MatrixStudio/Codeforces-Python-Submissions

500

A

New Year Transportation

PROGRAMMING

1,000

[ "dfs and similar", "graphs", "implementation" ]

null

New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells. So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals. Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.

The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to. The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.

If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".

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

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

In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4. In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.

500

[ { "input": "8 4\n1 2 1 2 1 2 1", "output": "YES" }, { "input": "8 5\n1 2 1 2 1 1 1", "output": "NO" }, { "input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1", "output": "YES" }, { "input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1", "output": "YES" }, { "input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1", "output": "NO" }, { "input": "10 3\n8 3 5 4 2 3 2 2 1", "output": "NO" }, { "input": "10 9\n8 3 5 4 2 3 2 2 1", "output": "YES" }, { "input": "3 2\n1 1", "output": "YES" }, { "input": "3 2\n2 1", "output": "NO" }, { "input": "4 2\n2 1 1", "output": "NO" }, { "input": "4 4\n2 2 1", "output": "YES" }, { "input": "8 8\n1 2 1 2 1 2 1", "output": "YES" }, { "input": "3 3\n1 1", "output": "YES" }, { "input": "8 8\n1 2 1 2 1 1 1", "output": "YES" }, { "input": "3 3\n2 1", "output": "YES" }, { "input": "4 4\n1 1 1", "output": "YES" }, { "input": "8 8\n1 1 1 1 1 1 1", "output": "YES" }, { "input": "5 5\n1 1 1 1", "output": "YES" } ]

1,661,294,528

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

12

92

3,686,400

i=input().split() n=int(i[0]) t=int(i[1]) i=input().split() ai=list(map(lambda x:int(x),i)) n-=2 i=0 fl=0 while(i<=n): if i==t-1: fl=1 break i=i+ai[i] if fl==1: print('YES') else: print('NO')

Title: New Year Transportation Time Limit: None seconds Memory Limit: None megabytes Problem Description: New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells. So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals. Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. Input Specification: The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to. The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. Output Specification: If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". Demo Input: ['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4. In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.

```python i=input().split() n=int(i[0]) t=int(i[1]) i=input().split() ai=list(map(lambda x:int(x),i)) n-=2 i=0 fl=0 while(i<=n): if i==t-1: fl=1 break i=i+ai[i] if fl==1: print('YES') else: print('NO') ```

0

22

A

Second Order Statistics

PROGRAMMING

800

[ "brute force" ]

A. Second Order Statistics

2

256

Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.

The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.

If the given sequence has the second order statistics, output this order statistics, otherwise output NO.

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

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

none

0

[ { "input": "4\n1 2 2 -4", "output": "1" }, { "input": "5\n1 2 3 1 1", "output": "2" }, { "input": "1\n28", "output": "NO" }, { "input": "2\n-28 12", "output": "12" }, { "input": "3\n-83 40 -80", "output": "-80" }, { "input": "8\n93 77 -92 26 21 -48 53 91", "output": "-48" }, { "input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49", "output": "-86" }, { "input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94", "output": "-98" }, { "input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46", "output": "-91" }, { "input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89", "output": "-98" }, { "input": "2\n-1 -1", "output": "NO" }, { "input": "3\n-2 -2 -2", "output": "NO" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "NO" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100", "output": "100" }, { "input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47", "output": "40" }, { "input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90", "output": "-64" }, { "input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93", "output": "-93" }, { "input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83", "output": "-86" }, { "input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43", "output": "-95" }, { "input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92", "output": "-92" }, { "input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12", "output": "-98" }, { "input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100", "output": "-98" }, { "input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92", "output": "-98" }, { "input": "3\n1 2 3", "output": "2" }, { "input": "3\n1 3 2", "output": "2" }, { "input": "3\n2 1 3", "output": "2" }, { "input": "3\n2 3 1", "output": "2" }, { "input": "3\n3 1 2", "output": "2" }, { "input": "3\n3 2 1", "output": "2" }, { "input": "9\n99 99 99 99 100 100 100 100 100", "output": "100" }, { "input": "5\n-100 -100 -100 -100 -100", "output": "NO" } ]

1,656,518,205

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

2

62

0

n=int(input()) m=input().split(" ") min=m[0] my_list=[] for i in m: if i not in my_list: my_list.append(i) for count in range(len(my_list)): min=my_list[count] loc=count for c in range(count+1,len(my_list)): if my_list[c]<min: min=my_list[c] loc=c temp=my_list[count] my_list[count]=my_list[loc] my_list[loc]=temp print(my_list[1])

Title: Second Order Statistics Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. Output Specification: If the given sequence has the second order statistics, output this order statistics, otherwise output NO. Demo Input: ['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n'] Demo Output: ['1\n', '2\n'] Note: none

```python n=int(input()) m=input().split(" ") min=m[0] my_list=[] for i in m: if i not in my_list: my_list.append(i) for count in range(len(my_list)): min=my_list[count] loc=count for c in range(count+1,len(my_list)): if my_list[c]<min: min=my_list[c] loc=c temp=my_list[count] my_list[count]=my_list[loc] my_list[loc]=temp print(my_list[1]) ```

-1

353

A

Domino

PROGRAMMING

1,200

[ "implementation", "math" ]

null

Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even. To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.

The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.

Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.

[ "2\n4 2\n6 4\n", "1\n2 3\n", "3\n1 4\n2 3\n4 4\n" ]

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

In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything. In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd. In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.

500

[ { "input": "2\n4 2\n6 4", "output": "0" }, { "input": "1\n2 3", "output": "-1" }, { "input": "3\n1 4\n2 3\n4 4", "output": "1" }, { "input": "5\n5 4\n5 4\n1 5\n5 5\n3 3", "output": "1" }, { "input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n6 2\n4 3\n2 1\n6 2\n6 5\n4 5\n2 4\n1 4", "output": "-1" }, { "input": "100\n2 3\n2 4\n3 3\n1 4\n5 2\n5 4\n6 6\n3 4\n1 1\n4 2\n5 1\n5 5\n5 3\n3 6\n4 1\n1 6\n1 1\n3 2\n4 5\n6 1\n6 4\n1 1\n3 4\n3 3\n2 2\n1 1\n4 4\n6 4\n3 2\n5 2\n6 4\n3 2\n3 5\n4 4\n1 4\n5 2\n3 4\n1 4\n2 2\n5 6\n3 5\n6 1\n5 5\n1 6\n6 3\n1 4\n1 5\n5 5\n4 1\n3 2\n4 1\n5 5\n5 5\n1 5\n1 2\n6 4\n1 3\n3 6\n4 3\n3 5\n6 4\n2 6\n5 5\n1 4\n2 2\n2 3\n5 1\n2 5\n1 2\n2 6\n5 5\n4 6\n1 4\n3 6\n2 3\n6 1\n6 5\n3 2\n6 4\n4 5\n4 5\n2 6\n1 3\n6 2\n1 2\n2 3\n4 3\n5 4\n3 4\n1 6\n6 6\n2 4\n4 1\n3 1\n2 6\n5 4\n1 2\n6 5\n3 6\n2 4", "output": "-1" }, { "input": "1\n2 4", "output": "0" }, { "input": "1\n1 1", "output": "-1" }, { "input": "1\n1 2", "output": "-1" }, { "input": "2\n1 1\n3 3", "output": "0" }, { "input": "2\n1 1\n2 2", "output": "-1" }, { "input": "2\n1 1\n1 2", "output": "-1" }, { "input": "5\n1 2\n6 6\n1 1\n3 3\n6 1", "output": "1" }, { "input": "5\n5 4\n2 6\n6 2\n1 4\n6 2", "output": "0" }, { "input": "10\n4 1\n3 2\n1 2\n2 6\n3 5\n2 1\n5 2\n4 6\n5 6\n3 1", "output": "0" }, { "input": "10\n6 1\n4 4\n2 6\n6 5\n3 6\n6 3\n2 4\n5 1\n1 6\n1 5", "output": "-1" }, { "input": "15\n1 2\n5 1\n6 4\n5 1\n1 6\n2 6\n3 1\n6 4\n3 1\n2 1\n6 4\n3 5\n6 2\n1 6\n1 1", "output": "1" }, { "input": "15\n3 3\n2 1\n5 4\n3 3\n5 3\n5 4\n2 5\n1 3\n3 2\n3 3\n3 5\n2 5\n4 1\n2 3\n5 4", "output": "-1" }, { "input": "20\n1 5\n6 4\n4 3\n6 2\n1 1\n1 5\n6 3\n2 3\n3 6\n3 6\n3 6\n2 5\n4 3\n4 6\n5 5\n4 6\n3 4\n4 2\n3 3\n5 2", "output": "0" }, { "input": "20\n2 1\n6 5\n3 1\n2 5\n3 5\n4 1\n1 1\n5 4\n5 1\n2 4\n1 5\n3 2\n1 2\n3 5\n5 2\n1 2\n1 3\n4 2\n2 3\n4 5", "output": "-1" }, { "input": "25\n4 1\n6 3\n1 3\n2 3\n2 4\n6 6\n4 2\n4 2\n1 5\n5 4\n1 2\n2 5\n3 6\n4 1\n3 4\n2 6\n6 1\n5 6\n6 6\n4 2\n1 5\n3 3\n3 3\n6 5\n1 4", "output": "-1" }, { "input": "25\n5 5\n4 3\n2 5\n4 3\n4 6\n4 2\n5 6\n2 1\n5 4\n6 6\n1 3\n1 4\n2 3\n5 6\n5 4\n5 6\n5 4\n6 3\n3 5\n1 3\n2 5\n2 2\n4 4\n2 1\n4 4", "output": "-1" }, { "input": "30\n3 5\n2 5\n1 6\n1 6\n2 4\n5 5\n5 4\n5 6\n5 4\n2 1\n2 4\n1 6\n3 5\n1 1\n3 6\n5 5\n1 6\n3 4\n1 4\n4 6\n2 1\n3 3\n1 3\n4 5\n1 4\n1 6\n2 1\n4 6\n3 5\n5 6", "output": "1" }, { "input": "30\n2 3\n3 1\n6 6\n1 3\n5 5\n3 6\n4 5\n2 1\n1 3\n2 3\n4 4\n2 4\n6 4\n2 4\n5 4\n2 1\n2 5\n2 5\n4 2\n1 4\n2 6\n3 2\n3 2\n6 6\n4 2\n3 4\n6 3\n6 6\n6 6\n5 5", "output": "1" }, { "input": "35\n6 1\n4 3\n1 2\n4 3\n6 4\n4 6\n3 1\n5 5\n3 4\n5 4\n4 6\n1 6\n2 4\n6 6\n5 4\n5 2\n1 3\n1 4\n3 5\n1 4\n2 3\n4 5\n4 3\n6 1\n5 3\n3 2\n5 6\n3 5\n6 5\n4 1\n1 3\n5 5\n4 6\n6 1\n1 3", "output": "1" }, { "input": "35\n4 3\n5 6\n4 5\n2 5\n6 6\n4 1\n2 2\n4 2\n3 4\n4 1\n6 6\n6 3\n1 5\n1 5\n5 6\n4 2\n4 6\n5 5\n2 2\n5 2\n1 2\n4 6\n6 6\n6 5\n2 1\n3 5\n2 5\n3 1\n5 3\n6 4\n4 6\n5 6\n5 1\n3 4\n3 5", "output": "1" }, { "input": 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2", "output": "-1" }, { "input": "85\n6 3\n4 1\n1 2\n3 5\n6 4\n6 2\n2 6\n1 2\n1 5\n6 2\n1 4\n6 6\n2 4\n4 6\n4 5\n1 6\n3 1\n2 5\n5 1\n5 2\n3 5\n1 1\n4 1\n2 3\n1 1\n3 3\n6 4\n1 4\n1 1\n3 6\n1 5\n1 6\n2 5\n2 2\n5 1\n6 6\n1 3\n1 5\n5 6\n4 5\n4 3\n5 5\n1 3\n6 3\n4 6\n2 4\n5 6\n6 2\n4 5\n1 4\n1 4\n6 5\n1 6\n6 1\n1 6\n5 5\n2 1\n5 2\n2 3\n1 6\n1 6\n1 6\n5 6\n2 4\n6 5\n6 5\n4 2\n5 4\n3 4\n4 3\n6 6\n3 3\n3 2\n3 6\n2 5\n2 1\n2 5\n3 4\n1 2\n5 4\n6 2\n5 1\n1 4\n3 4\n4 5", "output": "0" }, { "input": "85\n3 1\n3 2\n6 3\n1 3\n2 1\n3 6\n1 4\n2 5\n6 5\n1 6\n1 5\n1 1\n4 3\n3 5\n4 6\n3 2\n6 6\n4 4\n4 1\n5 5\n4 2\n6 2\n2 2\n4 5\n6 1\n3 4\n4 5\n3 5\n4 2\n3 5\n4 4\n3 1\n4 4\n6 4\n1 4\n5 5\n1 5\n2 2\n6 5\n5 6\n6 5\n3 2\n3 2\n6 1\n6 5\n2 1\n4 6\n2 1\n3 1\n5 6\n1 3\n5 4\n1 4\n1 4\n5 3\n2 3\n1 3\n2 2\n5 3\n2 3\n2 3\n1 3\n3 6\n4 4\n6 6\n6 2\n5 1\n5 5\n5 5\n1 2\n1 4\n2 4\n3 6\n4 6\n6 3\n6 4\n5 5\n3 2\n5 4\n5 4\n4 5\n6 4\n2 1\n5 2\n5 1", "output": "-1" }, { "input": "90\n5 2\n5 5\n5 1\n4 6\n4 3\n5 3\n5 6\n5 1\n3 4\n1 3\n4 2\n1 6\n6 4\n1 2\n6 1\n4 1\n6 2\n6 5\n6 2\n5 4\n3 6\n1 1\n5 5\n2 2\n1 6\n3 5\n6 5\n1 6\n1 5\n2 3\n2 6\n2 3\n3 3\n1 3\n5 1\n2 5\n3 6\n1 2\n4 4\n1 6\n2 3\n1 5\n2 5\n1 3\n2 2\n4 6\n3 6\n6 3\n1 2\n4 3\n4 5\n4 6\n3 2\n6 5\n6 2\n2 5\n2 4\n1 3\n1 6\n4 3\n1 3\n6 4\n4 6\n4 1\n1 1\n4 1\n4 4\n6 2\n6 5\n1 1\n2 2\n3 1\n1 4\n6 2\n5 2\n1 4\n1 3\n6 5\n3 2\n6 4\n3 4\n2 6\n2 2\n6 3\n4 6\n1 2\n4 2\n3 4\n2 3\n1 5", "output": "-1" }, { "input": "90\n1 4\n3 5\n4 2\n2 5\n4 3\n2 6\n2 6\n3 2\n4 4\n6 1\n4 3\n2 3\n5 3\n6 6\n2 2\n6 3\n4 1\n4 4\n5 6\n6 4\n4 2\n5 6\n4 6\n4 4\n6 4\n4 1\n5 3\n3 2\n4 4\n5 2\n5 4\n6 4\n1 2\n3 3\n3 4\n6 4\n1 6\n4 2\n3 2\n1 1\n2 2\n5 1\n6 6\n4 1\n5 2\n3 6\n2 1\n2 2\n4 6\n6 5\n4 4\n5 5\n5 6\n1 6\n1 4\n5 6\n3 6\n6 3\n5 6\n6 5\n5 1\n6 1\n6 6\n6 3\n1 5\n4 5\n3 1\n6 6\n3 4\n6 2\n1 4\n2 2\n3 2\n5 6\n2 4\n1 4\n6 3\n4 6\n1 4\n5 2\n1 2\n6 5\n1 5\n1 4\n4 2\n2 5\n3 2\n5 1\n5 4\n5 3", "output": "-1" }, { "input": "95\n4 3\n3 2\n5 5\n5 3\n1 6\n4 4\n5 5\n6 5\n3 5\n1 5\n4 2\n5 1\n1 2\n2 3\n6 4\n2 3\n6 3\n6 5\n5 6\n1 4\n2 6\n2 6\n2 5\n2 1\n3 1\n3 5\n2 2\n6 1\n2 4\n4 6\n6 6\n6 4\n3 2\n5 1\n4 3\n6 5\n2 3\n4 1\n2 5\n6 5\n6 5\n6 5\n5 1\n5 4\n4 6\n3 2\n2 5\n2 6\n4 6\n6 3\n6 4\n5 6\n4 6\n2 4\n3 4\n1 4\n2 4\n2 3\n5 6\n6 4\n3 1\n5 1\n3 6\n3 5\n2 6\n6 3\n4 3\n3 1\n6 1\n2 2\n6 3\n2 2\n2 2\n6 4\n6 1\n2 1\n5 6\n5 4\n5 2\n3 4\n3 6\n2 1\n1 6\n5 5\n2 6\n2 3\n3 6\n1 3\n1 5\n5 1\n1 2\n2 2\n5 3\n6 4\n4 5", "output": "0" }, { "input": "95\n4 5\n5 6\n3 2\n5 1\n4 3\n4 1\n6 1\n5 2\n2 4\n5 3\n2 3\n6 4\n4 1\n1 6\n2 6\n2 3\n4 6\n2 4\n3 4\n4 2\n5 5\n1 1\n1 5\n4 3\n4 5\n6 2\n6 1\n6 3\n5 5\n4 1\n5 1\n2 3\n5 1\n3 6\n6 6\n4 5\n4 4\n4 3\n1 6\n6 6\n4 6\n6 4\n1 2\n6 2\n4 6\n6 6\n5 5\n6 1\n5 2\n4 5\n6 6\n6 5\n4 4\n1 5\n4 6\n4 1\n3 6\n5 1\n3 1\n4 6\n4 5\n1 3\n5 4\n4 5\n2 2\n6 1\n5 2\n6 5\n2 2\n1 1\n6 3\n6 1\n2 6\n3 3\n2 1\n4 6\n2 4\n5 5\n5 2\n3 2\n1 2\n6 6\n6 2\n5 1\n2 6\n5 2\n2 2\n5 5\n3 5\n3 3\n2 6\n5 3\n4 3\n1 6\n5 4", "output": "-1" }, { "input": "100\n1 1\n3 5\n2 1\n1 2\n3 4\n5 6\n5 6\n6 1\n5 5\n2 4\n5 5\n5 6\n6 2\n6 6\n2 6\n1 4\n2 2\n3 2\n1 3\n5 5\n6 3\n5 6\n1 1\n1 2\n1 2\n2 1\n2 3\n1 6\n4 3\n1 1\n2 5\n2 4\n4 4\n1 5\n3 3\n6 1\n3 5\n1 1\n3 6\n3 1\n4 2\n4 3\n3 6\n6 6\n1 6\n6 2\n2 5\n5 4\n6 3\n1 4\n2 6\n6 2\n3 4\n6 1\n6 5\n4 6\n6 5\n4 4\n3 1\n6 3\n5 1\n2 4\n5 1\n1 2\n2 4\n2 1\n6 6\n5 3\n4 6\n6 3\n5 5\n3 3\n1 1\n6 5\n4 3\n2 6\n1 5\n3 5\n2 4\n4 5\n1 6\n2 3\n6 3\n5 5\n2 6\n2 6\n3 4\n3 2\n6 1\n3 4\n6 4\n3 3\n2 3\n5 1\n3 1\n6 2\n2 3\n6 4\n1 4\n1 2", "output": "-1" }, { "input": "100\n1 1\n5 5\n1 2\n5 3\n5 5\n2 2\n1 5\n3 4\n3 2\n1 3\n5 6\n4 5\n2 1\n5 5\n2 2\n1 6\n6 1\n5 1\n4 1\n4 6\n3 5\n6 1\n2 3\n5 6\n3 6\n2 3\n5 6\n1 6\n3 2\n2 2\n3 3\n6 5\n5 5\n1 4\n5 6\n6 4\n1 4\n1 2\n2 6\n3 2\n6 4\n5 3\n3 3\n6 4\n4 6\n2 2\n5 6\n5 1\n1 2\n3 4\n4 5\n1 1\n3 4\n5 2\n4 5\n3 3\n1 1\n3 4\n1 6\n2 4\n1 3\n3 2\n6 5\n1 6\n3 6\n2 3\n2 6\n5 1\n5 5\n5 6\n4 1\n6 2\n3 6\n5 3\n2 2\n2 4\n6 6\n3 6\n4 6\n2 5\n5 3\n1 2\n3 4\n3 4\n6 2\n2 4\n2 2\n4 6\n3 5\n4 2\n5 6\n4 2\n2 3\n6 2\n5 6\n2 1\n3 3\n6 6\n4 3\n4 2", "output": "1" }, { "input": "1\n2 2", "output": "0" }, { "input": "3\n2 4\n6 6\n3 3", "output": "-1" }, { "input": "2\n3 6\n4 1", "output": "1" }, { "input": "3\n1 1\n1 1\n3 3", "output": "-1" }, { "input": "3\n2 3\n1 1\n2 3", "output": "1" }, { "input": "3\n2 2\n2 1\n1 2", "output": "1" }, { "input": "3\n1 1\n1 1\n1 1", "output": "-1" } ]

1,685,441,200

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

7

62

0

t = int(input()) upper, lower = [], [] for i in range(t): u, d = map(int, input().split()) upper.append(u) lower.append(d) if sum(lower) % 2 != sum(upper) % 2: print(-1) exit() if sum(lower) % 2 == sum(upper) % 2 and sum(upper) % 2 == 0: print(0) exit() k = 0 for j in range(t): k += 1 if upper[j] % 2 == lower[j] % 2: continue if upper[j] % 2 == 1: if sum(upper) % 2 == 1: upper[j], lower[j] = lower[j], upper[j] else: if sum(lower) % 2 == 1: upper[j], lower[j] = lower[j], upper[j] if sum(lower) % 2 == sum(upper) % 2 and sum(upper) % 2 == 0: print(k) exit()

Title: Domino Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even. To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half. Output Specification: Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1. Demo Input: ['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n'] Demo Output: ['0\n', '-1\n', '1\n'] Note: In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything. In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd. In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.

```python t = int(input()) upper, lower = [], [] for i in range(t): u, d = map(int, input().split()) upper.append(u) lower.append(d) if sum(lower) % 2 != sum(upper) % 2: print(-1) exit() if sum(lower) % 2 == sum(upper) % 2 and sum(upper) % 2 == 0: print(0) exit() k = 0 for j in range(t): k += 1 if upper[j] % 2 == lower[j] % 2: continue if upper[j] % 2 == 1: if sum(upper) % 2 == 1: upper[j], lower[j] = lower[j], upper[j] else: if sum(lower) % 2 == 1: upper[j], lower[j] = lower[j], upper[j] if sum(lower) % 2 == sum(upper) % 2 and sum(upper) % 2 == 0: print(k) exit() ```

0

729

A

Interview with Oleg

PROGRAMMING

900

[ "implementation", "strings" ]

null

Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters. There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers. The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here. To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length. Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview. The second line contains the string *s* of length *n*, consisting of lowercase English letters.

Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.

[ "7\naogogob\n", "13\nogogmgogogogo\n", "9\nogoogoogo\n" ]

[ "a***b\n", "***gmg***\n", "*********\n" ]

The first sample contains one filler word ogogo, so the interview for printing is "a***b". The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".

500

[ { "input": "7\naogogob", "output": "a***b" }, { "input": "13\nogogmgogogogo", "output": "***gmg***" }, { "input": "9\nogoogoogo", "output": "*********" }, { "input": "32\nabcdefogoghijklmnogoopqrstuvwxyz", "output": "abcdef***ghijklmn***opqrstuvwxyz" }, { "input": "100\nggogogoooggogooggoggogggggogoogoggooooggooggoooggogoooggoggoogggoogoggogggoooggoggoggogggogoogggoooo", "output": "gg***oogg***oggoggoggggg******ggooooggooggooogg***ooggoggoogggo***ggogggoooggoggoggoggg***ogggoooo" }, { "input": "10\nogooggoggo", "output": "***oggoggo" }, { "input": "20\nooggooogooogooogooog", "output": "ooggoo***o***o***oog" }, { "input": "30\ngoggogoooggooggggoggoggoogoggo", "output": "gogg***ooggooggggoggoggo***ggo" }, { "input": "40\nogggogooggoogoogggogooogogggoogggooggooo", "output": "oggg***oggo***oggg***o***gggoogggooggooo" }, { "input": "50\noggggogoogggggggoogogggoooggooogoggogooogogggogooo", "output": "ogggg***ogggggggo***gggoooggoo***gg***o***ggg***oo" }, { "input": "60\nggoooogoggogooogogooggoogggggogogogggggogggogooogogogggogooo", "output": "ggooo***gg***o***oggooggggg***gggggoggg***o***ggg***oo" }, { "input": "70\ngogoooggggoggoggggggoggggoogooogogggggooogggogoogoogoggogggoggogoooooo", "output": "g***ooggggoggoggggggoggggo***o***gggggoooggg*********ggogggogg***ooooo" }, { "input": "80\nooogoggoooggogogoggooooogoogogooogoggggogggggogoogggooogooooooggoggoggoggogoooog", "output": "oo***ggooogg***ggoooo******o***ggggoggggg***ogggoo***oooooggoggoggogg***ooog" }, { "input": "90\nooogoggggooogoggggoooogggggooggoggoggooooooogggoggogggooggggoooooogoooogooggoooogggggooooo", "output": "oo***ggggoo***ggggoooogggggooggoggoggooooooogggoggogggooggggooooo***oo***oggoooogggggooooo" }, { "input": "100\ngooogoggooggggoggoggooooggogoogggoogogggoogogoggogogogoggogggggogggggoogggooogogoggoooggogoooooogogg", "output": "goo***ggooggggoggoggoooogg***ogggo***gggo***gg***ggogggggogggggoogggoo***ggooogg***oooo***gg" }, { "input": "100\ngoogoogggogoooooggoogooogoogoogogoooooogooogooggggoogoggogooogogogoogogooooggoggogoooogooooooggogogo", "output": "go***oggg***ooooggo***o*********oooo***o***oggggo***gg***o******oooggogg***oo***ooooogg***" }, { "input": "100\ngoogoggggogggoooggoogoogogooggoggooggggggogogggogogggoogogggoogoggoggogooogogoooogooggggogggogggoooo", "output": "go***ggggogggoooggo******oggoggoogggggg***ggg***gggo***gggo***ggogg***o***oo***oggggogggogggoooo" }, { "input": "100\nogogogogogoggogogogogogogoggogogogoogoggoggooggoggogoogoooogogoogggogogogogogoggogogogogogogogogogoe", "output": "***gg***gg******ggoggooggogg******oo***oggg***gg***e" }, { "input": "5\nogoga", "output": "***ga" }, { "input": "1\no", "output": "o" }, { "input": "100\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogog", "output": "***g" }, { "input": "99\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo", "output": "***" }, { "input": "5\nggggg", "output": "ggggg" }, { "input": "6\ngoogoo", "output": "go***o" }, { "input": "7\nooogooo", "output": "oo***oo" }, { "input": "8\ngggggggg", "output": "gggggggg" }, { "input": "9\nogggogggg", "output": "ogggogggg" }, { "input": "10\nogogoggogo", "output": "***gg***" }, { "input": "11\noooggooggog", "output": "oooggooggog" }, { "input": "12\nogggooooggog", "output": "ogggooooggog" }, { "input": "13\nogoggogogooog", "output": "***gg***oog" }, { "input": "15\nggooogoooggoggg", "output": "ggoo***ooggoggg" }, { "input": "14\noogooooggooggo", "output": "o***oooggooggo" }, { "input": "1\na", "output": "a" }, { "input": "1\ng", "output": "g" }, { "input": "1\nz", "output": "z" }, { "input": "46\nsdfkogogogogergergogogergerogogogogoergoegogog", "output": "sdfk***gergerg***gerger***ergoeg***g" }, { "input": "86\ngdfggogogodfogogdofgoodfogodgodgfogoogdodhbofgobgbgfogoogdfogogogogogoogggsgfosogodgon", "output": "gdfgg***df***gdofgoodf***dgodgf***ogdodhbofgobgbgf***ogdf***ogggsgfos***dgon" }, { "input": "83\nergegergoogogogerogoerorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgtogooggsdgogoog", "output": "ergegergo***ger***erorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgt***oggsdg***og" }, { "input": "2\nog", "output": "og" }, { "input": "4\nogog", "output": "***g" }, { "input": "4\nogoo", "output": "***o" }, { "input": "29\naaaogogogogogogogogogogogobbb", "output": "aaa***bbb" }, { "input": "11\nogogogogogo", "output": "***" }, { "input": "9\nogogogogo", "output": "***" }, { "input": "3\nogo", "output": "***" }, { "input": "15\nogogogogogogogo", "output": "***" }, { "input": "25\nogogogogogogogogogogogogo", "output": "***" } ]

1,569,599,382

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

3

139

0

x = int(input()) y = input() i = len(y) while i > 0: i = i - 1 if ("og"*i + "o") in y: y = y.replace(("og"*i + "o"), "***") print(y)

Title: Interview with Oleg Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters. There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers. The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here. To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length. Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview. The second line contains the string *s* of length *n*, consisting of lowercase English letters. Output Specification: Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. Demo Input: ['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n'] Demo Output: ['a***b\n', '***gmg***\n', '*********\n'] Note: The first sample contains one filler word ogogo, so the interview for printing is "a***b". The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".

