "input": "5 1 1\n5\n3 4", "output": "1" }, { "input": "3 2 1\n1 2\n2 1", "output": "1" }, { "input": "5 1 2\n2\n2 1\n2 3", "output": "2" }, { "input": "3 1 2\n2\n2 1\n2 3", "output": "2" }, { "input": "3 2 2\n2 3\n2 1\n2 3", "output": "2" }, { "input": "4 2 1\n1 2\n2 1", "output": "1" }, { "input": "4 1 1\n2\n2 3", "output": "1" }, { "input": "3 2 1\n1 3\n3 1", "output": "1" }, { "input": "10 1 3\n2\n2 1\n2 4\n9 10", "output": "2" }, { "input": "5 2 4\n3 5\n1 2\n4 2\n3 4\n3 5", "output": "3" }, { "input": "4 3 1\n1 2 3\n2 1", "output": "1" } ]

1,599,517,052

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

93

307,200

n, m, k = map(int, input().split()) m1 = list(map(int, input().split())) tmp = 1 tmp1 = 0 cond = 0 for i in range(k): l1, l2 = map(int, input().split()) if l1 == 1 or l1 == tmp: tmp1 = l1 cond = 1 tmp = l2 if l1 == tmp1 and l2 in m1 and cond==1: exit(print(l2)) print(tmp)

Title: Find The Bone Time Limit: None seconds Memory Limit: None megabytes Problem Description: Zane the wizard is going to perform a magic show shuffling the cups. There are *n* cups, numbered from 1 to *n*, placed along the *x*-axis on a table that has *m* holes on it. More precisely, cup *i* is on the table at the position *x*<==<=*i*. The problematic bone is initially at the position *x*<==<=1. Zane will confuse the audience by swapping the cups *k* times, the *i*-th time of which involves the cups at the positions *x*<==<=*u**i* and *x*<==<=*v**i*. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations. Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at *x*<==<=4 and the one at *x*<==<=6, they will not be at the position *x*<==<=5 at any moment during the operation. Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone. Input Specification: The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=106, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively. The second line contains *m* distinct integers *h*1,<=*h*2,<=...,<=*h**m* (1<=≤<=*h**i*<=≤<=*n*) — the positions along the *x*-axis where there is a hole on the table. Each of the next *k* lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the positions of the cups to be swapped. Output Specification: Print one integer — the final position along the *x*-axis of the bone. Demo Input: ['7 3 4\n3 4 6\n1 2\n2 5\n5 7\n7 1\n', '5 1 2\n2\n1 2\n2 4\n'] Demo Output: ['1', '2'] Note: In the first sample, after the operations, the bone becomes at *x* = 2, *x* = 5, *x* = 7, and *x* = 1, respectively. In the second sample, after the first operation, the bone becomes at *x* = 2, and falls into the hole onto the ground.

```python n, m, k = map(int, input().split()) m1 = list(map(int, input().split())) tmp = 1 tmp1 = 0 cond = 0 for i in range(k): l1, l2 = map(int, input().split()) if l1 == 1 or l1 == tmp: tmp1 = l1 cond = 1 tmp = l2 if l1 == tmp1 and l2 in m1 and cond==1: exit(print(l2)) print(tmp) ```

0

747

A

Display Size

PROGRAMMING

800

[ "brute force", "math" ]

null

A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels. Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that: - there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.

The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.

Print two integers — the number of rows and columns on the display.

[ "8\n", "64\n", "5\n", "999999\n" ]

[ "2 4\n", "8 8\n", "1 5\n", "999 1001\n" ]

In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels. In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels. In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.

500

[ { "input": "8", "output": "2 4" }, { "input": "64", "output": "8 8" }, { "input": "5", "output": "1 5" }, { "input": "999999", "output": "999 1001" }, { "input": "716539", "output": "97 7387" }, { "input": "1", "output": "1 1" }, { "input": "2", "output": "1 2" }, { "input": "3", "output": "1 3" }, { "input": "4", "output": "2 2" }, { "input": "6", "output": "2 3" }, { "input": "7", "output": "1 7" }, { "input": "9", "output": "3 3" }, { "input": "10", "output": "2 5" }, { "input": "11", "output": "1 11" }, { "input": "12", "output": "3 4" }, { "input": "15", "output": "3 5" }, { "input": "100", "output": "10 10" }, { "input": "101", "output": "1 101" }, { "input": "169", "output": "13 13" }, { "input": "179", "output": "1 179" }, { "input": "190", "output": "10 19" }, { "input": "1000", "output": "25 40" }, { "input": "10000", "output": "100 100" }, { "input": "10001", "output": "73 137" }, { "input": "100000", "output": "250 400" }, { "input": "100001", "output": "11 9091" }, { "input": "1000000", "output": "1000 1000" }, { "input": "999983", "output": "1 999983" }, { "input": "524288", "output": "512 1024" }, { "input": "954493", "output": "971 983" }, { "input": "966289", "output": "983 983" }, { "input": "944663", "output": "961 983" }, { "input": "912673", "output": "97 9409" }, { "input": "732641", "output": "679 1079" }, { "input": "232897", "output": "343 679" }, { "input": "16807", "output": "49 343" }, { "input": "999958", "output": "2 499979" }, { "input": "990151", "output": "1 990151" }, { "input": "997002", "output": "998 999" }, { "input": "20", "output": "4 5" }, { "input": "20261", "output": "1 20261" }, { "input": "999123", "output": "3 333041" }, { "input": "901841", "output": "1 901841" } ]

1,633,078,226

2,147,483,647

PyPy 3

OK

TESTS

43

109

20,172,800

a = int(input()) b = int(a ** 0.5) while a % b != 0: b -= 1 print(b, a // b)

Title: Display Size Time Limit: None seconds Memory Limit: None megabytes Problem Description: A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels. Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that: - there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible. Input Specification: The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have. Output Specification: Print two integers — the number of rows and columns on the display. Demo Input: ['8\n', '64\n', '5\n', '999999\n'] Demo Output: ['2 4\n', '8 8\n', '1 5\n', '999 1001\n'] Note: In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels. In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels. In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.

```python a = int(input()) b = int(a ** 0.5) while a % b != 0: b -= 1 print(b, a // b) ```

3

742

A

Arpa’s hard exam and Mehrdad’s naive cheat

PROGRAMMING

1,000

[ "implementation", "math", "number theory" ]

null

There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do. Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*. Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.

The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).

Print single integer — the last digit of 1378*n*.

[ "1\n", "2\n" ]

[ "8", "4" ]

In the first example, last digit of 13781 = 1378 is 8. In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.

500

[ { "input": "1", "output": "8" }, { "input": "2", "output": "4" }, { "input": "1000", "output": "6" }, { "input": "3", "output": "2" }, { "input": "4", "output": "6" }, { "input": "1000000000", "output": "6" }, { "input": "5", "output": "8" }, { "input": "6", "output": "4" }, { "input": "999999999", "output": "2" }, { "input": "1378", "output": "4" }, { "input": "13781378", "output": "4" }, { "input": "51202278", "output": "4" }, { "input": "999999998", "output": "4" }, { "input": "999999997", "output": "8" }, { "input": "12193721", "output": "8" }, { "input": "0", "output": "1" }, { "input": "989898989", "output": "8" }, { "input": "7", "output": "2" }, { "input": "8", "output": "6" }, { "input": "9", "output": "8" }, { "input": "10", "output": "4" }, { "input": "11", "output": "2" }, { "input": "12", "output": "6" }, { "input": "13", "output": "8" }, { "input": "14", "output": "4" }, { "input": "15", "output": "2" }, { "input": "16", "output": "6" }, { "input": "999999996", "output": "6" }, { "input": "999999995", "output": "2" }, { "input": "999999994", "output": "4" } ]

1,680,338,687

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

5

1,000

8,704,000

y=int(input()) d=1378**y print(d%10)

Title: Arpa’s hard exam and Mehrdad’s naive cheat Time Limit: None seconds Memory Limit: None megabytes Problem Description: There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do. Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*. Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat. Input Specification: The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109). Output Specification: Print single integer — the last digit of 1378*n*. Demo Input: ['1\n', '2\n'] Demo Output: ['8', '4'] Note: In the first example, last digit of 13781 = 1378 is 8. In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.

```python y=int(input()) d=1378**y print(d%10) ```

0

961

B

Lecture Sleep

PROGRAMMING

1,200

[ "data structures", "dp", "implementation", "two pointers" ]

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" } ]

1,595,235,215

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

2

249

29,900,800

n, k = map(int,input().strip().split(" ")) s = list(map(int, input().strip().split(" "))) l = list(map(int, input().strip().split(" "))) sum = 0 start = 0 result = [] time = k i = 0 while i < len(s): if l[i] == 1 and start == 0: i += 1 else: if l[i] == 0 and start == 0: start = 1 sum += s[i] time -= 1 i += 1 else: if l[i] == 1 and start == 1: sum += s[i] i += 1 else: if l[i] == 0 and start == 1: if time > 0: sum += s[i] time -= 1 if time == 0: pass else: i += 1 if time == 0: start = 0 time = k result.append(sum) sum = 0 print(max(result))

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 n, k = map(int,input().strip().split(" ")) s = list(map(int, input().strip().split(" "))) l = list(map(int, input().strip().split(" "))) sum = 0 start = 0 result = [] time = k i = 0 while i < len(s): if l[i] == 1 and start == 0: i += 1 else: if l[i] == 0 and start == 0: start = 1 sum += s[i] time -= 1 i += 1 else: if l[i] == 1 and start == 1: sum += s[i] i += 1 else: if l[i] == 0 and start == 1: if time > 0: sum += s[i] time -= 1 if time == 0: pass else: i += 1 if time == 0: start = 0 time = k result.append(sum) sum = 0 print(max(result)) ```

-1

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,633,956,133

2,147,483,647

Python 3

OK

TESTS

81

124

6,963,200

n=int(input()) x=0 y=0 z=0 for i in range(n): m=[int(x) for x in input().split()] x+=m[0] y+=m[1] z+=m[2] if x==y==z==0: print('YES') else: print('NO')

Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none

```python n=int(input()) x=0 y=0 z=0 for i in range(n): m=[int(x) for x in input().split()] x+=m[0] y+=m[1] z+=m[2] if x==y==z==0: print('YES') else: print('NO') ```

3.95603

581

A

Vasya the Hipster

PROGRAMMING

800

[ "implementation", "math" ]

null

One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him?

The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.

Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.

[ "3 1\n", "2 3\n", "7 3\n" ]

[ "1 1\n", "2 0\n", "3 2\n" ]

In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.

500

[ { "input": "3 1", "output": "1 1" }, { "input": "2 3", "output": "2 0" }, { "input": "7 3", "output": "3 2" }, { "input": "100 100", "output": "100 0" }, { "input": "4 10", "output": "4 3" }, { "input": "6 10", "output": "6 2" }, { "input": "6 11", "output": "6 2" }, { "input": "10 40", "output": "10 15" }, { "input": "11 56", "output": "11 22" }, { "input": "34 30", "output": "30 2" }, { "input": "33 33", "output": "33 0" }, { "input": "100 45", "output": "45 27" }, { "input": "100 23", "output": "23 38" }, { "input": "45 12", "output": "12 16" }, { "input": "1 1", "output": "1 0" }, { "input": "1 100", "output": "1 49" }, { "input": "100 1", "output": "1 49" }, { "input": "68 59", "output": "59 4" }, { "input": "45 99", "output": "45 27" }, { "input": "99 100", "output": "99 0" }, { "input": "100 98", "output": "98 1" }, { "input": "59 12", "output": "12 23" }, { "input": "86 4", "output": "4 41" }, { "input": "68 21", "output": "21 23" }, { "input": "100 11", "output": "11 44" }, { "input": "100 10", "output": "10 45" }, { "input": "15 45", "output": "15 15" }, { "input": "11 32", "output": "11 10" }, { "input": "34 96", "output": "34 31" }, { "input": "89 89", "output": "89 0" } ]

1,667,914,670

2,147,483,647

PyPy 3-64

OK

TESTS

30

62

0

a,b = map(int,input().split()) ; x,y = min(a,b) , max(a,b) ; print(x,(y-x)//2)

Title: Vasya the Hipster Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him? Input Specification: The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. Output Specification: Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. Demo Input: ['3 1\n', '2 3\n', '7 3\n'] Demo Output: ['1 1\n', '2 0\n', '3 2\n'] Note: In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.

```python a,b = map(int,input().split()) ; x,y = min(a,b) , max(a,b) ; print(x,(y-x)//2) ```

3

119

A

Epic Game

PROGRAMMING

800

[ "implementation" ]

null

Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.

The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.

If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).

[ "3 5 9\n", "1 1 100\n" ]

[ "0", "1" ]

The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

500

[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]

1,649,657,340

2,147,483,647

PyPy 3-64

OK

TESTS

45

122

0

from math import gcd a,b,c=map(int,input().split()) x=True k=0 while x: if k%2==0: if c-gcd(a,c)>=0: c-=gcd(a,c) k+=1 else: print(1) x=False else: if c-gcd(b,c)>=0: c-=gcd(b,c) k+=1 else: print(0) x=False

Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

```python from math import gcd a,b,c=map(int,input().split()) x=True k=0 while x: if k%2==0: if c-gcd(a,c)>=0: c-=gcd(a,c) k+=1 else: print(1) x=False else: if c-gcd(b,c)>=0: c-=gcd(b,c) k+=1 else: print(0) x=False ```

3

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,518,364,679

2,147,483,647

Python 3

OK

TESTS

34

124

5,632,000

n = int(input()) record = [] for i in range(0,n): record.append(input()) team1 = record[0] exists = 0 for i in range(1,n): if(record[i]!=team1): team2 = record[i] exists = 1 break if(exists == 0): print(team1) else: score1 = 0 score2 = 0 for i in range(0,n): if(record[i] == team1): score1+=1 else: score2+=1 if(score1>score2): print(team1) else: print(team2)

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python n = int(input()) record = [] for i in range(0,n): record.append(input()) team1 = record[0] exists = 0 for i in range(1,n): if(record[i]!=team1): team2 = record[i] exists = 1 break if(exists == 0): print(team1) else: score1 = 0 score2 = 0 for i in range(0,n): if(record[i] == team1): score1+=1 else: score2+=1 if(score1>score2): print(team1) else: print(team2) ```

3.95851

628

F

Bear and Fair Set

PROGRAMMING

2,500

[ "flows", "graphs" ]

null

Limak is a grizzly bear. He is big and dreadful. You were chilling in the forest when you suddenly met him. It's very unfortunate for you. He will eat all your cookies unless you can demonstrate your mathematical skills. To test you, Limak is going to give you a puzzle to solve. It's a well-known fact that Limak, as every bear, owns a set of numbers. You know some information about the set: - The elements of the set are distinct positive integers. - The number of elements in the set is *n*. The number *n* is divisible by 5. - All elements are between 1 and *b*, inclusive: bears don't know numbers greater than *b*. - For each *r* in {0,<=1,<=2,<=3,<=4}, the set contains exactly elements that give remainder *r* when divided by 5. (That is, there are elements divisible by 5, elements of the form 5*k*<=+<=1, elements of the form 5*k*<=+<=2, and so on.) Limak smiles mysteriously and gives you *q* hints about his set. The *i*-th hint is the following sentence: "If you only look at elements that are between 1 and *upTo**i*, inclusive, you will find exactly *quantity**i* such elements in my set." In a moment Limak will tell you the actual puzzle, but something doesn't seem right... That smile was very strange. You start to think about a possible reason. Maybe Limak cheated you? Or is he a fair grizzly bear? Given *n*, *b*, *q* and hints, check whether Limak can be fair, i.e. there exists at least one set satisfying the given conditions. If it's possible then print ''fair". Otherwise, print ''unfair".

The first line contains three integers *n*, *b* and *q* (5<=≤<=*n*<=≤<=*b*<=≤<=104, 1<=≤<=*q*<=≤<=104, *n* divisible by 5) — the size of the set, the upper limit for numbers in the set and the number of hints. The next *q* lines describe the hints. The *i*-th of them contains two integers *upTo**i* and *quantity**i* (1<=≤<=*upTo**i*<=≤<=*b*, 0<=≤<=*quantity**i*<=≤<=*n*).

Print ''fair" if there exists at least one set that has all the required properties and matches all the given hints. Otherwise, print ''unfair".