```python x = int(input()) y = input() i = len(y) while i > 0: i = i - 1 if ("og"*i + "o") in y: y = y.replace(("og"*i + "o"), "***") print(y) ```

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" } ]

1,607,404,964

2,147,483,647

PyPy 3

OK

TESTS

30

310

0

string = input() lower_case = len([i for i in string if i.islower()]) upper_case = len([i for i in string if i.isupper()]) if lower_case >= upper_case: print(string.lower()) else: print(string.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 string = input() lower_case = len([i for i in string if i.islower()]) upper_case = len([i for i in string if i.isupper()]) if lower_case >= upper_case: print(string.lower()) else: print(string.upper()) ```

3.9225

845

A

Chess Tourney

PROGRAMMING

1,100

[ "implementation", "sortings" ]

null

Berland annual chess tournament is coming! Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil. Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins. Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win. After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random. Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100). The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000).

If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".

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

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

none

0

[ { "input": "2\n1 3 2 4", "output": "YES" }, { "input": "1\n3 3", "output": "NO" }, { "input": "5\n1 1 1 1 2 2 3 3 3 3", "output": "NO" }, { "input": "5\n1 1 1 1 1 2 2 2 2 2", "output": "YES" }, { "input": "10\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "1\n2 3", "output": "YES" }, { "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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "35\n919 240 231 858 456 891 959 965 758 30 431 73 505 694 874 543 975 445 16 147 904 690 940 278 562 127 724 314 30 233 389 442 353 652 581 383 340 445 487 283 85 845 578 946 228 557 906 572 919 388 686 181 958 955 736 438 991 170 632 593 475 264 178 344 159 414 739 590 348 884", "output": "YES" }, { "input": "5\n1 2 3 4 10 10 6 7 8 9", "output": "YES" }, { "input": "2\n1 1 1 2", "output": "NO" }, { "input": "2\n10 4 4 4", "output": "NO" }, { "input": "2\n2 3 3 3", "output": "NO" }, { "input": "4\n1 2 3 4 5 4 6 7", "output": "NO" }, { "input": "4\n2 5 4 5 8 3 1 5", "output": "YES" }, { "input": "4\n8 2 2 4 1 4 10 9", "output": "NO" }, { "input": "2\n3 8 10 2", "output": "YES" }, { "input": "3\n1 3 4 4 5 6", "output": "NO" }, { "input": "2\n3 3 3 4", "output": "NO" }, { "input": "2\n1 1 2 2", "output": "YES" }, { "input": "2\n1 1 3 3", "output": "YES" }, { "input": "2\n1 2 3 2", "output": "NO" }, { "input": "10\n1 2 7 3 9 4 1 5 10 3 6 1 10 7 8 5 7 6 1 4", "output": "NO" }, { "input": "3\n1 2 3 3 4 5", "output": "NO" }, { "input": "2\n2 2 1 1", "output": "YES" }, { "input": "7\n1 2 3 4 5 6 7 7 8 9 10 11 12 19", "output": "NO" }, { "input": "5\n1 2 3 4 5 3 3 5 6 7", "output": "YES" }, { "input": "4\n1 1 2 2 3 3 3 3", "output": "YES" }, { "input": "51\n576 377 63 938 667 992 959 997 476 94 652 272 108 410 543 456 942 800 917 163 931 584 357 890 895 318 544 179 268 130 649 916 581 350 573 223 495 26 377 695 114 587 380 424 744 434 332 249 318 522 908 815 313 384 981 773 585 747 376 812 538 525 997 896 859 599 437 163 878 14 224 733 369 741 473 178 153 678 12 894 630 921 505 635 128 404 64 499 208 325 343 996 970 39 380 80 12 756 580 57 934 224", "output": "YES" }, { "input": "3\n3 3 3 2 3 2", "output": "NO" }, { "input": "2\n5 3 3 6", "output": "YES" }, { "input": "2\n1 2 2 3", "output": "NO" }, { "input": "2\n1 3 2 2", "output": "NO" }, { "input": "2\n1 3 3 4", "output": "NO" }, { "input": "2\n1 2 2 2", "output": "NO" }, { "input": "3\n1 2 7 19 19 7", "output": "NO" }, { "input": "3\n1 2 3 3 5 6", "output": "NO" }, { "input": "2\n1 2 2 4", "output": "NO" }, { "input": "2\n6 6 5 5", "output": "YES" }, { "input": "2\n3 1 3 1", "output": "YES" }, { "input": "3\n1 2 3 3 1 1", "output": "YES" }, { "input": "3\n3 2 1 3 4 5", "output": "NO" }, { "input": "3\n4 5 6 4 2 1", "output": "NO" }, { "input": "3\n1 1 2 3 2 4", "output": "NO" }, { "input": "3\n100 99 1 1 1 1", "output": "NO" }, { "input": "3\n1 2 3 6 5 3", "output": "NO" }, { "input": "2\n2 2 1 2", "output": "NO" }, { "input": "4\n1 2 3 4 5 6 7 4", "output": "NO" }, { "input": "3\n1 2 3 1 1 1", "output": "NO" }, { "input": "3\n6 5 3 3 1 3", "output": "NO" }, { "input": "2\n1 2 1 2", "output": "YES" }, { "input": "3\n1 2 5 6 8 6", "output": "YES" }, { "input": "5\n1 2 3 4 5 3 3 3 3 3", "output": "NO" }, { "input": "2\n1 2 4 2", "output": "NO" }, { "input": "3\n7 7 4 5 319 19", "output": "NO" }, { "input": "3\n1 2 4 4 3 5", "output": "YES" }, { "input": "3\n3 2 3 4 5 2", "output": "NO" }, { "input": "5\n1 2 3 4 4 5 3 6 7 8", "output": "NO" }, { "input": "3\n3 3 4 4 5 1", "output": "YES" }, { "input": "2\n3 4 3 3", "output": "NO" }, { "input": "2\n2 5 4 4", "output": "NO" }, { "input": "5\n1 2 3 3 4 5 6 7 8 4", "output": "NO" }, { "input": "3\n1 2 3 3 5 5", "output": "NO" }, { "input": "2\n3 4 4 4", "output": "NO" }, { "input": "2\n1 4 5 4", "output": "NO" }, { "input": "2\n1 2 3 3", "output": "YES" }, { "input": "2\n1 1 2 1", "output": "NO" }, { "input": "4\n1 1 1 1 2 2 2 2", "output": "YES" }, { "input": "4\n1 2 3 5 6 7 8 5", "output": "NO" }, { "input": "2\n4 3 3 1", "output": "NO" }, { "input": "3\n3 1 2 4 3 5", "output": "NO" }, { "input": "3\n1 2 3 3 4 6", "output": "NO" }, { "input": "4\n2 2 2 4 5 5 5 5", "output": "YES" }, { "input": "2\n1 3 4 3", "output": "NO" }, { "input": "2\n3 3 2 3", "output": "NO" }, { "input": "2\n1 2 1 1", "output": "NO" }, { "input": "3\n1 3 4 4 2 5", "output": "YES" }, { "input": "4\n4 7 1 2 3 5 6 4", "output": "NO" }, { "input": "2\n3 2 2 2", "output": "NO" }, { "input": "1\n2 1", "output": "YES" }, { "input": "2\n3 3 1 2", "output": "YES" }, { "input": "1\n8 6", "output": "YES" }, { "input": "7\n6 7 6 7 3 1 9 4 6 10 8 2 5 7", "output": "NO" }, { "input": "2\n3 9 2 1", "output": "YES" }, { "input": "2\n3 3 3 3", "output": "NO" } ]

1,639,470,435

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

31

0

def main(): n=int(input()) a=[int(i) for i in input().split()] a.sort() if a[n/2]>a[(n/2) - 1]: print('YES') else: print('NO') if __name__=='__main__': main()

Title: Chess Tourney Time Limit: None seconds Memory Limit: None megabytes Problem Description: Berland annual chess tournament is coming! Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil. Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins. Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win. After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random. Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing? Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100). The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000). Output Specification: If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO". Demo Input: ['2\n1 3 2 4\n', '1\n3 3\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python def main(): n=int(input()) a=[int(i) for i in input().split()] a.sort() if a[n/2]>a[(n/2) - 1]: print('YES') else: print('NO') if __name__=='__main__': main() ```

-1

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!" } ]

1,646,716,411

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

2

46

0

for i in range(int(input())): sen = input() if (sen.endswith('miao.')) and (not sen.startswith('lala.')): print("Rainbow's") elif (sen.endswith('lala.')) and (not sen.startswith('miao.')): print("Freda's") else: print("OMG>.< I don't know!")

Title: Whose sentence is it? Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. Output Specification: For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< 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 for i in range(int(input())): sen = input() if (sen.endswith('miao.')) and (not sen.startswith('lala.')): print("Rainbow's") elif (sen.endswith('lala.')) and (not sen.startswith('miao.')): print("Freda's") else: print("OMG>.< I don't know!") ```

0

932

A

Palindromic Supersequence

PROGRAMMING

800

[ "constructive algorithms" ]

null

You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*. A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest". A palindrome is a string that reads the same forward or backward. The length of string *B* should be at most 104. It is guaranteed that there always exists such string. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.

First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.

Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.

[ "aba\n", "ab\n" ]

[ "aba", "aabaa" ]

In the first example, "aba" is a subsequence of "aba" which is a palindrome. In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.

500

[ { "input": "aba", "output": "abaaba" }, { "input": "ab", "output": "abba" }, { "input": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpa", "output": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpaapkxovfyaifzjikpcxhemrcpwhypaabnoeifgfygtiqvuoihntvuvbrlnkywutodwrmvgrumdjzqhfoenxphzrdgtwmljdm..." }, { "input": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadco", "output": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadcoocdafcilhkrgpvtsyrmzilywzgrtebnaldqqcmrfvsfehjkbetprwlvyscuxef..." }, { "input": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxar", "output": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxarraxluhthchsyjjpegbmmzflbvzbunpfhzruoghtlgtearurzueovxcivvfbqlssonqkjiybmbjbzpffixfcdwbrpwyhvzbdjmchzjfnrjdoupimgkyyhfgppveltacqczktdxkawpzdmkmyikyjtqznvjdnkkmikemtyjsgdlheyjltiwcxpfvxqtxwwvgkjcrzbkxkandsekqwxpequ..." }, { "input": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjg", "output": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjggjferujlznjmyalmuspib..." }, { "input": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgm", "output": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgmmgszzmttxopywvrbvmgavsgwqergikztssmlnktgpolmcbrropknnkeiytztyrlqthshvbosemmjbzpsvcytuzynrqygcfoqsywdkrvjftdagjdgjyynfkcdkmwaqhzfewry..." }, { "input": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyj", "output": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyjjycqccvmgjgqjfbxjzrdgbsqyseehcjsyxlbrettpfcjdvjerwvaixetfjxjxkdwcwlmwqkrgrhwzopjnilmonwtxsdonvyuazaqghlplvlisqdqbpvxkjpovqohbarttqoqm..." }, { "input": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgd", "output": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgddgmsixjxsmwpfzyxrnkjqrbrfyprvlvbegdmgsidepusixegsdukxmudmjuqqzkefrzymgkluqqrigtpmznhvswwhqbiiizogjofdvfqdrfabswiclfpkiermnozbdmxieoxornesifoiuxgffjelsvcoofviyorrpwkffvvfjaubzeuzqbdhfjgujqoejsktzfzaypaxipweiawzojidcustlgyfhkbfgwshbdkzneilwwqemqblw..." }, { "input": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohlig", "output": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohliggilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlw..." }, { "input": "abcab", "output": "abcabbacba" }, { "input": "baaaaaaa", "output": "baaaaaaaaaaaaaab" }, { "input": "baaaaaa", "output": "baaaaaaaaaaaab" }, { "input": "baaaaaaaaa", "output": "baaaaaaaaaaaaaaaaaab" }, { "input": "baaaaaaaa", "output": "baaaaaaaaaaaaaaaab" } ]

1,518,705,519

219

Python 3

WRONG_ANSWER

PRETESTS

1

61

5,632,000

read=lambda:map(int,input().split()) s=input() print(s+s)

Title: Palindromic Supersequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*. A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest". A palindrome is a string that reads the same forward or backward. The length of string *B* should be at most 104. It is guaranteed that there always exists such string. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. Input Specification: First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*. Output Specification: Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them. Demo Input: ['aba\n', 'ab\n'] Demo Output: ['aba', 'aabaa'] Note: In the first example, "aba" is a subsequence of "aba" which is a palindrome. In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.

```python read=lambda:map(int,input().split()) s=input() print(s+s) ```

0

none

0

[ "none" ]

null

Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*. Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array. Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.

The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array. Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105). Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). The numbers in the lines are separated by single spaces.

On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

[ "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n", "1 1 1\n1\n1 1 1\n1 1\n", "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n" ]

[ "9 18 17\n", "2\n", "5 18 31 20\n" ]

none

0

[ { "input": "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3", "output": "9 18 17" }, { "input": "1 1 1\n1\n1 1 1\n1 1", "output": "2" }, { "input": "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3", "output": "5 18 31 20" }, { "input": "1 1 1\n0\n1 1 0\n1 1", "output": "0" } ]

1,675,447,669

2,147,483,647

PyPy 3-64

MEMORY_LIMIT_EXCEEDED

TESTS

10

483

268,390,400

import sys input = sys.stdin.readline n, m, k = map(int, input().split()) arr = list(map(int, input().split())) oper = [] queries = [] while (m > 0): oper.append(tuple(map(int, input().split()))) m -= 1 while (k > 0): queries.append(tuple(map(int, input().split()))) k -= 1 ap_arr = [] ap_arr.append(arr) new_arr = arr.copy() for l,r,d in oper: new_arr = new_arr[:l-1] + [i+d for i in new_arr[l-1:r]] + new_arr[r:] ap_arr.append(new_arr) for xi, yi in queries: add = [a_i - b_i for a_i, b_i in zip(ap_arr[yi], ap_arr[xi-1])] arr = [sum(x) for x in zip(arr,add)] print(*arr, sep = ' ') # for xi, yi in queries: # for l, r, d in oper: # arr[l-1:r] = [i + d for i in arr[l-1:r]]

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*. Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array. Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg. Input Specification: The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array. Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105). Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). The numbers in the lines are separated by single spaces. Output Specification: On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier. Demo Input: ['3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n', '1 1 1\n1\n1 1 1\n1 1\n', '4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n'] Demo Output: ['9 18 17\n', '2\n', '5 18 31 20\n'] Note: none

```python import sys input = sys.stdin.readline n, m, k = map(int, input().split()) arr = list(map(int, input().split())) oper = [] queries = [] while (m > 0): oper.append(tuple(map(int, input().split()))) m -= 1 while (k > 0): queries.append(tuple(map(int, input().split()))) k -= 1 ap_arr = [] ap_arr.append(arr) new_arr = arr.copy() for l,r,d in oper: new_arr = new_arr[:l-1] + [i+d for i in new_arr[l-1:r]] + new_arr[r:] ap_arr.append(new_arr) for xi, yi in queries: add = [a_i - b_i for a_i, b_i in zip(ap_arr[yi], ap_arr[xi-1])] arr = [sum(x) for x in zip(arr,add)] print(*arr, sep = ' ') # for xi, yi in queries: # for l, r, d in oper: # arr[l-1:r] = [i + d for i in arr[l-1:r]] ```

0

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..." } ]

1,619,673,075

2,147,483,647

Python 3

OK

TESTS

21

154

0

n = int(input()) i, j = 1, n * n for __ in range(n): for ___ in range(n // 2): print(str(i) + " " + str(j), end=" ") i, j = i + 1, j - 1 print()

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()) i, j = 1, n * n for __ in range(n): for ___ in range(n // 2): print(str(i) + " " + str(j), end=" ") i, j = i + 1, j - 1 print() ```

3

499

B

Lecture

PROGRAMMING

1,000

[ "implementation", "strings" ]

null

You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.

The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.

Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.

[ "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n", "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n" ]

[ "codeforces round letter round\n", "hbnyiyc joll joll un joll\n" ]

none

500

[ { "input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest", "output": "codeforces round letter round" }, { "input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll", "output": "hbnyiyc joll joll un joll" }, { "input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b", "output": "f jpggvr b hnxsd b" }, { "input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb", "output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs" }, { "input": "1 1\namit am\namit", "output": "am" }, { "input": "1 1\na c\na", "output": "a" } ]

1,664,470,759

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

30

0

s = list(map(int, input().split(" "))) n, m = s[0], s[1] names = [] for i in range(m): s1 = input().split(" ") a, b = s1[0], s1[1] if(len(a)<=len(b)): names.append([a, a]) else: names.append([a, b]) s2 = input().split(" ") resList=[] for i in s2: for j in range(len(names)): if i==names[j][0]: resList.append(names[j][1]) print(resList)

Title: Lecture Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes. Input Specification: The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters. Output Specification: Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input. Demo Input: ['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n'] Demo Output: ['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n'] Note: none

```python s = list(map(int, input().split(" "))) n, m = s[0], s[1] names = [] for i in range(m): s1 = input().split(" ") a, b = s1[0], s1[1] if(len(a)<=len(b)): names.append([a, a]) else: names.append([a, b]) s2 = input().split(" ") resList=[] for i in s2: for j in range(len(names)): if i==names[j][0]: resList.append(names[j][1]) print(resList) ```

0

452

A

Eevee

PROGRAMMING

1,000

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

null

You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon. You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.

First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string. Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).

Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).

[ "7\nj......\n", "7\n...feon\n", "7\n.l.r.o.\n" ]

[ "jolteon\n", "leafeon\n", "flareon\n" ]

Here's a set of names in a form you can paste into your solution: ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] {"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}

500

[ { "input": "7\n...feon", "output": "leafeon" }, { "input": "7\n.l.r.o.", "output": "flareon" }, { "input": "6\n.s..o.", "output": "espeon" }, { "input": "7\nglaceon", "output": "glaceon" }, { "input": "8\n.a.o.e.n", "output": "vaporeon" }, { "input": "7\n.laceon", "output": "glaceon" }, { "input": "7\n..lveon", "output": "sylveon" }, { "input": "7\n.l.ceon", "output": "glaceon" }, { "input": "7\n..areon", "output": "flareon" } ]

1,652,807,631

2,147,483,647

PyPy 3-64

RUNTIME_ERROR

TESTS

1

77

31,744,000

a = ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] n = int(input()) b = [i for i in input()] for u in a: k = 0 for i in range(len(b)): if b[i] != '.' and u[i] != b[i]: k += 1 if k == 0: print(u)

Title: Eevee Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon. You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it. Input Specification: First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string. Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword). Output Specification: Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter). Demo Input: ['7\nj......\n', '7\n...feon\n', '7\n.l.r.o.\n'] Demo Output: ['jolteon\n', 'leafeon\n', 'flareon\n'] Note: Here's a set of names in a form you can paste into your solution: ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] {"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}

```python a = ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] n = int(input()) b = [i for i in input()] for u in a: k = 0 for i in range(len(b)): if b[i] != '.' and u[i] != b[i]: k += 1 if k == 0: print(u) ```

-1

0

none

0

[ "none" ]

null

Fox Ciel has *n* boxes in her room. They have the same size and weight, but they might have different strength. The *i*-th box can hold at most *x**i* boxes on its top (we'll call *x**i* the strength of the box). Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile. Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than *x**i* boxes on the top of *i*-th box. What is the minimal number of piles she needs to construct?

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100).

Output a single integer — the minimal possible number of piles.

[ "3\n0 0 10\n", "5\n0 1 2 3 4\n", "4\n0 0 0 0\n", "9\n0 1 0 2 0 1 1 2 10\n" ]

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

In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2. In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).

0

[ { "input": "3\n0 0 10", "output": "2" }, { "input": "5\n0 1 2 3 4", "output": "1" }, { "input": "4\n0 0 0 0", "output": "4" }, { "input": "9\n0 1 0 2 0 1 1 2 10", "output": "3" }, { "input": "1\n0", "output": "1" }, { "input": "2\n0 0", "output": "2" }, { "input": "2\n0 1", "output": "1" }, { "input": "2\n100 99", "output": "1" }, { "input": "9\n0 1 1 0 2 0 3 45 4", "output": "3" }, { "input": "10\n1 1 1 1 2 2 2 2 2 2", "output": "4" }, { "input": "100\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50", "output": "2" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "100" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "1" }, { "input": "11\n71 34 31 71 42 38 64 60 36 76 67", "output": "1" }, { "input": "39\n54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54", "output": "1" }, { "input": "59\n61 33 84 76 56 47 70 94 46 77 95 85 35 90 83 62 48 74 36 74 83 97 62 92 95 75 70 82 94 67 82 42 78 70 50 73 80 76 94 83 96 80 80 88 91 79 83 54 38 90 33 93 53 33 86 95 48 34 46", "output": "1" }, { "input": "87\n52 63 93 90 50 35 67 66 46 89 43 64 33 88 34 80 69 59 75 55 55 68 66 83 46 33 72 36 73 34 54 85 52 87 67 68 47 95 52 78 92 58 71 66 84 61 36 77 69 44 84 70 71 55 43 91 33 65 77 34 43 59 83 70 95 38 92 92 74 53 66 65 81 45 55 89 49 52 43 69 78 41 37 79 63 70 67", "output": "1" }, { "input": "15\n20 69 36 63 40 40 52 42 20 43 59 68 64 49 47", "output": "1" }, { "input": "39\n40 20 49 35 80 18 20 75 39 62 43 59 46 37 58 52 67 16 34 65 32 75 59 42 59 41 68 21 41 61 66 19 34 63 19 63 78 62 24", "output": "1" }, { "input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "18" }, { "input": "46\n14 13 13 10 13 15 8 8 12 9 11 15 8 10 13 8 12 13 11 8 12 15 12 15 11 13 12 9 13 12 10 8 13 15 9 15 8 13 11 8 9 9 9 8 11 8", "output": "3" }, { "input": "70\n6 1 4 1 1 6 5 2 5 1 1 5 2 1 2 4 1 1 1 2 4 5 2 1 6 6 5 2 1 4 3 1 4 3 6 5 2 1 3 4 4 1 4 5 6 2 1 2 4 4 5 3 6 1 1 2 2 1 5 6 1 6 3 1 4 4 2 3 1 4", "output": "11" }, { "input": "94\n11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11", "output": "8" }, { "input": "18\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "9" }, { "input": "46\n14 8 7 4 8 7 8 8 12 9 9 12 9 12 14 8 10 14 14 6 9 11 7 14 14 13 11 4 13 13 11 13 9 10 10 12 10 8 12 10 13 10 7 13 14 6", "output": "4" }, { "input": "74\n4 4 5 5 5 5 5 5 6 6 5 4 4 4 3 3 5 4 5 3 4 4 5 6 3 3 5 4 4 5 4 3 5 5 4 4 3 5 6 4 3 6 6 3 4 5 4 4 3 3 3 6 3 5 6 5 5 5 5 3 6 4 5 4 4 6 6 3 4 5 6 6 6 6", "output": "11" }, { "input": "100\n48 35 44 37 35 42 42 39 49 53 35 55 41 42 42 39 43 49 46 54 48 39 42 53 55 39 56 43 43 38 48 40 54 36 48 55 46 40 41 39 45 56 38 40 47 46 45 46 53 51 38 41 54 35 35 47 42 43 54 54 39 44 49 41 37 49 36 37 37 49 53 44 47 37 55 49 45 40 35 51 44 40 42 35 46 48 53 48 35 38 42 36 54 46 44 47 41 40 41 42", "output": "2" }, { "input": "100\n34 3 37 35 40 44 38 46 13 31 12 23 26 40 26 18 28 36 5 21 2 4 10 29 3 46 38 41 37 28 44 14 39 10 35 17 24 28 38 16 29 6 2 42 47 34 43 2 43 46 7 16 16 43 33 32 20 47 8 48 32 4 45 38 15 7 25 25 19 41 20 35 16 2 31 5 31 25 27 3 45 29 32 36 9 47 39 35 9 21 32 17 21 41 29 48 11 40 5 25", "output": "3" }, { "input": "100\n2 4 5 5 0 5 3 0 3 0 5 3 4 1 0 3 0 5 5 0 4 3 3 3 0 2 1 2 2 4 4 2 4 0 1 3 4 1 4 2 5 3 5 2 3 0 1 2 5 5 2 0 4 2 5 1 0 0 4 0 1 2 0 1 2 4 1 4 5 3 4 5 5 1 0 0 3 1 4 0 4 5 1 3 3 0 4 2 0 4 5 2 3 0 5 1 4 4 1 0", "output": "21" }, { "input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "17" }, { "input": "100\n1 1 1 2 2 2 2 2 2 1 1 1 2 0 2 2 0 0 0 0 0 2 0 0 2 2 1 0 2 0 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 0 1 2 2 0 2 2 2 2 1 0 1 0 0 0 2 0 2 0 1 1 0 2 2 2 2 1 1 1 2 1 1 2 1 1 1 2 1 0 2 1 0 1 2 0 1 1 2 0 0 1 1 0 1 1", "output": "34" }, { "input": "100\n0 3 1 0 3 2 1 2 2 1 2 1 3 2 1 2 1 3 2 0 0 2 3 0 0 2 1 2 2 3 1 2 2 2 0 3 3 2 0 0 1 0 1 2 3 1 0 3 3 3 0 2 1 3 0 1 3 2 2 2 2 3 3 2 0 2 0 1 0 1 3 0 1 2 0 1 3 2 0 3 1 1 2 3 1 3 1 0 3 0 3 0 2 1 1 1 2 2 0 1", "output": "26" }, { "input": "100\n1 0 2 2 2 2 1 0 1 2 2 2 0 1 0 1 2 1 2 1 0 1 2 2 2 1 0 1 0 2 1 2 0 2 1 1 2 1 1 0 1 2 1 1 2 1 1 0 2 2 0 0 1 2 0 2 0 0 1 1 0 0 2 1 2 1 0 2 2 2 2 2 2 1 2 0 1 2 1 2 1 0 1 0 1 0 1 1 0 2 1 0 0 1 2 2 1 0 0 1", "output": "34" }, { "input": "100\n3 4 4 4 3 3 3 3 3 4 4 4 3 3 3 4 3 4 4 4 3 4 3 4 3 4 3 3 4 4 3 4 4 3 4 4 4 4 4 3 4 3 3 3 4 3 3 4 3 4 3 4 3 3 4 4 4 3 3 3 3 3 4 4 3 4 4 3 4 3 3 3 4 4 3 3 3 3 3 4 3 4 4 3 3 4 3 4 3 4 4 4 3 3 3 4 4 4 4 3", "output": "20" }, { "input": "100\n8 7 9 10 2 7 8 11 11 4 7 10 2 5 8 9 10 3 9 4 10 5 5 6 3 8 8 9 6 9 5 5 4 11 4 2 11 8 3 5 6 6 11 9 8 11 9 8 3 3 8 9 8 9 4 8 6 11 4 4 4 9 7 5 3 4 11 3 9 11 8 10 3 5 5 7 6 9 4 5 2 11 3 6 2 10 9 4 6 10 5 11 8 10 10 8 9 8 5 3", "output": "9" }, { "input": "5\n4 1 1 1 1", "output": "2" } ]

1,497,131,835

2,147,483,647

Python 3

OK

TESTS

36

62

0

n = int(input()) nums = sorted(list(map(int, input().split(' ')))) v = [] b = [0] * 105 cnt = 0 k = 0 while cnt < n: k += 1 for i in range(n): if nums[i] >= len(v) and b[i] == 0: v.append(nums[i]) b[i] = 1 cnt += 1 #print(list(reversed(v))) v = [] print(k)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Fox Ciel has *n* boxes in her room. They have the same size and weight, but they might have different strength. The *i*-th box can hold at most *x**i* boxes on its top (we'll call *x**i* the strength of the box). Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile. Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than *x**i* boxes on the top of *i*-th box. What is the minimal number of piles she needs to construct? Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100). Output Specification: Output a single integer — the minimal possible number of piles. Demo Input: ['3\n0 0 10\n', '5\n0 1 2 3 4\n', '4\n0 0 0 0\n', '9\n0 1 0 2 0 1 1 2 10\n'] Demo Output: ['2\n', '1\n', '4\n', '3\n'] Note: In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2. In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).

```python n = int(input()) nums = sorted(list(map(int, input().split(' ')))) v = [] b = [0] * 105 cnt = 0 k = 0 while cnt < n: k += 1 for i in range(n): if nums[i] >= len(v) and b[i] == 0: v.append(nums[i]) b[i] = 1 cnt += 1 #print(list(reversed(v))) v = [] print(k) ```

3

550

A

Two Substrings

PROGRAMMING

1,500

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

null

You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).

The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters.

Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.

[ "ABA\n", "BACFAB\n", "AXBYBXA\n" ]

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

In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".

1,000

[ { "input": "ABA", "output": "NO" }, { "input": "BACFAB", "output": "YES" }, { "input": "AXBYBXA", "output": "NO" }, { "input": "ABABAB", "output": "YES" }, { "input": "BBBBBBBBBB", "output": "NO" }, { "input": "ABBA", "output": "YES" }, { "input": "ABAXXXAB", "output": "YES" }, { "input": "TESTABAXXABTEST", "output": "YES" }, { "input": "A", "output": "NO" }, { "input": "B", "output": "NO" }, { "input": "X", "output": "NO" }, { "input": "BA", "output": "NO" }, { "input": "AB", "output": "NO" }, { "input": "AA", "output": "NO" }, { "input": "BB", "output": "NO" }, { "input": "BAB", "output": "NO" }, { "input": "AAB", "output": "NO" }, { "input": "BAA", "output": "NO" }, { "input": "ABB", "output": "NO" }, { "input": "BBA", "output": "NO" }, { "input": "AAA", "output": "NO" }, { "input": "BBB", "output": "NO" }, { "input": "AXBXBXA", "output": "NO" }, { "input": "SKDSKDJABSDBADKFJDK", "output": "YES" }, { "input": "ABAXXBBXXAA", "output": "NO" }, { "input": "ABAB", "output": "NO" }, { "input": "BABA", "output": "NO" }, { "input": "AAAB", "output": "NO" }, { "input": "AAAA", "output": "NO" }, { "input": "AABA", "output": "NO" }, { "input": "ABAA", "output": "NO" }, { "input": "BAAA", "output": "NO" }, { "input": "AABB", "output": "NO" }, { "input": "BAAB", "output": "YES" }, { "input": "BBAA", "output": "NO" }, { "input": "BBBA", "output": "NO" }, { "input": "BBAB", "output": "NO" }, { "input": "BABB", "output": "NO" }, { "input": "ABBB", "output": "NO" }, { "input": "BBBB", "output": "NO" }, { "input": "BABAB", "output": "YES" }, { "input": "ABABA", "output": "YES" }, { "input": "AAABAAACBBBC", "output": "NO" }, { "input": "AABBBACBBBBBBAACBCCACBBAABBBBBCAACABAACABCACCBCBCCCBCBCABCBBCCBCBBAACBACAABACBBCACCBCCACCABBCBABBAAC", "output": "YES" }, { "input": "CBBABDDBBADAC", "output": "YES" }, { "input": "ABYFAB", "output": "NO" }, { "input": "BABXXAB", "output": "YES" }, { "input": "ABAXAB", "output": "YES" }, { "input": "ABABXAB", "output": "YES" }, { "input": "ABXABXABXABXABXBAXBAXBAXBA", "output": "YES" }, { "input": "QQQQQQQQQABABQQQQQQQQ", "output": "NO" } ]

1,695,142,945

2,147,483,647

PyPy 3-64

OK

TESTS

85

62

0

s = input() a = s.find("AB") b = s[a + 2:].find("BA") c = s.find("BA") d = s[c + 2:].find("AB") print("YES" if a >= 0 and b >= 0 or c >= 0 and d >= 0 else "NO")

Title: Two Substrings Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). Input Specification: The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters. Output Specification: Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. Demo Input: ['ABA\n', 'BACFAB\n', 'AXBYBXA\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n'] Note: In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".

```python s = input() a = s.find("AB") b = s[a + 2:].find("BA") c = s.find("BA") d = s[c + 2:].find("AB") print("YES" if a >= 0 and b >= 0 or c >= 0 and d >= 0 else "NO") ```

3

750

A

New Year and Hurry

PROGRAMMING

800

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

null

Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party?

The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.

Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.

[ "3 222\n", "4 190\n", "7 1\n" ]

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

In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

500

[ { "input": "3 222", "output": "2" }, { "input": "4 190", "output": "4" }, { "input": "7 1", "output": "7" }, { "input": "10 135", "output": "6" }, { "input": "10 136", "output": "5" }, { "input": "1 1", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "10 240", "output": "0" }, { "input": "9 240", "output": "0" }, { "input": "9 1", "output": "9" }, { "input": "9 235", "output": "1" }, { "input": "9 236", "output": "0" }, { "input": "5 225", "output": "2" }, { "input": "5 226", "output": "1" }, { "input": "4 210", "output": "3" }, { "input": "4 211", "output": "2" }, { "input": "4 191", "output": "3" }, { "input": "10 165", "output": "5" }, { "input": "10 166", "output": "4" }, { "input": "8 100", "output": "7" }, { "input": "8 101", "output": "6" }, { "input": "8 60", "output": "8" }, { "input": "8 61", "output": "7" }, { "input": "10 15", "output": "9" }, { "input": "10 16", "output": "8" }, { "input": "4 100", "output": "4" }, { "input": "4 101", "output": "4" }, { "input": "7 167", "output": "4" }, { "input": "10 164", "output": "5" }, { "input": "9 170", "output": "4" }, { "input": "8 160", "output": "5" }, { "input": "1 100", "output": "1" }, { "input": "8 123", "output": "6" }, { "input": "2 99", "output": "2" }, { "input": "10 88", "output": "7" }, { "input": "1 235", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "4 240", "output": "0" }, { "input": "1 55", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "3 240", "output": "0" }, { "input": "10 240", "output": "0" }, { "input": "2 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "9 1", "output": "9" }, { "input": "2 236", "output": "0" }, { "input": "10 2", "output": "9" }, { "input": "3 239", "output": "0" }, { "input": "1 237", "output": "0" }, { "input": "7 8", "output": "7" }, { "input": "10 235", "output": "1" } ]

1,688,313,639

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

30

0

n, k = map(int,input().split()) c = [] m = 240 rest = m - k for i in range(n+1): c.append(i*5) tim = (sum(c)) if tim <= rest: print(n) else: while rest > 0: n1 = 0 for x in c: rest = rest - x if rest > 0: n1 += 1 print(n1 - 1)

Title: New Year and Hurry Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party? Input Specification: The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house. Output Specification: Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier. Demo Input: ['3 222\n', '4 190\n', '7 1\n'] Demo Output: ['2\n', '4\n', '7\n'] Note: In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

```python n, k = map(int,input().split()) c = [] m = 240 rest = m - k for i in range(n+1): c.append(i*5) tim = (sum(c)) if tim <= rest: print(n) else: while rest > 0: n1 = 0 for x in c: rest = rest - x if rest > 0: n1 += 1 print(n1 - 1) ```

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" } ]

1,605,520,003

2,147,483,647

Python 3

OK

TESTS

32

218

307,200

n=int(input()) l=list(map(int,input().split())) odd=0 odd_list=[] even=0 p=0 even_list=[] for i in l: if i%2==0: even+=1 even_list.append(i) else: odd+=1 odd_list.append(i) if odd>even: p=even_list[0] print(l.index(p)+1) else: p=odd_list[0] print(l.index(p)+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 n=int(input()) l=list(map(int,input().split())) odd=0 odd_list=[] even=0 p=0 even_list=[] for i in l: if i%2==0: even+=1 even_list.append(i) else: odd+=1 odd_list.append(i) if odd>even: p=even_list[0] print(l.index(p)+1) else: p=odd_list[0] print(l.index(p)+1) ```

3.944928

750

A

New Year and Hurry

PROGRAMMING

800

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

null

Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party?

The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.

Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.

[ "3 222\n", "4 190\n", "7 1\n" ]

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

In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

500

[ { "input": "3 222", "output": "2" }, { "input": "4 190", "output": "4" }, { "input": "7 1", "output": "7" }, { "input": "10 135", "output": "6" }, { "input": "10 136", "output": "5" }, { "input": "1 1", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "10 240", "output": "0" }, { "input": "9 240", "output": "0" }, { "input": "9 1", "output": "9" }, { "input": "9 235", "output": "1" }, { "input": "9 236", "output": "0" }, { "input": "5 225", "output": "2" }, { "input": "5 226", "output": "1" }, { "input": "4 210", "output": "3" }, { "input": "4 211", "output": "2" }, { "input": "4 191", "output": "3" }, { "input": "10 165", "output": "5" }, { "input": "10 166", "output": "4" }, { "input": "8 100", "output": "7" }, { "input": "8 101", "output": "6" }, { "input": "8 60", "output": "8" }, { "input": "8 61", "output": "7" }, { "input": "10 15", "output": "9" }, { "input": "10 16", "output": "8" }, { "input": "4 100", "output": "4" }, { "input": "4 101", "output": "4" }, { "input": "7 167", "output": "4" }, { "input": "10 164", "output": "5" }, { "input": "9 170", "output": "4" }, { "input": "8 160", "output": "5" }, { "input": "1 100", "output": "1" }, { "input": "8 123", "output": "6" }, { "input": "2 99", "output": "2" }, { "input": "10 88", "output": "7" }, { "input": "1 235", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "4 240", "output": "0" }, { "input": "1 55", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "3 240", "output": "0" }, { "input": "10 240", "output": "0" }, { "input": "2 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "9 1", "output": "9" }, { "input": "2 236", "output": "0" }, { "input": "10 2", "output": "9" }, { "input": "3 239", "output": "0" }, { "input": "1 237", "output": "0" }, { "input": "7 8", "output": "7" }, { "input": "10 235", "output": "1" } ]

1,692,189,405

2,147,483,647

PyPy 3-64

OK

TESTS

52

62

0

n, k = map(int, input().split()) time = 4 * 60 - k for i in range(1, n + 1): time -= 5 * i if time < 0: print(i - 1) break else: print(n)

Title: New Year and Hurry Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party? Input Specification: The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house. Output Specification: Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier. Demo Input: ['3 222\n', '4 190\n', '7 1\n'] Demo Output: ['2\n', '4\n', '7\n'] Note: In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

```python n, k = map(int, input().split()) time = 4 * 60 - k for i in range(1, n + 1): time -= 5 * i if time < 0: print(i - 1) break else: print(n) ```

3

608

B

Hamming Distance Sum

PROGRAMMING

1,500

[ "combinatorics", "strings" ]

null

Genos needs your help. He was asked to solve the following programming problem by Saitama: The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2. Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.

The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000). The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000). Both strings are guaranteed to consist of characters '0' and '1' only.

Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.

[ "01\n00111\n", "0011\n0110\n" ]

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

For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3. The second sample case is described in the statement.

1,000

[ { "input": "01\n00111", "output": "3" }, { "input": "0011\n0110", "output": "2" }, { "input": "0\n0", "output": "0" }, { "input": "1\n0", "output": "1" }, { "input": "0\n1", "output": "1" }, { "input": "1\n1", "output": "0" }, { "input": "1001101001101110101101000\n01111000010011111111110010001101000100011110101111", "output": "321" }, { "input": "1110010001000101001011111\n00011011000000100001010000010100110011010001111010", "output": "316" } ]

1,453,626,150

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

62

0

#! /usr/bin/env python # -*- coding: utf-8 -*- # vim:fenc=utf-8 # # Copyright © 2016 missingdays <missingdays@missingdays> # # Distributed under terms of the MIT license. """ """ s1 = input() s2 = input() zeros = 0 ones = 0 i = 0 j = -1 l1 = len(s1) l2 = len(s2) answ = 0 while j < l2 - l1: j += 1 if s2[j] == "0": zeros += 1 else: ones += 1 while j < l2: if s1[i] == "0": answ += ones else: answ += zeros if j == l2 - 1: break j += 1 i += 1 if s2[i] == "0" and s2[j] == "1": zeros -= 1 ones += 1 elif s2[i] == "1" and s2[j] == "0": zeros += 1 ones -= 1 print(answ)

Title: Hamming Distance Sum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Genos needs your help. He was asked to solve the following programming problem by Saitama: The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2. Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. Input Specification: The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000). The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000). Both strings are guaranteed to consist of characters '0' and '1' only. Output Specification: Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. Demo Input: ['01\n00111\n', '0011\n0110\n'] Demo Output: ['3\n', '2\n'] Note: For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3. The second sample case is described in the statement.

```python #! /usr/bin/env python # -*- coding: utf-8 -*- # vim:fenc=utf-8 # # Copyright © 2016 missingdays <missingdays@missingdays> # # Distributed under terms of the MIT license. """ """ s1 = input() s2 = input() zeros = 0 ones = 0 i = 0 j = -1 l1 = len(s1) l2 = len(s2) answ = 0 while j < l2 - l1: j += 1 if s2[j] == "0": zeros += 1 else: ones += 1 while j < l2: if s1[i] == "0": answ += ones else: answ += zeros if j == l2 - 1: break j += 1 i += 1 if s2[i] == "0" and s2[j] == "1": zeros -= 1 ones += 1 elif s2[i] == "1" and s2[j] == "0": zeros += 1 ones -= 1 print(answ) ```

0

938

A

Word Correction

PROGRAMMING

800

[ "implementation" ]

null

Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange. Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct. You are given a word *s*. Can you predict what will it become after correction? In this problem letters a, e, i, o, u and y are considered to be vowels.

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction. The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction.

Output the word *s* after the correction.

[ "5\nweird\n", "4\nword\n", "5\naaeaa\n" ]

[ "werd\n", "word\n", "a\n" ]

Explanations of the examples: 1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.

0

[ { "input": "5\nweird", "output": "werd" }, { "input": "4\nword", "output": "word" }, { "input": "5\naaeaa", "output": "a" }, { "input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw", "output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw" }, { "input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" }, { "input": "12\nmmmmmmmmmmmm", "output": "mmmmmmmmmmmm" }, { "input": "18\nyaywptqwuyiqypwoyw", "output": "ywptqwuqypwow" }, { "input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" }, { "input": "13\nmmmmmmmmmmmmm", "output": "mmmmmmmmmmmmm" }, { "input": "10\nmmmmmmmmmm", "output": "mmmmmmmmmm" }, { "input": "11\nmmmmmmmmmmm", "output": "mmmmmmmmmmm" }, { "input": "15\nmmmmmmmmmmmmmmm", "output": "mmmmmmmmmmmmmmm" }, { "input": "1\na", "output": "a" }, { "input": "14\nmmmmmmmmmmmmmm", "output": "mmmmmmmmmmmmmm" }, { "input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm", "output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm" }, { "input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" }, { "input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" }, { "input": "2\naa", "output": "a" }, { "input": "18\niuiuqpyyaoaetiwliu", "output": "iqpytiwli" }, { "input": "5\nxxxxx", "output": "xxxxx" }, { "input": "6\nxxxahg", "output": "xxxahg" }, { "input": "3\nzcv", "output": "zcv" }, { "input": "4\naepo", "output": "apo" }, { "input": "5\nqqqqq", "output": "qqqqq" }, { "input": "6\naaaaaa", "output": "a" }, { "input": "4\naeta", "output": "ata" }, { "input": "20\nttyttlwaoieulyiluuri", "output": "ttyttlwalyluri" }, { "input": "1\nb", "output": "b" }, { "input": "3\nanc", "output": "anc" }, { "input": "1\ne", "output": "e" }, { "input": "3\naie", "output": "a" }, { "input": "3\nvio", "output": "vi" }, { "input": "2\nea", "output": "e" }, { "input": "3\nuas", "output": "us" }, { "input": "2\nba", "output": "ba" }, { "input": "2\naq", "output": "aq" }, { "input": "2\nya", "output": "y" }, { "input": "2\nou", "output": "o" }, { "input": "2\nbb", "output": "bb" }, { "input": "7\nayylmao", "output": "alma" }, { "input": "2\nab", "output": "ab" }, { "input": "19\nyuouiyaoiiweqrryqqp", "output": "yweqrryqqp" }, { "input": "25\niqypwqpriiioetiuqqqttouei", "output": "iqypwqpritiqqqtto" }, { "input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz", "output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz" }, { "input": "17\naccccccccccccccca", "output": "accccccccccccccca" }, { "input": "5\nababa", "output": "ababa" }, { "input": "10\naaaaaaaaaa", "output": "a" }, { "input": "22\naaaaabbbbboyoyoyoyoyac", "output": "abbbbboc" }, { "input": "7\nmahmoud", "output": "mahmod" } ]

1,594,042,025

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

3

124

20,172,800

n = int ( input ()) s = input () l = set ([ 'a' , 'i' , 'e' , 'o' , 'u' ]) p = [ '=' ] for i in s : if i in l and p [- 1 ] in l : continue p . append ( i ) print ( '' . join ( p [ 1 :]))

Title: Word Correction Time Limit: None seconds Memory Limit: None megabytes Problem Description: Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange. Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct. You are given a word *s*. Can you predict what will it become after correction? In this problem letters a, e, i, o, u and y are considered to be vowels. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction. The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction. Output Specification: Output the word *s* after the correction. Demo Input: ['5\nweird\n', '4\nword\n', '5\naaeaa\n'] Demo Output: ['werd\n', 'word\n', 'a\n'] Note: Explanations of the examples: 1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.

```python n = int ( input ()) s = input () l = set ([ 'a' , 'i' , 'e' , 'o' , 'u' ]) p = [ '=' ] for i in s : if i in l and p [- 1 ] in l : continue p . append ( i ) print ( '' . join ( p [ 1 :])) ```

0

849

A

Odds and Ends

PROGRAMMING

1,000

[ "implementation" ]

null

Where do odds begin, and where do they end? Where does hope emerge, and will they ever break? Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers. A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not.

The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence.

Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise. You can output each letter in any case (upper or lower).

[ "3\n1 3 5\n", "5\n1 0 1 5 1\n", "3\n4 3 1\n", "4\n3 9 9 3\n" ]

[ "Yes\n", "Yes\n", "No\n", "No\n" ]

In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met. In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}. In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met. In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

500

[ { "input": "3\n1 3 5", "output": "Yes" }, { "input": "5\n1 0 1 5 1", "output": "Yes" }, { "input": "3\n4 3 1", "output": "No" }, { "input": "4\n3 9 9 3", "output": "No" }, { "input": "1\n1", "output": "Yes" }, { "input": "5\n100 99 100 99 99", "output": "No" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "No" }, { "input": "1\n0", "output": "No" }, { "input": "2\n1 1", "output": "No" }, { "input": "2\n10 10", "output": "No" }, { "input": "2\n54 21", "output": "No" }, { "input": "5\n0 0 0 0 0", "output": "No" }, { "input": "5\n67 92 0 26 43", "output": "Yes" }, { "input": "15\n45 52 35 80 68 80 93 57 47 32 69 23 63 90 43", "output": "Yes" }, { "input": "15\n81 28 0 82 71 64 63 89 87 92 38 30 76 72 36", "output": "No" }, { "input": "50\n49 32 17 59 77 98 65 50 85 10 40 84 65 34 52 25 1 31 61 45 48 24 41 14 76 12 33 76 44 86 53 33 92 58 63 93 50 24 31 79 67 50 72 93 2 38 32 14 87 99", "output": "No" }, { "input": "55\n65 69 53 66 11 100 68 44 43 17 6 66 24 2 6 6 61 72 91 53 93 61 52 96 56 42 6 8 79 49 76 36 83 58 8 43 2 90 71 49 80 21 75 13 76 54 95 61 58 82 40 33 73 61 46", "output": "No" }, { "input": "99\n73 89 51 85 42 67 22 80 75 3 90 0 52 100 90 48 7 15 41 1 54 2 23 62 86 68 2 87 57 12 45 34 68 54 36 49 27 46 22 70 95 90 57 91 90 79 48 89 67 92 28 27 25 37 73 66 13 89 7 99 62 53 48 24 73 82 62 88 26 39 21 86 50 95 26 27 60 6 56 14 27 90 55 80 97 18 37 36 70 2 28 53 36 77 39 79 82 42 69", "output": "Yes" }, { "input": "99\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99", "output": "Yes" }, { "input": "100\n61 63 34 45 20 91 31 28 40 27 94 1 73 5 69 10 56 94 80 23 79 99 59 58 13 56 91 59 77 78 88 72 80 72 70 71 63 60 41 41 41 27 83 10 43 14 35 48 0 78 69 29 63 33 42 67 1 74 51 46 79 41 37 61 16 29 82 28 22 14 64 49 86 92 82 55 54 24 75 58 95 31 3 34 26 23 78 91 49 6 30 57 27 69 29 54 42 0 61 83", "output": "No" }, { "input": "6\n1 2 2 2 2 1", "output": "No" }, { "input": "3\n1 2 1", "output": "Yes" }, { "input": "4\n1 3 2 3", "output": "No" }, { "input": "6\n1 1 1 1 1 1", "output": "No" }, { "input": "6\n1 1 0 0 1 1", "output": "No" }, { "input": "4\n1 4 9 3", "output": "No" }, { "input": "4\n1 0 1 1", "output": "No" }, { "input": "10\n1 0 0 1 1 1 1 1 1 1", "output": "No" }, { "input": "10\n9 2 5 7 8 3 1 9 4 9", "output": "No" }, { "input": "99\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 2", "output": "No" }, { "input": "6\n1 2 1 2 2 1", "output": "No" }, { "input": "6\n1 0 1 0 0 1", "output": "No" }, { "input": "4\n1 3 4 7", "output": "No" }, { "input": "8\n1 1 1 2 1 1 1 1", "output": "No" }, { "input": "3\n1 1 2", "output": "No" }, { "input": "5\n1 2 1 2 1", "output": "Yes" }, { "input": "5\n5 4 4 2 1", "output": "Yes" }, { "input": "6\n1 3 3 3 3 1", "output": "No" }, { "input": "7\n1 2 1 2 2 2 1", "output": "Yes" }, { "input": "4\n1 2 2 1", "output": "No" }, { "input": "6\n1 2 3 4 6 5", "output": "No" }, { "input": "5\n1 1 2 2 2", "output": "No" }, { "input": "5\n1 0 0 1 1", "output": "Yes" }, { "input": "3\n1 2 4", "output": "No" }, { "input": "3\n1 0 2", "output": "No" }, { "input": "5\n1 1 1 0 1", "output": "Yes" }, { "input": "4\n3 9 2 3", "output": "No" }, { "input": "6\n1 1 1 4 4 1", "output": "No" }, { "input": "6\n1 2 3 5 6 7", "output": "No" }, { "input": "6\n1 1 1 2 2 1", "output": "No" }, { "input": "6\n1 1 1 0 0 1", "output": "No" }, { "input": "5\n1 2 2 5 5", "output": "Yes" }, { "input": "5\n1 3 2 4 5", "output": "Yes" }, { "input": "8\n1 2 3 5 7 8 8 5", "output": "No" }, { "input": "10\n1 1 1 2 1 1 1 1 1 1", "output": "No" }, { "input": "4\n1 0 0 1", "output": "No" }, { "input": "7\n1 0 1 1 0 0 1", "output": "Yes" }, { "input": "7\n1 4 5 7 6 6 3", "output": "Yes" }, { "input": "4\n2 2 2 2", "output": "No" }, { "input": "5\n2 3 4 5 6", "output": "No" }, { "input": "4\n1 1 2 1", "output": "No" }, { "input": "3\n1 2 3", "output": "Yes" }, { "input": "6\n1 3 3 2 2 3", "output": "No" }, { "input": "4\n1 1 2 3", "output": "No" }, { "input": "4\n1 2 3 5", "output": "No" }, { "input": "5\n3 4 4 3 3", "output": "Yes" }, { "input": "4\n3 2 2 3", "output": "No" }, { "input": "6\n1 1 1 1 2 1", "output": "No" }, { "input": "6\n1 1 2 2 1 1", "output": "No" }, { "input": "10\n3 4 2 4 3 2 2 4 4 3", "output": "No" }, { "input": "7\n1 2 4 3 2 4 5", "output": "Yes" }, { "input": "28\n75 51 25 52 13 7 34 29 5 59 68 56 13 2 9 37 59 83 18 32 36 30 20 43 92 76 78 67", "output": "No" }, { "input": "79\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18", "output": "No" }, { "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": "No" } ]

1,595,868,452

2,147,483,647

Python 3

OK

TESTS

74

124

6,758,400

n = int(input()) arr = list(map(int,input().split())) if n%2!=0 and arr[0]%2!=0 and arr[-1]%2!=0: print("Yes") else: print("No")

Title: Odds and Ends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Where do odds begin, and where do they end? Where does hope emerge, and will they ever break? Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers. A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not. Input Specification: The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence. Output Specification: Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise. You can output each letter in any case (upper or lower). Demo Input: ['3\n1 3 5\n', '5\n1 0 1 5 1\n', '3\n4 3 1\n', '4\n3 9 9 3\n'] Demo Output: ['Yes\n', 'Yes\n', 'No\n', 'No\n'] Note: In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met. In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}. In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met. In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

```python n = int(input()) arr = list(map(int,input().split())) if n%2!=0 and arr[0]%2!=0 and arr[-1]%2!=0: print("Yes") else: print("No") ```

3

574

A

Bear and Elections

PROGRAMMING

1,200

[ "greedy", "implementation" ]

null

Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland. There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate. Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?

The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1. Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.

Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.

[ "5\n5 1 11 2 8\n", "4\n1 8 8 8\n", "2\n7 6\n" ]

[ "4\n", "6\n", "0\n" ]

In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8. In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6. In the third sample Limak is a winner without bribing any citizen.

500

[ { "input": "5\n5 1 11 2 8", "output": "4" }, { "input": "4\n1 8 8 8", "output": "6" }, { "input": "2\n7 6", "output": "0" }, { "input": "2\n1 1", "output": "1" }, { "input": "10\n100 200 57 99 1 1000 200 200 200 500", "output": "451" }, { "input": "16\n7 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "932" }, { "input": "100\n47 64 68 61 68 66 69 61 69 65 69 63 62 60 68 65 64 65 65 62 63 68 60 70 63 63 65 67 70 69 68 69 61 65 63 60 60 65 61 60 70 66 66 65 62 60 65 68 61 62 67 64 66 65 67 68 60 69 70 63 65 62 64 65 67 67 69 68 66 69 70 67 65 70 60 66 70 67 67 64 69 69 66 68 60 64 62 62 68 69 67 69 60 70 69 68 62 63 68 66", "output": "23" }, { "input": "2\n96 97", "output": "1" }, { "input": "2\n1000 1000", "output": "1" }, { "input": "3\n999 1000 1000", "output": "2" }, { "input": "3\n1 2 3", "output": "2" }, { "input": "7\n10 940 926 990 946 980 985", "output": "817" }, { "input": "10\n5 3 4 5 5 2 1 8 4 1", "output": "2" }, { "input": "15\n17 15 17 16 13 17 13 16 14 14 17 17 13 15 17", "output": "1" }, { "input": "20\n90 5 62 9 50 7 14 43 44 44 56 13 71 22 43 35 52 60 73 54", "output": "0" }, { "input": "30\n27 85 49 7 77 38 4 68 23 28 81 100 40 9 78 38 1 60 60 49 98 44 45 92 46 39 98 24 37 39", "output": "58" }, { "input": "51\n90 47 100 12 21 96 2 68 84 60 2 9 33 8 45 13 59 50 100 93 22 97 4 81 51 2 3 78 19 16 25 63 52 34 79 32 34 87 7 42 96 93 30 33 33 43 69 8 63 58 57", "output": "8" }, { "input": "77\n1000 2 2 3 1 1 1 3 3 2 1 1 3 2 2 2 3 2 3 1 3 1 1 2 2 2 3 1 1 2 2 2 3 2 1 3 3 1 2 3 3 3 2 1 3 2 1 3 3 2 3 3 2 1 3 1 1 1 2 3 2 3 1 3 1 2 1 2 2 2 1 2 2 3 2 2 2", "output": "0" }, { "input": "91\n3 92 89 83 85 80 91 94 95 82 92 95 80 88 90 85 81 90 87 86 94 88 90 87 88 82 95 84 84 93 83 95 91 85 89 88 88 85 87 90 93 80 89 95 94 92 93 86 83 82 86 84 91 80 90 95 84 86 84 85 84 92 82 84 83 91 87 95 94 95 90 95 86 92 86 80 95 86 88 80 82 87 84 83 91 93 81 81 91 89 88", "output": "89" }, { "input": "100\n1 3 71 47 64 82 58 61 61 35 52 36 57 62 63 54 52 21 78 100 24 94 4 80 99 62 43 72 21 70 90 4 23 14 72 4 76 49 71 96 96 99 78 7 32 11 14 61 19 69 1 68 100 77 86 54 14 86 47 53 30 88 67 66 61 70 17 63 40 5 99 53 38 31 91 18 41 5 77 61 53 30 87 21 23 54 52 17 23 75 58 99 99 63 20 1 78 72 28 11", "output": "90" }, { "input": "100\n1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "99" }, { "input": "94\n3 100 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 100 99 100 99 100 100 100 99 100 99 100 99 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 99 99 100 99 100 99 99 99 100 100 99 100 100 99 99 100 100 100 99 100 99 99 99 99 99 100 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 99 100 100", "output": "97" }, { "input": "97\n99 99 98 98 100 98 99 99 98 100 100 100 99 99 100 99 99 98 99 99 98 98 98 100 100 99 98 99 100 98 99 98 98 100 98 99 100 98 98 99 98 98 99 98 100 99 99 99 99 98 98 98 100 99 100 100 99 99 100 99 99 98 98 98 100 100 98 100 100 99 98 99 100 98 98 98 98 99 99 98 98 99 100 100 98 98 99 98 99 100 98 99 100 98 99 99 100", "output": "2" }, { "input": "100\n100 55 70 81 73 51 6 75 45 85 33 61 98 63 11 59 1 8 14 28 78 74 44 80 7 69 7 5 90 73 43 78 64 64 43 92 59 70 80 19 33 39 31 70 38 85 24 23 86 79 98 56 92 63 92 4 36 8 79 74 2 81 54 13 69 44 49 63 17 76 78 99 42 36 47 71 19 90 9 58 83 53 27 2 35 51 65 59 90 51 74 87 84 48 98 44 84 100 84 93", "output": "1" }, { "input": "100\n100 637 498 246 615 901 724 673 793 33 282 908 477 185 185 969 34 859 90 70 107 492 227 918 919 131 620 182 802 703 779 184 403 891 448 499 628 553 905 392 70 396 8 575 66 908 992 496 792 174 667 355 836 610 855 377 244 827 836 808 667 354 800 114 746 556 75 894 162 367 99 718 394 273 833 776 151 433 315 470 759 12 552 613 85 793 775 649 225 86 296 624 557 201 209 595 697 527 282 168", "output": "749" }, { "input": "100\n107 172 549 883 564 56 399 970 173 990 224 217 601 381 948 631 159 958 512 136 61 584 633 202 652 355 26 723 663 237 410 721 688 552 699 24 748 186 461 88 34 243 872 205 471 298 654 693 244 33 359 533 471 116 386 653 654 887 531 303 335 829 319 340 827 89 602 191 422 289 361 200 593 421 592 402 256 813 606 589 741 9 148 893 3 142 50 169 219 360 642 45 810 818 507 624 561 743 303 111", "output": "729" }, { "input": "90\n670 694 651 729 579 539 568 551 707 638 604 544 502 531 775 805 558 655 506 729 802 778 653 737 591 770 594 535 588 604 658 713 779 705 504 563 513 651 529 572 505 553 515 750 621 574 727 774 714 725 665 798 670 747 751 635 755 798 635 717 583 682 517 546 740 802 743 507 658 700 645 671 533 594 506 633 768 584 672 666 703 522 530 501 592 528 678 708 619 786", "output": "111" }, { "input": "90\n10 265 429 431 343 305 806 746 284 313 503 221 594 351 83 653 232 431 427 610 458 88 255 215 529 205 492 549 55 694 535 104 45 327 816 432 595 549 454 141 216 557 250 415 531 494 190 749 718 380 78 447 784 347 196 814 16 780 262 462 776 315 160 307 593 694 692 41 528 725 376 777 337 44 438 630 345 502 384 184 742 429 570 361 394 267 820 778 662 377", "output": "714" }, { "input": "95\n800 280 176 472 587 763 588 838 760 378 667 231 566 278 713 305 354 815 140 220 188 409 109 180 251 268 474 590 853 143 235 691 313 785 386 92 783 471 43 342 718 592 678 404 256 362 239 504 163 85 521 81 356 73 754 589 380 159 196 862 838 509 149 42 366 630 467 292 698 123 187 796 576 37 689 800 186 518 488 432 159 860 349 799 282 304 880 283 23 312 55 507 734 370 490", "output": "52" }, { "input": "100\n95 88 84 85 74 97 100 52 91 94 62 66 90 56 86 66 95 73 79 68 54 67 99 52 82 62 81 71 93 85 72 72 95 52 72 63 57 90 92 89 88 77 84 78 95 59 72 86 98 64 89 64 80 70 54 93 88 86 79 78 94 64 89 66 50 90 54 82 52 96 99 54 81 66 83 79 69 80 51 73 81 69 93 82 76 52 58 87 93 92 52 67 78 63 63 87 77 95 58 78", "output": "4" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "1" }, { "input": "100\n999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "2" }, { "input": "100\n901 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "99" }, { "input": "4\n8 5 11 3", "output": "2" }, { "input": "3\n3 10 1", "output": "4" } ]

1,616,215,117

2,147,483,647

PyPy 3

OK

TESTS

35

109

1,536,000

n = int(input()) a = list(map(int, input().split())) mx = max(a) for i in range(a[0], mx + 2): need = i - a[0] changes = sum([x - i + 1 for x in a[1:] if x >= i]) if changes <= need: print(need) exit(0)

Title: Bear and Elections Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland. There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate. Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe? Input Specification: The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1. Note that after bribing number of votes for some candidate might be zero or might be greater than 1000. Output Specification: Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate. Demo Input: ['5\n5 1 11 2 8\n', '4\n1 8 8 8\n', '2\n7 6\n'] Demo Output: ['4\n', '6\n', '0\n'] Note: In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8. In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6. In the third sample Limak is a winner without bribing any citizen.