[ "10 20 1\n10 10\n", "10 20 3\n15 10\n5 0\n10 5\n", "10 20 2\n15 3\n20 10\n" ]

[ "fair\n", "fair\n", "unfair\n" ]

In the first example there is only one set satisfying all conditions: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. In the second example also there is only one set satisfying all conditions: {6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. Easy to see that there is no set satisfying all conditions from the third example. So Limak lied to you :-(

0

[ { "input": "10 20 1\n10 10", "output": "fair" }, { "input": "10 20 3\n15 10\n5 0\n10 5", "output": "fair" }, { "input": "10 20 2\n15 3\n20 10", "output": "unfair" }, { "input": "15 27 2\n6 4\n23 5", "output": "unfair" }, { "input": "50 7654 4\n1273 11\n6327 38\n1244 3\n5208 22", "output": "fair" }, { "input": "50 7654 4\n2899 15\n3848 26\n2718 12\n5511 36", "output": "fair" }, { "input": "50 7654 4\n4881 20\n4957 6\n4764 50\n944 44", "output": "unfair" }, { "input": "50 6457 1\n945 41", "output": "fair" }, { "input": "500 5000 5\n1289 221\n694 178\n2179 454\n160 11\n1398 232", "output": "fair" }, { "input": "500 5000 10\n2905 421\n573 82\n1602 205\n4523 491\n970 100\n3810 453\n2553 418\n2033 364\n1664 245\n1924 311", "output": "fair" }, { "input": "500 5000 20\n875 49\n73 16\n2405 136\n811 33\n2477 140\n3475 303\n4640 496\n4025 369\n4482 440\n3475 272\n3594 346\n3945 368\n3807 346\n2605 159\n4045 382\n2861 270\n4488 448\n1894 61\n2388 113\n4071 383", "output": "unfair" }, { "input": "500 5000 50\n687 73\n3816 389\n4333 436\n1660 177\n2238 231\n2936 312\n899 96\n541 55\n4218 425\n4512 457\n1302 132\n2322 239\n688 73\n4423 449\n2765 284\n3755 382\n4192 422\n2718 277\n2254 231\n1354 140\n4891 490\n2722 277\n344 35\n4774 479\n988 101\n2530 256\n3679 375\n3258 341\n1870 201\n1391 146\n643 68\n1040 105\n2607 266\n906 97\n4790 480\n2390 245\n3101 326\n2616 267\n1064 107\n1091 110\n1735 187\n2434 247\n3887 397\n1335 137\n2073 219\n450 45\n480 47\n3519 359\n157 16\n4316 434", "output": "fair" }, { "input": "15 40 3\n2 0\n13 9\n4 1", "output": "fair" }, { "input": "15 41 3\n16 8\n14 2\n40 9", "output": "unfair" }, { "input": "15 40 3\n8 0\n38 14\n28 9", "output": "fair" }, { "input": "15 40 3\n1 9\n24 0\n35 7", "output": "unfair" }, { "input": "15 40 2\n23 4\n36 7", "output": "unfair" }, { "input": "15 41 2\n19 12\n2 0", "output": "fair" }, { "input": "15 40 2\n35 13\n36 14", "output": "fair" }, { "input": "15 40 2\n15 4\n24 4", "output": "fair" }, { "input": "5 6 2\n4 4\n5 4", "output": "unfair" }, { "input": "10 20 4\n3 3\n5 3\n9 7\n17 7", "output": "unfair" }, { "input": "5 10 3\n2 1\n5 4\n7 4", "output": "unfair" }, { "input": "5 30 10\n1 1\n5 1\n6 2\n10 2\n11 3\n15 3\n16 4\n20 4\n21 5\n30 5", "output": "unfair" }, { "input": "10 13 2\n3 3\n5 4", "output": "unfair" }, { "input": "5 10 3\n1 1\n5 1\n9 5", "output": "unfair" }, { "input": "10 14 2\n4 4\n5 4", "output": "unfair" } ]

1,689,173,111

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

46

0

print("_RANDOM_GUESS_1689173111.00439")# 1689173111.00441

Title: Bear and Fair Set Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is a grizzly bear. He is big and dreadful. You were chilling in the forest when you suddenly met him. It's very unfortunate for you. He will eat all your cookies unless you can demonstrate your mathematical skills. To test you, Limak is going to give you a puzzle to solve. It's a well-known fact that Limak, as every bear, owns a set of numbers. You know some information about the set: - The elements of the set are distinct positive integers. - The number of elements in the set is *n*. The number *n* is divisible by 5. - All elements are between 1 and *b*, inclusive: bears don't know numbers greater than *b*. - For each *r* in {0,<=1,<=2,<=3,<=4}, the set contains exactly elements that give remainder *r* when divided by 5. (That is, there are elements divisible by 5, elements of the form 5*k*<=+<=1, elements of the form 5*k*<=+<=2, and so on.) Limak smiles mysteriously and gives you *q* hints about his set. The *i*-th hint is the following sentence: "If you only look at elements that are between 1 and *upTo**i*, inclusive, you will find exactly *quantity**i* such elements in my set." In a moment Limak will tell you the actual puzzle, but something doesn't seem right... That smile was very strange. You start to think about a possible reason. Maybe Limak cheated you? Or is he a fair grizzly bear? Given *n*, *b*, *q* and hints, check whether Limak can be fair, i.e. there exists at least one set satisfying the given conditions. If it's possible then print ''fair". Otherwise, print ''unfair". Input Specification: The first line contains three integers *n*, *b* and *q* (5<=≤<=*n*<=≤<=*b*<=≤<=104, 1<=≤<=*q*<=≤<=104, *n* divisible by 5) — the size of the set, the upper limit for numbers in the set and the number of hints. The next *q* lines describe the hints. The *i*-th of them contains two integers *upTo**i* and *quantity**i* (1<=≤<=*upTo**i*<=≤<=*b*, 0<=≤<=*quantity**i*<=≤<=*n*). Output Specification: Print ''fair" if there exists at least one set that has all the required properties and matches all the given hints. Otherwise, print ''unfair". Demo Input: ['10 20 1\n10 10\n', '10 20 3\n15 10\n5 0\n10 5\n', '10 20 2\n15 3\n20 10\n'] Demo Output: ['fair\n', 'fair\n', 'unfair\n'] Note: In the first example there is only one set satisfying all conditions: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. In the second example also there is only one set satisfying all conditions: {6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. Easy to see that there is no set satisfying all conditions from the third example. So Limak lied to you :-(

```python print("_RANDOM_GUESS_1689173111.00439")# 1689173111.00441 ```

0

70

A

Cookies

PROGRAMMING

1,300

[ "math" ]

A. Cookies

1

256

Fangy collects cookies. Once he decided to take a box and put cookies into it in some way. If we take a square *k*<=×<=*k* in size, divided into blocks 1<=×<=1 in size and paint there the main diagonal together with cells, which lie above it, then the painted area will be equal to the area occupied by one cookie *k* in size. Fangy also has a box with a square base 2*n*<=×<=2*n*, divided into blocks 1<=×<=1 in size. In a box the cookies should not overlap, and they should not be turned over or rotated. See cookies of sizes 2 and 4 respectively on the figure: To stack the cookies the little walrus uses the following algorithm. He takes out of the repository the largest cookie which can fit in some place in the box and puts it there. Everything could be perfect but alas, in the repository the little walrus has infinitely many cookies of size 2 and larger, and there are no cookies of size 1, therefore, empty cells will remain in the box. Fangy wants to know how many empty cells will be left in the end.

The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1000).

Print the single number, equal to the number of empty cells in the box. The answer should be printed modulo 106<=+<=3.

[ "3\n" ]

[ "9" ]

If the box possesses the base of 23 × 23 (as in the example), then the cookies will be put there in the following manner:

500

[ { "input": "3", "output": "9" }, { "input": "1", "output": "1" }, { "input": "2", "output": "3" }, { "input": "4", "output": "27" }, { "input": "6", "output": "243" }, { "input": "11", "output": "59049" }, { "input": "14", "output": "594320" }, { "input": "15", "output": "782957" }, { "input": "7", "output": "729" }, { "input": "0", "output": "1" }, { "input": "1000", "output": "691074" }, { "input": "657", "output": "874011" }, { "input": "561", "output": "842553" }, { "input": "823", "output": "858672" }, { "input": "850", "output": "557186" }, { "input": "298", "output": "999535" }, { "input": "262", "output": "946384" }, { "input": "910", "output": "678945" }, { "input": "617", "output": "247876" }, { "input": "857", "output": "562128" }, { "input": "69", "output": "327984" }, { "input": "589", "output": "889192" }, { "input": "928", "output": "794863" }, { "input": "696", "output": "695035" }, { "input": "226", "output": "376094" } ]

1,643,501,344

2,147,483,647

Python 3

OK

TESTS

25

46

0

mod=1000003 c=int(input()) ans=1 i = 2 while i<=c: ans=(ans%mod + (ans%mod * 2)%mod)%mod i+=1 print(ans)

Title: Cookies Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Fangy collects cookies. Once he decided to take a box and put cookies into it in some way. If we take a square *k*<=×<=*k* in size, divided into blocks 1<=×<=1 in size and paint there the main diagonal together with cells, which lie above it, then the painted area will be equal to the area occupied by one cookie *k* in size. Fangy also has a box with a square base 2*n*<=×<=2*n*, divided into blocks 1<=×<=1 in size. In a box the cookies should not overlap, and they should not be turned over or rotated. See cookies of sizes 2 and 4 respectively on the figure: To stack the cookies the little walrus uses the following algorithm. He takes out of the repository the largest cookie which can fit in some place in the box and puts it there. Everything could be perfect but alas, in the repository the little walrus has infinitely many cookies of size 2 and larger, and there are no cookies of size 1, therefore, empty cells will remain in the box. Fangy wants to know how many empty cells will be left in the end. Input Specification: The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1000). Output Specification: Print the single number, equal to the number of empty cells in the box. The answer should be printed modulo 106<=+<=3. Demo Input: ['3\n'] Demo Output: ['9'] Note: If the box possesses the base of 23 × 23 (as in the example), then the cookies will be put there in the following manner:

```python mod=1000003 c=int(input()) ans=1 i = 2 while i<=c: ans=(ans%mod + (ans%mod * 2)%mod)%mod i+=1 print(ans) ```

3.977

231

A

Team

PROGRAMMING

800

[ "brute force", "greedy" ]

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" } ]

1,696,482,648

2,147,483,647

Python 3

OK

TESTS

21

92

0

import sys if __name__ == '__main__': n = int(sys.stdin.readline()) count = 0 for _ in range(n): conf = [int(i) for i in sys.stdin.readline().strip().split()] if sum(conf) >= 2: count += 1 print(count)

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 import sys if __name__ == '__main__': n = int(sys.stdin.readline()) count = 0 for _ in range(n): conf = [int(i) for i in sys.stdin.readline().strip().split()] if sum(conf) >= 2: count += 1 print(count) ```

3

766

B

Mahmoud and a Triangle

PROGRAMMING

1,000

[ "constructive algorithms", "geometry", "greedy", "math", "number theory", "sortings" ]

null

Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle. Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.

In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.

[ "5\n1 5 3 2 4\n", "3\n4 1 2\n" ]

[ "YES\n", "NO\n" ]

For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

1,000

[ { "input": "5\n1 5 3 2 4", "output": "YES" }, { "input": "3\n4 1 2", "output": "NO" }, { "input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576", "output": "NO" }, { "input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761", "output": "YES" }, { "input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974", "output": "YES" }, { "input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291", "output": "NO" }, { "input": "3\n1 1000000000 1000000000", "output": "YES" }, { "input": "4\n1 1000000000 1000000000 1000000000", "output": "YES" }, { "input": "3\n1 1000000000 1", "output": "NO" }, { "input": "5\n1 2 3 5 2", "output": "YES" }, { "input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431", "output": "NO" }, { "input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159", "output": "YES" }, { "input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023", "output": "YES" }, { "input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015", "output": "YES" }, { "input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14", "output": "YES" }, { "input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368", "output": "NO" }, { "input": "3\n1 1000000000 999999999", "output": "NO" }, { "input": "5\n1 1 1 1 1", "output": "YES" }, { "input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000", "output": "NO" }, { "input": "5\n2 3 4 10 20", "output": "YES" }, { "input": "6\n18 23 40 80 160 161", "output": "YES" }, { "input": "4\n5 6 7 888", "output": "YES" }, { "input": "9\n1 1 2 2 4 5 10 10 20", "output": "YES" }, { "input": "7\n3 150 900 4 500 1500 5", "output": "YES" }, { "input": "3\n2 2 3", "output": "YES" }, { "input": "7\n1 2 100 200 250 1000000 2000000", "output": "YES" }, { "input": "8\n2 3 5 5 5 6 6 13", "output": "YES" }, { "input": "3\n2 3 4", "output": "YES" }, { "input": "6\n1 1 1 4 5 100", "output": "YES" }, { "input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377", "output": "YES" }, { "input": "4\n2 3 4 8", "output": "YES" }, { "input": "3\n5 6 7", "output": "YES" }, { "input": "5\n1 4 5 6 1000000", "output": "YES" }, { "input": "4\n5 6 7 20", "output": "YES" }, { "input": "6\n1 1 1 1 1 65", "output": "YES" }, { "input": "4\n3 4 5 100", "output": "YES" }, { "input": "3\n2 4 5", "output": "YES" }, { "input": "7\n1 1 1 1 1 10 1000", "output": "YES" }, { "input": "4\n1 1 2 3", "output": "NO" }, { "input": "11\n1 2 5 6 7 8 9 17 18 19 100", "output": "YES" }, { "input": "4\n5 16 20 200", "output": "YES" }, { "input": "5\n17 6 3 3 1", "output": "YES" }, { "input": "3\n1 1 1", "output": "YES" }, { "input": "6\n1 1 1 2 3 5", "output": "YES" }, { "input": "4\n2 4 6 6", "output": "YES" }, { "input": "9\n1 2 4 4 4 4 7 8 20", "output": "YES" }, { "input": "9\n1 1 2 5 5 5 10 10 20", "output": "YES" }, { "input": "7\n3 150 600 4 1700 6000 5", "output": "YES" }, { "input": "5\n5761 20966 27841 28800 29399", "output": "YES" }, { "input": "9\n1 2 3 6 7 10 11 12 24", "output": "YES" }, { "input": "4\n1 2 1 1", "output": "YES" }, { "input": "5\n1 1 2 3 4", "output": "YES" } ]

1,675,765,157

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

30

0

# LUOGU_RID: 101648785 n, *a = map(int, open(0).read().split()) a.sort() print(a[-1] + a[-2] > a[0] and 'YES' or 'NO')

Title: Mahmoud and a Triangle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle. Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. Input Specification: The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. Output Specification: In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. Demo Input: ['5\n1 5 3 2 4\n', '3\n4 1 2\n'] Demo Output: ['YES\n', 'NO\n'] Note: For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

```python # LUOGU_RID: 101648785 n, *a = map(int, open(0).read().split()) a.sort() print(a[-1] + a[-2] > a[0] and 'YES' or 'NO') ```

0

266

A

Stones on the Table

PROGRAMMING

800

[ "implementation" ]

null

There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.

The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table. The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.

Print a single integer — the answer to the problem.

[ "3\nRRG\n", "5\nRRRRR\n", "4\nBRBG\n" ]

[ "1\n", "4\n", "0\n" ]

none

500

[ { "input": "3\nRRG", "output": "1" }, { "input": "5\nRRRRR", "output": "4" }, { "input": "4\nBRBG", "output": "0" }, { "input": "1\nB", "output": "0" }, { "input": "2\nBG", "output": "0" }, { "input": "3\nBGB", "output": "0" }, { "input": "4\nRBBR", "output": "1" }, { "input": "5\nRGGBG", "output": "1" }, { "input": "10\nGGBRBRGGRB", "output": "2" }, { "input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB", "output": "18" }, { "input": "15\nBRRBRGGBBRRRRGR", "output": "6" }, { "input": "20\nRRGBBRBRGRGBBGGRGRRR", "output": "6" }, { "input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB", "output": "6" }, { "input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB", "output": "9" }, { "input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB", "output": "14" }, { "input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG", "output": "20" }, { "input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG", "output": "11" }, { "input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG", "output": "17" }, { "input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB", "output": "16" }, { "input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG", "output": "19" }, { "input": "10\nGRRBRBRBGR", "output": "1" }, { "input": "10\nBRBGBGRRBR", "output": "1" }, { "input": "20\nGBGBGGRRRRGRBBGRGRGR", "output": "5" }, { "input": "20\nRRGGRBBGBBRBGRRBRRBG", "output": "6" }, { "input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB", "output": "8" }, { "input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB", "output": "11" }, { "input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB", "output": "9" }, { "input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG", "output": "13" }, { "input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG", "output": "13" }, { "input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB", "output": "12" }, { "input": "2\nBB", "output": "1" }, { "input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "49" }, { "input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "47" }, { "input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR", "output": "43" } ]

1,698,396,665

2,147,483,647

PyPy 3-64

OK

TESTS

34

122

0

num = int(input()) user = list(input()) counter = 0 length = len(user) for x in range(0, length - 1): if user[x] == user[x + 1]: counter = counter + 1 print(counter)

Title: Stones on the Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table. The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. Output Specification: Print a single integer — the answer to the problem. Demo Input: ['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n'] Demo Output: ['1\n', '4\n', '0\n'] Note: none

```python num = int(input()) user = list(input()) counter = 0 length = len(user) for x in range(0, length - 1): if user[x] == user[x + 1]: counter = counter + 1 print(counter) ```

3

41

A

Translation

PROGRAMMING

800

[ "implementation", "strings" ]

A. Translation

2

256

The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.