```python n = int(input()) a = list(map(int, input().split())) mx = max(a) for i in range(a[0], mx + 2): need = i - a[0] changes = sum([x - i + 1 for x in a[1:] if x >= i]) if changes <= need: print(need) exit(0) ```

3

621

A

Wet Shark and Odd and Even

PROGRAMMING

900

[ "implementation" ]

null

Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.

The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

Print the maximum possible even sum that can be obtained if we use some of the given integers.

[ "3\n1 2 3\n", "5\n999999999 999999999 999999999 999999999 999999999\n" ]

[ "6", "3999999996" ]

In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

500

[ { "input": "3\n1 2 3", "output": "6" }, { "input": "5\n999999999 999999999 999999999 999999999 999999999", "output": "3999999996" }, { "input": "1\n1", "output": "0" }, { "input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98", "output": "870" }, { "input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25", "output": "840" }, { "input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44", "output": "922" }, { "input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59", "output": "674" }, { "input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82", "output": "848" }, { "input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93", "output": "902" }, { "input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20", "output": "798" }, { "input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31", "output": "632" }, { "input": "1\n4", "output": "4" }, { "input": "2\n1 4", "output": "4" }, { "input": "3\n1 2 4", "output": "6" }, { "input": "2\n9 3", "output": "12" }, { "input": "2\n1000000000 1001", "output": "1000000000" }, { "input": "3\n1 8 4", "output": "12" }, { "input": "3\n7 4 4", "output": "8" }, { "input": "5\n2 3 4 5 3", "output": "14" }, { "input": "2\n4 5", "output": "4" }, { "input": "3\n2 4 5", "output": "6" }, { "input": "3\n2 2 3", "output": "4" }, { "input": "2\n2 3", "output": "2" }, { "input": "4\n2 3 7 7", "output": "16" }, { "input": "2\n999999999 2", "output": "2" }, { "input": "2\n2 5", "output": "2" }, { "input": "3\n5 3 1", "output": "8" }, { "input": "4\n3 2 5 7", "output": "14" } ]

1,598,207,415

2,147,483,647

Python 3

OK

TESTS

87

202

15,462,400

n=int(input()) a=list(map(int,input().split())) s=sum(a) ans,even,odd=0,0,0 a.sort() for i in range(n): if a[i]%2!=0: even+=1 odd+=1 if even==1: ans=a[i] if odd%2==0: print(s) else: print(s-ans)

Title: Wet Shark and Odd and Even Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. Input Specification: The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. Output Specification: Print the maximum possible even sum that can be obtained if we use some of the given integers. Demo Input: ['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n'] Demo Output: ['6', '3999999996'] Note: In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

```python n=int(input()) a=list(map(int,input().split())) s=sum(a) ans,even,odd=0,0,0 a.sort() for i in range(n): if a[i]%2!=0: even+=1 odd+=1 if even==1: ans=a[i] if odd%2==0: print(s) else: print(s-ans) ```

3

946

A

Partition

PROGRAMMING

800

[ "greedy" ]

null

You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences. Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.

Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.

[ "3\n1 -2 0\n", "6\n16 23 16 15 42 8\n" ]

[ "3\n", "120\n" ]

In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3. In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.

0

[ { "input": "3\n1 -2 0", "output": "3" }, { "input": "6\n16 23 16 15 42 8", "output": "120" }, { "input": "1\n-1", "output": "1" }, { "input": "100\n-100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100", "output": "10000" }, { "input": "2\n-1 5", "output": "6" }, { "input": "3\n-2 0 1", "output": "3" }, { "input": "12\n-1 -2 -3 4 4 -6 -6 56 3 3 -3 3", "output": "94" }, { "input": "4\n1 -1 1 -1", "output": "4" }, { "input": "4\n100 -100 100 -100", "output": "400" }, { "input": "3\n-2 -5 10", "output": "17" }, { "input": "5\n1 -2 3 -4 5", "output": "15" }, { "input": "3\n-100 100 -100", "output": "300" }, { "input": "6\n1 -1 1 -1 1 -1", "output": "6" }, { "input": "6\n2 -2 2 -2 2 -2", "output": "12" }, { "input": "9\n12 93 -2 0 0 0 3 -3 -9", "output": "122" }, { "input": "6\n-1 2 4 -5 -3 55", "output": "70" }, { "input": "6\n-12 8 68 -53 1 -15", "output": "157" }, { "input": "2\n-2 1", "output": "3" }, { "input": "3\n100 -100 100", "output": "300" }, { "input": "5\n100 100 -1 -100 2", "output": "303" }, { "input": "6\n-5 -4 -3 -2 -1 0", "output": "15" }, { "input": "6\n4 4 4 -3 -3 2", "output": "20" }, { "input": "2\n-1 2", "output": "3" }, { "input": "1\n100", "output": "100" }, { "input": "5\n-1 -2 3 1 2", "output": "9" }, { "input": "5\n100 -100 100 -100 100", "output": "500" }, { "input": "5\n1 -1 1 -1 1", "output": "5" }, { "input": "4\n0 0 0 -1", "output": "1" }, { "input": "5\n100 -100 -1 2 100", "output": "303" }, { "input": "2\n75 0", "output": "75" }, { "input": "4\n55 56 -59 -58", "output": "228" }, { "input": "2\n9 71", "output": "80" }, { "input": "2\n9 70", "output": "79" }, { "input": "2\n9 69", "output": "78" }, { "input": "2\n100 -100", "output": "200" }, { "input": "4\n-9 4 -9 5", "output": "27" }, { "input": "42\n91 -27 -79 -56 80 -93 -23 10 80 94 61 -89 -64 81 34 99 31 -32 -69 92 79 -9 73 66 -8 64 99 99 58 -19 -40 21 1 -33 93 -23 -62 27 55 41 57 36", "output": "2348" }, { "input": "7\n-1 2 2 2 -1 2 -1", "output": "11" }, { "input": "6\n-12 8 17 -69 7 -88", "output": "201" }, { "input": "3\n1 -2 5", "output": "8" }, { "input": "6\n-2 3 -4 5 6 -1", "output": "21" }, { "input": "2\n-5 1", "output": "6" }, { "input": "4\n2 2 -2 4", "output": "10" }, { "input": "68\n21 47 -75 -25 64 83 83 -21 89 24 43 44 -35 34 -42 92 -96 -52 -66 64 14 -87 25 -61 -78 83 -96 -18 95 83 -93 -28 75 49 87 65 -93 -69 -2 95 -24 -36 -61 -71 88 -53 -93 -51 -81 -65 -53 -46 -56 6 65 58 19 100 57 61 -53 44 -58 48 -8 80 -88 72", "output": "3991" }, { "input": "5\n5 5 -10 -1 1", "output": "22" }, { "input": "3\n-1 2 3", "output": "6" }, { "input": "76\n57 -38 -48 -81 93 -32 96 55 -44 2 38 -46 42 64 71 -73 95 31 -39 -62 -1 75 -17 57 28 52 12 -11 82 -84 59 -86 73 -97 34 97 -57 -85 -6 39 -5 -54 95 24 -44 35 -18 9 91 7 -22 -61 -80 54 -40 74 -90 15 -97 66 -52 -49 -24 65 21 -93 -29 -24 -4 -1 76 -93 7 -55 -53 1", "output": "3787" }, { "input": "5\n-1 -2 1 2 3", "output": "9" }, { "input": "4\n2 2 -2 -2", "output": "8" }, { "input": "6\n100 -100 100 -100 100 -100", "output": "600" }, { "input": "100\n-59 -33 34 0 69 24 -22 58 62 -36 5 45 -19 -73 61 -9 95 42 -73 -64 91 -96 2 53 -8 82 -79 16 18 -5 -53 26 71 38 -31 12 -33 -1 -65 -6 3 -89 22 33 -27 -36 41 11 -47 -32 47 -56 -38 57 -63 -41 23 41 29 78 16 -65 90 -58 -12 6 -60 42 -36 -52 -54 -95 -10 29 70 50 -94 1 93 48 -71 -77 -16 54 56 -60 66 76 31 8 44 -61 -74 23 37 38 18 -18 29 41", "output": "4362" }, { "input": "2\n-1 1", "output": "2" }, { "input": "3\n1 -2 100", "output": "103" }, { "input": "5\n1 -2 3 1 2", "output": "9" }, { "input": "10\n100 -10 -100 10 10 10 10 10 10 10", "output": "280" }, { "input": "4\n2 0 -2 4", "output": "8" }, { "input": "4\n3 -3 1 -1", "output": "8" }, { "input": "3\n1 -1 1", "output": "3" }, { "input": "4\n2 5 -2 4", "output": "13" }, { "input": "2\n-2 2", "output": "4" }, { "input": "3\n1 -2 1", "output": "4" }, { "input": "5\n-1 -2 1 1 -1", "output": "6" }, { "input": "4\n-2 0 2 4", "output": "8" }, { "input": "8\n-42 7 87 -16 -5 65 -88 1", "output": "311" }, { "input": "3\n1 -3 4", "output": "8" }, { "input": "1\n1", "output": "1" }, { "input": "2\n0 1", "output": "1" }, { "input": "3\n-1 2 -1", "output": "4" }, { "input": "18\n-21 12 65 66 -24 62 82 35 -45 -47 28 37 5 -32 22 -14 -69 -95", "output": "761" }, { "input": "4\n-1 1 -1 1", "output": "4" }, { "input": "5\n-1 2 1 1 1", "output": "6" }, { "input": "3\n1 1 1", "output": "3" } ]

1,626,567,305

2,147,483,647

Python 3

OK

TESTS

72

77

6,758,400

n = int(input()) a = list(map(int, input().split())) b = [] c = [] for x in a: if (x >= 0): b.append(x) else: c.append(x) print(sum(b) - sum(c))

Title: Partition Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences. Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*? Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*. Output Specification: Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*. Demo Input: ['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n'] Demo Output: ['3\n', '120\n'] Note: In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3. In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.

```python n = int(input()) a = list(map(int, input().split())) b = [] c = [] for x in a: if (x >= 0): b.append(x) else: c.append(x) print(sum(b) - sum(c)) ```

3

991

A

If at first you don't succeed...

PROGRAMMING

1,000

[ "implementation" ]

null

Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam. Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group. Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?

The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).

If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam. If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.

[ "10 10 5 20\n", "2 2 0 4\n", "2 2 2 1\n" ]

[ "5", "-1", "-1" ]

The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam. In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible. The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.

500

[ { "input": "10 10 5 20", "output": "5" }, { "input": "2 2 0 4", "output": "-1" }, { "input": "2 2 2 1", "output": "-1" }, { "input": "98 98 97 100", "output": "1" }, { "input": "1 5 2 10", "output": "-1" }, { "input": "5 1 2 10", "output": "-1" }, { "input": "6 7 5 8", "output": "-1" }, { "input": "6 7 5 9", "output": "1" }, { "input": "6 7 5 7", "output": "-1" }, { "input": "50 50 1 100", "output": "1" }, { "input": "8 3 2 12", "output": "3" }, { "input": "10 19 6 25", "output": "2" }, { "input": "1 0 0 99", "output": "98" }, { "input": "0 1 0 98", "output": "97" }, { "input": "1 1 0 97", "output": "95" }, { "input": "1 1 1 96", "output": "95" }, { "input": "0 0 0 0", "output": "-1" }, { "input": "100 0 0 0", "output": "-1" }, { "input": "0 100 0 0", "output": "-1" }, { "input": "100 100 0 0", "output": "-1" }, { "input": "0 0 100 0", "output": "-1" }, { "input": "100 0 100 0", "output": "-1" }, { "input": "0 100 100 0", "output": "-1" }, { "input": "100 100 100 0", "output": "-1" }, { "input": "0 0 0 100", "output": "100" }, { "input": "100 0 0 100", "output": "-1" }, { "input": "0 100 0 100", "output": "-1" }, { "input": "100 100 0 100", "output": "-1" }, { "input": "0 0 100 100", "output": "-1" }, { "input": "100 0 100 100", "output": "-1" }, { "input": "0 100 100 100", "output": "-1" }, { "input": "100 100 100 100", "output": "-1" }, { "input": "10 45 7 52", "output": "4" }, { "input": "38 1 1 68", "output": "30" }, { "input": "8 45 2 67", "output": "16" }, { "input": "36 36 18 65", "output": "11" }, { "input": "10 30 8 59", "output": "27" }, { "input": "38 20 12 49", "output": "3" }, { "input": "8 19 4 38", "output": "15" }, { "input": "36 21 17 72", "output": "32" }, { "input": "14 12 12 89", "output": "75" }, { "input": "38 6 1 44", "output": "1" }, { "input": "13 4 6 82", "output": "-1" }, { "input": "5 3 17 56", "output": "-1" }, { "input": "38 5 29 90", "output": "-1" }, { "input": "22 36 18 55", "output": "15" }, { "input": "13 0 19 75", "output": "-1" }, { "input": "62 65 10 89", "output": "-1" }, { "input": "2 29 31 72", "output": "-1" }, { "input": "1 31 19 55", "output": "-1" }, { "input": "1 25 28 88", "output": "-1" }, { "input": "34 32 28 33", "output": "-1" }, { "input": "43 36 1 100", "output": "22" }, { "input": "16 39 55 70", "output": "-1" }, { "input": "2 3 0 91", "output": "86" }, { "input": "55 29 12 48", "output": "-1" }, { "input": "7 33 20 88", "output": "-1" }, { "input": "40 38 27 99", "output": "48" }, { "input": "18 28 14 84", "output": "52" }, { "input": "34 25 25 92", "output": "58" }, { "input": "4 24 5 76", "output": "-1" }, { "input": "5 22 16 96", "output": "-1" }, { "input": "1 1 0 4", "output": "2" }, { "input": "5 5 3 1", "output": "-1" }, { "input": "0 0 0 1", "output": "1" }, { "input": "2 3 0 8", "output": "3" }, { "input": "5 5 2 5", "output": "-1" }, { "input": "1 2 1 3", "output": "1" }, { "input": "3 0 0 4", "output": "1" }, { "input": "0 0 0 5", "output": "5" }, { "input": "5 5 0 3", "output": "-1" }, { "input": "5 6 1 7", "output": "-1" }, { "input": "10 10 10 11", "output": "1" }, { "input": "0 0 0 10", "output": "10" }, { "input": "5 15 5 30", "output": "15" }, { "input": "3 2 0 7", "output": "2" } ]

1,655,958,506

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

4

46

0

import sys input = sys.stdin.readline a, b, c, n = map(int, input().split()) ans = n - a - b + c if(ans > 0): print(ans) else: print(-1)

Title: If at first you don't succeed... Time Limit: None seconds Memory Limit: None megabytes Problem Description: Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam. Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group. Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? Input Specification: The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). Output Specification: If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam. If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. Demo Input: ['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n'] Demo Output: ['5', '-1', '-1'] Note: The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam. In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible. The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.

```python import sys input = sys.stdin.readline a, b, c, n = map(int, input().split()) ans = n - a - b + c if(ans > 0): print(ans) else: print(-1) ```

0

4

C

Registration System

PROGRAMMING

1,300

[ "data structures", "hashing", "implementation" ]

C. Registration system

5

64

A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle. Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database.

The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.

Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.

[ "4\nabacaba\nacaba\nabacaba\nacab\n", "6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n" ]

[ "OK\nOK\nabacaba1\nOK\n", "OK\nfirst1\nOK\nsecond1\nOK\nthird1\n" ]

none

0

[ { "input": "4\nabacaba\nacaba\nabacaba\nacab", "output": "OK\nOK\nabacaba1\nOK" }, { "input": "6\nfirst\nfirst\nsecond\nsecond\nthird\nthird", "output": "OK\nfirst1\nOK\nsecond1\nOK\nthird1" }, { "input": "1\nn", "output": "OK" }, { "input": "2\nu\nu", "output": "OK\nu1" }, { "input": "3\nb\nb\nb", "output": "OK\nb1\nb2" }, { "input": "2\nc\ncn", "output": "OK\nOK" }, { "input": "3\nvhn\nvhn\nh", "output": "OK\nvhn1\nOK" }, { "input": "4\nd\nhd\nd\nh", "output": "OK\nOK\nd1\nOK" }, { "input": "10\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp\nbhnqaptmp", "output": "OK\nbhnqaptmp1\nbhnqaptmp2\nbhnqaptmp3\nbhnqaptmp4\nbhnqaptmp5\nbhnqaptmp6\nbhnqaptmp7\nbhnqaptmp8\nbhnqaptmp9" }, { "input": "10\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\nfpqhfouqdldravpjttarh\njmvlplnrmba\nfpqhfouqdldravpjttarh\njmvlplnrmba\nfpqhfouqdldravpjttarh", "output": "OK\nfpqhfouqdldravpjttarh1\nfpqhfouqdldravpjttarh2\nfpqhfouqdldravpjttarh3\nfpqhfouqdldravpjttarh4\nfpqhfouqdldravpjttarh5\nOK\nfpqhfouqdldravpjttarh6\njmvlplnrmba1\nfpqhfouqdldravpjttarh7" }, { "input": "10\niwexcrupuubwzbooj\niwexcrupuubwzbooj\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\njzsyjnxttliyfpunxyhsouhunenzxedi\niwexcrupuubwzbooj\niwexcrupuubwzbooj\niwexcrupuubwzbooj", "output": "OK\niwexcrupuubwzbooj1\nOK\njzsyjnxttliyfpunxyhsouhunenzxedi1\njzsyjnxttliyfpunxyhsouhunenzxedi2\njzsyjnxttliyfpunxyhsouhunenzxedi3\njzsyjnxttliyfpunxyhsouhunenzxedi4\niwexcrupuubwzbooj2\niwexcrupuubwzbooj3\niwexcrupuubwzbooj4" }, { "input": "10\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzz", "output": "OK\nzzzzzzzzzzzzzzzzzzzzzzzzzzz1\nOK\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz1\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz3\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz4\nzzzzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzzzzzzzzzzzzzzzz3\nzzzzzzzzzzzzzzzzzzzzzzzzzzz4" }, { "input": "20\nzzzzzzzzz\nzzzzzzzzzzzzz\nz\nzzzzzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzz\nz\nzzzzzzzzz\nzzzzzzzzz\nzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzz", "output": "OK\nOK\nOK\nzzzzzzzzzzzzz1\nzzzzzzzzz1\nzzzzzzzzz2\nzzzzzzzzzzzzz2\nzzzzzzzzzzzzz3\nOK\nzzzzzzzzzzzzz4\nzzzzzzzzzzzzz5\nzzzzzzzzzzzzzzzzzzzzzzzz1\nzzzzzzzzzzzzzzzzzzzzzzzz2\nzzzzzzzzzzzzz6\nzzzzzzzzzzzzz7\nz1\nzzzzzzzzz3\nzzzzzzzzz4\nzzzzzzzzzzzzz8\nzzzzzzzzzzzzzzzzzzzzzzzz3" } ]

1,692,190,762

2,147,483,647

PyPy 3-64

OK

TESTS

25

2,370

23,347,200

import io,os from collections import * from math import * from string import * from itertools import * from statistics import * from fractions import * from functools import reduce from heapq import * #subsets here (; def f(s,n): return list(combinations(s,n)) def subsets(s): x=[] for i in range(0,len(s)+1): x.extend(f(s,i)) return x d={} for _ in range(int(input())): s=input() if s in d: print(s+str(d[s])) d[s]+=1 else: d[s]=1 print('OK')

Title: Registration System Time Limit: 5 seconds Memory Limit: 64 megabytes Problem Description: A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle. Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database. Input Specification: The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters. Output Specification: Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken. Demo Input: ['4\nabacaba\nacaba\nabacaba\nacab\n', '6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n'] Demo Output: ['OK\nOK\nabacaba1\nOK\n', 'OK\nfirst1\nOK\nsecond1\nOK\nthird1\n'] Note: none

```python import io,os from collections import * from math import * from string import * from itertools import * from statistics import * from fractions import * from functools import reduce from heapq import * #subsets here (; def f(s,n): return list(combinations(s,n)) def subsets(s): x=[] for i in range(0,len(s)+1): x.extend(f(s,i)) return x d={} for _ in range(int(input())): s=input() if s in d: print(s+str(d[s])) d[s]+=1 else: d[s]=1 print('OK') ```

3.58905

296

A

Yaroslav and Permutations

PROGRAMMING

1,100

[ "greedy", "math" ]

null

Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav.

The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.

In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.

[ "1\n1\n", "3\n1 1 2\n", "4\n7 7 7 7\n" ]

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

In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.

500

[ { "input": "1\n1", "output": "YES" }, { "input": "3\n1 1 2", "output": "YES" }, { "input": "4\n7 7 7 7", "output": "NO" }, { "input": "4\n479 170 465 146", "output": "YES" }, { "input": "5\n996 437 605 996 293", "output": "YES" }, { "input": "6\n727 539 896 668 36 896", "output": "YES" }, { "input": "7\n674 712 674 674 674 674 674", "output": "NO" }, { "input": "8\n742 742 742 742 742 289 742 742", "output": "NO" }, { "input": "9\n730 351 806 806 806 630 85 757 967", "output": "YES" }, { "input": "10\n324 539 83 440 834 640 440 440 440 440", "output": "YES" }, { "input": "7\n925 830 925 98 987 162 356", "output": "YES" }, { "input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808", "output": "YES" }, { "input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94", "output": "YES" }, { "input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402", "output": "NO" }, { "input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187", "output": "YES" }, { "input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103", "output": "YES" }, { "input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925", "output": "NO" }, { "input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353", "output": "YES" }, { "input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285", "output": "NO" }, { "input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99", "output": "YES" }, { "input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26", "output": "YES" }, { "input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516", "output": "YES" }, { "input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314", "output": "NO" }, { "input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800", "output": "NO" }, { "input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317", "output": "NO" }, { "input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784", "output": "YES" }, { "input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474", "output": "YES" }, { "input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499", "output": "NO" }, { "input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816", "output": "NO" }, { "input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167", "output": "YES" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "2\n1 1", "output": "NO" }, { "input": "1\n1000", "output": "YES" }, { "input": "12\n2 2 4 4 4 4 6 6 6 6 6 6", "output": "YES" } ]

1,634,030,229

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

92

6,758,400

n = int(input()) l = [int(i) for i in input().split()] ans = 'YES' for i in range(n): if l.count(i)>(n+1)/2: ans = 'NO' print(ans)

Title: Yaroslav and Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. Output Specification: In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. Demo Input: ['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.

```python n = int(input()) l = [int(i) for i in input().split()] ans = 'YES' for i in range(n): if l.count(i)>(n+1)/2: ans = 'NO' print(ans) ```

0

1

A

Theatre Square

PROGRAMMING

1,000

[ "math" ]

A. Theatre Square

1

256

Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).

Write the needed number of flagstones.

[ "6 6 4\n" ]

[ "4\n" ]

none

0

[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]

1,671,467,142

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

46

0

import math n, m, a = map(int,input().split()) s = (m*n) k = (a*a) print(math.ceil((s/k)*2))

Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none

```python import math n, m, a = map(int,input().split()) s = (m*n) k = (a*a) print(math.ceil((s/k)*2)) ```

0

305

B

Continued Fractions

PROGRAMMING

1,700

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

null

A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal.

The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction. The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Print "YES" if these fractions are equal and "NO" otherwise.

[ "9 4\n2\n2 4\n", "9 4\n3\n2 3 1\n", "9 4\n3\n1 2 4\n" ]

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

In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

1,000

[ { "input": "9 4\n2\n2 4", "output": "YES" }, { "input": "9 4\n3\n2 3 1", "output": "YES" }, { "input": "9 4\n3\n1 2 4", "output": "NO" }, { "input": "39088169 24157817\n36\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 2", "output": "YES" }, { "input": "39088169 24157817\n36\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 4", "output": "NO" }, { "input": "61305790721611591 37889062373143906\n80\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 4", "output": "NO" }, { "input": "61305790721611591 37889062373143906\n80\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 2", "output": "YES" }, { "input": "565049485241691020 228217260073568804\n40\n2 2 9 1 7 1 2 1 2 1 1 1 9 1 2 1 9 1 3 2 3 10 13 2 1 2 7 1 1 2 2 2 1 1 2 1 6 5 3 2", "output": "YES" }, { "input": "2 1\n4\n2 1 1 1", "output": "NO" }, { "input": "4 1\n2\n3 1", "output": "YES" }, { "input": "72723460248141 1597\n1\n45537545554", "output": "NO" }, { "input": "14930352 13\n6\n1148488 1 1 1 1 2", "output": "YES" }, { "input": "86267571272 102334155\n6\n842 1 841 1 842 145", "output": "NO" }, { "input": "72723460248141 121393\n7\n599074578 122 1 122 2 1 2", "output": "YES" }, { "input": "168455988218483660 53310571951833359\n32\n3 6 3 1 14 1 48 1 3 2 1 1 39 2 1 3 13 23 4 1 11 1 1 23 1 3 3 2 1 1 1 3", "output": "NO" }, { "input": "382460255113156464 275525972692563593\n37\n1 2 1 1 2 1 3 4 5 5 1 4 2 1 1 1 4 2 2 1 2 1 1 2 3 3 1 2 2 50 4 1 4 2 5 109 8", "output": "YES" }, { "input": "1000000000000000000 1\n1\n1000000000000000000", "output": "YES" }, { "input": "362912509915545727 266073193475139553\n30\n1 2 1 2 1 25 75 1 14 6 6 9 1 1 1 1 210 2 2 2 5 2 1 3 1 1 13 3 14 3", "output": "NO" }, { "input": "933329105990871495 607249523603826772\n33\n1 1 1 6 3 1 5 24 3 55 1 15 2 2 1 12 2 2 3 109 1 1 4 1 4 1 7 2 4 1 3 3 2", "output": "YES" }, { "input": "790637895857383456 679586240913926415\n40\n1 6 8 2 1 2 1 7 2 4 1 1 1 10 1 10 1 4 1 4 41 1 1 7 1 1 2 1 2 4 1 2 1 63 1 2 1 1 4 3", "output": "NO" }, { "input": "525403371166594848 423455864168639615\n38\n1 4 6 1 1 32 3 1 14 1 3 1 2 4 5 4 1 2 1 5 8 1 3 1 2 1 46 1 1 1 3 1 4 1 11 1 2 4", "output": "YES" }, { "input": "1 1\n1\n1", "output": "YES" }, { "input": "2 1\n2\n1 2", "output": "NO" }, { "input": "531983955813463755 371380136962341468\n38\n1 2 3 4 1 37 1 12 1 3 2 1 6 3 1 7 3 2 8 1 2 1 1 7 1 1 1 7 1 47 2 1 3 1 1 5 1 2", "output": "YES" }, { "input": "32951280099 987\n7\n33385288 1 5 1 5 1 6", "output": "YES" }, { "input": "6557470319842 86267571272\n6\n76 76 76 76 76 76", "output": "YES" }, { "input": "934648630114363087 6565775686518446\n31\n142 2 1 5 2 2 1 1 3 1 2 8 1 3 12 2 1 23 5 1 10 1 863 1 1 1 2 1 14 2 3", "output": "YES" }, { "input": "61305790721611591 37889062373143906\n81\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", "output": "YES" }, { "input": "4 1\n1\n4", "output": "YES" }, { "input": "500000000000000001 5\n2\n100000000000000000 5", "output": "YES" }, { "input": "1000000000000000000 3\n3\n3 4 5", "output": "NO" }, { "input": "822981258385599125 28316248989464296\n39\n29 15 1 1 1 4 4 4 1 3 1 5 12 1 1 1 1 1 6 5 2 1 11 1 1 26 1 2 2 2 14 1 1 1 3 2 4 1 1", "output": "NO" }, { "input": "823443107025550834 331822464812968648\n42\n2 2 13 14 4 4 1 1 1 1 2 1 1 1 1 113 1 1 8 1 1 1 1 2 2 1 15 1 5 1 1 2 1 1 1 14 4 3 1 5 1 1", "output": "NO" }, { "input": "226137305050296073 27076290603746056\n30\n8 2 1 5 3 67 2 1 6 1 2 1 5 1 11 8 43 2 1 7 1 95 2 3 1 11 5 2 1 1", "output": "NO" }, { "input": "524928871965838747 313083111434773473\n35\n1 1 2 10 1 4 12 3 28 1 23 1 1 1 4 1 4 3 1 3 2 3 1 4 3 1 3 2 3 11 21 1 35 1 1", "output": "NO" }, { "input": "633468529243155234 4\n90\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", "output": "NO" }, { "input": "742143496299253703 2\n90\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", "output": "NO" }, { "input": "550736960584023286 3\n90\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", "output": "NO" }, { "input": "2 1\n10\n99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999", "output": "NO" }, { "input": "262882295792523313 105000000000078855\n1\n105000000000078855", "output": "NO" }, { "input": "990130967049151695 166430169817556175\n1\n564668656008429569", "output": "NO" }, { "input": "9 4\n2\n2 3", "output": "NO" }, { "input": "529824479480396864 4705882352941177\n2\n80000000000000007 80000000000000009", "output": "NO" }, { "input": "985625905209512860 565433601688714177\n10\n6423 24947 27507 13031 16414 29169 901 32592 18763 1656", "output": "NO" }, { "input": "913255926290448385 4400000000\n2\n4400000000 4400000000", "output": "NO" }, { "input": "7 2\n2\n2 1", "output": "NO" }, { "input": "10 3\n1\n3", "output": "NO" }, { "input": "4 2\n1\n2", "output": "YES" }, { "input": "1337 42\n1\n31", "output": "NO" } ]

1,381,065,932

2,147,483,647

Python 3

OK

TESTS

49

124

0

p, q = list(map(int, input().split())) n = int(input()) a = list(map(int, input().split())) fract1 = [] x, y = p, q while y > 0: if y == 1: fract1 += [x - 1, 1] else: fract1 += [x // y] x, y = y, x % y fract2 = [] x, y = p, q while y > 0: fract2 += [x // y] x, y = y, x % y # print(fract1) # print(fract2) if fract1 == a or fract2 == a: print("YES") else: print("NO")

Title: Continued Fractions Time Limit: None seconds Memory Limit: None megabytes Problem Description: A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal. Input Specification: The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction. The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print "YES" if these fractions are equal and "NO" otherwise. Demo Input: ['9 4\n2\n2 4\n', '9 4\n3\n2 3 1\n', '9 4\n3\n1 2 4\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

```python p, q = list(map(int, input().split())) n = int(input()) a = list(map(int, input().split())) fract1 = [] x, y = p, q while y > 0: if y == 1: fract1 += [x - 1, 1] else: fract1 += [x // y] x, y = y, x % y fract2 = [] x, y = p, q while y > 0: fract2 += [x // y] x, y = y, x % y # print(fract1) # print(fract2) if fract1 == a or fract2 == a: print("YES") else: print("NO") ```

3

844

B

Rectangles

PROGRAMMING

1,300

[ "combinatorics", "math" ]

null

You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that: 1. All cells in a set have the same color. 1. Every two cells in a set share row or column.

The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly. The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.

Output single integer — the number of non-empty sets from the problem description.

[ "1 1\n0\n", "2 3\n1 0 1\n0 1 0\n" ]

[ "1\n", "8\n" ]

In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.

1,000

[ { "input": "1 1\n0", "output": "1" }, { "input": "2 3\n1 0 1\n0 1 0", "output": "8" }, { "input": "2 2\n1 1\n1 1", "output": "8" }, { "input": "1 10\n0 0 0 0 0 0 0 0 0 0", "output": "1023" }, { "input": "11 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1", "output": "2047" }, { "input": "10 11\n1 1 0 1 1 0 0 0 1 0 0\n1 0 0 1 1 1 0 0 1 1 0\n0 0 1 0 1 1 0 1 0 1 1\n0 1 1 1 0 1 0 1 0 0 0\n1 1 1 1 1 1 1 0 1 0 0\n1 1 0 1 1 1 1 0 0 1 1\n1 0 1 0 1 0 0 1 1 1 0\n1 1 0 0 0 0 0 1 0 1 1\n1 1 0 1 1 1 0 0 1 1 0\n1 0 1 1 0 0 1 0 0 1 1", "output": "2444" }, { "input": "50 1\n0\n1\n0\n1\n0\n1\n0\n1\n1\n1\n0\n0\n1\n0\n0\n1\n1\n1\n1\n0\n1\n1\n0\n1\n1\n1\n0\n1\n0\n0\n0\n1\n1\n0\n1\n1\n0\n1\n0\n1\n0\n0\n1\n0\n0\n0\n1\n1\n0\n1", "output": "142606334" }, { "input": "1 50\n0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1", "output": "142606334" }, { "input": "2 20\n0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0", "output": "589853" }, { "input": "5 5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "285" }, { "input": "6 6\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1", "output": "720" }, { "input": "21 2\n0 1\n1 1\n0 1\n0 0\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": "1310745" }, { "input": "3 15\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 1 0 1 0 0 0 0 0 1 0\n1 0 0 1 0 0 0 0 0 0 0 0 1 0 1", "output": "22587" }, { "input": "10 11\n0 1 0 0 0 0 0 0 0 0 0\n0 1 0 1 0 0 1 0 0 0 0\n0 0 0 0 0 0 1 1 1 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 1 0 0 0 0 1 0\n0 0 0 0 0 0 1 0 0 0 0\n0 0 0 0 0 0 0 0 0 1 0\n0 0 1 0 0 0 1 1 0 0 0\n0 0 0 0 0 0 0 0 1 0 0\n0 0 1 0 1 0 0 0 0 1 1", "output": "12047" }, { "input": "14 15\n0 1 0 0 0 0 0 0 1 0 0 0 1 0 1\n0 0 0 1 1 1 1 0 1 0 0 1 1 0 0\n1 0 0 0 0 1 1 0 0 0 0 0 0 0 0\n0 1 0 0 0 1 0 1 1 0 0 1 0 0 0\n0 0 1 1 0 1 0 1 0 1 1 0 1 0 0\n0 0 0 1 1 0 0 0 0 0 1 1 0 1 0\n0 0 1 0 0 0 0 0 0 1 0 0 1 1 0\n1 1 0 0 0 1 0 0 0 0 0 0 1 1 0\n0 0 0 0 1 0 1 1 1 0 0 0 1 0 1\n1 0 1 1 0 1 0 0 1 0 0 1 1 1 0\n1 0 0 0 0 1 0 0 0 0 0 1 0 0 0\n0 0 0 1 0 1 0 0 0 0 1 0 0 0 1\n0 0 1 0 1 0 0 0 1 1 1 1 0 0 0\n0 0 0 0 0 0 0 0 0 0 0 1 0 0 0", "output": "53166" }, { "input": "1 50\n0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0", "output": "1099511628798" }, { "input": "50 1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1", "output": "35184372088862" }, { "input": "1 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "1125899906842623" }, { "input": "5 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\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\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\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\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "5629499534214415" }, { "input": "32 2\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": "8589934622" }, { "input": "1 50\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "562949953421312" }, { "input": "50 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1", "output": "1125899906842623" }, { "input": "1 49\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", "output": "562949953421311" }, { "input": "2 50\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "2251799813685296" }, { "input": "3 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\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\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "3377699720528069" }, { "input": "1 50\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "1125899906842623" }, { "input": "1 40\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", "output": "1099511627775" }, { "input": "1 33\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", "output": "8589934591" }, { "input": "2 40\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "2199023255590" }, { "input": "1 35\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", "output": "34359738367" }, { "input": "50 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0", "output": "1125899906842623" } ]

1,662,776,833

2,147,483,647

Python 3

OK

TESTS

55

46

0

n, m = map(int, input().split()) table = [] sets = 0 for _ in range(n): table.append(list(map(int, input().split()))) for i in range(n): ones = table[i].count(1) zeros = table[i].count(0) sets += (2**ones - 1) + (2**zeros -1) for i in range(m): ones = 0 zeros = 0 for j in range(n): ones += (table[j][i] == 1) zeros += (table[j][i] == 0) sets += (ones > 1)*(2**ones - 1 - ones) + (zeros > 1)*(2**zeros -1 - zeros) print(sets)

Title: Rectangles Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that: 1. All cells in a set have the same color. 1. Every two cells in a set share row or column. Input Specification: The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly. The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black. Output Specification: Output single integer — the number of non-empty sets from the problem description. Demo Input: ['1 1\n0\n', '2 3\n1 0 1\n0 1 0\n'] Demo Output: ['1\n', '8\n'] Note: In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.

```python n, m = map(int, input().split()) table = [] sets = 0 for _ in range(n): table.append(list(map(int, input().split()))) for i in range(n): ones = table[i].count(1) zeros = table[i].count(0) sets += (2**ones - 1) + (2**zeros -1) for i in range(m): ones = 0 zeros = 0 for j in range(n): ones += (table[j][i] == 1) zeros += (table[j][i] == 0) sets += (ones > 1)*(2**ones - 1 - ones) + (zeros > 1)*(2**zeros -1 - zeros) print(sets) ```

3

743

C

Vladik and fractions

PROGRAMMING

1,500

[ "brute force", "constructive algorithms", "math", "number theory" ]

null

Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer *n* he can represent fraction as a sum of three distinct positive fractions in form . Help Vladik with that, i.e for a given *n* find three distinct positive integers *x*, *y* and *z* such that . Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109. If there is no such answer, print -1.

The single line contains single integer *n* (1<=≤<=*n*<=≤<=104).

If the answer exists, print 3 distinct numbers *x*, *y* and *z* (1<=≤<=*x*,<=*y*,<=*z*<=≤<=109, *x*<=≠<=*y*, *x*<=≠<=*z*, *y*<=≠<=*z*). Otherwise print -1. If there are multiple answers, print any of them.

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

[ "2 7 42\n", "7 8 56\n" ]

none

1,250

[ { "input": "3", "output": "2 7 42" }, { "input": "7", "output": "7 8 56" }, { "input": "2", "output": "2 3 6" }, { "input": "5", "output": "5 6 30" }, { "input": "4", "output": "4 5 20" }, { "input": "7", "output": "7 8 56" }, { "input": "82", "output": "82 83 6806" }, { "input": "56", "output": "56 57 3192" }, { "input": "30", "output": "30 31 930" }, { "input": "79", "output": "79 80 6320" }, { "input": "28", "output": "28 29 812" }, { "input": "4116", "output": "4116 4117 16945572" }, { "input": "1", "output": "-1" }, { "input": "6491", "output": "6491 6492 42139572" }, { "input": "8865", "output": "8865 8866 78597090" }, { "input": "1239", "output": "1239 1240 1536360" }, { "input": "3614", "output": "3614 3615 13064610" }, { "input": "5988", "output": "5988 5989 35862132" }, { "input": "8363", "output": "8363 8364 69948132" }, { "input": "737", "output": "737 738 543906" }, { "input": "3112", "output": "3112 3113 9687656" }, { "input": "9562", "output": "9562 9563 91441406" }, { "input": "1936", "output": "1936 1937 3750032" }, { "input": "4311", "output": "4311 4312 18589032" }, { "input": "6685", "output": "6685 6686 44695910" }, { "input": "9060", "output": "9060 9061 82092660" }, { "input": "1434", "output": "1434 1435 2057790" }, { "input": "3809", "output": "3809 3810 14512290" }, { "input": "6183", "output": "6183 6184 38235672" }, { "input": "8558", "output": "8558 8559 73247922" }, { "input": "932", "output": "932 933 869556" }, { "input": "7274", "output": "7274 7275 52918350" }, { "input": "9648", "output": "9648 9649 93093552" }, { "input": "2023", "output": "2023 2024 4094552" }, { "input": "4397", "output": "4397 4398 19338006" }, { "input": "6772", "output": "6772 6773 45866756" }, { "input": "9146", "output": "9146 9147 83658462" }, { "input": "1521", "output": "1521 1522 2314962" }, { "input": "4938", "output": "4938 4939 24388782" }, { "input": "6143", "output": "6143 6144 37742592" }, { "input": "6230", "output": "6230 6231 38819130" }, { "input": "10000", "output": "10000 10001 100010000" }, { "input": "1", "output": "-1" }, { "input": "101", "output": "101 102 10302" }, { "input": "9999", "output": "9999 10000 99990000" }, { "input": "512", "output": "512 513 262656" }, { "input": "511", "output": "511 512 261632" }, { "input": "513", "output": "513 514 263682" }, { "input": "9973", "output": "9973 9974 99470702" }, { "input": "9972", "output": "9972 9973 99450756" }, { "input": "9974", "output": "9974 9975 99490650" }, { "input": "9409", "output": "9409 9410 88538690" }, { "input": "9408", "output": "9408 9409 88519872" }, { "input": "9410", "output": "9410 9411 88557510" }, { "input": "719", "output": "719 720 517680" }, { "input": "720", "output": "720 721 519120" }, { "input": "721", "output": "721 722 520562" }, { "input": "9796", "output": "9796 9797 95971412" }, { "input": "9797", "output": "9797 9798 95991006" }, { "input": "9798", "output": "9798 9799 96010602" }, { "input": "4999", "output": "4999 5000 24995000" }, { "input": "9998", "output": "9998 9999 99970002" } ]

1,481,765,519

2,147,483,647

Python 3

OK

TESTS

62

93

0

n = int(input()) x, y, z = n, n+1, n*(n+1) if n != 1: print(x, y, z) else: print(-1)

Title: Vladik and fractions Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer *n* he can represent fraction as a sum of three distinct positive fractions in form . Help Vladik with that, i.e for a given *n* find three distinct positive integers *x*, *y* and *z* such that . Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109. If there is no such answer, print -1. Input Specification: The single line contains single integer *n* (1<=≤<=*n*<=≤<=104). Output Specification: If the answer exists, print 3 distinct numbers *x*, *y* and *z* (1<=≤<=*x*,<=*y*,<=*z*<=≤<=109, *x*<=≠<=*y*, *x*<=≠<=*z*, *y*<=≠<=*z*). Otherwise print -1. If there are multiple answers, print any of them. Demo Input: ['3\n', '7\n'] Demo Output: ['2 7 42\n', '7 8 56\n'] Note: none

```python n = int(input()) x, y, z = n, n+1, n*(n+1) if n != 1: print(x, y, z) else: print(-1) ```

3

733

A

Grasshopper And the String

PROGRAMMING

1,000

[ "implementation" ]

null

One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump. Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability. The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.

The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.

Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.

[ "ABABBBACFEYUKOTT\n", "AAA\n" ]

[ "4", "1" ]

none

500

[ { "input": "ABABBBACFEYUKOTT", "output": "4" }, { "input": "AAA", "output": "1" }, { "input": "A", "output": "1" }, { "input": "B", "output": "2" }, { "input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU", "output": "39" }, { "input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI", "output": "1" }, { "input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC", "output": "85" }, { "input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ", "output": "18" }, { "input": "PKLKBWTXVJ", "output": "11" }, { "input": "CFHFPTGMOKXVLJJZJDQW", "output": "12" }, { "input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX", "output": "9" }, { "input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY", "output": "4" }, { "input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD", "output": "101" }, { "input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC", "output": "76" }, { "input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC", "output": "45" }, { "input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT", "output": "48" }, { "input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX", "output": "47" }, { "input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA", "output": "65" }, { "input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS", "output": "28" }, { "input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW", "output": "35" }, { "input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL", "output": "30" }, { "input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL", "output": "19" }, { "input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB", "output": "30" }, { "input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB", "output": "34" }, { "input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR", "output": "17" }, { "input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT", "output": "15" }, { "input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB", "output": "9" }, { "input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW", "output": "5" }, { "input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA", "output": "4" }, { "input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE", "output": "5" }, { "input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA", "output": "5" }, { "input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO", "output": "3" }, { "input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY", "output": "3" }, { "input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY", "output": "2" }, { "input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE", "output": "2" }, { "input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII", "output": "2" }, { "input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE", "output": "2" }, { "input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU", "output": "1" }, { "input": "ABBABBB", "output": "4" }, { "input": "ABCD", "output": "4" }, { "input": "XXYC", "output": "3" }, { "input": "YYY", "output": "1" }, { "input": "ABABBBBBBB", "output": "8" }, { "input": "YYYY", "output": "1" }, { "input": "YYYYY", "output": "1" }, { "input": "AXXX", "output": "4" }, { "input": "YYYYYYY", "output": "1" }, { "input": "BYYBBB", "output": "4" }, { "input": "YYYYYYYYY", "output": "1" }, { "input": "CAAAAA", "output": "2" }, { "input": "CCCACCCC", "output": "5" }, { "input": "ABABBBACFEYUKOTTTT", "output": "5" }, { "input": "AABBYYYYYYYY", "output": "3" }, { "input": "BYBACYC", "output": "2" }, { "input": "Y", "output": "1" }, { "input": "ABBBBBB", "output": "7" }, { "input": "BACDYDI", "output": "3" }, { "input": "XEXXXXXXXXXXXXXXX", "output": "16" }, { "input": "TTYTT", "output": "3" }, { "input": "AAYBC", "output": "3" }, { "input": "ABABBBACFEYUKOTTTTT", "output": "6" }, { "input": "YYAYY", "output": "1" }, { "input": "YZZY", "output": "3" }, { "input": "YY", "output": "1" }, { "input": "ZZYZZ", "output": "3" }, { "input": "YBBBY", "output": "4" }, { "input": "BBBACCCCCCC", "output": "8" }, { "input": "YBBBBY", "output": "5" }, { "input": "YYYYYYYYYY", "output": "1" }, { "input": "ABABBBBBBBBBBBB", "output": "13" } ]

1,643,522,811

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

3

30

0

s = input() vowels = ['A', 'E', 'I', 'O', 'U', 'Y'] jumps = [] jump = 0 for i in s: if i in vowels: jump+=1 jumps.append(jump) jump = 0 if i not in vowels: jump+=1 print(max(jumps))

Title: Grasshopper And the String Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump. Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability. The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'. Input Specification: The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100. Output Specification: Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels. Demo Input: ['ABABBBACFEYUKOTT\n', 'AAA\n'] Demo Output: ['4', '1'] Note: none

```python s = input() vowels = ['A', 'E', 'I', 'O', 'U', 'Y'] jumps = [] jump = 0 for i in s: if i in vowels: jump+=1 jumps.append(jump) jump = 0 if i not in vowels: jump+=1 print(max(jumps)) ```

-1

129

A

Cookies

PROGRAMMING

900

[ "implementation" ]

null

Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?

The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.

Print in the only line the only number — the sought number of ways. If there are no such ways print 0.

[ "1\n1\n", "10\n1 2 2 3 4 4 4 2 2 2\n", "11\n2 2 2 2 2 2 2 2 2 2 99\n" ]

[ "1\n", "8\n", "1\n" ]

In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.

500

[ { "input": "1\n1", "output": "1" }, { "input": "10\n1 2 2 3 4 4 4 2 2 2", "output": "8" }, { "input": "11\n2 2 2 2 2 2 2 2 2 2 99", "output": "1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n2 2", "output": "2" }, { "input": "2\n1 2", "output": "1" }, { "input": "7\n7 7 7 7 7 7 7", "output": "7" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "4" }, { "input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2", "output": "50" }, { "input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99", "output": "49" }, { "input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15", "output": "50" }, { "input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38", "output": "7" }, { "input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87", "output": "39" }, { "input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53", "output": "37" }, { "input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32", "output": "51" }, { "input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80", "output": "21" }, { "input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81", "output": "37" }, { "input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59", "output": "28" }, { "input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94", "output": "47" }, { "input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28", "output": "15" }, { "input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42", "output": "11" }, { "input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35", "output": "33" }, { "input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22", "output": "37" }, { "input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14", "output": "15" }, { "input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77", "output": "35" }, { "input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75", "output": "42" }, { "input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26", "output": "23" }, { "input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78", "output": "15" }, { "input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78", "output": "13" }, { "input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51", "output": "17" }, { "input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59", "output": "45" }, { "input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43", "output": "53" }, { "input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8", "output": "40" }, { "input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54", "output": "53" }, { "input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77", "output": "55" }, { "input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66", "output": "52" }, { "input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89", "output": "51" }, { "input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2", "output": "53" }, { "input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23", "output": "48" }, { "input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25", "output": "48" }, { "input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71", "output": "58" }, { "input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45", "output": "53" }, { "input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17", "output": "54" }, { "input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52", "output": "49" }, { "input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31", "output": "44" }, { "input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74", "output": "46" }, { "input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75", "output": "53" }, { "input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37", "output": "43" }, { "input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82", "output": "51" }, { "input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5", "output": "47" }, { "input": "1\n2", "output": "1" }, { "input": "5\n1 1 3 2 2", "output": "3" } ]

1,667,849,445

2,147,483,647

Python 3

OK

TESTS

52

92

0

n = int(input()) a = list(map(int, input().split())) counte = 0 counto = 0 b = sum(a) % 2 print(sum(i % 2 == b for i in a))

Title: Cookies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. Output Specification: Print in the only line the only number — the sought number of ways. If there are no such ways print 0. Demo Input: ['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n'] Demo Output: ['1\n', '8\n', '1\n'] Note: In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.

```python n = int(input()) a = list(map(int, input().split())) counte = 0 counto = 0 b = sum(a) % 2 print(sum(i % 2 == b for i in a)) ```

3

909

B

Segments

PROGRAMMING

1,300

[ "constructive algorithms", "math" ]

null

You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them. You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis. Find the minimal number of layers you have to use for the given *N*.

The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100).

Output a single integer - the minimal number of layers required to draw the segments for the given *N*.

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

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

As an example, here are the segments and their optimal arrangement into layers for *N* = 4.

1,000

[ { "input": "2", "output": "2" }, { "input": "3", "output": "4" }, { "input": "4", "output": "6" }, { "input": "21", "output": "121" }, { "input": "100", "output": "2550" }, { "input": "1", "output": "1" }, { "input": "5", "output": "9" }, { "input": "6", "output": "12" }, { "input": "7", "output": "16" }, { "input": "8", "output": "20" }, { "input": "9", "output": "25" }, { "input": "10", "output": "30" }, { "input": "11", "output": "36" }, { "input": "12", "output": "42" }, { "input": "13", "output": "49" }, { "input": "14", "output": "56" }, { "input": "15", "output": "64" }, { "input": "16", "output": "72" }, { "input": "17", "output": "81" }, { "input": "18", "output": "90" }, { "input": "19", "output": "100" }, { "input": "20", "output": "110" }, { "input": "22", "output": "132" }, { "input": "23", "output": "144" }, { "input": "24", "output": "156" }, { "input": "25", "output": "169" }, { "input": "26", "output": "182" }, { "input": "27", "output": "196" }, { "input": "28", "output": "210" }, { "input": "29", "output": "225" }, { "input": "30", "output": "240" }, { "input": "31", "output": "256" }, { "input": "32", "output": "272" }, { "input": "33", "output": "289" }, { "input": "34", "output": "306" }, { "input": "35", "output": "324" }, { "input": "36", "output": "342" }, { "input": "37", "output": "361" }, { "input": "38", "output": "380" }, { "input": "39", "output": "400" }, { "input": "40", "output": "420" }, { "input": "41", "output": "441" }, { "input": "42", "output": "462" }, { "input": "43", "output": "484" }, { "input": "44", "output": "506" }, { "input": "45", "output": "529" }, { "input": "46", "output": "552" }, { "input": "47", "output": "576" }, { "input": "48", "output": "600" }, { "input": "49", "output": "625" }, { "input": "50", "output": "650" }, { "input": "51", "output": "676" }, { "input": "52", "output": "702" }, { "input": "53", "output": "729" }, { "input": "54", "output": "756" }, { "input": "55", "output": "784" }, { "input": "56", "output": "812" }, { "input": "57", "output": "841" }, { "input": "58", "output": "870" }, { "input": "59", "output": "900" }, { "input": "60", "output": "930" }, { "input": "61", "output": "961" }, { "input": "62", "output": "992" }, { "input": "63", "output": "1024" }, { "input": "64", "output": "1056" }, { "input": "65", "output": "1089" }, { "input": "66", "output": "1122" }, { "input": "67", "output": "1156" }, { "input": "68", "output": "1190" }, { "input": "69", "output": "1225" }, { "input": "70", "output": "1260" }, { "input": "71", "output": "1296" }, { "input": "72", "output": "1332" }, { "input": "73", "output": "1369" }, { "input": "74", "output": "1406" }, { "input": "75", "output": "1444" }, { "input": "76", "output": "1482" }, { "input": "77", "output": "1521" }, { "input": "78", "output": "1560" }, { "input": "79", "output": "1600" }, { "input": "80", "output": "1640" }, { "input": "81", "output": "1681" }, { "input": "82", "output": "1722" }, { "input": "83", "output": "1764" }, { "input": "84", "output": "1806" }, { "input": "85", "output": "1849" }, { "input": "86", "output": "1892" }, { "input": "87", "output": "1936" }, { "input": "88", "output": "1980" }, { "input": "89", "output": "2025" }, { "input": "90", "output": "2070" }, { "input": "91", "output": "2116" }, { "input": "92", "output": "2162" }, { "input": "93", "output": "2209" }, { "input": "94", "output": "2256" }, { "input": "95", "output": "2304" }, { "input": "96", "output": "2352" }, { "input": "97", "output": "2401" }, { "input": "98", "output": "2450" }, { "input": "99", "output": "2500" }, { "input": "1", "output": "1" }, { "input": "5", "output": "9" } ]

1,514,395,131

2,631

Python 3

WRONG_ANSWER

PRETESTS

1

46

5,529,600

#!/usr/bin/env python import sys def Rint(): return map(int, input().split()) a = [0,1] def f(n): return 1+2*((n+1)//2-(0 if n%2 else 1))+(0 if n%2 else 1)+a[n-2] def main(): for line in sys.stdin: n = int(line) if n==1: print(1) continue for i in range(2,n+1): a.append(f(i)) print(a[n]) main()

Title: Segments Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them. You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis. Find the minimal number of layers you have to use for the given *N*. Input Specification: The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100). Output Specification: Output a single integer - the minimal number of layers required to draw the segments for the given *N*. Demo Input: ['2\n', '3\n', '4\n'] Demo Output: ['2\n', '4\n', '6\n'] Note: As an example, here are the segments and their optimal arrangement into layers for *N* = 4.

```python #!/usr/bin/env python import sys def Rint(): return map(int, input().split()) a = [0,1] def f(n): return 1+2*((n+1)//2-(0 if n%2 else 1))+(0 if n%2 else 1)+a[n-2] def main(): for line in sys.stdin: n = int(line) if n==1: print(1) continue for i in range(2,n+1): a.append(f(i)) print(a[n]) main() ```

0

166

E

Tetrahedron

PROGRAMMING

1,500

[ "dp", "math", "matrices" ]

null

You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly. An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place. You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).

The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.

Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).