The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.

If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.

[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]

[ "YES\n", "NO\n", "NO\n" ]

none

500

[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]

1,696,593,835

2,147,483,647

Python 3

OK

TESTS

40

92

0

s = input() t = input() ls = list(s) lt = list(t) lt.reverse() if ls==lt : print('YES') else: print("NO")

Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none

```python s = input() t = input() ls = list(s) lt = list(t) lt.reverse() if ls==lt : print('YES') else: print("NO") ```

3.977

807

A

Is it rated?

PROGRAMMING

900

[ "implementation", "sortings" ]

null

Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.

If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".

[ "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n", "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n", "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n" ]

[ "rated\n", "unrated\n", "maybe\n" ]

In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

500

[ { "input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated" }, { "input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n1 1\n1 1", "output": "maybe" }, { "input": "2\n4126 4126\n4126 4126", "output": "maybe" }, { "input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423", "output": "rated" }, { "input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110", "output": "unrated" }, { "input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143", "output": "maybe" }, { "input": "2\n3936 3936\n2967 2967", "output": "maybe" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 1\n1 2", "output": "rated" }, { "input": "2\n2967 2967\n3936 3936", "output": "unrated" }, { "input": "3\n1200 1200\n1200 1200\n1300 1300", "output": "unrated" }, { "input": "3\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "3\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "2\n3 2\n3 2", "output": "rated" }, { "input": "3\n5 5\n4 4\n3 4", "output": "rated" }, { "input": "3\n200 200\n200 200\n300 300", "output": "unrated" }, { "input": "3\n1 1\n2 2\n3 3", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699", "output": "maybe" }, { "input": "2\n10 10\n8 8", "output": "maybe" }, { "input": "3\n1500 1500\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "3\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n100 100\n100 100\n70 70\n80 80", "output": "unrated" }, { "input": "2\n1 2\n2 1", "output": "rated" }, { "input": "3\n5 5\n4 3\n3 3", "output": "rated" }, { "input": "3\n1600 1650\n1500 1550\n1400 1450", "output": "rated" }, { "input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700", "output": "unrated" }, { "input": "2\n1600 1600\n1400 1400", "output": "maybe" }, { "input": "2\n3 1\n9 8", "output": "rated" }, { "input": "2\n2 1\n1 1", "output": "rated" }, { "input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670", "output": "unrated" }, { "input": "2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n10 11\n5 4", "output": "rated" }, { "input": "2\n15 14\n13 12", "output": "rated" }, { "input": "2\n2 1\n2 2", "output": "rated" }, { "input": "3\n2670 2670\n3670 3670\n4106 4106", "output": "unrated" }, { "input": "3\n4 5\n3 3\n2 2", "output": "rated" }, { "input": "2\n10 9\n10 10", "output": "rated" }, { "input": "3\n1011 1011\n1011 999\n2200 2100", "output": "rated" }, { "input": "2\n3 3\n5 5", "output": "unrated" }, { "input": "2\n1500 1500\n3000 2000", "output": "rated" }, { "input": "2\n5 6\n5 5", "output": "rated" }, { "input": "3\n2000 2000\n1500 1501\n500 500", "output": "rated" }, { "input": "2\n2 3\n2 2", "output": "rated" }, { "input": "2\n3 3\n2 2", "output": "maybe" }, { "input": "2\n1 2\n1 1", "output": "rated" }, { "input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n15 14\n14 13", "output": "rated" }, { "input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900", "output": "unrated" }, { "input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884", "output": "rated" }, { "input": "2\n100 99\n100 100", "output": "rated" }, { "input": "4\n2 2\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "3\n100 101\n100 100\n100 100", "output": "rated" }, { "input": "4\n1000 1001\n900 900\n950 950\n890 890", "output": "rated" }, { "input": "2\n2 3\n1 1", "output": "rated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n3 2\n2 2", "output": "rated" }, { "input": "2\n3 2\n3 3", "output": "rated" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "3\n3 2\n3 3\n3 3", "output": "rated" }, { "input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "3\n1000 1000\n500 500\n400 300", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000", "output": "unrated" }, { "input": "2\n1 1\n2 3", "output": "rated" }, { "input": "2\n6 2\n6 2", "output": "rated" }, { "input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246", "output": "unrated" }, { "input": "2\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699", "output": "maybe" }, { "input": "2\n20 30\n10 5", "output": "rated" }, { "input": "3\n1 1\n2 2\n1 1", "output": "unrated" }, { "input": "2\n1 2\n3 3", "output": "rated" }, { "input": "5\n5 5\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 2\n2 1", "output": "rated" }, { "input": "2\n100 100\n90 89", "output": "rated" }, { "input": "2\n1000 900\n2000 2000", "output": "rated" }, { "input": "2\n50 10\n10 50", "output": "rated" }, { "input": "2\n200 200\n100 100", "output": "maybe" }, { "input": "3\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "3\n1000 1000\n300 300\n100 100", "output": "maybe" }, { "input": "4\n2 2\n2 2\n3 3\n4 4", "output": "unrated" }, { "input": "2\n5 3\n6 3", "output": "rated" }, { "input": "2\n1200 1100\n1200 1000", "output": "rated" }, { "input": "2\n5 5\n4 4", "output": "maybe" }, { "input": "2\n5 5\n3 3", "output": "maybe" }, { "input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100", "output": "unrated" }, { "input": "5\n10 10\n9 9\n8 8\n7 7\n6 6", "output": "maybe" }, { "input": "3\n1000 1000\n300 300\n10 10", "output": "maybe" }, { "input": "5\n6 6\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "2\n3 3\n1 1", "output": "maybe" }, { "input": "4\n2 2\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n1000 1000\n700 700", "output": "maybe" }, { "input": "2\n4 3\n5 3", "output": "rated" }, { "input": "2\n1000 1000\n1100 1100", "output": "unrated" }, { "input": "4\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "3\n1 1\n2 3\n2 2", "output": "rated" }, { "input": "2\n1 2\n1 3", "output": "rated" }, { "input": "2\n3 3\n1 2", "output": "rated" }, { "input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "5\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "2\n10 10\n1 2", "output": "rated" }, { "input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900", "output": "unrated" }, { "input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699", "output": "unrated" }, { "input": "2\n100 100\n110 110", "output": "unrated" }, { "input": "3\n3 3\n3 3\n4 4", "output": "unrated" }, { "input": "3\n3 3\n3 2\n4 4", "output": "rated" }, { "input": "3\n5 2\n4 4\n3 3", "output": "rated" }, { "input": "4\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n1 1\n3 2", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699", "output": "unrated" }, { "input": "3\n3 3\n3 3\n3 4", "output": "rated" }, { "input": "3\n1 2\n2 2\n3 3", "output": "rated" }, { "input": "3\n1 2\n1 2\n1 2", "output": "rated" }, { "input": "2\n2 1\n2 1", "output": "rated" }, { "input": "2\n1 2\n3 4", "output": "rated" }, { "input": "2\n3 2\n2 3", "output": "rated" }, { "input": "3\n1500 1500\n1600 1600\n1600 1600", "output": "unrated" }, { "input": "3\n1 1\n3 3\n4 4", "output": "unrated" }, { "input": "3\n1 1\n2 2\n2 2", "output": "unrated" }, { "input": "2\n10 12\n8 8", "output": "rated" }, { "input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500", "output": "unrated" }, { "input": "2\n1 2\n2 2", "output": "rated" }, { "input": "3\n1500 1400\n1200 1200\n1100 1100", "output": "rated" }, { "input": "2\n10 12\n10 10", "output": "rated" }, { "input": "3\n1500 1500\n1400 1400\n1300 1300", "output": "maybe" }, { "input": "3\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "3\n2 6\n3 5\n4 4", "output": "rated" }, { "input": "2\n5 6\n4 6", "output": "rated" }, { "input": "4\n10 10\n10 10\n7 7\n8 8", "output": "unrated" }, { "input": "2\n4 4\n3 3", "output": "maybe" } ]

1,494,712,919

2,147,483,647

Python 3

OK

TESTS

150

62

0

import sys if __name__ == '__main__': n = int(input()) before = [] after = [] for i in range(0, n): x, y = [int(num) for num in input().split()] before.append(x) after.append(y) for i in range(0, n): if before[i] != after[i]: print("rated") sys.exit() for i in range(0, n - 1): if before[i + 1] > before[i]: print("unrated") sys.exit() print("maybe")

Title: Is it rated? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. Output Specification: If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". Demo Input: ['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n'] Demo Output: ['rated\n', 'unrated\n', 'maybe\n'] Note: In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

```python import sys if __name__ == '__main__': n = int(input()) before = [] after = [] for i in range(0, n): x, y = [int(num) for num in input().split()] before.append(x) after.append(y) for i in range(0, n): if before[i] != after[i]: print("rated") sys.exit() for i in range(0, n - 1): if before[i + 1] > before[i]: print("unrated") sys.exit() print("maybe") ```

3

9

C

Hexadecimal's Numbers

PROGRAMMING

1,200

[ "brute force", "implementation", "math" ]

C. Hexadecimal's Numbers

1

64

One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy. But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).

Output the only number — answer to the problem.

[ "10\n" ]

[ "2" ]

For *n* = 10 the answer includes numbers 1 and 10.

0

[ { "input": "10", "output": "2" }, { "input": "20", "output": "3" }, { "input": "72", "output": "3" }, { "input": "99", "output": "3" }, { "input": "100", "output": "4" }, { "input": "101", "output": "5" }, { "input": "102", "output": "5" }, { "input": "111", "output": "7" }, { "input": "112", "output": "7" }, { "input": "745", "output": "7" }, { "input": "23536", "output": "31" }, { "input": "1", "output": "1" }, { "input": "1010011", "output": "83" }, { "input": "312410141", "output": "511" }, { "input": "1000000000", "output": "512" }, { "input": "999999999", "output": "511" }, { "input": "111111111", "output": "511" }, { "input": "101010101", "output": "341" }, { "input": "121212121", "output": "511" }, { "input": "106341103", "output": "383" }, { "input": "901556123", "output": "511" }, { "input": "832513432", "output": "511" }, { "input": "3", "output": "1" }, { "input": "732875234", "output": "511" }, { "input": "7", "output": "1" }, { "input": "9", "output": "1" }, { "input": "2", "output": "1" }, { "input": "11", "output": "3" }, { "input": "12", "output": "3" }, { "input": "13", "output": "3" }, { "input": "101020101", "output": "351" }, { "input": "111100100", "output": "484" }, { "input": "110110101", "output": "437" }, { "input": "100111001", "output": "313" }, { "input": "100100", "output": "36" }, { "input": "110100102", "output": "421" } ]

1,622,470,739

2,147,483,647

PyPy 3

COMPILATION_ERROR

TESTS

0

#include<bits/stdc++.h> #define int long long using namespace std; #define pr pair<int,int> #define vect vector <int> v1 #define pb push_back #define wgr vector<pair<int,int> > #define lowestbit(x) __builtin_ffs(x) //returns 1+index of(lowest bit) #define highestbit(x) __builtin_clz(x) //returns number of 0's to left of MST #define is(x) cerr << #x << " is " << x << endl; #define sortv sort(v1.begin(),v1.end()); #define loop(i,a,b) for(int i=(a);i<(b);i++) #define inputarr loop(i,0,n) cin>>ar[i]; int mod=1e9+7; int inf=1e18; int mx=-1; int mn=1e18; int calpow(int a, int b, int md) { int mul = 1; while (b > 0) { if (b & 1) { mul = ((a % md) * (mul % md)) % md; b--; } else { a = ((a % md) * (a % md)) % md; b /= 2; } } return mul; } void solve() { int i,j,n,t,q,a,b,m,k,ct=0,sum=0,flag=0; cin>>n; j=31-highestbit(n); cout<<(calpow(2,j,inf)-1); } signed main() { //freopen("lineup.in","r",stdin); //freopen("lineup.out","w",stdout); ios::sync_with_stdio(false); cin.tie(NULL); // WATCH FOR SAME VARIABLE USAGE CLEARLY!!! //MAKE SURE TO MAKE A DUPLICATE OF ANYTHING IF PERFORMING OPERATION DIRECTLY ON SOMETHING !!! int t=1; //cin>>t; while(t--) { solve(); cout<< "\n" ; } cerr<< endl << clock() * 1000.0 / CLOCKS_PER_SEC <<'m'<<'s'; return 0; }

Title: Hexadecimal's Numbers Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy. But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully. Input Specification: Input data contains the only number *n* (1<=≤<=*n*<=≤<=109). Output Specification: Output the only number — answer to the problem. Demo Input: ['10\n'] Demo Output: ['2'] Note: For *n* = 10 the answer includes numbers 1 and 10.

```python #include<bits/stdc++.h> #define int long long using namespace std; #define pr pair<int,int> #define vect vector <int> v1 #define pb push_back #define wgr vector<pair<int,int> > #define lowestbit(x) __builtin_ffs(x) //returns 1+index of(lowest bit) #define highestbit(x) __builtin_clz(x) //returns number of 0's to left of MST #define is(x) cerr << #x << " is " << x << endl; #define sortv sort(v1.begin(),v1.end()); #define loop(i,a,b) for(int i=(a);i<(b);i++) #define inputarr loop(i,0,n) cin>>ar[i]; int mod=1e9+7; int inf=1e18; int mx=-1; int mn=1e18; int calpow(int a, int b, int md) { int mul = 1; while (b > 0) { if (b & 1) { mul = ((a % md) * (mul % md)) % md; b--; } else { a = ((a % md) * (a % md)) % md; b /= 2; } } return mul; } void solve() { int i,j,n,t,q,a,b,m,k,ct=0,sum=0,flag=0; cin>>n; j=31-highestbit(n); cout<<(calpow(2,j,inf)-1); } signed main() { //freopen("lineup.in","r",stdin); //freopen("lineup.out","w",stdout); ios::sync_with_stdio(false); cin.tie(NULL); // WATCH FOR SAME VARIABLE USAGE CLEARLY!!! //MAKE SURE TO MAKE A DUPLICATE OF ANYTHING IF PERFORMING OPERATION DIRECTLY ON SOMETHING !!! int t=1; //cin>>t; while(t--) { solve(); cout<< "\n" ; } cerr<< endl << clock() * 1000.0 / CLOCKS_PER_SEC <<'m'<<'s'; return 0; } ```

-1

441

B

Valera and Fruits

PROGRAMMING

1,400

[ "greedy", "implementation" ]

null

Valera loves his garden, where *n* fruit trees grow. This year he will enjoy a great harvest! On the *i*-th tree *b**i* fruit grow, they will ripen on a day number *a**i*. Unfortunately, the fruit on the tree get withered, so they can only be collected on day *a**i* and day *a**i*<=+<=1 (all fruits that are not collected in these two days, become unfit to eat). Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than *v* fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?

The first line contains two space-separated integers *n* and *v* (1<=≤<=*n*,<=*v*<=≤<=3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day. Next *n* lines contain the description of trees in the garden. The *i*-th line contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=3000) — the day the fruits ripen on the *i*-th tree and the number of fruits on the *i*-th tree.

Print a single integer — the maximum number of fruit that Valera can collect.

[ "2 3\n1 5\n2 3\n", "5 10\n3 20\n2 20\n1 20\n4 20\n5 20\n" ]

[ "8\n", "60\n" ]

In the first sample, in order to obtain the optimal answer, you should act as follows. - On the first day collect 3 fruits from the 1-st tree. - On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree. - On the third day collect the remaining fruits from the 2-nd tree. In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.

1,000

[ { "input": "2 3\n1 5\n2 3", "output": "8" }, { "input": "5 10\n3 20\n2 20\n1 20\n4 20\n5 20", "output": "60" }, { "input": "10 3000\n1 2522\n4 445\n8 1629\n5 772\n9 2497\n6 81\n3 426\n7 1447\n2 575\n10 202", "output": "10596" }, { "input": "5 3000\n5 772\n1 2522\n2 575\n4 445\n3 426", "output": "4740" }, { "input": "2 1500\n2 575\n1 2522", "output": "3097" }, { "input": "12 2856\n9 2728\n8 417\n3 1857\n10 1932\n1 775\n12 982\n9 1447\n1 426\n7 2918\n11 2522\n10 2497\n9 772", "output": "18465" }, { "input": "24 1524\n16 934\n23 1940\n21 1447\n20 417\n24 1340\n22 1932\n13 775\n19 2918\n12 2355\n9 593\n11 2676\n3 1857\n16 868\n13 426\n18 1679\n22 991\n9 2728\n10 2497\n16 1221\n9 772\n23 2522\n24 982\n12 1431\n18 757", "output": "25893" }, { "input": "1 10\n3000 30", "output": "20" }, { "input": "2 1\n30 3\n31 2", "output": "3" }, { "input": "4 2061\n1 426\n3 2522\n1 772\n1 1447", "output": "5167" }, { "input": "2 1\n1 1\n1 1", "output": "2" }, { "input": "1 10\n3000 20", "output": "20" }, { "input": "1 1000\n3000 2000", "output": "2000" }, { "input": "2 100\n3000 100\n3000 100", "output": "200" }, { "input": "2 3\n1 6\n3 6", "output": "12" }, { "input": "1 40\n3000 42", "output": "42" }, { "input": "1 100\n3000 200", "output": "200" }, { "input": "1 50\n3000 100", "output": "100" }, { "input": "1 1\n3000 2", "output": "2" }, { "input": "2 3000\n3000 3000\n3000 3000", "output": "6000" }, { "input": "2 2\n2999 3\n3000 2", "output": "5" }, { "input": "1 2\n3000 3", "output": "3" }, { "input": "2 5\n2999 10\n3000 5", "output": "15" }, { "input": "1 3\n5 3", "output": "3" }, { "input": "2 1000\n2999 2000\n3000 1000", "output": "3000" }, { "input": "1 5\n3000 10", "output": "10" }, { "input": "1 10\n3000 15", "output": "15" }, { "input": "5 1\n10 100\n2698 100\n200 100\n3000 100\n1500 100", "output": "10" }, { "input": "1 1\n3000 3000", "output": "2" }, { "input": "2 10\n2999 100\n3000 100", "output": "30" }, { "input": "1 10\n3000 100", "output": "20" } ]

1,586,997,295

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

77

204,800

""" int(input()) map(int,input().split()) list(map(int,input().split())) """ n,v=map(int,input().split()) N=[0 for i in range(3002)] for i in range(n): x,y=map(int,input().split()) N[x]+=y cost=0 cur=0 for i in range(3002): if(N[i]+cur>v): cost+=v if(cur>v): cur=0 else: cur=max(0,(N[i]-cur)) else: cost+=(N[i]+cur) cur=0 print(cost)

Title: Valera and Fruits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera loves his garden, where *n* fruit trees grow. This year he will enjoy a great harvest! On the *i*-th tree *b**i* fruit grow, they will ripen on a day number *a**i*. Unfortunately, the fruit on the tree get withered, so they can only be collected on day *a**i* and day *a**i*<=+<=1 (all fruits that are not collected in these two days, become unfit to eat). Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than *v* fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well? Input Specification: The first line contains two space-separated integers *n* and *v* (1<=≤<=*n*,<=*v*<=≤<=3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day. Next *n* lines contain the description of trees in the garden. The *i*-th line contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=3000) — the day the fruits ripen on the *i*-th tree and the number of fruits on the *i*-th tree. Output Specification: Print a single integer — the maximum number of fruit that Valera can collect. Demo Input: ['2 3\n1 5\n2 3\n', '5 10\n3 20\n2 20\n1 20\n4 20\n5 20\n'] Demo Output: ['8\n', '60\n'] Note: In the first sample, in order to obtain the optimal answer, you should act as follows. - On the first day collect 3 fruits from the 1-st tree. - On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree. - On the third day collect the remaining fruits from the 2-nd tree. In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.

```python """ int(input()) map(int,input().split()) list(map(int,input().split())) """ n,v=map(int,input().split()) N=[0 for i in range(3002)] for i in range(n): x,y=map(int,input().split()) N[x]+=y cost=0 cur=0 for i in range(3002): if(N[i]+cur>v): cost+=v if(cur>v): cur=0 else: cur=max(0,(N[i]-cur)) else: cost+=(N[i]+cur) cur=0 print(cost) ```

0

339

B

Xenia and Ringroad

PROGRAMMING

1,000

[ "implementation" ]

null

Xenia lives in a city that has *n* houses built along the main ringroad. The ringroad houses are numbered 1 through *n* in the clockwise order. The ringroad traffic is one way and also is clockwise. Xenia has recently moved into the ringroad house number 1. As a result, she's got *m* things to do. In order to complete the *i*-th task, she needs to be in the house number *a**i* and complete all tasks with numbers less than *i*. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time.

The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105). The second line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=*n*). Note that Xenia can have multiple consecutive tasks in one house.

Print a single integer — the time Xenia needs to complete all tasks. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

[ "4 3\n3 2 3\n", "4 3\n2 3 3\n" ]

[ "6\n", "2\n" ]

In the first test example the sequence of Xenia's moves along the ringroad looks as follows: 1 → 2 → 3 → 4 → 1 → 2 → 3. This is optimal sequence. So, she needs 6 time units.

1,000

[ { "input": "4 3\n3 2 3", "output": "6" }, { "input": "4 3\n2 3 3", "output": "2" }, { "input": "2 2\n1 1", "output": "0" }, { "input": "2 2\n1 2", "output": "1" }, { "input": "2 2\n1 2", "output": "1" }, { "input": "100 100\n56 46 1 47 5 86 45 35 81 1 31 70 67 70 62 99 100 47 44 33 78 35 32 37 92 12 95 18 3 22 54 24 22 90 25 22 78 88 51 92 46 84 15 29 28 40 8 5 93 68 77 47 45 76 85 39 84 94 52 69 93 64 31 60 99 17 51 59 62 37 46 47 86 60 88 14 68 22 47 93 50 10 55 87 46 50 43 63 44 43 61 65 91 43 33 97 67 57 66 70", "output": "4869" }, { "input": "78 58\n23 14 73 45 47 14 27 59 65 39 15 23 5 1 50 37 3 51 46 69 75 65 45 68 48 59 77 39 53 21 72 33 46 32 34 5 69 55 56 53 47 31 32 5 42 23 76 15 2 77 65 24 16 68 61 28 55 10", "output": "2505" }, { "input": "14 54\n9 13 14 9 5 12 4 7 3 14 5 12 13 1 1 11 10 2 7 9 5 2 2 8 10 7 3 9 5 11 2 2 6 12 11 5 4 11 11 6 2 11 14 13 8 7 13 9 4 9 11 3 7 13", "output": "362" }, { "input": "100 100\n48 73 63 16 49 88 36 17 66 6 87 13 94 52 58 70 71 52 7 70 25 42 24 36 57 9 79 26 75 39 13 14 38 26 33 66 88 28 75 98 53 48 67 54 63 25 69 87 88 32 72 17 36 35 29 67 74 89 70 47 20 90 78 13 94 57 32 73 29 74 45 78 85 64 81 56 12 65 19 67 34 86 55 71 41 33 76 13 100 47 44 76 86 78 37 15 26 98 83 98", "output": "4997" }, { "input": "99 100\n88 65 10 91 18 35 58 49 42 2 22 57 74 31 53 24 27 93 45 4 71 2 69 39 21 90 97 89 45 73 20 45 82 98 35 90 37 76 68 26 21 65 95 63 24 74 50 59 3 93 65 6 30 37 62 71 18 88 40 12 56 40 89 56 38 71 90 41 97 43 44 23 19 22 10 80 3 24 32 85 26 65 70 60 76 85 66 68 74 11 64 88 12 63 16 15 79 57 93 58", "output": "4809" }, { "input": "65 100\n53 14 5 10 32 60 31 52 52 56 38 6 8 17 52 23 59 3 18 28 15 2 46 26 8 2 40 6 58 30 28 46 49 23 47 24 9 53 3 47 55 12 36 49 12 24 54 55 58 7 50 42 15 4 58 49 34 40 19 4 59 19 31 17 35 65 36 50 45 5 33 11 29 52 55 40 48 11 32 41 31 7 46 55 32 41 56 51 39 13 5 59 58 34 38 50 55 10 43 30", "output": "3149" }, { "input": "10 100\n7 6 2 10 7 2 3 8 10 4 6 1 4 5 7 10 1 2 3 5 4 10 8 2 3 3 6 8 3 9 4 1 9 10 1 2 5 1 8 8 5 9 2 8 1 2 3 2 1 10 10 7 1 3 2 2 7 1 6 6 6 9 2 3 1 7 2 2 9 7 3 3 2 10 7 4 7 3 3 3 2 4 4 2 2 8 4 1 10 10 5 10 6 10 6 10 3 10 8 9", "output": "428" }, { "input": "2 100\n1 1 2 2 2 2 1 2 1 2 2 2 1 1 2 2 2 2 1 1 2 1 2 2 1 1 2 2 2 1 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 1 1 2 1 1 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 2 2 2 1 1 2 2 1 1 2 2 1 1 2 1 1 1 1 2 2 1 1 1 2 1 1 1 1 1 1 1 2", "output": "47" }, { "input": "67 100\n49 5 25 48 37 55 5 33 14 30 59 28 57 46 45 32 47 22 40 28 58 34 27 29 4 52 63 44 31 65 42 61 11 17 32 17 18 1 12 33 38 11 59 46 43 55 23 30 23 2 42 21 45 51 30 19 35 15 10 30 13 21 32 34 33 3 5 59 23 63 6 9 20 43 64 3 42 41 40 4 14 20 40 33 25 44 1 2 50 46 13 10 3 20 22 64 28 42 58 30", "output": "3245" }, { "input": "100 100\n81 62 26 90 7 87 60 35 75 81 54 94 53 71 64 80 58 83 53 70 40 96 87 50 37 63 5 77 83 75 53 28 22 98 52 45 72 70 62 21 1 99 37 93 16 98 68 27 96 37 21 83 92 23 62 96 85 31 85 66 38 77 45 16 90 86 6 21 75 42 18 60 2 17 51 61 94 68 97 79 17 59 66 14 87 78 70 40 63 78 58 22 87 7 90 17 20 97 9 31", "output": "5530" }, { "input": "2 1\n1", "output": "0" }, { "input": "2 10\n1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "2 10\n1 1 1 2 2 2 1 1 1 2", "output": "3" }, { "input": "3 6\n1 2 3 3 2 1", "output": "6" } ]

1,697,517,507

2,147,483,647

Python 3

OK

TESTS

29

154

13,721,600

# https://codeforces.com/problemset/problem/339/B n, m = [int(x) for x in input().split()] a = [int(x) for x in input().split()] c = 0 prev = -1 for e in a: if e < prev: c += n prev = e print(c + a[-1] -1)

Title: Xenia and Ringroad Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia lives in a city that has *n* houses built along the main ringroad. The ringroad houses are numbered 1 through *n* in the clockwise order. The ringroad traffic is one way and also is clockwise. Xenia has recently moved into the ringroad house number 1. As a result, she's got *m* things to do. In order to complete the *i*-th task, she needs to be in the house number *a**i* and complete all tasks with numbers less than *i*. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time. Input Specification: The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105). The second line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=*n*). Note that Xenia can have multiple consecutive tasks in one house. Output Specification: Print a single integer — the time Xenia needs to complete all tasks. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['4 3\n3 2 3\n', '4 3\n2 3 3\n'] Demo Output: ['6\n', '2\n'] Note: In the first test example the sequence of Xenia's moves along the ringroad looks as follows: 1 → 2 → 3 → 4 → 1 → 2 → 3. This is optimal sequence. So, she needs 6 time units.

```python # https://codeforces.com/problemset/problem/339/B n, m = [int(x) for x in input().split()] a = [int(x) for x in input().split()] c = 0 prev = -1 for e in a: if e < prev: c += n prev = e print(c + a[-1] -1) ```

3

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" } ]

1,672,425,476

2,147,483,647

Python 3

OK

TESTS

20

46

0

b,c,a=map(int,input().split()) print(-b//a*(-c//a))

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 b,c,a=map(int,input().split()) print(-b//a*(-c//a)) ```

3.977

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" } ]

1,629,980,688

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

25

122

6,758,400

n, m = map(int, input().split()) if min(n, m) != 1: if n*m % 5 != 0: print(round(m * n / 2)) else: print(int(m * n / 2)) else: print(int(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 n, m = map(int, input().split()) if min(n, m) != 1: if n*m % 5 != 0: print(round(m * n / 2)) else: print(int(m * n / 2)) else: print(int(m * n / 2)) ```

0

2

A

Winner

PROGRAMMING

1,500

[ "hashing", "implementation" ]

A. Winner

1

64

The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.

The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.

Print the name of the winner.

[ "3\nmike 3\nandrew 5\nmike 2\n", "3\nandrew 3\nandrew 2\nmike 5\n" ]

[ "andrew\n", "andrew\n" ]

none

0

[ { "input": "3\nmike 3\nandrew 5\nmike 2", "output": "andrew" }, { "input": "3\nandrew 3\nandrew 2\nmike 5", "output": "andrew" }, { "input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303", "output": "kaxqybeultn" }, { "input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734", "output": "ksjuuerbnlklcfdjeyq" }, { "input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879", "output": "fcgslzkicjrpbqaifgweyzreajjfdo" }, { "input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735", "output": "aawtvezfntstrcpgbzjbf" }, { "input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158", "output": "ivhgbxiv" } ]

1,545,232,679

2,147,483,647

PyPy 3

OK

TESTS

20

280

307,200

# -*- coding: utf-8 -*- """ Created on Wed Dec 19 22:45:37 2018 @author: Lenovo """ x = [] y = {} n = int(input()) for i in range(n): a, b = input().split() b = int(b) y[a] = y.get(a, 0) + b x.append([y[a], a]) m = max(y.values()) for n, a in x: if n >= m and y[a] >= m: print(a) break

Title: Winner Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. Input Specification: The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. Output Specification: Print the name of the winner. Demo Input: ['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n'] Demo Output: ['andrew\n', 'andrew\n'] Note: none

```python # -*- coding: utf-8 -*- """ Created on Wed Dec 19 22:45:37 2018 @author: Lenovo """ x = [] y = {} n = int(input()) for i in range(n): a, b = input().split() b = int(b) y[a] = y.get(a, 0) + b x.append([y[a], a]) m = max(y.values()) for n, a in x: if n >= m and y[a] >= m: print(a) break ```

3.857711

439

A

Devu, the Singer and Churu, the Joker

PROGRAMMING

900

[ "greedy", "implementation" ]

null

Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited. Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly. The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly. People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest. You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions: - The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible. If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.

The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).

If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.

[ "3 30\n2 2 1\n", "3 20\n2 1 1\n" ]

[ "5\n", "-1\n" ]

Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way: - First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes. Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes. Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.