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

[ "3\n", "21\n" ]

The required paths in the first sample are: - *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*

1,000

[ { "input": "2", "output": "3" }, { "input": "4", "output": "21" }, { "input": "1", "output": "0" }, { "input": "3", "output": "6" }, { "input": "5", "output": "60" }, { "input": "6", "output": "183" }, { "input": "7", "output": "546" }, { "input": "8", "output": "1641" }, { "input": "9", "output": "4920" }, { "input": "10", "output": "14763" }, { "input": "15", "output": "3587226" }, { "input": "30", "output": "782663359" }, { "input": "10000000", "output": "192336614" }, { "input": "100", "output": "721510432" }, { "input": "300", "output": "327873818" }, { "input": "900", "output": "295068084" }, { "input": "1500", "output": "451187545" }, { "input": "3000", "output": "645417275" }, { "input": "5000", "output": "755610910" }, { "input": "10000", "output": "723907367" }, { "input": "50000", "output": "969527595" }, { "input": "90000", "output": "548978368" }, { "input": "99999", "output": "909741855" }, { "input": "100000", "output": "729225554" }, { "input": "300000", "output": "93822635" }, { "input": "800000", "output": "178940616" }, { "input": "1000000", "output": "266233856" }, { "input": "4000000", "output": "882155933" }, { "input": "9000000", "output": "295060537" }, { "input": "9999999", "output": "730778875" }, { "input": "1000000", "output": "266233856" }, { "input": "9999999", "output": "730778875" }, { "input": "10000000", "output": "192336614" }, { "input": "9999998", "output": "576926295" }, { "input": "30", "output": "782663359" }, { "input": "31", "output": "347990060" } ]

1,639,244,125

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

12

2,000

8,704,000

import math mod=10**9+7 n=int(input()) d = 0 abc = 3 for i in range(2,n+1): d,abc = abc, 3*d+2*abc print(d%mod)

Title: Tetrahedron Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly. An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place. You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7). Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path. Output Specification: Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7). Demo Input: ['2\n', '4\n'] Demo Output: ['3\n', '21\n'] Note: The required paths in the first sample are: - *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*

```python import math mod=10**9+7 n=int(input()) d = 0 abc = 3 for i in range(2,n+1): d,abc = abc, 3*d+2*abc print(d%mod) ```

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" } ]

1,700,125,369

2,147,483,647

Python 3

OK

TESTS

48

46

0

a,b,c=map(int,(input().split())) mid=(a+b+c)//3 da=(abs(a-mid)) db=abs(b-mid) dc=abs(c-mid) a1=[0]*3 a1[0]=min(a,b,c) a1[2]=max(a,b,c) a1[1]=(a+b+c)-(a1[0]+a1[2]) ma=(abs(a-a1[1])) mb=abs(b-a1[1]) mc=abs(c-a1[1]) ans=min((da+db+dc),(ma+mb+mc)) print(ans)

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())) mid=(a+b+c)//3 da=(abs(a-mid)) db=abs(b-mid) dc=abs(c-mid) a1=[0]*3 a1[0]=min(a,b,c) a1[2]=max(a,b,c) a1[1]=(a+b+c)-(a1[0]+a1[2]) ma=(abs(a-a1[1])) mb=abs(b-a1[1]) mc=abs(c-a1[1]) ans=min((da+db+dc),(ma+mb+mc)) print(ans) ```

3

264

A

Escape from Stones

PROGRAMMING

1,200

[ "constructive algorithms", "data structures", "implementation", "two pointers" ]

null

Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.

The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".

Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.

[ "llrlr\n", "rrlll\n", "lrlrr\n" ]

[ "3\n5\n4\n2\n1\n", "1\n2\n5\n4\n3\n", "2\n4\n5\n3\n1\n" ]

In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.

500

[ { "input": "llrlr", "output": "3\n5\n4\n2\n1" }, { "input": "rrlll", "output": "1\n2\n5\n4\n3" }, { "input": "lrlrr", "output": "2\n4\n5\n3\n1" }, { "input": "lllrlrllrl", "output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1" }, { "input": "llrlrrrlrr", "output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1" }, { "input": "rlrrrllrrr", "output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2" }, { "input": "lrrlrrllrrrrllllllrr", "output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1" }, { "input": "rlrrrlrrrllrrllrlrll", "output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2" }, { "input": "lllrrlrlrllrrrrrllrl", "output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1" }, { "input": "rrrllrrrlllrlllrlrrr", "output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4" }, { "input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl", "output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3" }, { "input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll", "output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1" }, { "input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr", "output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1" }, { "input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl", "output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1" }, { "input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl", "output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1" }, { "input": "l", "output": "1" }, { "input": "r", "output": "1" } ]

1,598,524,594

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

30

2,000

32,051,200

s=input() l=[] v=[] for i in range(len(s)): if s[i]=='l': l.append(i+1) else: v.append(i+1) for i in range(len(v)): print(v[i]) for i in range(len(l)): print(l[len(l)-i-1])

Title: Escape from Stones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls. Input Specification: The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r". Output Specification: Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left. Demo Input: ['llrlr\n', 'rrlll\n', 'lrlrr\n'] Demo Output: ['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n'] Note: In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.

```python s=input() l=[] v=[] for i in range(len(s)): if s[i]=='l': l.append(i+1) else: v.append(i+1) for i in range(len(v)): print(v[i]) for i in range(len(l)): print(l[len(l)-i-1]) ```

0

469

A

I Wanna Be the Guy

PROGRAMMING

800

[ "greedy", "implementation" ]

null

There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game. Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?

The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100). The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.

If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).

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

[ "I become the guy.\n", "Oh, my keyboard!\n" ]

In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both. In the second sample, no one can pass level 4.

500

[ { "input": "4\n3 1 2 3\n2 2 4", "output": "I become the guy." }, { "input": "4\n3 1 2 3\n2 2 3", "output": "Oh, my keyboard!" }, { "input": "10\n5 8 6 1 5 4\n6 1 3 2 9 4 6", "output": "Oh, my keyboard!" }, { "input": "10\n8 8 10 7 3 1 4 2 6\n8 9 5 10 3 7 2 4 8", "output": "I become the guy." }, { "input": "10\n9 6 1 8 3 9 7 5 10 4\n7 1 3 2 7 6 9 5", "output": "I become the guy." }, { "input": "100\n75 83 69 73 30 76 37 48 14 41 42 21 35 15 50 61 86 85 46 3 31 13 78 10 2 44 80 95 56 82 38 75 77 4 99 9 84 53 12 11 36 74 39 72 43 89 57 28 54 1 51 66 27 22 93 59 68 88 91 29 7 20 63 8 52 23 64 58 100 79 65 49 96 71 33 45\n83 50 89 73 34 28 99 67 77 44 19 60 68 42 8 27 94 85 14 39 17 78 24 21 29 63 92 32 86 22 71 81 31 82 65 48 80 59 98 3 70 55 37 12 15 72 47 9 11 33 16 7 91 74 13 64 38 84 6 61 93 90 45 69 1 54 52 100 57 10 35 49 53 75 76 43 62 5 4 18 36 96 79 23", "output": "Oh, my keyboard!" }, { "input": "1\n1 1\n1 1", "output": "I become the guy." }, { "input": "1\n0\n1 1", "output": "I become the guy." }, { "input": "1\n1 1\n0", "output": "I become the guy." }, { "input": "1\n0\n0", "output": "Oh, my keyboard!" }, { "input": "100\n0\n0", "output": "Oh, my keyboard!" }, { "input": "100\n44 71 70 55 49 43 16 53 7 95 58 56 38 76 67 94 20 73 29 90 25 30 8 84 5 14 77 52 99 91 66 24 39 37 22 44 78 12 63 59 32 51 15 82 34\n56 17 10 96 80 69 13 81 31 57 4 48 68 89 50 45 3 33 36 2 72 100 64 87 21 75 54 74 92 65 23 40 97 61 18 28 98 93 35 83 9 79 46 27 41 62 88 6 47 60 86 26 42 85 19 1 11", "output": "I become the guy." }, { "input": "100\n78 63 59 39 11 58 4 2 80 69 22 95 90 26 65 16 30 100 66 99 67 79 54 12 23 28 45 56 70 74 60 82 73 91 68 43 92 75 51 21 17 97 86 44 62 47 85 78 72 64 50 81 71 5 57 13 31 76 87 9 49 96 25 42 19 35 88 53 7 83 38 27 29 41 89 93 10 84 18\n78 1 16 53 72 99 9 36 59 49 75 77 94 79 35 4 92 42 82 83 76 97 20 68 55 47 65 50 14 30 13 67 98 8 7 40 64 32 87 10 33 90 93 18 26 71 17 46 24 28 89 58 37 91 39 34 25 48 84 31 96 95 80 88 3 51 62 52 85 61 12 15 27 6 45 38 2 22 60", "output": "I become the guy." }, { "input": "2\n2 2 1\n0", "output": "I become the guy." }, { "input": "2\n1 2\n2 1 2", "output": "I become the guy." }, { "input": "80\n57 40 1 47 36 69 24 76 5 72 26 4 29 62 6 60 3 70 8 64 18 37 16 14 13 21 25 7 66 68 44 74 61 39 38 33 15 63 34 65 10 23 56 51 80 58 49 75 71 12 50 57 2 30 54 27 17 52\n61 22 67 15 28 41 26 1 80 44 3 38 18 37 79 57 11 7 65 34 9 36 40 5 48 29 64 31 51 63 27 4 50 13 24 32 58 23 19 46 8 73 39 2 21 56 77 53 59 78 43 12 55 45 30 74 33 68 42 47 17 54", "output": "Oh, my keyboard!" }, { "input": "100\n78 87 96 18 73 32 38 44 29 64 40 70 47 91 60 69 24 1 5 34 92 94 99 22 83 65 14 68 15 20 74 31 39 100 42 4 97 46 25 6 8 56 79 9 71 35 54 19 59 93 58 62 10 85 57 45 33 7 86 81 30 98 26 61 84 41 23 28 88 36 66 51 80 53 37 63 43 95 75\n76 81 53 15 26 37 31 62 24 87 41 39 75 86 46 76 34 4 51 5 45 65 67 48 68 23 71 27 94 47 16 17 9 96 84 89 88 100 18 52 69 42 6 92 7 64 49 12 98 28 21 99 25 55 44 40 82 19 36 30 77 90 14 43 50 3 13 95 78 35 20 54 58 11 2 1 33", "output": "Oh, my keyboard!" }, { "input": "100\n77 55 26 98 13 91 78 60 23 76 12 11 36 62 84 80 18 1 68 92 81 67 19 4 2 10 17 77 96 63 15 69 46 97 82 42 83 59 50 72 14 40 89 9 52 29 56 31 74 39 45 85 22 99 44 65 95 6 90 38 54 32 49 34 3 70 75 33 94 53 21 71 5 66 73 41 100 24\n69 76 93 5 24 57 59 6 81 4 30 12 44 15 67 45 73 3 16 8 47 95 20 64 68 85 54 17 90 86 66 58 13 37 42 51 35 32 1 28 43 80 7 14 48 19 62 55 2 91 25 49 27 26 38 79 89 99 22 60 75 53 88 82 34 21 87 71 72 61", "output": "I become the guy." }, { "input": "100\n74 96 32 63 12 69 72 99 15 22 1 41 79 77 71 31 20 28 75 73 85 37 38 59 42 100 86 89 55 87 68 4 24 57 52 8 92 27 56 98 95 58 34 9 45 14 11 36 66 76 61 19 25 23 78 49 90 26 80 43 70 13 65 10 5 74 81 21 44 60 97 3 47 93 6\n64 68 21 27 16 91 23 22 33 12 71 88 90 50 62 43 28 29 57 59 5 74 10 95 35 1 67 93 36 32 86 40 6 64 78 46 89 15 84 53 18 30 17 85 2 3 47 92 25 48 76 51 20 82 52 83 99 63 80 11 94 54 39 7 58", "output": "I become the guy." }, { "input": "100\n75 11 98 44 47 88 94 23 78 59 70 2 43 39 34 63 71 19 42 61 30 74 14 77 97 53 92 60 67 36 37 13 6 86 62 46 41 3 25 93 7 12 27 48 55 49 31 35 51 10 57 54 95 82 28 90 73 26 17 50 81 56 20 87 40 85 72 64 99 29 91 5 80 18 24 52\n72 93 59 5 88 47 9 58 48 1 43 50 100 87 61 91 45 98 99 56 25 84 53 73 78 54 63 38 37 2 77 95 89 85 4 90 10 33 12 22 74 32 34 70 71 52 96 57 15 66 31 27 75 8 21 39 62 44 67 94 81 68 14 19 36 28 11 79 16 65 46 83 76", "output": "Oh, my keyboard!" }, { "input": "3\n1 2\n2 2 3", "output": "Oh, my keyboard!" }, { "input": "4\n1 2\n3 1 3 4", "output": "I become the guy." }, { "input": "6\n2 1 2\n3 4 5 6", "output": "Oh, my keyboard!" } ]

1,698,918,828

2,147,483,647

PyPy 3-64

OK

TESTS

29

61

0

n = int(input()) p1 = list(map(int, input().split())) p2 = list(map(int, input().split())) print("Oh, my keyboard!" if {*range(1, n + 1)} - (set(p1[1:]).union(set(p2[1:]))) else "I become the guy.")

Title: I Wanna Be the Guy Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game. Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other? Input Specification: The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100). The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*. Output Specification: If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes). Demo Input: ['4\n3 1 2 3\n2 2 4\n', '4\n3 1 2 3\n2 2 3\n'] Demo Output: ['I become the guy.\n', 'Oh, my keyboard!\n'] Note: In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both. In the second sample, no one can pass level 4.

```python n = int(input()) p1 = list(map(int, input().split())) p2 = list(map(int, input().split())) print("Oh, my keyboard!" if {*range(1, n + 1)} - (set(p1[1:]).union(set(p2[1:]))) else "I become the guy.") ```

3

518

B

Tanya and Postcard

PROGRAMMING

1,400

[ "greedy", "implementation", "strings" ]

null

Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string *s* of length *n*, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string *s*. The newspaper contains string *t*, consisting of uppercase and lowercase English letters. We know that the length of string *t* greater or equal to the length of the string *s*. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some *n* letters out of the newspaper and make a message of length exactly *n*, so that it looked as much as possible like *s*. If the letter in some position has correct value and correct letter case (in the string *s* and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message.

The first line contains line *s* (1<=≤<=|*s*|<=≤<=2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line *t* (|*s*|<=≤<=|*t*|<=≤<=2·105), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |*a*| means the length of the string *a*.

Print two integers separated by a space: - the first number is the number of times Tanya shouts "YAY!" while making the message, - the second number is the number of times Tanya says "WHOOPS" while making the message.

[ "AbC\nDCbA\n", "ABC\nabc\n", "abacaba\nAbaCaBA\n" ]

[ "3 0\n", "0 3\n", "3 4\n" ]

none

1,000

[ { "input": "AbC\nDCbA", "output": "3 0" }, { "input": "ABC\nabc", "output": "0 3" }, { "input": "abacaba\nAbaCaBA", "output": "3 4" }, { "input": "zzzzz\nZZZZZ", "output": "0 5" }, { "input": "zzzZZZ\nZZZzzZ", "output": "5 1" }, { "input": "abcdefghijklmnopqrstuvwxyz\nABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "0 26" }, { "input": "abcdefghijklmnopqrstuvwxyz\nqrsimtabuvzhnwcdefgjklxyop", "output": "26 0" }, { "input": "l\nFPbAVjsMpPDTLkfwNYFmBDHPTDSWSOUlrBHYJHPM", "output": "1 0" }, { "input": "ncMeXssLHS\nuwyeMcaFatpInZVdEYpwJQSnVxLK", "output": "6 1" }, { "input": "DpiNBmCRFWxpdbfGOzvvOcemjructoAdEwegTvbVbfWWRPGyEAxGdDRWVlqNyGWMWHMrHAIZpyxvgaflrsVZhhZRouvpxrKXFZam\nwwPLFtNfPtJXvMLuHjKfYyaRhreNSWSzOvDpqHCGcqllACNPGHxReeFUCmAqIKXYytsSQwIxJzNiiUtgebVuwRmWpRALLyKAzyDPvgIGxALSaeeTIqm", "output": "66 12" }, { "input": "CCAE\ndcecc", "output": "0 3" }, { "input": "Dccb\nbeeeb", "output": "1 0" }, { "input": "Adc\neadeabcad", "output": "2 1" }, { "input": "DBAdeb\ndeeabcddadaa", "output": "3 2" }, { "input": "EDCED\neebeacdba", "output": "0 4" }, { "input": "CdAbD\ndecbde", "output": "2 2" }, { "input": "a\nB", "output": "0 0" }, { "input": "r\nqA", "output": "0 0" } ]

1,624,811,508

2,147,483,647

PyPy 3

OK

TESTS

49

202

4,300,800

#This code is contributed by Siddharth from bisect import * import math from collections import * from heapq import * from itertools import * inf=10**18 mod=10**9+7 # ---------------------------------------------------------Code--------------------------------------------------------- s=input() t=input() cnt1=Counter(s) cnt2=Counter(t) # firstly checking similar elements yay=0 for i in s: if cnt1[i]<=cnt2[i]: yay+=cnt1[i] cnt2[i]-=cnt1[i] cnt1[i]=0 else: yay+=cnt2[i] cnt1[i]-=cnt2[i] cnt2[i]=0 #now calculating the number of whoops whoops=0 for i in cnt1: if cnt1[i]: if i.isupper(): temp=min(cnt1[i],cnt2[chr(ord(i)+32)]) cnt2[chr(ord(i)+32)]-=temp else: temp = min(cnt1[i], cnt2[chr(ord(i) - 32)]) cnt2[chr(ord(i) - 32)] -= temp whoops+=temp print(yay,whoops)

Title: Tanya and Postcard Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string *s* of length *n*, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string *s*. The newspaper contains string *t*, consisting of uppercase and lowercase English letters. We know that the length of string *t* greater or equal to the length of the string *s*. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some *n* letters out of the newspaper and make a message of length exactly *n*, so that it looked as much as possible like *s*. If the letter in some position has correct value and correct letter case (in the string *s* and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. Input Specification: The first line contains line *s* (1<=≤<=|*s*|<=≤<=2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line *t* (|*s*|<=≤<=|*t*|<=≤<=2·105), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |*a*| means the length of the string *a*. Output Specification: Print two integers separated by a space: - the first number is the number of times Tanya shouts "YAY!" while making the message, - the second number is the number of times Tanya says "WHOOPS" while making the message. Demo Input: ['AbC\nDCbA\n', 'ABC\nabc\n', 'abacaba\nAbaCaBA\n'] Demo Output: ['3 0\n', '0 3\n', '3 4\n'] Note: none

```python #This code is contributed by Siddharth from bisect import * import math from collections import * from heapq import * from itertools import * inf=10**18 mod=10**9+7 # ---------------------------------------------------------Code--------------------------------------------------------- s=input() t=input() cnt1=Counter(s) cnt2=Counter(t) # firstly checking similar elements yay=0 for i in s: if cnt1[i]<=cnt2[i]: yay+=cnt1[i] cnt2[i]-=cnt1[i] cnt1[i]=0 else: yay+=cnt2[i] cnt1[i]-=cnt2[i] cnt2[i]=0 #now calculating the number of whoops whoops=0 for i in cnt1: if cnt1[i]: if i.isupper(): temp=min(cnt1[i],cnt2[chr(ord(i)+32)]) cnt2[chr(ord(i)+32)]-=temp else: temp = min(cnt1[i], cnt2[chr(ord(i) - 32)]) cnt2[chr(ord(i) - 32)] -= temp whoops+=temp print(yay,whoops) ```

3

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" } ]

1,678,813,041

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

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

-1

825

D

Suitable Replacement

PROGRAMMING

1,500

[ "binary search", "greedy", "implementation" ]

null

You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters. Suitability of string *s* is calculated by following metric: Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences. You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal.

The first line contains string *s* (1<=≤<=|*s*|<=≤<=106). The second line contains string *t* (1<=≤<=|*t*|<=≤<=106).

Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal. If there are multiple strings with maximal suitability then print any of them.

[ "?aa?\nab\n", "??b?\nza\n", "abcd\nabacaba\n" ]

[ "baab\n", "azbz\n", "abcd\n" ]

In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2. In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them. In the third example there are no '?' characters and the suitability of the string is 0.

0

[ { "input": "?aa?\nab", "output": "baab" }, { "input": "??b?\nza", "output": "azbz" }, { "input": "abcd\nabacaba", "output": "abcd" }, { "input": "mqwstphetbfrsyxuzdww\nrutseqtsbh", "output": "mqwstphetbfrsyxuzdww" }, { "input": "????????????????????\nxwkxsxlrre", "output": "eekkllrrrrsswwxxxxxx" }, { "input": "fzjqgrgzzzghrwgwhfjq\nggwfrjzzqh", "output": "fzjqgrgzzzghrwgwhfjq" }, { "input": "ng?\nvu", "output": "ngz" }, { "input": "???\nb", "output": "bbb" }, { "input": "??a?\na", "output": "aaaa" }, { "input": "c?aa???a?cdcc??\nabcadbaccb", "output": "cbaabbzazcdcczz" }, { "input": "cc?cdb?????b?cc?b?\ncabdacd", "output": "ccacdbaaaddbdcczbz" }, { "input": "ccaaa?bc?baccbccccbca?cccab??aacccbaac?bccacc?accccaccbcbbcbc\ncbaaacccc", "output": "ccaaaabcabaccbccccbcaacccabaaaacccbaaczbccacczaccccaccbcbbcbc" }, { "input": "aacaacbabc?bcca?ccca\nbc", "output": "aacaacbabcbbccabccca" } ]

1,670,373,087

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

4

46

614,400

from collections import Counter import sys def debug(*arg, sep=" ", end="\n"): print(*arg, sep=sep, end=end, file=sys.stderr) def z_func(S): n = len(S) z = [0] * n L = R = 0 for i in range(1, n): if i >= R: L = R = i while R < n and S[R] == S[R-L]: R += 1 z[i] = R - L continue if z[i-L] < R - i: z[i] = z[i-L] continue L = i while R < n and S[R] == S[R-L]: R += 1 z[i] = R - L return z def main(): S = input() T = input() if len(S) < len(T): print(S.replace("?", "r")) return valid = 0 for z in z_func(T): if z != 0: break valid += 1 T = T[:valid] blank = [] cnt = Counter(S) for _ in range(len(S) // len(T)): for t in T: if cnt[t] > 0: cnt[t] -= 1 else: blank.append(t) blank.reverse() for s in S: if s == "?": if blank: print(blank.pop(), end="") else: print("r", end="") else: print(s, end="") print() main()

Title: Suitable Replacement Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters. Suitability of string *s* is calculated by following metric: Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences. You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal. Input Specification: The first line contains string *s* (1<=≤<=|*s*|<=≤<=106). The second line contains string *t* (1<=≤<=|*t*|<=≤<=106). Output Specification: Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal. If there are multiple strings with maximal suitability then print any of them. Demo Input: ['?aa?\nab\n', '??b?\nza\n', 'abcd\nabacaba\n'] Demo Output: ['baab\n', 'azbz\n', 'abcd\n'] Note: In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2. In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them. In the third example there are no '?' characters and the suitability of the string is 0.

```python from collections import Counter import sys def debug(*arg, sep=" ", end="\n"): print(*arg, sep=sep, end=end, file=sys.stderr) def z_func(S): n = len(S) z = [0] * n L = R = 0 for i in range(1, n): if i >= R: L = R = i while R < n and S[R] == S[R-L]: R += 1 z[i] = R - L continue if z[i-L] < R - i: z[i] = z[i-L] continue L = i while R < n and S[R] == S[R-L]: R += 1 z[i] = R - L return z def main(): S = input() T = input() if len(S) < len(T): print(S.replace("?", "r")) return valid = 0 for z in z_func(T): if z != 0: break valid += 1 T = T[:valid] blank = [] cnt = Counter(S) for _ in range(len(S) // len(T)): for t in T: if cnt[t] > 0: cnt[t] -= 1 else: blank.append(t) blank.reverse() for s in S: if s == "?": if blank: print(blank.pop(), end="") else: print("r", end="") else: print(s, end="") print() main() ```

0

690

A1

Collective Mindsets (easy)

PROGRAMMING

1,100

[]

null

Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage. They will be *N* guests tonight: *N*<=-<=1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to *N*<=-<=1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, *N*. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure: The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.) You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order: 1. survive the event (they experienced death already once and know it is no fun), 1. get as many brains as possible. Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself. What is the smallest number of brains that have to be in the chest for this to be possible?

The only line of input contains one integer: *N*, the number of attendees (1<=≤<=*N*<=≤<=109).

Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home.

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

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

0

[ { "input": "1", "output": "1" }, { "input": "4", "output": "2" }, { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "5", "output": "3" }, { "input": "6", "output": "3" }, { "input": "7", "output": "4" }, { "input": "8", "output": "4" }, { "input": "9", "output": "5" }, { "input": "10", "output": "5" }, { "input": "11", "output": "6" }, { "input": "12", "output": "6" }, { "input": "13", "output": "7" }, { "input": "14", "output": "7" }, { "input": "15", "output": "8" }, { "input": "16", "output": "8" }, { "input": "17", "output": "9" }, { "input": "18", "output": "9" }, { "input": "19", "output": "10" }, { "input": "20", "output": "10" }, { "input": "100", "output": "50" }, { "input": "9999", "output": "5000" }, { "input": "21736", "output": "10868" }, { "input": "873467", "output": "436734" }, { "input": "4124980", "output": "2062490" }, { "input": "536870910", "output": "268435455" }, { "input": "536870912", "output": "268435456" }, { "input": "876543210", "output": "438271605" }, { "input": "987654321", "output": "493827161" }, { "input": "1000000000", "output": "500000000" } ]

1,489,426,354

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

31

4,608,000

input = input() result = input/2 print(result)

Title: Collective Mindsets (easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tonight is brain dinner night and all zombies will gather together to scarf down some delicious brains. The artful Heidi plans to crash the party, incognito, disguised as one of them. Her objective is to get away with at least one brain, so she can analyze the zombies' mindset back home and gain a strategic advantage. They will be *N* guests tonight: *N*<=-<=1 real zombies and a fake one, our Heidi. The living-dead love hierarchies as much as they love brains: each one has a unique rank in the range 1 to *N*<=-<=1, and Heidi, who still appears slightly different from the others, is attributed the highest rank, *N*. Tonight there will be a chest with brains on display and every attendee sees how many there are. These will then be split among the attendees according to the following procedure: The zombie of the highest rank makes a suggestion on who gets how many brains (every brain is an indivisible entity). A vote follows. If at least half of the attendees accept the offer, the brains are shared in the suggested way and the feast begins. But if majority is not reached, then the highest-ranked zombie is killed, and the next zombie in hierarchy has to make a suggestion. If he is killed too, then the third highest-ranked makes one, etc. (It's enough to have exactly half of the votes – in case of a tie, the vote of the highest-ranked alive zombie counts twice, and he will of course vote in favor of his own suggestion in order to stay alive.) You should know that zombies are very greedy and sly, and they know this too – basically all zombie brains are alike. Consequently, a zombie will never accept an offer which is suboptimal for him. That is, if an offer is not strictly better than a potential later offer, he will vote against it. And make no mistake: while zombies may normally seem rather dull, tonight their intellects are perfect. Each zombie's priorities for tonight are, in descending order: 1. survive the event (they experienced death already once and know it is no fun), 1. get as many brains as possible. Heidi goes first and must make an offer which at least half of the attendees will accept, and which allocates at least one brain for Heidi herself. What is the smallest number of brains that have to be in the chest for this to be possible? Input Specification: The only line of input contains one integer: *N*, the number of attendees (1<=≤<=*N*<=≤<=109). Output Specification: Output one integer: the smallest number of brains in the chest which allows Heidi to take one brain home. Demo Input: ['1\n', '4\n'] Demo Output: ['1\n', '2\n']

```python input = input() result = input/2 print(result) ```

-1

411

A

Password Check

PROGRAMMING

800

[ "*special", "implementation" ]

null

You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check. Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions: - the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit. You are given a password. Please implement the automatic check of its complexity for company Q.

The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".

If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).