500

[ { "input": "3 30\n2 2 1", "output": "5" }, { "input": "3 20\n2 1 1", "output": "-1" }, { "input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1", "output": "1943" }, { "input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1", "output": "1891" }, { "input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2", "output": "1688" }, { "input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12", "output": "1391" }, { "input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32", "output": "368" }, { "input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1", "output": "435" }, { "input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100", "output": "326" }, { "input": "45 894\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 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1", "output": "139" }, { "input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86", "output": "514" }, { "input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1", "output": "586" }, { "input": "81 2200\n1 59 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1", "output": "384" }, { "input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1", "output": "174" }, { "input": "81 4131\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 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "807" }, { "input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96", "output": "490" }, { "input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1", "output": "284" }, { "input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1", "output": "1589" }, { "input": "10 100\n1 1 1 1 1 1 1 1 1 1", "output": "18" }, { "input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99", "output": "1017" }, { "input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96", "output": "-1" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 6\n1", "output": "1" }, { "input": "1 5\n1", "output": "0" }, { "input": "1 3\n4", "output": "-1" }, { "input": "3 24\n2 1 2", "output": "-1" } ]

1,570,893,673

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

93

0

i=lambda:map(int,input().split()) n,d=i() d-=sum(i()) print([-1,d//5][d%5==0])

Title: Devu, the Singer and Churu, the Joker Time Limit: None seconds Memory Limit: None megabytes Problem Description: Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited. Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly. The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly. People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest. You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions: - The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible. If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. Input Specification: The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). Output Specification: If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. Demo Input: ['3 30\n2 2 1\n', '3 20\n2 1 1\n'] Demo Output: ['5\n', '-1\n'] Note: Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way: - First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes. Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes. Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.

```python i=lambda:map(int,input().split()) n,d=i() d-=sum(i()) print([-1,d//5][d%5==0]) ```

0

245

E

Mishap in Club

PROGRAMMING

1,400

[ "greedy", "implementation" ]

null

Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times.

The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive.

Print the sought minimum number of people

[ "+-+-+\n", "---" ]

[ "1\n", "3" ]

none

0

[ { "input": "+-+-+", "output": "1" }, { "input": "---", "output": "3" }, { "input": "-", "output": "1" }, { "input": "--", "output": "2" }, { "input": "---", "output": "3" }, { "input": "----", "output": "4" }, { "input": "---+", "output": "3" }, { "input": "--+-", "output": "2" }, { "input": "--++", "output": "2" }, { "input": "-+--", "output": "2" }, { "input": "-++", "output": "2" }, { "input": "-++-", "output": "2" }, { "input": "+", "output": "1" }, { "input": "+-", "output": "1" }, { "input": "+--", "output": "2" }, { "input": "+--+", "output": "2" }, { "input": "++--", "output": "2" }, { "input": "-+++--+-++--+-+--+-+", "output": "3" }, { "input": "++-++--+++++-+++++---+++-++-++-", "output": "12" }, { "input": "----+-+--++---++---++-+-----+--", "output": "11" }, { "input": "-+++---+++++++++++++-++-++++++-++-+-+++-", "output": "22" }, { "input": "++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++", "output": "300" }, { "input": "------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------", "output": "300" }, { "input": "++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++", "output": "298" }, { "input": "++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++", "output": "296" }, { "input": "+++++++++++++++++++++++++++++++++++++++++-++++++++-++++++++++++-+++++++++++++++++++++++++++++++++++++++++++++++++++++-+++++++++++++++++++++++++-++++++++++++++++++++++++-++++-+++++++++++++-++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-+++-+++++++++++++++++", "output": "280" }, { "input": "++++-+---+++--++++-++-++++++++-++-++++-++++++-+-+-+++--+-++++++-+++-++-+++-++++-++++-+-+----+++++---++++-+---+++--+++++-+++-+-++++++----+--+++++++++-+--+++-+-+-++++++--+-+-+-+-++--+-+-----++++++-+++-++--+++++++++---+-++++++-++-++++-+--+-++-++++-+-+--++-+--+++-+-++++++++++++-+++-+----++++++++--+-+-++", "output": "100" }, { "input": "+----++-----+----+++--++---+++--+-++++-++---++++++--++++--++-++--++--++----++++---+--+++----++--++--++--+--++++++++--++--+++----+++----++----++-+--+---+--+-++--+--+--+-+--+---++-+-++--+++++-++------+++-++--+--+--+++++++--++-+--+-+--++++-++--+---+-+-++-+-++----+-++++++-+++--+----++-+--++-----+++-++-+", "output": "15" }, { "input": "-+++----+-++--+-+----+--+++++----+---+-++-+---+++--+---++-+-----+----+------+--+----++-++-----+++--+---+-+-----++++------+--+-----++---+---+---+-++------++++--+-+-------------+---+--+-+--------++---+-++---+-----+++--+---+-++-+---+-+---+++--++-----++------+----+---+---+--+-+-++-+---++--------+----++", "output": "103" }, { "input": "----------+-----------------------------------------------------------+-+-------------+--------------------------------------------------------------------------------+--+-----+-+-------------------------------------------+-----------------------------------------------------------------+-----------", "output": "280" } ]

1,568,426,359

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

6

248

0

s = input() ans = 1 max_ans = 1 for i in range(1, len(s)): if len(s) == 1: print(1) else: if s[i] == s[i - 1]: ans += 1 elif s[i] != s[i - 1]: if ans > max_ans: max_ans = ans ans = 1 print(ans)

Title: Mishap in Club Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. Input Specification: The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. Output Specification: Print the sought minimum number of people Demo Input: ['+-+-+\n', '---'] Demo Output: ['1\n', '3'] Note: none

```python s = input() ans = 1 max_ans = 1 for i in range(1, len(s)): if len(s) == 1: print(1) else: if s[i] == s[i - 1]: ans += 1 elif s[i] != s[i - 1]: if ans > max_ans: max_ans = ans ans = 1 print(ans) ```

0

710

B

Optimal Point on a Line

PROGRAMMING

1,400

[ "brute force", "sortings" ]

null

You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.

The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line. The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.

Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.

[ "4\n1 2 3 4\n" ]

[ "2\n" ]

none

0

[ { "input": "4\n1 2 3 4", "output": "2" }, { "input": "5\n-1 -10 2 6 7", "output": "2" }, { "input": "10\n-68 10 87 22 30 89 82 -97 -52 25", "output": "22" }, { "input": "100\n457 827 807 17 871 935 907 -415 536 170 551 -988 865 758 -457 -892 -875 -488 684 19 0 555 -807 -624 -239 826 318 811 20 -732 -91 460 551 -610 555 -493 -154 442 -141 946 -913 -104 704 -380 699 32 106 -455 -518 214 -464 -861 243 -798 -472 559 529 -844 -32 871 -459 236 387 626 -318 -580 -611 -842 790 486 64 951 81 78 -693 403 -731 309 678 696 891 846 -106 918 212 -44 994 606 -829 -454 243 -477 -402 -818 -819 -310 -837 -209 736 424", "output": "64" }, { "input": "2\n-1 0", "output": "-1" }, { "input": "48\n-777 -767 -764 -713 -688 -682 -606 -586 -585 -483 -465 -440 -433 -397 -390 -377 -299 -252 -159 -147 -96 -29 -15 15 52 109 124 129 142 218 231 314 320 339 442 496 505 548 575 576 594 624 694 827 891 979 981 981", "output": "15" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 1 1 1 1 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "1" }, { "input": "4\n-1 -1 0 1", "output": "-1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 1000000000", "output": "0" }, { "input": "2\n1 -1", "output": "-1" }, { "input": "2\n100 50", "output": "50" }, { "input": "2\n1 2", "output": "1" }, { "input": "1\n10", "output": "10" }, { "input": "3\n606194955 -856471310 117647402", "output": "117647402" }, { "input": "2\n615002717 -843553590", "output": "-843553590" }, { "input": "2\n-1 2", "output": "-1" }, { "input": "1\n0", "output": "0" }, { "input": "1\n2", "output": "2" }, { "input": "5\n-638512131 348325781 -550537933 -618161835 -567935532", "output": "-567935532" }, { "input": "1\n120", "output": "120" }, { "input": "2\n-1000000000 1000000000", "output": "-1000000000" }, { "input": "1\n618309368", "output": "618309368" } ]

1,486,934,063

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

31

4,608,000

n = int(input()) numbers = list(map(int, input().split())) numbers.sort() ls = 0 rs = sum(numbers) midiff = 99999999999999 distanceindex = 0 for i in range(n): rks = rs - ((n-i) * numbers[i]) ls += numbers[i] if abs(rks - ls) < midiff: midiff = abs(rks - ls) distanceindex = i print(rks, ls ,midiff) print(numbers[distanceindex])

Title: Optimal Point on a Line Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line. The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points. Output Specification: Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer. Demo Input: ['4\n1 2 3 4\n'] Demo Output: ['2\n'] Note: none

```python n = int(input()) numbers = list(map(int, input().split())) numbers.sort() ls = 0 rs = sum(numbers) midiff = 99999999999999 distanceindex = 0 for i in range(n): rks = rs - ((n-i) * numbers[i]) ls += numbers[i] if abs(rks - ls) < midiff: midiff = abs(rks - ls) distanceindex = i print(rks, ls ,midiff) print(numbers[distanceindex]) ```

0

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,694,533,080

2,147,483,647

Python 3

OK

TESTS

81

92

0

from sys import stdin input = stdin.readline if __name__ == "__main__": n = int(input()) data = [[0]*n, [0]*n, [0]*n] for i in range(n): data[0][i], data[1][i], data[2][i] = map(int, input().split()) print(["NO", "YES"][sum(data[0]) == sum(data[1]) == sum(data[2]) == 0])

Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none

```python from sys import stdin input = stdin.readline if __name__ == "__main__": n = int(input()) data = [[0]*n, [0]*n, [0]*n] for i in range(n): data[0][i], data[1][i], data[2][i] = map(int, input().split()) print(["NO", "YES"][sum(data[0]) == sum(data[1]) == sum(data[2]) == 0]) ```

3.977

447

A

DZY Loves Hash

PROGRAMMING

800

[ "implementation" ]

null

DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*. However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.

The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109).

Output a single integer — the answer to the problem.

[ "10 5\n0\n21\n53\n41\n53\n", "5 5\n0\n1\n2\n3\n4\n" ]

[ "4\n", "-1\n" ]

none

500

[ { "input": "10 5\n0\n21\n53\n41\n53", "output": "4" }, { "input": "5 5\n0\n1\n2\n3\n4", "output": "-1" }, { "input": "10 6\n811966798\n734823552\n790326404\n929189974\n414343256\n560346537", "output": "4" }, { "input": "2 2\n788371161\n801743052", "output": "-1" }, { "input": "10 6\n812796223\n122860157\n199259103\n597650585\n447742024\n521549402", "output": "3" }, { "input": "300 2\n822454942\n119374431", "output": "-1" }, { "input": "300 2\n823284367\n507345500", "output": "-1" }, { "input": "2 5\n791579811\n35613889\n997079893\n878677665\n693717467", "output": "2" }, { "input": "20 5\n793926268\n28931770\n842870287\n974950617\n859404206", "output": "-1" }, { "input": "100 15\n805069249\n778178198\n633897389\n844316223\n146759898\n870527016\n82668754\n42060733\n943602929\n979451110\n746979598\n47406033\n607284062\n850870259\n229415316", "output": "5" }, { "input": "100 15\n806204335\n189490323\n718805086\n716787474\n262315718\n822030596\n894644222\n724054623\n141384399\n579354205\n192622443\n672556242\n97417563\n243354557\n208957882", "output": "8" }, { "input": "100 15\n807033760\n577461392\n275221433\n532633429\n295714486\n783298996\n255799943\n99107143\n729119412\n59302896\n37640015\n313610861\n630550567\n534283052\n681062462", "output": "8" }, { "input": "100 15\n808103310\n136224397\n360129131\n405104681\n263786657\n734802577\n67808179\n928584682\n926900882\n511722343\n483348395\n938695534\n120684068\n74152694\n808088675", "output": "9" }, { "input": "2 2\n2\n2", "output": "2" }, { "input": "300 2\n0\n300", "output": "2" }, { "input": "2 2\n0\n0", "output": "2" } ]

1,461,062,925

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

46

4,608,000

p, n = map(int, input().split()) s = set() for i in range(n): x = int(input()) if x in s: print(i) break else: s.add(x)

Title: DZY Loves Hash Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*. However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1. Input Specification: The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109). Output Specification: Output a single integer — the answer to the problem. Demo Input: ['10 5\n0\n21\n53\n41\n53\n', '5 5\n0\n1\n2\n3\n4\n'] Demo Output: ['4\n', '-1\n'] Note: none

```python p, n = map(int, input().split()) s = set() for i in range(n): x = int(input()) if x in s: print(i) break else: s.add(x) ```

0

119

A

Epic Game

PROGRAMMING

800

[ "implementation" ]

null

Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.

The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.

If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).

[ "3 5 9\n", "1 1 100\n" ]

[ "0", "1" ]

The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

500

[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]

1,557,304,887

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

186

0

from math import gcd a,b,n=map(int,input().split()) Turn = 1 while True: yy=gcd(a,n) if n-yy>=0: Turn*=-1 n-=yy else: break print(0 if Turn == -1 else 1)

Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

```python from math import gcd a,b,n=map(int,input().split()) Turn = 1 while True: yy=gcd(a,n) if n-yy>=0: Turn*=-1 n-=yy else: break print(0 if Turn == -1 else 1) ```

0

784

A

Numbers Joke

PROGRAMMING

1,700

[ "*special" ]

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" } ]

1,491,253,180

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

46

4,608,000

x = int(input(x)) print(27)

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 x = int(input(x)) print(27) ```

-1

92

A

Chips

PROGRAMMING

800

[ "implementation", "math" ]

A. Chips

2

256

There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*. The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.

The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.

Print the number of chips the presenter ended up with.

[ "4 11\n", "17 107\n", "3 8\n" ]

[ "0\n", "2\n", "1\n" ]

In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes. In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.

500

[ { "input": "4 11", "output": "0" }, { "input": "17 107", "output": "2" }, { "input": "3 8", "output": "1" }, { "input": "46 7262", "output": "35" }, { "input": "32 6864", "output": "0" }, { "input": "36 6218", "output": "14" }, { "input": "25 9712", "output": "11" }, { "input": "9 7601", "output": "5" }, { "input": "1 9058", "output": "0" }, { "input": "29 7772", "output": "26" }, { "input": "45 9465", "output": "14" }, { "input": "46 866", "output": "5" }, { "input": "29 1241", "output": "20" }, { "input": "17 4248", "output": "12" }, { "input": "20 8082", "output": "11" }, { "input": "50 9555", "output": "0" }, { "input": "4 7455", "output": "2" }, { "input": "36 880", "output": "4" }, { "input": "24 7440", "output": "9" }, { "input": "44 7888", "output": "12" }, { "input": "1 1", "output": "0" }, { "input": "50 10000", "output": "40" }, { "input": "1 10000", "output": "0" }, { "input": "50 1", "output": "0" }, { "input": "50 50", "output": "5" } ]

1,614,376,323

2,147,483,647

Python 3

OK

TESTS

25

124

0

n,m = map(int,input().split()) i=1 while(i<=m): m-=i i+=1 if i==n+1: i=1 print(m)

Title: Chips Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*. The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. Output Specification: Print the number of chips the presenter ended up with. Demo Input: ['4 11\n', '17 107\n', '3 8\n'] Demo Output: ['0\n', '2\n', '1\n'] Note: In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes. In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.

```python n,m = map(int,input().split()) i=1 while(i<=m): m-=i i+=1 if i==n+1: i=1 print(m) ```

3.969

381

A

Sereja and Dima

PROGRAMMING

800

[ "greedy", "implementation", "two pointers" ]

null

Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.

The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.

On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.

[ "4\n4 1 2 10\n", "7\n1 2 3 4 5 6 7\n" ]

[ "12 5\n", "16 12\n" ]

In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

500

[ { "input": "4\n4 1 2 10", "output": "12 5" }, { "input": "7\n1 2 3 4 5 6 7", "output": "16 12" }, { "input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13", "output": "613 418" }, { "input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24", "output": "644 500" }, { "input": "1\n3", "output": "3 0" }, { "input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646", "output": "6848 6568" }, { "input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727", "output": "9562 9561" }, { "input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12", "output": "315 315" }, { "input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304", "output": "3238 2222" }, { "input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325", "output": "5246 4864" }, { "input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71", "output": "8147 7807" }, { "input": "1\n1", "output": "1 0" } ]

1,696,482,917

2,147,483,647

Python 3

OK

TESTS

34

46

0

n=int(input()) a=list(map(int,input().split())) sc1=0 sc2=0 p1=0 p2=n-1 for i in range(n): if(i%2==0): if (a[p1]>a[p2]): sc1+=a[p1] p1+=1 elif(a[p1]<a[p2]): sc1+=a[p2] p2-=1 else: sc1+=a[p1] else: if (a[p1]>a[p2]): sc2+=a[p1] p1+=1 elif(a[p1]<a[p2]): sc2+=a[p2] p2-=1 else: sc2+=a[p1] print(sc1,sc2)

Title: Sereja and Dima Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. Output Specification: On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. Demo Input: ['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n'] Demo Output: ['12 5\n', '16 12\n'] Note: In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

```python n=int(input()) a=list(map(int,input().split())) sc1=0 sc2=0 p1=0 p2=n-1 for i in range(n): if(i%2==0): if (a[p1]>a[p2]): sc1+=a[p1] p1+=1 elif(a[p1]<a[p2]): sc1+=a[p2] p2-=1 else: sc1+=a[p1] else: if (a[p1]>a[p2]): sc2+=a[p1] p1+=1 elif(a[p1]<a[p2]): sc2+=a[p2] p2-=1 else: sc2+=a[p1] print(sc1,sc2) ```

3

633

C

Spy Syndrome 2

PROGRAMMING

1,900

[ "data structures", "dp", "hashing", "implementation", "sortings", "string suffix structures", "strings" ]

null

After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique. For a given sentence, the cipher is processed as: 1. Convert all letters of the sentence to lowercase. 1. Reverse each of the words of the sentence individually. 1. Remove all the spaces in the sentence. For example, when this cipher is applied to the sentence Kira is childish and he hates losing the resulting string is ariksihsidlihcdnaehsetahgnisol Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10<=000) — the length of the ciphered text. The second line consists of *n* lowercase English letters — the ciphered text *t*. The third line contains a single integer *m* (1<=≤<=*m*<=≤<=100<=000) — the number of words which will be considered while deciphering the text. Each of the next *m* lines contains a non-empty word *w**i* (|*w**i*|<=≤<=1<=000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1<=000<=000.

Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.

[ "30\nariksihsidlihcdnaehsetahgnisol\n10\nKira\nhates\nis\nhe\nlosing\ndeath\nchildish\nL\nand\nNote\n", "12\niherehtolleh\n5\nHI\nHo\nthere\nHeLLo\nhello\n" ]

[ "Kira is childish and he hates losing \n", "HI there HeLLo \n" ]

In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.

1,500

[ { "input": "30\nariksihsidlihcdnaehsetahgnisol\n10\nKira\nhates\nis\nhe\nlosing\ndeath\nchildish\nL\nand\nNote", "output": "Kira is childish and he hates losing " }, { "input": "12\niherehtolleh\n5\nHI\nHo\nthere\nHeLLo\nhello", "output": "HI there HeLLo " }, { "input": "71\nbaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n8\na\naa\naaa\naaaa\naaaaa\naaaaaa\naaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab " }, { "input": "11\naaaaaaaaaaa\n3\naaaa\naaaaa\naaaaaa", "output": "aaaaaa aaaaa " }, { "input": "5\ncbaed\n6\nbc\nea\nab\ncd\nabc\nde", "output": "abc de " }, { "input": "5\nlehet\n3\nel\nhel\nte", "output": "hel te " }, { "input": "3\naab\n2\na\nbaa", "output": "baa " }, { "input": "3\naab\n2\naa\nbaa", "output": "baa " }, { "input": "9\naaaaaaaaa\n2\naa\naaa", "output": "aaa aa aa aa " }, { "input": "7\nabababc\n2\nba\ncba", "output": "ba ba cba " }, { "input": "7\nkirkirk\n2\nrik\nkrik", "output": "rik krik " } ]

1,456,512,442

5,542

Python 3

WRONG_ANSWER

PRETESTS

0

46

0

lol = input() data = input() right_data = '' for l in data: right_data = l + right_data # print(right_data) N = int(input()) for i in range(N): t = input() if t.lower() in right_data: print(t, end='')

Title: Spy Syndrome 2 Time Limit: None seconds Memory Limit: None megabytes Problem Description: After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique. For a given sentence, the cipher is processed as: 1. Convert all letters of the sentence to lowercase. 1. Reverse each of the words of the sentence individually. 1. Remove all the spaces in the sentence. For example, when this cipher is applied to the sentence Kira is childish and he hates losing the resulting string is ariksihsidlihcdnaehsetahgnisol Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10<=000) — the length of the ciphered text. The second line consists of *n* lowercase English letters — the ciphered text *t*. The third line contains a single integer *m* (1<=≤<=*m*<=≤<=100<=000) — the number of words which will be considered while deciphering the text. Each of the next *m* lines contains a non-empty word *w**i* (|*w**i*|<=≤<=1<=000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1<=000<=000. Output Specification: Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those. Demo Input: ['30\nariksihsidlihcdnaehsetahgnisol\n10\nKira\nhates\nis\nhe\nlosing\ndeath\nchildish\nL\nand\nNote\n', '12\niherehtolleh\n5\nHI\nHo\nthere\nHeLLo\nhello\n'] Demo Output: ['Kira is childish and he hates losing \n', 'HI there HeLLo \n'] Note: In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.

```python lol = input() data = input() right_data = '' for l in data: right_data = l + right_data # print(right_data) N = int(input()) for i in range(N): t = input() if t.lower() in right_data: print(t, end='') ```

0

231

A

Team

PROGRAMMING

800

[ "brute force", "greedy" ]

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" } ]

1,697,695,076

2,147,483,647

Python 3

OK

TESTS

21

92

0

n=int(input()) c=0 for i in range (0,n): #print("sakib") x,y,z=map(int,input().split()) sum=x+y+z if sum>1: c+=1 print(c)

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 i in range (0,n): #print("sakib") x,y,z=map(int,input().split()) sum=x+y+z if sum>1: c+=1 print(c) ```

3

864

B

Polycarp and Letters

PROGRAMMING

1,000

[ "brute force", "implementation", "strings" ]

null

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters.

Print maximum number of elements in pretty set of positions for string *s*.

[ "11\naaaaBaabAbA\n", "12\nzACaAbbaazzC\n", "3\nABC\n" ]

[ "2\n", "3\n", "0\n" ]

In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.

1,000

[ { "input": "11\naaaaBaabAbA", "output": "2" }, { "input": "12\nzACaAbbaazzC", "output": "3" }, { "input": "3\nABC", "output": "0" }, { "input": "1\na", "output": "1" }, { "input": "2\naz", "output": "2" }, { "input": "200\nXbTJZqcbpYuZQEoUrbxlPXAPCtVLrRExpQzxzqzcqsqzsiisswqitswzCtJQxOavicSdBIodideVRKHPojCNHmbnrLgwJlwOpyrJJIhrUePszxSjJGeUgTtOfewPQnPVWhZAtogRPrJLwyShNQaeNsvrJwjuuBOMPCeSckBMISQzGngfOmeyfDObncyeNsihYVtQbSEh", "output": "8" }, { "input": "2\nAZ", "output": "0" }, { "input": "28\nAabcBabcCBNMaaaaabbbbbcccccc", "output": "3" }, { "input": "200\nrsgraosldglhdoorwhkrsehjpuxrjuwgeanjgezhekprzarelduuaxdnspzjuooguuwnzkowkuhzduakdrzpnslauejhrrkalwpurpuuswdgeadlhjwzjgegwpknepazwwleulppwrlgrgedlwdzuodzropsrrkxusjnuzshdkjrxxpgzanzdrpnggdwxarpwohxdepJ", "output": "17" }, { "input": "1\nk", "output": "1" }, { "input": "1\nH", "output": "0" }, { "input": "2\nzG", "output": "1" }, { "input": "2\ngg", "output": "1" }, { "input": "2\nai", "output": "2" }, { "input": "20\npEjVrKWLIFCZjIHgggVU", "output": "1" }, { "input": "20\niFSiiigiYFSKmDnMGcgM", "output": "2" }, { "input": "20\nedxedxxxCQiIVmYEUtLi", "output": "3" }, { "input": "20\nprnchweyabjvzkoqiltm", "output": "20" }, { "input": "35\nQLDZNKFXKVSVLUVHRTDPQYMSTDXBELXBOTS", "output": "0" }, { "input": "35\nbvZWiitgxodztelnYUyljYGnCoWluXTvBLp", "output": "10" }, { "input": "35\nBTexnaeplecllxwlanarpcollawHLVMHIIF", "output": "10" }, { "input": "35\nhhwxqysolegsthsvfcqiryenbujbrrScobu", "output": "20" }, { "input": "26\npbgfqosklxjuzmdheyvawrictn", "output": "26" }, { "input": "100\nchMRWwymTDuZDZuSTvUmmuxvSscnTasyjlwwodhzcoifeahnbmcifyeobbydwparebduoLDCgHlOsPtVRbYGGQXfnkdvrWKIwCRl", "output": "20" }, { "input": "100\nhXYLXKUMBrGkjqQJTGbGWAfmztqqapdbjbhcualhypgnaieKXmhzGMnqXVlcPesskfaEVgvWQTTShRRnEtFahWDyuBzySMpugxCM", "output": "19" }, { "input": "100\nucOgELrgjMrFOgtHzqgvUgtHngKJxdMFKBjfcCppciqmGZXXoiSZibgpadshyljqrwxbomzeutvnhTLGVckZUmyiFPLlwuLBFito", "output": "23" }, { "input": "200\nWTCKAKLVGXSYFVMVJDUYERXNMVNTGWXUGRFCGMYXJQGLODYZTUIDENHYEGFKXFIEUILAMESAXAWZXVCZPJPEYUXBITHMTZOTMKWITGRSFHODKVJHPAHVVWTCTHIVAWAREQXWMPUWQSTPPJFHKGKELBTPUYDAVIUMGASPUEDIODRYXIWCORHOSLIBLOZUNJPHHMXEXOAY", "output": "0" }, { "input": "200\neLCCuYMPPwQoNlCpPOtKWJaQJmWfHeZCKiMSpILHSKjFOYGpRMzMCfMXdDuQdBGNsCNrHIVJzEFfBZcNMwNcFjOFVJvEtUQmLbFNKVHgNDyFkFVQhUTUQDgXhMjJZgFSSiHhMKuTgZQYJqAqKBpHoHddddddddddddddddXSSYNKNnRrKuOjAVKZlRLzCjExPdHaDHBT", "output": "1" }, { "input": "200\nitSYxgOLlwOoAkkkkkzzzzzzzzkzkzkzkkkkkzkzzkzUDJSKybRPBvaIDsNuWImPJvrHkKiMeYukWmtHtgZSyQsgYanZvXNbKXBlFLSUcqRnGWSriAvKxsTkDJfROqaKdzXhvJsPEDATueCraWOGEvRDWjPwXuiNpWsEnCuhDcKWOQxjBkdBqmFatWFkgKsbZuLtRGtY", "output": "2" }, { "input": "200\noggqoqqogoqoggggoggqgooqggogogooogqqgggoqgggqoqogogggogggqgooqgqggqqqoqgqgoooqgqogqoggoqqgqoqgoooqoogooqoogqoqoqqgoqgoqgggogqqqoqoggoqoqqoqggqoggooqqqoqggoggqqqqqqqqqgogqgggggooogogqgggqogqgoqoqogoooq", "output": "3" }, { "input": "200\nCtclUtUnmqFniaLqGRmMoUMeLyFfAgWxIZxdrBarcRQprSOGcdUYsmDbooSuOvBLgrYlgaIjJtFgcxJKHGkCXpYfVKmUbouuIqGstFrrwJzYQqjjqqppqqqqqpqqqjpjjpjqjXRYkfPhGAatOigFuItkKxkjCBLdiNMVGjmdWNMgOOvmaJEdGsWNoaERrINNKqKeQajv", "output": "3" }, { "input": "200\nmeZNrhqtSTSmktGQnnNOTcnyAMTKSixxKQKiagrMqRYBqgbRlsbJhvtNeHVUuMCyZLCnsIixRYrYEAkfQOxSVqXkrPqeCZQksInzRsRKBgvIqlGVPxPQnypknSXjgMjsjElcqGsaJRbegJVAKtWcHoOnzHqzhoKReqBBsOhZYLaYJhmqOMQsizdCsQfjUDHcTtHoeYwu", "output": "4" }, { "input": "200\nvFAYTHJLZaivWzSYmiuDBDUFACDSVbkImnVaXBpCgrbgmTfXKJfoglIkZxWPSeVSFPnHZDNUAqLyhjLXSuAqGLskBlDxjxGPJyGdwzlPfIekwsblIrkxzfhJeNoHywdfAGlJzqXOfQaKceSqViVFTRJEGfACnsFeSFpOYisIHJciqTMNAmgeXeublTvfWoPnddtvKIyF", "output": "6" }, { "input": "200\ngnDdkqJjYvduVYDSsswZDvoCouyaYZTfhmpSakERWLhufZtthWsfbQdTGwhKYjEcrqWBOyxBbiFhdLlIjChLOPiOpYmcrJgDtXsJfmHtLrabyGKOfHQRukEtTzwoqBHfmyVXPebfcpGQacLkGWFwerszjdHpTBXGssYXmGHlcCBgBXyGJqxbVhvDffLyCrZnxonABEXV", "output": "7" }, { "input": "200\nBmggKNRZBXPtJqlJaXLdKKQLDJvXpDuQGupiRQfDwCJCJvAlDDGpPZNOvXkrdKOFOEFBVfrsZjWyHPoKGzXmTAyPJGEmxCyCXpeAdTwbrMtWLmlmGNqxvuxmqpmtpuhrmxxtrquSLFYVlnSYgRJDYHWgHBbziBLZRwCIJNvbtsEdLLxmTbnjkoqSPAuzEeTYLlmejOUH", "output": "9" }, { "input": "200\nMkuxcDWdcnqsrlTsejehQKrTwoOBRCUAywqSnZkDLRmVBDVoOqdZHbrInQQyeRFAjiYYmHGrBbWgWstCPfLPRdNVDXBdqFJsGQfSXbufsiogybEhKDlWfPazIuhpONwGzZWaQNwVnmhTqWdewaklgjwaumXYDGwjSeEcYXjkVtLiYSWULEnTFukIlWQGWsXwWRMJGTcI", "output": "10" }, { "input": "200\nOgMBgYeuMJdjPtLybvwmGDrQEOhliaabEtwulzNEjsfnaznXUMoBbbxkLEwSQzcLrlJdjJCLGVNBxorghPxTYCoqniySJMcilpsqpBAbqdzqRUDVaYOgqGhGrxlIJkyYgkOdTUgRZwpgIkeZFXojLXpDilzirHVVadiHaMrxhzodzpdvhvrzdzxbhmhdpxqqpoDegfFQ", "output": "11" }, { "input": "200\nOLaJOtwultZLiZPSYAVGIbYvbIuZkqFZXwfsqpsavCDmBMStAuUFLBVknWDXNzmiuUYIsUMGxtoadWlPYPqvqSvpYdOiJRxFzGGnnmstniltvitnrmyrblnqyruylummmlsqtqitlbulvtuitiqimuintbimqyurviuntqnnvslynlNYMpYVKYwKVTbIUVdlNGrcFZON", "output": "12" }, { "input": "200\nGAcmlaqfjSAQLvXlkhxujXgSbxdFAwnoxDuldDvYmpUhTWJdcEQSdARLrozJzIgFVCkzPUztWIpaGfiKeqzoXinEjVuoKqyBHmtFjBWcRdBmyjviNlGAIkpikjAimmBgayfphrstfbjexjbttzfzfzaysxfyrjazfhtpghnbbeffjhxrjxpttesgzrnrfbgzzsRsCgmz", "output": "15" }, { "input": "200\nYRvIopNqSTYDhViTqCLMwEbTTIdHkoeuBmAJWhgtOgVxlcHSsavDNzMfpwTghkBvYEtCYQxicLUxdgAcaCzOOgbQYsfnaTXFlFxbeEiGwdNvxwHzkTdKtWlqzalwniDDBDipkxfflpaqkfkgfezbkxdvzemlfohwtgytzzywmwhvzUgPlPdeAVqTPAUZbogQheRXetvT", "output": "20" }, { "input": "200\nNcYVomemswLCUqVRSDKHCknlBmqeSWhVyRzQrnZaOANnTGqsRFMjpczllcEVebqpxdavzppvztxsnfmtcharzqlginndyjkawzurqkxJLXiXKNZTIIxhSQghDpjwzatEqnLMTLxwoEKpHytvWkKFDUcZjLShCiVdocxRvvJtbXHCDGpJvMwRKWLhcTFtswdLUHkbhfau", "output": "25" }, { "input": "200\nDxNZuvkTkQEqdWIkLzcKAwfqvZQiptnTazydSCTIfGjDhLMrlPZiKEsqIdDhgKPAlEvXyzNwWtYorotgkcwydpabjqnzubaksdchucxtkmjzfretdmvlxgklyvicrtftvztsbiUaQorfNIYUOdwQDRsKpxLUiLknbLbinilpPXPTTwLAnXVpMHBaAcKWgDBeOFabPtXU", "output": "26" }, { "input": "4\nabbc", "output": "3" }, { "input": "3\naaa", "output": "1" }, { "input": "3\naba", "output": "2" }, { "input": "3\nabb", "output": "2" }, { "input": "3\nbba", "output": "2" }, { "input": "3\nAaa", "output": "1" }, { "input": "3\nAba", "output": "2" }, { "input": "3\naBa", "output": "1" }, { "input": "3\naAa", "output": "1" }, { "input": "3\naAb", "output": "1" }, { "input": "3\nAaA", "output": "1" }, { "input": "5\naBacd", "output": "3" }, { "input": "5\naAabc", "output": "3" } ]

1,539,246,968

2,147,483,647

Python 3

OK

TESTS

55

124

0

##a = list(map(int, input().split())) ##print(' '.join(map(str, res))) n = int(input()) s = input() res = 0 p = 0 while p < n: q = p while q < n and s[q].islower(): q += 1 if q-p > 0: h = dict() for i in range(p, q): if s[i] in h.keys(): h[s[i]] += 1 else: h.update({s[i]:1}) res = max(res, len(h)) p = q+1 print(res)

Title: Polycarp and Letters Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters. Output Specification: Print maximum number of elements in pretty set of positions for string *s*. Demo Input: ['11\naaaaBaabAbA\n', '12\nzACaAbbaazzC\n', '3\nABC\n'] Demo Output: ['2\n', '3\n', '0\n'] Note: In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.