[ "abacaba\n", "X12345\n", "CONTEST_is_STARTED!!11\n" ]

[ "Too weak\n", "Too weak\n", "Correct\n" ]

none

0

[ { "input": "abacaba", "output": "Too weak" }, { "input": "X12345", "output": "Too weak" }, { "input": "CONTEST_is_STARTED!!11", "output": "Correct" }, { "input": "1zA__", "output": "Correct" }, { "input": "1zA_", "output": "Too weak" }, { "input": "zA___", "output": "Too weak" }, { "input": "1A___", "output": "Too weak" }, { "input": "z1___", "output": "Too weak" }, { "input": "0", "output": "Too weak" }, { "input": "_", "output": "Too weak" }, { "input": "a", "output": "Too weak" }, { "input": "D", "output": "Too weak" }, { "input": "_", "output": "Too weak" }, { "input": "?", "output": "Too weak" }, { "input": "?", "output": "Too weak" }, { "input": "._,.!.,...?_,!.", "output": "Too weak" }, { "input": "!_?_,?,?.,.,_!!!.!,.__,?!!,_!,?_,!??,?!..._!?_,?_!,?_.,._,,_.,.", "output": "Too weak" }, { "input": "?..!.,,?,__.,...????_???__!,?...?.,,,,___!,.!,_,,_,??!_?_,!!?_!_??.?,.!!?_?_.,!", "output": "Too weak" }, { "input": "XZX", "output": "Too weak" }, { "input": "R", "output": "Too weak" }, { "input": "H.FZ", "output": "Too weak" }, { "input": "KSHMICWPK,LSBM_JVZ!IPDYDG_GOPCHXFJTKJBIFY,FPHMY,CB?PZEAG..,X,.GFHPIDBB,IQ?MZ", "output": "Too weak" }, { "input": "EFHI,,Y?HMMUI,,FJGAY?FYPBJQMYM!DZHLFCTFWT?JOPDW,S_!OR?ATT?RWFBMAAKUHIDMHSD?LCZQY!UD_CGYGBAIRDPICYS", "output": "Too weak" }, { "input": "T,NDMUYCCXH_L_FJHMCCAGX_XSCPGOUZSY?D?CNDSYRITYS,VAT!PJVKNTBMXGGRYKACLYU.RJQ_?UWKXYIDE_AE", "output": "Too weak" }, { "input": "y", "output": "Too weak" }, { "input": "qgw", "output": "Too weak" }, { "input": "g", "output": "Too weak" }, { "input": "loaray", "output": "Too weak" }, { "input": "d_iymyvxolmjayhwpedocopqwmy.oalrdg!_n?.lrxpamhygps?kkzxydsbcaihfs.j?eu!oszjsy.vzu?!vs.bprz_j", "output": "Too weak" }, { "input": "txguglvclyillwnono", "output": "Too weak" }, { "input": "FwX", "output": "Too weak" }, { "input": "Zi", "output": "Too weak" }, { "input": "PodE", "output": "Too weak" }, { "input": "SdoOuJ?nj_wJyf", "output": "Too weak" }, { "input": "MhnfZjsUyXYw?f?ubKA", "output": "Too weak" }, { "input": "CpWxDVzwHfYFfoXNtXMFuAZr", "output": "Too weak" }, { "input": "9.,0", "output": "Too weak" }, { "input": "5,8", "output": "Too weak" }, { "input": "7", "output": "Too weak" }, { "input": "34__39_02!,!,82!129!2!566", "output": "Too weak" }, { "input": "96156027.65935663!_87!,44,..7914_!0_1,.4!!62!.8350!17_282!!9.2584,!!7__51.526.7", "output": "Too weak" }, { "input": "90328_", "output": "Too weak" }, { "input": "B9", "output": "Too weak" }, { "input": "P1H", "output": "Too weak" }, { "input": "J2", "output": "Too weak" }, { "input": "M6BCAKW!85OSYX1D?.53KDXP42F", "output": "Too weak" }, { "input": "C672F429Y8X6XU7S,.K9111UD3232YXT81S4!729ER7DZ.J7U1R_7VG6.FQO,LDH", "output": "Too weak" }, { "input": "W2PI__!.O91H8OFY6AB__R30L9XOU8800?ZUD84L5KT99818NFNE35V.8LJJ5P2MM.B6B", "output": "Too weak" }, { "input": "z1", "output": "Too weak" }, { "input": "p1j", "output": "Too weak" }, { "input": "j9", "output": "Too weak" }, { "input": "v8eycoylzv0qkix5mfs_nhkn6k!?ovrk9!b69zy!4frc?k", "output": "Too weak" }, { "input": "l4!m_44kpw8.jg!?oh,?y5oraw1tg7_x1.osl0!ny?_aihzhtt0e2!mr92tnk0es!1f,9he40_usa6c50l", "output": "Too weak" }, { "input": "d4r!ak.igzhnu!boghwd6jl", "output": "Too weak" }, { "input": "It0", "output": "Too weak" }, { "input": "Yb1x", "output": "Too weak" }, { "input": "Qf7", "output": "Too weak" }, { "input": "Vu7jQU8.!FvHBYTsDp6AphaGfnEmySP9te", "output": "Correct" }, { "input": "Ka4hGE,vkvNQbNolnfwp", "output": "Correct" }, { "input": "Ee9oluD?amNItsjeQVtOjwj4w_ALCRh7F3eaZah", "output": "Correct" }, { "input": "Um3Fj?QLhNuRE_Gx0cjMLOkGCm", "output": "Correct" }, { "input": "Oq2LYmV9HmlaW", "output": "Correct" }, { "input": "Cq7r3Wrb.lDb_0wsf7!ruUUGSf08RkxD?VsBEDdyE?SHK73TFFy0f8gmcATqGafgTv8OOg8or2HyMPIPiQ2Hsx8q5rn3_WZe", "output": "Correct" }, { "input": "Wx4p1fOrEMDlQpTlIx0p.1cnFD7BnX2K8?_dNLh4cQBx_Zqsv83BnL5hGKNcBE9g3QB,!fmSvgBeQ_qiH7", "output": "Correct" }, { "input": "k673,", "output": "Too weak" }, { "input": "LzuYQ", "output": "Too weak" }, { "input": "Pasq!", "output": "Too weak" }, { "input": "x5hve", "output": "Too weak" }, { "input": "b27fk", "output": "Too weak" }, { "input": "h6y1l", "output": "Too weak" }, { "input": "i9nij", "output": "Too weak" }, { "input": "Gf5Q6", "output": "Correct" }, { "input": "Uf24o", "output": "Correct" }, { "input": "Oj9vu", "output": "Correct" }, { "input": "c7jqaudcqmv8o7zvb5x_gp6zcgl6nwr7tz5or!28.tj8s1m2.wxz5a4id03!rq07?662vy.7.p5?vk2f2mc7ag8q3861rgd0rmbr", "output": "Too weak" }, { "input": "i6a.,8jb,n0kv4.1!7h?p.96pnhhgy6cl7dg7e4o6o384ys3z.t71kkq,,w,oqi4?u,,m5!rzu6wym_4hm,ohjy!.vvksl?pt,,1", "output": "Too weak" }, { "input": "M10V_MN_1K8YX2LA!89EYV7!5V9?,.IDHDP6JEC.OGLY.180LMZ6KW3Z5E17IT94ZNHS!79GN09Q6LH0,F3AYNKP?KM,QP_?XRD6", "output": "Too weak" }, { "input": "Hi7zYuVXCPhaho68YgCMzzgLILM6toQTJq8akMqqrnUn6ZCD36iA1yVVpvlsIiMpCu!1QZd4ycIrQ5Kcrhk5k0jTrwdAAEEP_T2f", "output": "Correct" }, { "input": "Bk2Q38vDSW5JqYu.077iYC.9YoiPc!Dh6FJWOVze6?YXiFjPNa4F1RG?154m9mY2jQobBnbxM,cDV8l1UX1?v?p.tTYIyJO!NYmE", "output": "Correct" }, { "input": "Ro1HcZ.piN,JRR88DLh,WtW!pbFM076?wCSbqfK7N2s5zUySFBtzk7HV,BxHXR0zALAr016z5jvvB.WUdEcKgYFav5TygwHQC..C", "output": "Correct" }, { "input": "!?.,_", "output": "Too weak" } ]

1,399,822,023

2,147,483,647

Python 3

OK

TESTS

81

77

0

s = input() if s.lower() != s and s.upper() != s and len([x for x in s if x.isdigit() == True]) != 0 and len(s) >= 5: print('Correct') else: print('Too weak')

Title: Password Check Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check. Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions: - the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit. You are given a password. Please implement the automatic check of its complexity for company Q. Input Specification: The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_". Output Specification: If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes). Demo Input: ['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n'] Demo Output: ['Too weak\n', 'Too weak\n', 'Correct\n'] Note: none

```python s = input() if s.lower() != s and s.upper() != s and len([x for x in s if x.isdigit() == True]) != 0 and len(s) >= 5: print('Correct') else: print('Too weak') ```

3

918

A

Eleven

PROGRAMMING

800

[ "brute force", "implementation" ]

null

Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters. Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where - *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2). As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.

The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).

Print Eleven's new name on the first and only line of output.

[ "8\n", "15\n" ]

[ "OOOoOooO\n", "OOOoOooOooooOoo\n" ]

none

500

[ { "input": "8", "output": "OOOoOooO" }, { "input": "15", "output": "OOOoOooOooooOoo" }, { "input": "85", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo" }, { "input": "381", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo" }, { "input": "805", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "1000", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "1", "output": "O" }, { "input": "2", "output": "OO" }, { "input": "3", "output": "OOO" }, { "input": "5", "output": "OOOoO" }, { "input": "17", "output": "OOOoOooOooooOoooo" }, { "input": "49", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo" }, { "input": "256", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo" }, { "input": "512", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "933", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "61", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo" }, { "input": "781", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "999", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." } ]

1,629,580,649

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

7

92

6,963,200

num=int(input()) frst= 0 scnd= 1 i=0 lst=[] if(num==1): print('O') else: while (i < num): if (i <= 1): fibo = i lst.append(fibo) else: fibo = frst + scnd frst = scnd scnd = fibo lst.append(fibo) i = i + 1 lst2 = [] for i in range(num + 1): if i in lst: lst2.append('O') else: lst2.append('o') lst3 = [] for i in range(1, len(lst2)): el = lst2[i] lst3.append(el) z = ''.join([str(elem) for elem in lst3]) print(z)

Title: Eleven Time Limit: None seconds Memory Limit: None megabytes Problem Description: Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters. Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where - *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2). As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name. Input Specification: The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000). Output Specification: Print Eleven's new name on the first and only line of output. Demo Input: ['8\n', '15\n'] Demo Output: ['OOOoOooO\n', 'OOOoOooOooooOoo\n'] Note: none

```python num=int(input()) frst= 0 scnd= 1 i=0 lst=[] if(num==1): print('O') else: while (i < num): if (i <= 1): fibo = i lst.append(fibo) else: fibo = frst + scnd frst = scnd scnd = fibo lst.append(fibo) i = i + 1 lst2 = [] for i in range(num + 1): if i in lst: lst2.append('O') else: lst2.append('o') lst3 = [] for i in range(1, len(lst2)): el = lst2[i] lst3.append(el) z = ''.join([str(elem) for elem in lst3]) print(z) ```

0

583

A

Asphalting Roads

PROGRAMMING

1,000

[ "implementation" ]

null

City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them. Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted. Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads. According to the schedule of road works tell in which days at least one road will be asphalted.

The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city. Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.

In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.

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

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

In the sample the brigade acts like that: 1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.

500

[ { "input": "2\n1 1\n1 2\n2 1\n2 2", "output": "1 4 " }, { "input": "1\n1 1", "output": "1 " }, { "input": "2\n1 1\n2 2\n1 2\n2 1", "output": "1 2 " }, { "input": "2\n1 2\n2 2\n2 1\n1 1", "output": "1 3 " }, { "input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1", "output": "1 4 5 " }, { "input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3", "output": "1 2 6 " }, { "input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3", "output": "1 3 5 14 " }, { "input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4", "output": "1 2 9 12 " }, { "input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8", "output": "1 2 4 9 10 14 16 32 56 " }, { "input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5", "output": "1 6 11 18 28 36 39 56 " }, { "input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8", "output": "1 2 3 4 5 6 7 8 9 " } ]

1,447,511,062

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

46

102,400

n = int(input()) v = [True] * n h = [True] * n ans = [] for i in range(1, n * n + 1): a, b = map(lambda x: int(x) - 1, input().split()) if v[a] and h[b]: ans.append(i) v[a] = False h[a] = False print(' '.join(map(str, ans)))

Title: Asphalting Roads Time Limit: None seconds Memory Limit: None megabytes Problem Description: City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them. Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted. Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads. According to the schedule of road works tell in which days at least one road will be asphalted. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city. Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. Output Specification: In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. Demo Input: ['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n'] Demo Output: ['1 4 \n', '1 \n'] Note: In the sample the brigade acts like that: 1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.

```python n = int(input()) v = [True] * n h = [True] * n ans = [] for i in range(1, n * n + 1): a, b = map(lambda x: int(x) - 1, input().split()) if v[a] and h[b]: ans.append(i) v[a] = False h[a] = False print(' '.join(map(str, ans))) ```

0

447

B

DZY Loves Strings

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?

The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.

Print a single integer — the largest possible value of the resulting string DZY could get.

[ "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n" ]

[ "41\n" ]

In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

1,000

[ { "input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "41" }, { "input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453", "output": "29978" }, { "input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314", "output": "1762894" }, { "input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585", "output": "2960349" }, { "input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933", "output": "99140444" }, { "input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686", "output": "301124161" }, { "input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495", "output": "30125295" }, { "input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165", "output": "136422" }, { "input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348", "output": "2578628" }, { "input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338", "output": "9168707" }, { "input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657", "output": "144901921" }, { "input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860", "output": "809931" }, { "input": "a\n0\n5 1 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": "5" }, { "input": "lol\n3\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", "output": "21" } ]

1,427,435,967

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

46

0

# coding: utf-8 s = input() k = int(input()) w = [int(i) for i in input().split()] li = [] for ch in s: li.append(w[ord(ch)-ord('a')]) li.extend([max(w)]*k) li.sort() ans = 0 for i in range(0,len(li)): ans += (i+1)*li[i] print(ans)

Title: DZY Loves Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get? Input Specification: The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000. Output Specification: Print a single integer — the largest possible value of the resulting string DZY could get. Demo Input: ['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n'] Demo Output: ['41\n'] Note: In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

```python # coding: utf-8 s = input() k = int(input()) w = [int(i) for i in input().split()] li = [] for ch in s: li.append(w[ord(ch)-ord('a')]) li.extend([max(w)]*k) li.sort() ans = 0 for i in range(0,len(li)): ans += (i+1)*li[i] print(ans) ```

0

489

C

Given Length and Sum of Digits...

PROGRAMMING

1,400

[ "dp", "greedy", "implementation" ]

null

You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.

The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.

In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).

[ "2 15\n", "3 0\n" ]

[ "69 96\n", "-1 -1\n" ]

none

1,500

[ { "input": "2 15", "output": "69 96" }, { "input": "3 0", "output": "-1 -1" }, { "input": "2 1", "output": "10 10" }, { "input": "3 10", "output": "109 910" }, { "input": "100 100", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000099999999999 9999999999910000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "1 900", "output": "-1 -1" }, { "input": "1 9", "output": "9 9" }, { "input": "1 0", "output": "0 0" }, { "input": "1 1", "output": "1 1" }, { "input": "1 2", "output": "2 2" }, { "input": "1 8", "output": "8 8" }, { "input": "1 10", "output": "-1 -1" }, { "input": "1 11", "output": "-1 -1" }, { "input": "2 0", "output": "-1 -1" }, { "input": "2 1", "output": "10 10" }, { "input": "2 2", "output": "11 20" }, { "input": "2 8", "output": "17 80" }, { "input": "2 10", "output": "19 91" }, { "input": "2 11", "output": "29 92" }, { "input": "2 16", "output": "79 97" }, { "input": "2 17", "output": "89 98" }, { "input": "2 18", "output": "99 99" }, { "input": "2 19", "output": "-1 -1" }, { "input": "2 20", "output": "-1 -1" }, { "input": "2 900", "output": "-1 -1" }, { "input": "3 1", "output": "100 100" }, { "input": "3 2", "output": "101 200" }, { "input": "3 3", "output": "102 300" }, { "input": "3 9", "output": "108 900" }, { "input": "3 10", "output": "109 910" }, { "input": "3 20", "output": "299 992" }, { "input": "3 21", "output": "399 993" }, { "input": "3 26", "output": "899 998" }, { "input": "3 27", "output": "999 999" }, { "input": "3 28", "output": "-1 -1" }, { "input": "3 100", "output": "-1 -1" }, { "input": "100 0", "output": "-1 -1" }, { "input": "100 1", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 2", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 9", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 10", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009 9100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 11", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000019 9200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 296", "output": "1000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999999999999 9999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 297", "output": "1000000000000000000000000000000000000000000000000000000000000000000899999999999999999999999999999999 9999999999999999999999999999999990000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 298", "output": "1000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999 9999999999999999999999999999999991000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 299", "output": "1000000000000000000000000000000000000000000000000000000000000000001999999999999999999999999999999999 9999999999999999999999999999999992000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 300", "output": "1000000000000000000000000000000000000000000000000000000000000000002999999999999999999999999999999999 9999999999999999999999999999999993000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 301", "output": "1000000000000000000000000000000000000000000000000000000000000000003999999999999999999999999999999999 9999999999999999999999999999999994000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 895", "output": "4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994" }, { "input": "100 896", "output": "5999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995" }, { "input": "100 897", "output": "6999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996" }, { "input": "100 898", "output": "7999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997" }, { "input": "100 899", "output": "8999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998" }, { "input": "100 900", "output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999" }, { "input": "99 900", "output": "-1 -1" }, { "input": "99 891", "output": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999" }, { "input": "99 892", "output": "-1 -1" }, { "input": "96 364", "output": "100000000000000000000000000000000000000000000000000000039999999999999999999999999999999999999999 999999999999999999999999999999999999999940000000000000000000000000000000000000000000000000000000" }, { "input": "97 206", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000079999999999999999999999 9999999999999999999999800000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "98 250", "output": "10000000000000000000000000000000000000000000000000000000000000000000006999999999999999999999999999 99999999999999999999999999970000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "99 96", "output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000059999999999 999999999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "100 215", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999 9999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "5 18", "output": "10089 99000" }, { "input": "2 19", "output": "-1 -1" }, { "input": "3 3", "output": "102 300" } ]

1,678,616,649

2,147,483,647

Python 3

OK

TESTS

65

46

0

m,s=map(int,input().split()) if s>9*m or (s<1and m!=1):print(-1,-1) else:print(int(10**(m-1)+((s-1)%9+1)*10**((s-1)//9)-1),10**m-10**(m-(s-1)//9-1)*(9-(s-1)%9))

Title: Given Length and Sum of Digits... Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes. Input Specification: The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers. Output Specification: In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes). Demo Input: ['2 15\n', '3 0\n'] Demo Output: ['69 96\n', '-1 -1\n'] Note: none

```python m,s=map(int,input().split()) if s>9*m or (s<1and m!=1):print(-1,-1) else:print(int(10**(m-1)+((s-1)%9+1)*10**((s-1)//9)-1),10**m-10**(m-(s-1)//9-1)*(9-(s-1)%9)) ```

3

937

A

Olympiad

PROGRAMMING

800

[ "implementation", "sortings" ]

null

The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score.

Print a single integer — the desired number of ways.

[ "4\n1 3 3 2\n", "3\n1 1 1\n", "4\n42 0 0 42\n" ]

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

There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.

500

[ { "input": "4\n1 3 3 2", "output": "3" }, { "input": "3\n1 1 1", "output": "1" }, { "input": "4\n42 0 0 42", "output": "1" }, { "input": "10\n1 0 1 0 1 0 0 0 0 1", "output": "1" }, { "input": "10\n572 471 540 163 50 30 561 510 43 200", "output": "10" }, { "input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600", "output": "94" }, { "input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600", "output": "1" }, { "input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12", "output": "13" }, { "input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391", "output": "33" }, { "input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3", "output": "3" }, { "input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55", "output": "93" }, { "input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13", "output": "13" }, { "input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1", "output": "1" }, { "input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37", "output": "61" }, { "input": "2\n0 1", "output": "1" }, { "input": "2\n0 600", "output": "1" }, { "input": "4\n1 1 1 2", "output": "2" }, { "input": "4\n0 0 1 2", "output": "2" }, { "input": "1\n5", "output": "1" }, { "input": "2\n0 5", "output": "1" }, { "input": "5\n1 0 0 1 2", "output": "2" } ]

1,558,191,208

2,147,483,647

PyPy 3

OK

TESTS

21

124

0

n = int(input()) a = input().split() b = [] for i in range(n): a[i] = int(a[i]) true = True if a[i] == 0: true = False for j in range(len(b)): if a[i] == b[j]: true = False if not true: break if true: b += [a[i]] print(len(b))

Title: Olympiad Time Limit: None seconds Memory Limit: None megabytes Problem Description: The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score. Output Specification: Print a single integer — the desired number of ways. Demo Input: ['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n'] Demo Output: ['3\n', '1\n', '1\n'] Note: There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.

```python n = int(input()) a = input().split() b = [] for i in range(n): a[i] = int(a[i]) true = True if a[i] == 0: true = False for j in range(len(b)): if a[i] == b[j]: true = False if not true: break if true: b += [a[i]] print(len(b)) ```

3

355

A

Vasya and Digital Root

PROGRAMMING

1,100

[ "constructive algorithms", "implementation" ]

null

Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you. Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to: 1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10. For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3. Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000). Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist.

The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9).

In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist. The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes.

[ "4 4\n", "5 1\n", "1 0\n" ]

[ "5881\n", "36172\n", "0\n" ]

For the first test sample *dr*(5881) = *dr*(22) = 4. For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1.

500

[ { "input": "4 4", "output": "5881" }, { "input": "5 1", "output": "36172" }, { "input": "1 0", "output": "0" }, { "input": "8 7", "output": "49722154" }, { "input": "487 0", "output": "No solution" }, { "input": "1000 5", "output": "8541939554067890866522280268745476436249986028349767396372181155840878549622667946850256234534972693110974918858266403731194206972478044933297639886527448596769215803533001453375065914421371731616055420973164037664278812596299678416020519508892847037891229851414508562230407367486468987019052183250172396304562086008837592345867873765321840214188417303688776985319268802181355472294386101622570417737061113209187893810568585166094583478900129912239498334853726870963804475563182775380744565964067602555515611220..." }, { "input": "22 9", "output": "1583569962049529809017" }, { "input": "1 1", "output": "1" }, { "input": "1 9", "output": "9" }, { "input": "13 5", "output": "1381199538344" }, { "input": "100 4", "output": "6334594910586850938286642284598905674550356974741186703111536643493065423553455569335256292313330478" }, { "input": "123 6", "output": "928024873067884441426263446866614165147002631091527531801777528825238463822318502518751375671158771476735217071878592158343" }, { "input": "1000 1", "output": "8286301124628812353504240076754144327937426329149605334362213339655339076564408659154706137278060590992944494591503606137350736487608756923833530346502466262820452589925067370165968733865814927433418675056573256434073937686361155637721866942352171450747045834987797118866710087297111065178077368748085213082452303815796793489599773148508108295035303578345492871662297456131736137780231762177312635688688714815857818196180724774924848693916003108422682889382923194020205691379066085156078824413573001257245677878..." }, { "input": "2 0", "output": "No solution" }, { "input": "734 9", "output": 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"8948712698877635315965401396781625629035528026219922557326466734622505808493494219669540192792500692387387200924494124219975316993592377253517258369463882172533672158172302951620486954085880606055358057621166471042557383036617719864238933843342304818076672889894622975857106353903546493307325157763617269195570831067239463586554245706859061059010215520785892192165179414199200952765077228454366556358805840526959104496983177402562569401945586771345953166346316987259989961516385311376707950154520512125143059966..." }, { "input": "960 6", "output": "7291446744949293530598651243840704118065364362420848463900543089429494124955982767375712583398715647208330285855809398453361266463042342917037983463277320070057956978767965467358862479678812136400444143489366786562672928727263605336304125661306952421127807059398289873947797478996261161224877129724944902005212399176751167053423457968483336961277157597291131065544350665072172392437597673561840137077580044817979332136630042866681837917711758227272499659805765131669208911408670581308412686469802437930679571593..." } ]

1,626,595,646

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

1

77

20,172,800

k, d = [int(x) for x in input().split()] if k>0 and d==1: print("No solution") else: print(d, end='') for i in range(1,k): print(0, end='')

Title: Vasya and Digital Root Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you. Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to: 1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10. For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3. Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000). Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist. Input Specification: The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9). Output Specification: In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist. The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes. Demo Input: ['4 4\n', '5 1\n', '1 0\n'] Demo Output: ['5881\n', '36172\n', '0\n'] Note: For the first test sample *dr*(5881) = *dr*(22) = 4. For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1.

```python k, d = [int(x) for x in input().split()] if k>0 and d==1: print("No solution") else: print(d, end='') for i in range(1,k): print(0, end='') ```

0

813

B

The Golden Age

PROGRAMMING

1,800

[ "brute force", "math" ]

null

Unlucky year in Berland is such a year that its number *n* can be represented as *n*<==<=*x**a*<=+<=*y**b*, where *a* and *b* are non-negative integer numbers. For example, if *x*<==<=2 and *y*<==<=3 then the years 4 and 17 are unlucky (4<==<=20<=+<=31, 17<==<=23<=+<=32<==<=24<=+<=30) and year 18 isn't unlucky as there is no such representation for it. Such interval of years that there are no unlucky years in it is called The Golden Age. You should write a program which will find maximum length of The Golden Age which starts no earlier than the year *l* and ends no later than the year *r*. If all years in the interval [*l*,<=*r*] are unlucky then the answer is 0.

The first line contains four integer numbers *x*, *y*, *l* and *r* (2<=≤<=*x*,<=*y*<=≤<=1018, 1<=≤<=*l*<=≤<=*r*<=≤<=1018).

Print the maximum length of The Golden Age within the interval [*l*,<=*r*]. If all years in the interval [*l*,<=*r*] are unlucky then print 0.

[ "2 3 1 10\n", "3 5 10 22\n", "2 3 3 5\n" ]

[ "1\n", "8\n", "0\n" ]

In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8]. In the second example the longest Golden Age is the interval [15, 22].

0

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"output": "12884901886" }, { "input": "100914030314340517 512922595840756536 812829791042966971 966156272123068006", "output": "153326481080101036" }, { "input": "288230376151711744 288230376151711744 1 1000000000000000000", "output": "423539247696576512" }, { "input": "6 9 681 750", "output": "49" }, { "input": "880356874212472951 178538501711453307 162918237570625233 224969951233811739", "output": "46431449522358431" }, { "input": "2 7 405373082004080437 771991379629433514", "output": "153172782079203571" }, { "input": "10 11 10 11", "output": "1" } ]

1,622,432,212

1,072

PyPy 3

WRONG_ANSWER

TESTS

18

202

6,451,200

import sys #import random from bisect import bisect_left as lb from collections import deque #sys.setrecursionlimit(10**8) from queue import PriorityQueue as pq from math import * input_ = lambda: sys.stdin.readline().strip("\r\n") ii = lambda : int(input_()) il = lambda : list(map(int, input_().split())) ilf = lambda : list(map(float, input_().split())) ip = lambda : input_() fi = lambda : float(input_()) ap = lambda ab,bc,cd : ab[bc].append(cd) li = lambda : list(input_()) pr = lambda x : print(x) prinT = lambda x : print(x) f = lambda : sys.stdout.flush() inv =lambda x:pow(x,mod-2,mod) mod = 10**9 + 7 x,y,l,r = il() a = [] for i in range (70) : for j in range (70) : a.append(x**i + y**j) a.sort() ans = 0 for i in range (len(a)-1) : ans = max(ans , min(r,a[i+1]-1) - max(l,a[i]+1)+1) print(ans)

Title: The Golden Age Time Limit: None seconds Memory Limit: None megabytes Problem Description: Unlucky year in Berland is such a year that its number *n* can be represented as *n*<==<=*x**a*<=+<=*y**b*, where *a* and *b* are non-negative integer numbers. For example, if *x*<==<=2 and *y*<==<=3 then the years 4 and 17 are unlucky (4<==<=20<=+<=31, 17<==<=23<=+<=32<==<=24<=+<=30) and year 18 isn't unlucky as there is no such representation for it. Such interval of years that there are no unlucky years in it is called The Golden Age. You should write a program which will find maximum length of The Golden Age which starts no earlier than the year *l* and ends no later than the year *r*. If all years in the interval [*l*,<=*r*] are unlucky then the answer is 0. Input Specification: The first line contains four integer numbers *x*, *y*, *l* and *r* (2<=≤<=*x*,<=*y*<=≤<=1018, 1<=≤<=*l*<=≤<=*r*<=≤<=1018). Output Specification: Print the maximum length of The Golden Age within the interval [*l*,<=*r*]. If all years in the interval [*l*,<=*r*] are unlucky then print 0. Demo Input: ['2 3 1 10\n', '3 5 10 22\n', '2 3 3 5\n'] Demo Output: ['1\n', '8\n', '0\n'] Note: In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8]. In the second example the longest Golden Age is the interval [15, 22].

```python import sys #import random from bisect import bisect_left as lb from collections import deque #sys.setrecursionlimit(10**8) from queue import PriorityQueue as pq from math import * input_ = lambda: sys.stdin.readline().strip("\r\n") ii = lambda : int(input_()) il = lambda : list(map(int, input_().split())) ilf = lambda : list(map(float, input_().split())) ip = lambda : input_() fi = lambda : float(input_()) ap = lambda ab,bc,cd : ab[bc].append(cd) li = lambda : list(input_()) pr = lambda x : print(x) prinT = lambda x : print(x) f = lambda : sys.stdout.flush() inv =lambda x:pow(x,mod-2,mod) mod = 10**9 + 7 x,y,l,r = il() a = [] for i in range (70) : for j in range (70) : a.append(x**i + y**j) a.sort() ans = 0 for i in range (len(a)-1) : ans = max(ans , min(r,a[i+1]-1) - max(l,a[i]+1)+1) print(ans) ```

0

755

A

PolandBall and Hypothesis

PROGRAMMING

800

[ "brute force", "graphs", "math", "number theory" ]

null

PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number". Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.