```python ##a = list(map(int, input().split())) ##print(' '.join(map(str, res))) n = int(input()) s = input() res = 0 p = 0 while p < n: q = p while q < n and s[q].islower(): q += 1 if q-p > 0: h = dict() for i in range(p, q): if s[i] in h.keys(): h[s[i]] += 1 else: h.update({s[i]:1}) res = max(res, len(h)) p = q+1 print(res) ```

3

437

A

The Child and Homework

PROGRAMMING

1,300

[ "implementation" ]

null

Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct. Fortunately the child knows how to solve such complicated test. The child will follow the algorithm: - If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice). You are given a multiple-choice questions, can you predict child's choose?

The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length. Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".

Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).

[ "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n", "A.ab\nB.abcde\nC.ab\nD.abc\n", "A.c\nB.cc\nC.c\nD.c\n" ]

[ "D\n", "C\n", "B\n" ]

In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D. In the second sample, no choice is great, so the child will choose the luckiest choice C. In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.

500

[ { "input": "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute", "output": "D" }, { "input": "A.ab\nB.abcde\nC.ab\nD.abc", "output": "C" }, { "input": "A.c\nB.cc\nC.c\nD.c", "output": "B" }, { "input": "A.He_nan_de_yang_guang_zhao_yao_zhe_wo_men_mei_guo_ren_lian_shang_dou_xiao_kai_yan_wahaaaaaaaaaaaaaaaa\nB.Li_bai_li_bai_fei_liu_zhi_xia_san_qian_chi_yi_si_yin_he_luo_jiu_tian_li_bai_li_bai_li_bai_li_bai_shi\nC.Peng_yu_xiang_shi_zai_tai_shen_le_jian_zhi_jiu_shi_ye_jie_du_liu_a_si_mi_da_zhen_shi_tai_shen_le_a_a\nD.Wo_huo_le_si_shi_er_nian_zhen_de_shi_cong_lai_ye_mei_you_jian_guo_zhe_me_biao_zhun_de_yi_bai_ge_zi_a", "output": "C" }, { "input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____", "output": "C" }, { "input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_", "output": "D" }, { "input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____", "output": "C" }, { "input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_", "output": "D" }, { "input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h", "output": "A" }, { "input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_", "output": "C" }, { "input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg", "output": "B" }, { "input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___", "output": "D" }, { "input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS", "output": "C" }, { "input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_", "output": "C" }, { "input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi", "output": "C" }, { "input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__", "output": "C" }, { "input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h", "output": "A" }, { "input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_", "output": "C" }, { "input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg", "output": "B" }, { "input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___", "output": "D" }, { "input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS", "output": "C" }, { "input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_", "output": "C" }, { "input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi", "output": "C" }, { "input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__", "output": "C" }, { "input": "A.aaaaaa\nB.aaa\nC.aaa\nD.aaa", "output": "A" }, { "input": "A.aaa\nB.aaaaaa\nC.aaaaaa\nD.aaaaaa", "output": "A" }, { "input": "A.a\nB.b\nC.c\nD.d", "output": "C" }, { "input": "A._\nB.__\nC.____\nD.________", "output": "C" }, { "input": "A.____\nB.________\nC.________\nD._______", "output": "C" }, { "input": "A.h\nB.asdf\nC.asqw\nD.qwertasdfg", "output": "C" }, { "input": "A.aa\nB.aaaaa\nC.aaaaaa\nD.aaaaaaaaaaaaa", "output": "C" }, { "input": "A.ccc\nB.ccccccc\nC.ccc\nD.c", "output": "C" }, { "input": "A.c\nB.ccc\nC.cccccccccccccccccc\nD.cccccc", "output": "C" }, { "input": "A.aa\nB.bb\nC.cc\nD.ddd", "output": "C" }, { "input": "A.QW\nB.WERT\nC.QWER\nD.QWERTYUI", "output": "C" } ]

1,608,617,187

2,147,483,647

PyPy 3

COMPILATION_ERROR

TESTS

0

#include <iostream> #include <vector> #include <set> #include <algorithm> #include <map> #include <queue> #include <string> using namespace std; int main(){ int n; cin>>n; vector<int> arr(n); for(int &x : arr){ cin>>x; } int x=*min_element(arr.begin(), arr.end()); vector<int> pos; vector<int> ano=arr; for(int i=0;i<n;i++){ ano.push_back(arr[i]); } for(int i=0;i<ano.size();i++){ if(ano[i]==x){ pos.push_back(i); } } int max_dist=0; for(int i=1;i<pos.size();i++){ max_dist=max(max_dist, pos[i]-pos[i-1]-1); } cout<<x*n+max_dist<<endl; }

Title: The Child and Homework Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct. Fortunately the child knows how to solve such complicated test. The child will follow the algorithm: - If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice). You are given a multiple-choice questions, can you predict child's choose? Input Specification: The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length. Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_". Output Specification: Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes). Demo Input: ['A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n', 'A.ab\nB.abcde\nC.ab\nD.abc\n', 'A.c\nB.cc\nC.c\nD.c\n'] Demo Output: ['D\n', 'C\n', 'B\n'] Note: In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D. In the second sample, no choice is great, so the child will choose the luckiest choice C. In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.

```python #include <iostream> #include <vector> #include <set> #include <algorithm> #include <map> #include <queue> #include <string> using namespace std; int main(){ int n; cin>>n; vector<int> arr(n); for(int &x : arr){ cin>>x; } int x=*min_element(arr.begin(), arr.end()); vector<int> pos; vector<int> ano=arr; for(int i=0;i<n;i++){ ano.push_back(arr[i]); } for(int i=0;i<ano.size();i++){ if(ano[i]==x){ pos.push_back(i); } } int max_dist=0; for(int i=1;i<pos.size();i++){ max_dist=max(max_dist, pos[i]-pos[i-1]-1); } cout<<x*n+max_dist<<endl; } ```

-1

0

none

0

[ "none" ]

null

Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.

You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.

Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.

[ "fixprefixsuffix\n", "abcdabc\n" ]

[ "fix", "Just a legend" ]

none

0

[ { "input": "fixprefixsuffix", "output": "fix" }, { "input": "abcdabc", "output": "Just a legend" }, { "input": "qwertyqwertyqwerty", "output": "qwerty" }, { "input": "papapapap", "output": "papap" }, { "input": "aaaaaaaaaa", "output": "aaaaaaaa" }, { "input": "ghbdtn", "output": "Just a legend" }, { "input": "a", "output": "Just a legend" }, { "input": "aa", "output": "Just a legend" }, { "input": "ab", "output": "Just a legend" }, { "input": "aaa", "output": "a" }, { "input": "aba", "output": "Just a legend" }, { "input": "aab", "output": "Just a legend" }, { "input": "abb", "output": "Just a legend" }, { "input": "abc", "output": "Just a legend" }, { "input": "aaabaabaaaaab", "output": "Just a legend" }, { "input": "aabaaabaaaaab", "output": "aab" }, { "input": "aaabaaaabab", "output": "Just a legend" }, { "input": "abcabcabcabcabc", "output": "abcabcabc" }, { "input": "aaaaabaaaa", "output": "aaaa" }, { "input": "aaaabaaaaaaa", "output": "aaaa" }, { "input": "ghghghgxghghghg", "output": "ghghg" }, { "input": "kincenvizh", "output": "Just a legend" }, { "input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw", "output": "Just a legend" }, { "input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx", "output": "Just a legend" }, { "input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb", "output": "nfbdzgdlbjhrlvfryyjb" }, { "input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp", "output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp" }, { "input": "abcazabca", "output": "a" } ]

1,464,934,276

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

92

4,505,600

def failurefunc(s): ls = len(s) F = [0] * ls i = 1 j = 0 while i < ls: #print (i,j,s[i],s[j]) if s[i] == s[j]: F[i] = j+1 i += 1 j += 1 elif j > 0: j = F[j-1] else: F[i] = 0 i += 1 return F s = input() a = {0} F = failurefunc(s) print(F) k = F[len(s)-1] while k>0: a.add(k) k = F[k-1] l = 0 for i in range(len(s)): f = F[i] if f in a and f > l: l = f if l: print(s[:l]) else: print('Just a legend')

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. Input Specification: You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. Output Specification: Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. Demo Input: ['fixprefixsuffix\n', 'abcdabc\n'] Demo Output: ['fix', 'Just a legend'] Note: none

```python def failurefunc(s): ls = len(s) F = [0] * ls i = 1 j = 0 while i < ls: #print (i,j,s[i],s[j]) if s[i] == s[j]: F[i] = j+1 i += 1 j += 1 elif j > 0: j = F[j-1] else: F[i] = 0 i += 1 return F s = input() a = {0} F = failurefunc(s) print(F) k = F[len(s)-1] while k>0: a.add(k) k = F[k-1] l = 0 for i in range(len(s)): f = F[i] if f in a and f > l: l = f if l: print(s[:l]) else: print('Just a legend') ```

0

none

0

[ "none" ]

null

The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the road there are *n* friends, and *i*-th of them is standing at the point *x**i* meters and can move with any speed no greater than *v**i* meters per second in any of the two directions along the road: south or north. You are to compute the minimum time needed to gather all the *n* friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate.

The first line contains single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of friends. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) — the current coordinates of the friends, in meters. The third line contains *n* integers *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — the maximum speeds of the friends, in meters per second.

Print the minimum time (in seconds) needed for all the *n* friends to meet at some point on the road. Your answer will be considered correct, if its absolute or relative error isn't greater than 10<=-<=6. Formally, let your answer be *a*, while jury's answer be *b*. Your answer will be considered correct if holds.

[ "3\n7 1 3\n1 2 1\n", "4\n5 10 3 2\n2 3 2 4\n" ]

[ "2.000000000000\n", "1.400000000000\n" ]

In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.

0

[ { "input": "3\n7 1 3\n1 2 1", "output": "2.000000000000" }, { "input": "4\n5 10 3 2\n2 3 2 4", "output": "1.400000000000" }, { "input": "3\n1 1000000000 2\n1 2 1000000000", "output": "333333332.999999999971" }, { "input": "2\n4 5\n10 8", "output": "0.055555555556" }, { "input": "4\n14 12 10 17\n8 6 5 10", "output": "0.466666666667" }, { "input": "5\n1 15 61 29 43\n15 11 19 19 19", "output": "1.764705882353" }, { "input": "10\n20 11 17 38 15 27 2 40 24 37\n22 30 22 30 28 16 7 20 22 13", "output": "1.750000000000" }, { "input": "2\n1000000000 1000000000\n1 1", "output": "0.000000000000" }, { "input": "3\n1 1 1\n1 1 1", "output": "0.000000000000" } ]

1,628,017,531

31

Python 3

WRONG_ANSWER

TESTS

2

46

6,963,200

n = int(input()) positions = list(map(int, input().split())) speeds = list(map(int, input().split())) upperBound = 10 ** 6 lowerBound = 0 time = (upperBound + lowerBound) / 2 while upperBound - lowerBound > 10 ** -7: time = (upperBound + lowerBound) / 2 maxL = float("-inf") minR = float("inf") for friend in range(len(positions)): pos = positions[friend] speed = speeds[friend] L = pos - speed * time R = pos + speed * time if L > maxL: maxL = L if R < minR: minR = R if maxL <= minR: upperBound = time else: lowerBound = time print(round(time, 6))

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the road there are *n* friends, and *i*-th of them is standing at the point *x**i* meters and can move with any speed no greater than *v**i* meters per second in any of the two directions along the road: south or north. You are to compute the minimum time needed to gather all the *n* friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate. Input Specification: The first line contains single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of friends. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) — the current coordinates of the friends, in meters. The third line contains *n* integers *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — the maximum speeds of the friends, in meters per second. Output Specification: Print the minimum time (in seconds) needed for all the *n* friends to meet at some point on the road. Your answer will be considered correct, if its absolute or relative error isn't greater than 10<=-<=6. Formally, let your answer be *a*, while jury's answer be *b*. Your answer will be considered correct if holds. Demo Input: ['3\n7 1 3\n1 2 1\n', '4\n5 10 3 2\n2 3 2 4\n'] Demo Output: ['2.000000000000\n', '1.400000000000\n'] Note: In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.

```python n = int(input()) positions = list(map(int, input().split())) speeds = list(map(int, input().split())) upperBound = 10 ** 6 lowerBound = 0 time = (upperBound + lowerBound) / 2 while upperBound - lowerBound > 10 ** -7: time = (upperBound + lowerBound) / 2 maxL = float("-inf") minR = float("inf") for friend in range(len(positions)): pos = positions[friend] speed = speeds[friend] L = pos - speed * time R = pos + speed * time if L > maxL: maxL = L if R < minR: minR = R if maxL <= minR: upperBound = time else: lowerBound = time print(round(time, 6)) ```

0

723

A

The New Year: Meeting Friends

PROGRAMMING

800

[ "implementation", "math", "sortings" ]

null

There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer.

The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.

Print one integer — the minimum total distance the friends need to travel in order to meet together.

[ "7 1 4\n", "30 20 10\n" ]

[ "6\n", "20\n" ]

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

500

[ { "input": "7 1 4", "output": "6" }, { "input": "30 20 10", "output": "20" }, { "input": "1 4 100", "output": "99" }, { "input": "100 1 91", "output": "99" }, { "input": "1 45 100", "output": "99" }, { "input": "1 2 3", "output": "2" }, { "input": "71 85 88", "output": "17" }, { "input": "30 38 99", "output": "69" }, { "input": "23 82 95", "output": "72" }, { "input": "22 41 47", "output": "25" }, { "input": "9 94 77", "output": "85" }, { "input": "1 53 51", "output": "52" }, { "input": "25 97 93", "output": "72" }, { "input": "42 53 51", "output": "11" }, { "input": "81 96 94", "output": "15" }, { "input": "21 5 93", "output": "88" }, { "input": "50 13 75", "output": "62" }, { "input": "41 28 98", "output": "70" }, { "input": "69 46 82", "output": "36" }, { "input": "87 28 89", "output": "61" }, { "input": "44 45 40", "output": "5" }, { "input": "86 97 68", "output": "29" }, { "input": "43 92 30", "output": "62" }, { "input": "16 70 1", "output": "69" }, { "input": "40 46 19", "output": "27" }, { "input": "71 38 56", "output": "33" }, { "input": "82 21 80", "output": "61" }, { "input": "75 8 35", "output": "67" }, { "input": "75 24 28", "output": "51" }, { "input": "78 23 56", "output": "55" }, { "input": "85 31 10", "output": "75" }, { "input": "76 50 9", "output": "67" }, { "input": "95 37 34", "output": "61" }, { "input": "84 61 35", "output": "49" }, { "input": "87 85 37", "output": "50" }, { "input": "1 3 2", "output": "2" }, { "input": "4 2 6", "output": "4" }, { "input": "6 9 3", "output": "6" }, { "input": "12 4 8", "output": "8" }, { "input": "15 10 5", "output": "10" }, { "input": "1 50 17", "output": "49" }, { "input": "10 5 15", "output": "10" }, { "input": "8 1 9", "output": "8" }, { "input": "3 5 4", "output": "2" }, { "input": "2 1 3", "output": "2" }, { "input": "1 8 2", "output": "7" }, { "input": "1 100 2", "output": "99" }, { "input": "1 4 6", "output": "5" } ]

1,682,418,331

2,147,483,647

PyPy 3-64

OK

TESTS

48

93

0

a,b,c=map(int, input().split()) x=a+b+c-max(a,b,c)-min(a,b,c) print(abs(a-x)+abs(b-x)+abs(c-x))

Title: The New Year: Meeting Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer. Input Specification: The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. Output Specification: Print one integer — the minimum total distance the friends need to travel in order to meet together. Demo Input: ['7 1 4\n', '30 20 10\n'] Demo Output: ['6\n', '20\n'] Note: In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

```python a,b,c=map(int, input().split()) x=a+b+c-max(a,b,c)-min(a,b,c) print(abs(a-x)+abs(b-x)+abs(c-x)) ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,663,667,028

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

80

92

0

n=int(input()) b=0 for i in range(n): a=input().split() for m in a: b+=int(m) if b==0: print('YES') else: print('NO')

Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none

```python n=int(input()) b=0 for i in range(n): a=input().split() for m in a: b+=int(m) if b==0: print('YES') else: print('NO') ```

0

236

A

Boy or Girl

PROGRAMMING

800

[ "brute force", "implementation", "strings" ]

null

Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network. But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names. This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.

The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.

If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).

[ "wjmzbmr\n", "xiaodao\n", "sevenkplus\n" ]

[ "CHAT WITH HER!\n", "IGNORE HIM!\n", "CHAT WITH HER!\n" ]

For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".

500

[ { "input": "wjmzbmr", "output": "CHAT WITH HER!" }, { "input": "xiaodao", "output": "IGNORE HIM!" }, { "input": "sevenkplus", "output": "CHAT WITH HER!" }, { "input": "pezu", "output": "CHAT WITH HER!" }, { "input": "wnemlgppy", "output": "CHAT WITH HER!" }, { "input": "zcinitufxoldnokacdvtmdohsfdjepyfioyvclhmujiqwvmudbfjzxjfqqxjmoiyxrfsbvseawwoyynn", "output": "IGNORE HIM!" }, { "input": "qsxxuoynwtebujwpxwpajitiwxaxwgbcylxneqiebzfphugwkftpaikixmumkhfbjiswmvzbtiyifbx", "output": "CHAT WITH HER!" }, { "input": "qwbdfzfylckctudyjlyrtmvbidfatdoqfmrfshsqqmhzohhsczscvwzpwyoyswhktjlykumhvaounpzwpxcspxwlgt", "output": "IGNORE HIM!" }, { "input": "nuezoadauueermoeaabjrkxttkatspjsjegjcjcdmcxgodowzbwuqncfbeqlhkk", "output": "IGNORE HIM!" }, { "input": "lggvdmulrsvtuagoavstuyufhypdxfomjlzpnduulukszqnnwfvxbvxyzmleocmofwclmzz", "output": "IGNORE HIM!" }, { "input": "tgcdptnkc", "output": "IGNORE HIM!" }, { "input": "wvfgnfrzabgibzxhzsojskmnlmrokydjoexnvi", "output": "IGNORE HIM!" }, { "input": "sxtburpzskucowowebgrbovhadrrayamuwypmmxhscrujkmcgvyinp", "output": "IGNORE HIM!" }, { "input": "pjqxhvxkyeqqvyuujxhmbspatvrckhhkfloottuybjivkkhpyivcighxumavrxzxslfpggnwbtalmhysyfllznphzia", "output": "IGNORE HIM!" }, { "input": "fpellxwskyekoyvrfnuf", "output": "CHAT WITH HER!" }, { "input": "xninyvkuvakfbs", "output": "IGNORE HIM!" }, { "input": "vnxhrweyvhqufpfywdwftoyrfgrhxuamqhblkvdpxmgvphcbeeqbqssresjifwyzgfhurmamhkwupymuomak", "output": "CHAT WITH HER!" }, { "input": "kmsk", "output": "IGNORE HIM!" }, { "input": "lqonogasrkzhryjxppjyriyfxmdfubieglthyswz", "output": "CHAT WITH HER!" }, { "input": "ndormkufcrkxlihdhmcehzoimcfhqsmombnfjrlcalffq", "output": "CHAT WITH HER!" }, { "input": "zqzlnnuwcfufwujygtczfakhcpqbtxtejrbgoodychepzdphdahtxyfpmlrycyicqthsgm", "output": "IGNORE HIM!" }, { "input": "ppcpbnhwoizajrl", "output": "IGNORE HIM!" }, { "input": "sgubujztzwkzvztitssxxxwzanfmddfqvv", "output": "CHAT WITH HER!" }, { "input": "ptkyaxycecpbrjnvxcjtbqiocqcswnmicxbvhdsptbxyxswbw", "output": "IGNORE HIM!" }, { "input": "yhbtzfppwcycxqjpqdfmjnhwaogyuaxamwxpnrdrnqsgdyfvxu", "output": "CHAT WITH HER!" }, { "input": "ojjvpnkrxibyevxk", "output": "CHAT WITH HER!" }, { "input": "wjweqcrqfuollfvfbiyriijovweg", "output": "IGNORE HIM!" }, { "input": "hkdbykboclchfdsuovvpknwqr", "output": "IGNORE HIM!" }, { "input": "stjvyfrfowopwfjdveduedqylerqugykyu", "output": "IGNORE HIM!" }, { "input": "rafcaanqytfclvfdegak", "output": "CHAT WITH HER!" }, { "input": "xczn", "output": "CHAT WITH HER!" }, { "input": "arcoaeozyeawbveoxpmafxxzdjldsielp", "output": "IGNORE HIM!" }, { "input": "smdfafbyehdylhaleevhoggiurdgeleaxkeqdixyfztkuqsculgslheqfafxyghyuibdgiuwrdxfcitojxika", "output": "CHAT WITH HER!" }, { "input": "vbpfgjqnhfazmvtkpjrdasfhsuxnpiepxfrzvoh", "output": "CHAT WITH HER!" }, { "input": "dbdokywnpqnotfrhdbrzmuyoxfdtrgrzcccninbtmoqvxfatcqg", "output": "CHAT WITH HER!" }, { "input": "udlpagtpq", "output": "CHAT WITH HER!" }, { "input": "zjurevbytijifnpfuyswfchdzelxheboruwjqijxcucylysmwtiqsqqhktexcynquvcwhbjsipy", "output": "CHAT WITH HER!" }, { "input": "qagzrqjomdwhagkhrjahhxkieijyten", "output": "CHAT WITH HER!" }, { "input": "achhcfjnnfwgoufxamcqrsontgjjhgyfzuhklkmiwybnrlsvblnsrjqdytglipxsulpnphpjpoewvlusalsgovwnsngb", "output": "CHAT WITH HER!" }, { "input": "qbkjsdwpahdbbohggbclfcufqelnojoehsxxkr", "output": "CHAT WITH HER!" }, { "input": "cpvftiwgyvnlmbkadiafddpgfpvhqqvuehkypqjsoibpiudfvpkhzlfrykc", "output": "IGNORE HIM!" }, { "input": "lnpdosnceumubvk", "output": "IGNORE HIM!" }, { "input": "efrk", "output": "CHAT WITH HER!" }, { "input": "temnownneghnrujforif", "output": "IGNORE HIM!" }, { "input": "ottnneymszwbumgobazfjyxewkjakglbfflsajuzescplpcxqta", "output": "IGNORE HIM!" }, { "input": "eswpaclodzcwhgixhpyzvhdwsgneqidanbzdzszquefh", "output": "IGNORE HIM!" }, { "input": "gwntwbpj", "output": "IGNORE HIM!" }, { "input": "wuqvlbblkddeindiiswsinkfrnkxghhwunzmmvyovpqapdfbolyim", "output": "IGNORE HIM!" }, { "input": "swdqsnzmzmsyvktukaoyqsqzgfmbzhezbfaqeywgwizrwjyzquaahucjchegknqaioliqd", "output": "CHAT WITH HER!" }, { "input": "vlhrpzezawyolhbmvxbwhtjustdbqggexmzxyieihjlelvwjosmkwesfjmramsikhkupzvfgezmrqzudjcalpjacmhykhgfhrjx", "output": "IGNORE HIM!" }, { "input": "lxxwbkrjgnqjwsnflfnsdyxihmlspgivirazsbveztnkuzpaxtygidniflyjheejelnjyjvgkgvdqks", "output": "CHAT WITH HER!" }, { "input": "wpxbxzfhtdecetpljcrvpjjnllosdqirnkzesiqeukbedkayqx", "output": "CHAT WITH HER!" }, { "input": "vmzxgacicvweclaodrunmjnfwtimceetsaoickarqyrkdghcmyjgmtgsqastcktyrjgvjqimdc", "output": "CHAT WITH HER!" }, { "input": "yzlzmesxdttfcztooypjztlgxwcr", "output": "IGNORE HIM!" }, { "input": "qpbjwzwgdzmeluheirjrvzrhbmagfsjdgvzgwumjtjzecsfkrfqjasssrhhtgdqqfydlmrktlgfc", "output": "IGNORE HIM!" }, { "input": "aqzftsvezdgouyrirsxpbuvdjupnzvbhguyayeqozfzymfnepvwgblqzvmxxkxcilmsjvcgyqykpoaktjvsxbygfgsalbjoq", "output": "CHAT WITH HER!" }, { "input": "znicjjgijhrbdlnwmtjgtdgziollrfxroabfhadygnomodaembllreorlyhnehijfyjbfxucazellblegyfrzuraogadj", "output": "IGNORE HIM!" }, { "input": "qordzrdiknsympdrkgapjxokbldorpnmnpucmwakklmqenpmkom", "output": "CHAT WITH HER!" }, { "input": "wqfldgihuxfktzanyycluzhtewmwvnawqlfoavuguhygqrrxtstxwouuzzsryjqtfqo", "output": "CHAT WITH HER!" }, { "input": "vujtrrpshinkskgyknlcfckmqdrwtklkzlyipmetjvaqxdsslkskschbalmdhzsdrrjmxdltbtnxbh", "output": "IGNORE HIM!" }, { "input": "zioixjibuhrzyrbzqcdjbbhhdmpgmqykixcxoqupggaqajuzonrpzihbsogjfsrrypbiphehonyhohsbybnnukqebopppa", "output": "CHAT WITH HER!" }, { "input": "oh", "output": "CHAT WITH HER!" }, { "input": "kxqthadqesbpgpsvpbcbznxpecqrzjoilpauttzlnxvaczcqwuri", "output": "IGNORE HIM!" }, { "input": "zwlunigqnhrwirkvufqwrnwcnkqqonebrwzcshcbqqwkjxhymjjeakuzjettebciadjlkbfp", "output": "CHAT WITH HER!" }, { "input": "fjuldpuejgmggvvigkwdyzytfxzwdlofrpifqpdnhfyroginqaufwgjcbgshyyruwhofctsdaisqpjxqjmtpp", "output": "CHAT WITH HER!" }, { "input": "xiwntnheuitbtqxrmzvxmieldudakogealwrpygbxsbluhsqhtwmdlpjwzyafckrqrdduonkgo", "output": "CHAT WITH HER!" }, { "input": "mnmbupgo", "output": "IGNORE HIM!" }, { "input": "mcjehdiygkbmrbfjqwpwxidbdfelifwhstaxdapigbymmsgrhnzsdjhsqchl", "output": "IGNORE HIM!" }, { "input": "yocxrzspinchmhtmqo", "output": "CHAT WITH HER!" }, { "input": "vasvvnpymtgjirnzuynluluvmgpquskuaafwogeztfnvybblajvuuvfomtifeuzpikjrolzeeoftv", "output": "CHAT WITH HER!" }, { "input": "ecsdicrznvglwggrdbrvehwzaenzjutjydhvimtqegweurpxtjkmpcznshtrvotkvrghxhacjkedidqqzrduzad", "output": "IGNORE HIM!" }, { "input": "ubvhyaebyxoghakajqrpqpctwbrfqzli", "output": "CHAT WITH HER!" }, { "input": "gogbxfeqylxoummvgxpkoqzsmobasesxbqjjktqbwqxeiaagnnhbvepbpy", "output": "IGNORE HIM!" }, { "input": "nheihhxkbbrmlpxpxbhnpofcjmxemyvqqdbanwd", "output": "IGNORE HIM!" }, { "input": "acrzbavz", "output": "CHAT WITH HER!" }, { "input": "drvzznznvrzskftnrhvvzxcalwutxmdza", "output": "IGNORE HIM!" }, { "input": "oacwxipdfcoabhkwxqdbtowiekpnflnqhlrkustgzryvws", "output": "CHAT WITH HER!" }, { "input": "tpnwfmfsibnccvdwjvzviyvjfljupinfigfunyff", "output": "CHAT WITH HER!" }, { "input": "gavaihhamfolcndgytcsgucqdqngxkrlovpthvteacmmthoglxu", "output": "CHAT WITH HER!" }, { "input": "hsfcfvameeupldgvchmogrvwxrvsmnwxxkxoawwodtsahqvehlcw", "output": "IGNORE HIM!" }, { "input": "sbkydrscoojychxchqsuciperfroumenelgiyiwlqfwximrgdbyvkmacy", "output": "CHAT WITH HER!" }, { "input": "rhh", "output": "CHAT WITH HER!" }, { "input": "zhdouqfmlkenjzdijxdfxnlegxeuvhelo", "output": "IGNORE HIM!" }, { "input": "yufkkfwyhhvcjntsgsvpzbhqtmtgyxifqoewmuplphykmptfdebjxuaxigomjtwgtljwdjhjernkitifbomifbhysnmadtnyn", "output": "CHAT WITH HER!" }, { "input": "urigreuzpxnej", "output": "CHAT WITH HER!" } ]

1,696,416,214

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

30

0

#蒋世刚2300016304 a=len(input()) if a%2!=0: print('IGNORE HIM!') else: print('CHAT WITH HER!')

Title: Boy or Girl Time Limit: None seconds Memory Limit: None megabytes Problem Description: Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network. But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names. This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method. Input Specification: The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters. Output Specification: If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes). Demo Input: ['wjmzbmr\n', 'xiaodao\n', 'sevenkplus\n'] Demo Output: ['CHAT WITH HER!\n', 'IGNORE HIM!\n', 'CHAT WITH HER!\n'] Note: For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".

```python #蒋世刚2300016304 a=len(input()) if a%2!=0: print('IGNORE HIM!') else: print('CHAT WITH HER!') ```

0

779

C

Dishonest Sellers

PROGRAMMING

1,200

[ "constructive algorithms", "greedy", "sortings" ]

null

Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*. Not all of sellers are honest, so now some products could be more expensive than after a week of discounts. Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.

In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now. The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now). The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).

Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.

[ "3 1\n5 4 6\n3 1 5\n", "5 3\n3 4 7 10 3\n4 5 5 12 5\n" ]

[ "10\n", "25\n" ]

In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10. In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.

1,000

[ { "input": "3 1\n5 4 6\n3 1 5", "output": "10" }, { "input": "5 3\n3 4 7 10 3\n4 5 5 12 5", "output": "25" }, { "input": "1 0\n9\n8", "output": "8" }, { "input": "2 0\n4 10\n1 2", "output": "3" }, { "input": "4 2\n19 5 17 13\n3 18 8 10", "output": "29" }, { "input": "5 3\n28 17 20 45 45\n39 12 41 27 9", "output": "101" }, { "input": "10 5\n87 96 19 81 10 88 7 49 36 21\n11 75 28 28 74 17 64 19 81 31", "output": "243" }, { "input": "50 45\n80 125 152 122 85 62 123 195 157 73 8 127 47 193 70 132 131 38 98 97 185 98 68 166 114 53 144 106 141 106 191 123 44 199 35 18 84 169 46 52 129 102 155 130 63 169 116 145 154 41\n155 187 79 106 136 173 17 146 9 47 173 21 37 183 110 74 185 14 82 156 147 25 162 195 26 25 22 80 171 180 7 92 53 18 100 17 159 184 37 127 116 192 163 49 94 95 118 102 8 30", "output": "4680" }, { "input": "5 2\n1 1 1 1 1\n2 2 2 2 2", "output": "5" }, { "input": "5 2\n2 4 4 4 4\n3 1 1 1 1", "output": "9" }, { "input": "6 3\n9 9 9 2 2 2\n6 6 6 1 1 1", "output": "24" }, { "input": "5 3\n10 17 23 26 35\n7 18 18 29 32", "output": "103" }, { "input": "2 1\n10 12\n5 6", "output": "16" } ]

1,661,753,837

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

7

233

16,179,200

n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] b = [int(i) for i in input().split()] q = sorted(zip(a, b), key = lambda x : x[0]-x[1]) ans, c = 0, 0 for i in range(n): if q[i][0]-q[i][1] <=0 : c+=1; ans += q[i][0] else : break #print(q) q = q[i:] for i in range(len(q)): if i < (max(0, k-c)) : ans += q[i][0] else : ans += q[i][1] #print(q) print(ans)

Title: Dishonest Sellers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*. Not all of sellers are honest, so now some products could be more expensive than after a week of discounts. Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. Input Specification: In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now. The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now). The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). Output Specification: Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. Demo Input: ['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n'] Demo Output: ['10\n', '25\n'] Note: In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10. In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.

```python n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] b = [int(i) for i in input().split()] q = sorted(zip(a, b), key = lambda x : x[0]-x[1]) ans, c = 0, 0 for i in range(n): if q[i][0]-q[i][1] <=0 : c+=1; ans += q[i][0] else : break #print(q) q = q[i:] for i in range(len(q)): if i < (max(0, k-c)) : ans += q[i][0] else : ans += q[i][1] #print(q) print(ans) ```

0