The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.

Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.

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

[ "1", "2" ]

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. For the first sample testcase, 3·1 + 1 = 4. We can output 1. In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.

500

[ { "input": "3", "output": "1" }, { "input": "4", "output": "2" }, { "input": "10", "output": "2" }, { "input": "153", "output": "1" }, { "input": "1000", "output": "1" }, { "input": "1", "output": "3" }, { "input": "2", "output": "4" }, { "input": "5", "output": "1" }, { "input": "6", "output": "4" }, { "input": "7", "output": "1" }, { "input": "8", "output": "1" }, { "input": "9", "output": "1" }, { "input": "11", "output": "1" }, { "input": "998", "output": "1" }, { "input": "996", "output": "3" }, { "input": "36", "output": "4" }, { "input": "210", "output": "4" }, { "input": "270", "output": "4" }, { "input": "306", "output": "4" }, { "input": "330", "output": "5" }, { "input": "336", "output": "4" }, { "input": "600", "output": "4" }, { "input": "726", "output": "4" }, { "input": "988", "output": "1" }, { "input": "12", "output": "2" }, { "input": "987", "output": "1" }, { "input": "13", "output": "1" }, { "input": "986", "output": "1" }, { "input": "14", "output": "1" }, { "input": "985", "output": "1" }, { "input": "15", "output": "1" }, { "input": "984", "output": "1" }, { "input": "16", "output": "2" }, { "input": "983", "output": "1" }, { "input": "17", "output": "1" }, { "input": "982", "output": "2" }, { "input": "18", "output": "3" }, { "input": "981", "output": "1" }, { "input": "19", "output": "1" }, { "input": "980", "output": "1" }, { "input": "20", "output": "1" }, { "input": "979", "output": "1" }, { "input": "21", "output": "1" }, { "input": "978", "output": "1" }, { "input": "22", "output": "2" }, { "input": "977", "output": "1" }, { "input": "23", "output": "1" }, { "input": "976", "output": "2" }, { "input": "24", "output": "1" }, { "input": "975", "output": "1" }, { "input": "25", "output": "1" }, { "input": "2", "output": "4" }, { "input": "6", "output": "4" }, { "input": "996", "output": "3" }, { "input": "8", "output": "1" }, { "input": "998", "output": "1" }, { "input": "330", "output": "5" }, { "input": "18", "output": "3" }, { "input": "12", "output": "2" }, { "input": "24", "output": "1" }, { "input": "999", "output": "1" }, { "input": "726", "output": "4" }, { "input": "876", "output": "3" } ]

1,641,296,598

2,147,483,647

Python 3

OK

TESTS

63

46

0

from math import sqrt n = int(input()) def solve( n ) : i = 1 def isPrime(n) : if n <= 2 : return True for i in range(2 , int(sqrt(n)) + 1) : if n % i == 0 : return False return True while isPrime(n * i + 1) : i += 1 return i print(solve( n ))

Title: PolandBall and Hypothesis Time Limit: None seconds Memory Limit: None megabytes Problem Description: PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number". Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*. Input Specification: The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis. Output Specification: Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists. Demo Input: ['3\n', '4\n'] Demo Output: ['1', '2'] Note: A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. For the first sample testcase, 3·1 + 1 = 4. We can output 1. In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.

```python from math import sqrt n = int(input()) def solve( n ) : i = 1 def isPrime(n) : if n <= 2 : return True for i in range(2 , int(sqrt(n)) + 1) : if n % i == 0 : return False return True while isPrime(n * i + 1) : i += 1 return i print(solve( n )) ```

3

975

A

Aramic script

PROGRAMMING

900

[ "implementation", "strings" ]

null

In Aramic language words can only represent objects. Words in Aramic have special properties: - A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root. You have an ancient script in Aramic. What is the number of different objects mentioned in the script?

The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script. The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$. It is guaranteed that all characters of the strings are small latin letters.

Output one integer — the number of different objects mentioned in the given ancient Aramic script.

[ "5\na aa aaa ab abb\n", "3\namer arem mrea\n" ]

[ "2", "1" ]

In the first test, there are two objects mentioned. The roots that represent them are "a","ab". In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

500

[ { "input": "5\na aa aaa ab abb", "output": "2" }, { "input": "3\namer arem mrea", "output": "1" }, { "input": "10\nbda bbb cda dca dda dcb bcd dcb ada ddd", "output": "6" }, { "input": "2\nfhjlqs aceginpr", "output": "2" }, { "input": "2\nbcdfghimn efghijlmo", "output": "2" } ]

1,542,468,225

2,147,483,647

PyPy 3

OK

TESTS

30

186

9,625,600

n =int(input()) l = list(map(str,input().split())) l = [frozenset(i) for i in l] s = set() for i in l: s.add(i) print(len(s))

Title: Aramic script Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Aramic language words can only represent objects. Words in Aramic have special properties: - A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root. You have an ancient script in Aramic. What is the number of different objects mentioned in the script? Input Specification: The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script. The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$. It is guaranteed that all characters of the strings are small latin letters. Output Specification: Output one integer — the number of different objects mentioned in the given ancient Aramic script. Demo Input: ['5\na aa aaa ab abb\n', '3\namer arem mrea\n'] Demo Output: ['2', '1'] Note: In the first test, there are two objects mentioned. The roots that represent them are "a","ab". In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

```python n =int(input()) l = list(map(str,input().split())) l = [frozenset(i) for i in l] s = set() for i in l: s.add(i) print(len(s)) ```

3

177

B1

Rectangular Game

PROGRAMMING

1,000

[ "number theory" ]

null

The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*. Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble. The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where: - *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1 The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.

The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has. The input limitations for getting 30 points are: - 2<=≤<=*n*<=≤<=50 The input limitations for getting 100 points are: - 2<=≤<=*n*<=≤<=109

Print a single number — the maximum possible result of the game.

[ "10\n", "8\n" ]

[ "16\n", "15\n" ]

Consider the first example (*c*1 = 10). The possible options for the game development are: - Arrange the pebbles in 10 rows, one pebble per row. Then *c*2 = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*3 = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*2 = 5, *c*3 = 1, and the game ends with the result of 16 — the maximum possible result.

30

[ { "input": "10", "output": "16" }, { "input": "8", "output": "15" }, { "input": "4", "output": "7" }, { "input": "36", "output": "67" }, { "input": "32", "output": "63" }, { "input": "46", "output": "70" }, { "input": "6", "output": "10" }, { "input": "13", "output": "14" }, { "input": "9", "output": "13" }, { "input": "18", "output": "31" }, { "input": "20", "output": "36" }, { "input": "37", "output": "38" }, { "input": "42", "output": "71" }, { "input": "50", "output": "81" }, { "input": "48", "output": "94" }, { "input": "33", "output": "45" }, { "input": "49", "output": "57" }, { "input": "34", "output": "52" }, { "input": "35", "output": "43" }, { "input": "37", "output": "38" }, { "input": "39", "output": "53" }, { "input": "42", "output": "71" }, { "input": "14", "output": "22" }, { "input": "2", "output": "3" }, { "input": "3", "output": "4" }, { "input": "5", "output": "6" } ]

1,569,833,830

2,147,483,647

Python 3

OK

TESTS1

26

218

0

def min_div(num): for i in range(2,int(num**0.5)+1): if num % i == 0: return i return 1 n = int(input()) count = n while min_div(n)!=1: n = n//min_div(n) count += n count += 1 print(count)

Title: Rectangular Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*. Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble. The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where: - *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1 The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game. Input Specification: The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has. The input limitations for getting 30 points are: - 2<=≤<=*n*<=≤<=50 The input limitations for getting 100 points are: - 2<=≤<=*n*<=≤<=109 Output Specification: Print a single number — the maximum possible result of the game. Demo Input: ['10\n', '8\n'] Demo Output: ['16\n', '15\n'] Note: Consider the first example (*c*1 = 10). The possible options for the game development are: - Arrange the pebbles in 10 rows, one pebble per row. Then *c*2 = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*3 = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*2 = 5, *c*3 = 1, and the game ends with the result of 16 — the maximum possible result.

```python def min_div(num): for i in range(2,int(num**0.5)+1): if num % i == 0: return i return 1 n = int(input()) count = n while min_div(n)!=1: n = n//min_div(n) count += n count += 1 print(count) ```

3

271

A

Beautiful Year

PROGRAMMING

800

[ "brute force" ]

null

It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits. Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.

The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.

Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.

[ "1987\n", "2013\n" ]

[ "2013\n", "2014\n" ]

none

500

[ { "input": "1987", "output": "2013" }, { "input": "2013", "output": "2014" }, { "input": "1000", "output": "1023" }, { "input": "1001", "output": "1023" }, { "input": "1234", "output": "1235" }, { "input": "5555", "output": "5601" }, { "input": "9000", "output": "9012" }, { "input": "1111", "output": "1203" }, { "input": "8999", "output": "9012" }, { "input": "4572", "output": "4573" }, { "input": "6666", "output": "6701" }, { "input": "2001", "output": "2013" }, { "input": "3000", "output": "3012" }, { "input": "7712", "output": "7801" }, { "input": "8088", "output": "8091" }, { "input": "1594", "output": "1596" }, { "input": "6016", "output": "6017" }, { "input": "8800", "output": "8901" }, { "input": "3331", "output": "3401" }, { "input": "5090", "output": "5091" }, { "input": "2342", "output": "2345" }, { "input": "2334", "output": "2340" }, { "input": "1123", "output": "1203" }, { "input": "8989", "output": "9012" }, { "input": "8977", "output": "9012" }, { "input": "6869", "output": "6870" }, { "input": "8999", "output": "9012" } ]

1,694,434,443

2,147,483,647

Python 3

OK

TESTS

27

92

0

def good(n): aa=[] re=0 for i in range(len(str(n))): a=n%10 aa.append(a) n=n//10 for j in range(len(aa)-1): for k in range(j+1): if aa[j+1]==aa[k]: re=1 if re==1: return(1) else: return(0) begin=int(input()) ans=begin+1 while good(ans)==1: ans+=1 print(ans)

Title: Beautiful Year Time Limit: None seconds Memory Limit: None megabytes Problem Description: It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits. Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. Input Specification: The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number. Output Specification: Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists. Demo Input: ['1987\n', '2013\n'] Demo Output: ['2013\n', '2014\n'] Note: none

```python def good(n): aa=[] re=0 for i in range(len(str(n))): a=n%10 aa.append(a) n=n//10 for j in range(len(aa)-1): for k in range(j+1): if aa[j+1]==aa[k]: re=1 if re==1: return(1) else: return(0) begin=int(input()) ans=begin+1 while good(ans)==1: ans+=1 print(ans) ```

3

913

A

Modular Exponentiation

PROGRAMMING

900

[ "implementation", "math" ]

null

The following problem is well-known: given integers *n* and *m*, calculate where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*. You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108). The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108).

Output a single integer — the value of .

[ "4\n42\n", "1\n58\n", "98765432\n23456789\n" ]

[ "10\n", "0\n", "23456789\n" ]

In the first example, the remainder of division of 42 by 24 = 16 is equal to 10. In the second example, 58 is divisible by 21 = 2 without remainder, and the answer is 0.

500

[ { "input": "4\n42", "output": "10" }, { "input": "1\n58", "output": "0" }, { "input": "98765432\n23456789", "output": "23456789" }, { "input": "8\n88127381", "output": "149" }, { "input": "32\n92831989", "output": "92831989" }, { "input": "92831989\n25", "output": "25" }, { "input": "100000000\n100000000", "output": "100000000" }, { "input": "7\n1234", "output": "82" }, { "input": "1\n1", "output": "1" }, { "input": "1\n100000000", "output": "0" }, { "input": "100000000\n1", "output": "1" }, { "input": "1\n2", "output": "0" }, { "input": "2\n1", "output": "1" }, { "input": "2\n2", "output": "2" }, { "input": "2\n3", "output": "3" }, { "input": "2\n4", "output": "0" }, { "input": "2\n5", "output": "1" }, { "input": "25\n33554432", "output": "0" }, { "input": "26\n33554432", "output": "33554432" }, { "input": "25\n67108864", "output": "0" }, { "input": "26\n67108864", "output": "0" }, { "input": "25\n92831989", "output": "25723125" }, { "input": "26\n92831989", "output": "25723125" }, { "input": "27\n92831989", "output": "92831989" }, { "input": "29\n92831989", "output": "92831989" }, { "input": "30\n92831989", "output": "92831989" }, { "input": "31\n92831989", "output": "92831989" }, { "input": "99812399\n93172892", "output": "93172892" }, { "input": "67108864\n12345678", "output": "12345678" }, { "input": "3\n98391849", "output": "1" }, { "input": "13\n1231232", "output": "2432" }, { "input": "18\n9842194", "output": "142866" }, { "input": "24\n71237177", "output": "4128313" }, { "input": "11\n65535", "output": "2047" }, { "input": "11\n65536", "output": "0" }, { "input": "12\n123", "output": "123" }, { "input": "123456\n12", "output": "12" }, { "input": "88\n888", "output": "888" }, { "input": "888\n88", "output": "88" }, { "input": "8888\n8888", "output": "8888" }, { "input": "64\n128", "output": "128" }, { "input": "128\n64", "output": "64" }, { "input": "60\n9999999", "output": "9999999" }, { "input": "200\n100000", "output": "100000" }, { "input": "3\n8", "output": "0" }, { "input": "10000000\n25024886", "output": "25024886" }, { "input": "100\n100000", "output": "100000" }, { "input": "99999999\n100000000", "output": "100000000" }, { "input": "70\n12345678", "output": "12345678" } ]

1,624,691,863

2,147,483,647

PyPy 3

OK

TESTS

49

109

13,107,200

n=int(input()) m=int(input()) n=2**n print(int(m%n))

Title: Modular Exponentiation Time Limit: None seconds Memory Limit: None megabytes Problem Description: The following problem is well-known: given integers *n* and *m*, calculate where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*. You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108). The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108). Output Specification: Output a single integer — the value of . Demo Input: ['4\n42\n', '1\n58\n', '98765432\n23456789\n'] Demo Output: ['10\n', '0\n', '23456789\n'] Note: In the first example, the remainder of division of 42 by 24 = 16 is equal to 10. In the second example, 58 is divisible by 21 = 2 without remainder, and the answer is 0.

```python n=int(input()) m=int(input()) n=2**n print(int(m%n)) ```

3

1,007

E

Mini Metro

PROGRAMMING

3,400

[ "dp" ]

null

In a simplified version of a "Mini Metro" game, there is only one subway line, and all the trains go in the same direction. There are $n$ stations on the line, $a_i$ people are waiting for the train at the $i$-th station at the beginning of the game. The game starts at the beginning of the $0$-th hour. At the end of each hour (couple minutes before the end of the hour), $b_i$ people instantly arrive to the $i$-th station. If at some moment, the number of people at the $i$-th station is larger than $c_i$, you lose. A player has several trains which he can appoint to some hours. The capacity of each train is $k$ passengers. In the middle of the appointed hour, the train goes from the $1$-st to the $n$-th station, taking as many people at each station as it can accommodate. A train can not take people from the $i$-th station if there are people at the $i-1$-th station. If multiple trains are appointed to the same hour, their capacities are being added up and they are moving together. The player wants to stay in the game for $t$ hours. Determine the minimum number of trains he will need for it.

The first line contains three integers $n$, $t$, and $k$ ($1 \leq n, t \leq 200, 1 \leq k \leq 10^9$) — the number of stations on the line, hours we want to survive, and capacity of each train respectively. Each of the next $n$ lines contains three integers $a_i$, $b_i$, and $c_i$ ($0 \leq a_i, b_i \leq c_i \leq 10^9$) — number of people at the $i$-th station in the beginning of the game, number of people arriving to $i$-th station in the end of each hour and maximum number of people at the $i$-th station allowed respectively.

Output a single integer number — the answer to the problem.

[ "3 3 10\n2 4 10\n3 3 9\n4 2 8\n", "4 10 5\n1 1 1\n1 0 1\n0 5 8\n2 7 100\n" ]

[ "2\n", "12\n" ]

<img class="tex-graphics" src="https://espresso.codeforces.com/bfa11d535d9fc44e73f6f8280d06436e4e327753.png" style="max-width: 100.0%;max-height: 100.0%;"/> Let's look at the sample. There are three stations, on the first, there are initially 2 people, 3 people on the second, and 4 people on the third. Maximal capacities of the stations are 10, 9, and 8 respectively. One of the winning strategies is to appoint two trains to the first and the third hours. Then on the first hour, the train takes all of the people from the stations, and at the end of the hour, 4 people arrive at the first station, 3 on the second, and 2 on the third. In the second hour there are no trains appointed, and at the end of it, the same amount of people are arriving again. In the third hour, the train first takes 8 people from the first station, and when it arrives at the second station, it takes only 2 people because it can accommodate no more than 10 people. Then it passes by the third station because it is already full. After it, people arrive at the stations once more, and the game ends. As there was no such moment when the number of people at a station exceeded maximal capacity, we won using two trains.

2,500

[]

1,689,847,741

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

31

0

n, t, k = map(int, input().split()) stations = [] for _ in range(n): a, b, c = map(int, input().split()) stations.append((a, b, c)) trains_needed = 0 for i in range(n): while stations[i][0] > 0: # Check if the current station can accommodate a full train if stations[i][0] >= k: stations[i] = (stations[i][0] - k, stations[i][1], stations[i][2]) else: # If the current station doesn't have enough passengers for a full train, check the next station j = i + 1 while j < n and stations[j][0] <= k - stations[i][0]: stations[j] = (stations[j][0] - (k - stations[i][0]), stations[j][1], stations[j][2]) i = j j += 1 # If no next station can be reached, increment the number of trains needed if j == n: trains_needed += 1 break # Increment the number of trains needed for the remaining passengers at the current station if stations[i][0] > 0: trains_needed += 1 print(trains_needed)

Title: Mini Metro Time Limit: None seconds Memory Limit: None megabytes Problem Description: In a simplified version of a "Mini Metro" game, there is only one subway line, and all the trains go in the same direction. There are $n$ stations on the line, $a_i$ people are waiting for the train at the $i$-th station at the beginning of the game. The game starts at the beginning of the $0$-th hour. At the end of each hour (couple minutes before the end of the hour), $b_i$ people instantly arrive to the $i$-th station. If at some moment, the number of people at the $i$-th station is larger than $c_i$, you lose. A player has several trains which he can appoint to some hours. The capacity of each train is $k$ passengers. In the middle of the appointed hour, the train goes from the $1$-st to the $n$-th station, taking as many people at each station as it can accommodate. A train can not take people from the $i$-th station if there are people at the $i-1$-th station. If multiple trains are appointed to the same hour, their capacities are being added up and they are moving together. The player wants to stay in the game for $t$ hours. Determine the minimum number of trains he will need for it. Input Specification: The first line contains three integers $n$, $t$, and $k$ ($1 \leq n, t \leq 200, 1 \leq k \leq 10^9$) — the number of stations on the line, hours we want to survive, and capacity of each train respectively. Each of the next $n$ lines contains three integers $a_i$, $b_i$, and $c_i$ ($0 \leq a_i, b_i \leq c_i \leq 10^9$) — number of people at the $i$-th station in the beginning of the game, number of people arriving to $i$-th station in the end of each hour and maximum number of people at the $i$-th station allowed respectively. Output Specification: Output a single integer number — the answer to the problem. Demo Input: ['3 3 10\n2 4 10\n3 3 9\n4 2 8\n', '4 10 5\n1 1 1\n1 0 1\n0 5 8\n2 7 100\n'] Demo Output: ['2\n', '12\n'] Note: <img class="tex-graphics" src="https://espresso.codeforces.com/bfa11d535d9fc44e73f6f8280d06436e4e327753.png" style="max-width: 100.0%;max-height: 100.0%;"/> Let's look at the sample. There are three stations, on the first, there are initially 2 people, 3 people on the second, and 4 people on the third. Maximal capacities of the stations are 10, 9, and 8 respectively. One of the winning strategies is to appoint two trains to the first and the third hours. Then on the first hour, the train takes all of the people from the stations, and at the end of the hour, 4 people arrive at the first station, 3 on the second, and 2 on the third. In the second hour there are no trains appointed, and at the end of it, the same amount of people are arriving again. In the third hour, the train first takes 8 people from the first station, and when it arrives at the second station, it takes only 2 people because it can accommodate no more than 10 people. Then it passes by the third station because it is already full. After it, people arrive at the stations once more, and the game ends. As there was no such moment when the number of people at a station exceeded maximal capacity, we won using two trains.

```python n, t, k = map(int, input().split()) stations = [] for _ in range(n): a, b, c = map(int, input().split()) stations.append((a, b, c)) trains_needed = 0 for i in range(n): while stations[i][0] > 0: # Check if the current station can accommodate a full train if stations[i][0] >= k: stations[i] = (stations[i][0] - k, stations[i][1], stations[i][2]) else: # If the current station doesn't have enough passengers for a full train, check the next station j = i + 1 while j < n and stations[j][0] <= k - stations[i][0]: stations[j] = (stations[j][0] - (k - stations[i][0]), stations[j][1], stations[j][2]) i = j j += 1 # If no next station can be reached, increment the number of trains needed if j == n: trains_needed += 1 break # Increment the number of trains needed for the remaining passengers at the current station if stations[i][0] > 0: trains_needed += 1 print(trains_needed) ```

0

427

A

Police Recruits

PROGRAMMING

800

[ "implementation" ]

null

The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups. Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime. If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated. Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.

The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers. If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.

Print a single integer, the number of crimes which will go untreated.

[ "3\n-1 -1 1\n", "8\n1 -1 1 -1 -1 1 1 1\n", "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n" ]

[ "2\n", "1\n", "8\n" ]

Lets consider the second example: 1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired. The answer is one, as one crime (on step 5) will go untreated.

500

[ { "input": "3\n-1 -1 1", "output": "2" }, { "input": "8\n1 -1 1 -1 -1 1 1 1", "output": "1" }, { "input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1", "output": "8" }, { "input": "7\n-1 -1 1 1 -1 -1 1", "output": "2" }, { "input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5", "output": "5" }, { "input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -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": "13" }, { "input": "3\n-1 5 4", "output": "1" }, { "input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1", "output": "5" }, { "input": "1\n2", "output": "0" }, { "input": "1\n-1", "output": "1" }, { "input": "2\n1 -1", "output": "0" }, { "input": "2\n-1 1", "output": "1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n-1 -1", "output": "2" }, { "input": "4\n10 -1 -1 -1", "output": "0" }, { "input": "1\n1", "output": "0" }, { "input": "4\n-1 -1 1 1", "output": "2" } ]

1,686,852,132

2,147,483,647

PyPy 3-64

RUNTIME_ERROR

TESTS

0

77

2,867,200

n = input() a = list(map(int, input().split(' '))) men = 0 crimes = 0 unscrimes = 0 for i in range(n): if a[i] > 0: men += a[i] elif a[i] < 0: if men > 0 : if crimes < men: men = men - crimes crimes = 0 elif crimes > men: men = 0 unscrimes = unscrimes + crimes - men elif men == 0: unscrimes += crimes print(unscrimes)

Title: Police Recruits Time Limit: None seconds Memory Limit: None megabytes Problem Description: The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups. Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime. If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated. Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. Input Specification: The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers. If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. Output Specification: Print a single integer, the number of crimes which will go untreated. Demo Input: ['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n'] Demo Output: ['2\n', '1\n', '8\n'] Note: Lets consider the second example: 1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired. The answer is one, as one crime (on step 5) will go untreated.

```python n = input() a = list(map(int, input().split(' '))) men = 0 crimes = 0 unscrimes = 0 for i in range(n): if a[i] > 0: men += a[i] elif a[i] < 0: if men > 0 : if crimes < men: men = men - crimes crimes = 0 elif crimes > men: men = 0 unscrimes = unscrimes + crimes - men elif men == 0: unscrimes += crimes print(unscrimes) ```

-1

876

A

Trip For Meal

PROGRAMMING

900

[ "math" ]

null

Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters. For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store). Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.

First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits. Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses. Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses. Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.

Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.

[ "3\n2\n3\n1\n", "1\n2\n3\n5\n" ]

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

In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3. In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.

500

[ { "input": "3\n2\n3\n1", "output": "3" }, { "input": "1\n2\n3\n5", "output": "0" }, { "input": "10\n1\n8\n3", "output": "9" }, { "input": "7\n10\n5\n6", "output": "30" }, { "input": "9\n9\n7\n5", "output": "42" }, { "input": "9\n37\n85\n76", "output": "296" }, { "input": "76\n46\n77\n11", "output": "860" }, { "input": "80\n42\n1\n37", "output": "79" }, { "input": "8\n80\n55\n1", "output": "61" }, { "input": "10\n13\n72\n17", "output": "117" }, { "input": "9\n24\n1\n63", "output": "8" }, { "input": "65\n5\n8\n7", "output": "320" }, { "input": "56\n8\n9\n3", "output": "170" }, { "input": "59\n8\n1\n2", "output": "58" }, { "input": "75\n50\n50\n5", "output": "415" }, { "input": "75\n54\n76\n66", "output": "3996" }, { "input": "73\n71\n69\n66", "output": "4755" }, { "input": "83\n58\n88\n16", "output": "1354" }, { "input": "74\n31\n11\n79", "output": "803" }, { "input": "62\n27\n16\n72", "output": "976" }, { "input": "72\n95\n27\n9", "output": "657" }, { "input": "1\n2\n2\n1", "output": "0" }, { "input": "1\n1\n1\n1", "output": "0" }, { "input": "1\n1\n1\n99", "output": "0" }, { "input": "100\n100\n100\n100", "output": "9900" }, { "input": "2\n1\n1\n3", "output": "1" }, { "input": "1\n3\n2\n1", "output": "0" }, { "input": "1\n5\n6\n1", "output": "0" }, { "input": "1\n2\n6\n1", "output": "0" }, { "input": "1\n30\n20\n1", "output": "0" }, { "input": "1\n15\n13\n11", "output": "0" }, { "input": "1\n100\n50\n1", "output": "0" }, { "input": "1\n2\n3\n1", "output": "0" }, { "input": "5\n5\n5\n5", "output": "20" }, { "input": "1\n100\n50\n5", "output": "0" }, { "input": "1\n5\n7\n3", "output": "0" }, { "input": "1\n3\n4\n2", "output": "0" }, { "input": "3\n3\n3\n3", "output": "6" }, { "input": "1\n99\n98\n1", "output": "0" }, { "input": "5\n3\n3\n1", "output": "6" }, { "input": "2\n1\n1\n1", "output": "1" }, { "input": "2\n1\n7\n8", "output": "1" }, { "input": "2\n2\n2\n1", "output": "2" }, { "input": "2\n2\n3\n1", "output": "2" }, { "input": "1\n5\n6\n2", "output": "0" }, { "input": "100\n1\n1\n100", "output": "99" }, { "input": "2\n3\n2\n1", "output": "2" }, { "input": "1\n99\n98\n97", "output": "0" } ]

1,613,472,678

2,147,483,647

PyPy 3

OK

TESTS

48

108

0

n=int(input()) a=int(input()) b=int(input()) c=int(input()) d=min(a,b) if n==1 or min(d,c)==a or min(d,c)==b: print(d*(n-1)) else: print(d+(c*(n-2)))

Title: Trip For Meal Time Limit: None seconds Memory Limit: None megabytes Problem Description: Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters. For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store). Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance. Input Specification: First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits. Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses. Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses. Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses. Output Specification: Output one number — minimum distance in meters Winnie must go through to have a meal *n* times. Demo Input: ['3\n2\n3\n1\n', '1\n2\n3\n5\n'] Demo Output: ['3\n', '0\n'] Note: In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3. In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.

```python n=int(input()) a=int(input()) b=int(input()) c=int(input()) d=min(a,b) if n==1 or min(d,c)==a or min(d,c)==b: print(d*(n-1)) else: print(d+(c*(n-2))) ```

3

315

B

Sereja and Array

PROGRAMMING

1,200

[ "implementation" ]

null

Sereja has got an array, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n*. Sereja is an active boy, so he is now going to complete *m* operations. Each operation will have one of the three forms: 1. Make *v**i*-th array element equal to *x**i*. In other words, perform the assignment *a**v**i*<==<=*x**i*. 1. Increase each array element by *y**i*. In other words, perform *n* assignments *a**i*<==<=*a**i*<=+<=*y**i* (1<=≤<=*i*<=≤<=*n*). 1. Take a piece of paper and write out the *q**i*-th array element. That is, the element *a**q**i*. Help Sereja, complete all his operations.

The first line contains integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the original array. Next *m* lines describe operations, the *i*-th line describes the *i*-th operation. The first number in the *i*-th line is integer *t**i* (1<=≤<=*t**i*<=≤<=3) that represents the operation type. If *t**i*<==<=1, then it is followed by two integers *v**i* and *x**i*, (1<=≤<=*v**i*<=≤<=*n*,<=1<=≤<=*x**i*<=≤<=109). If *t**i*<==<=2, then it is followed by integer *y**i* (1<=≤<=*y**i*<=≤<=104). And if *t**i*<==<=3, then it is followed by integer *q**i* (1<=≤<=*q**i*<=≤<=*n*).

For each third type operation print value *a**q**i*. Print the values in the order, in which the corresponding queries follow in the input.

[ "10 11\n1 2 3 4 5 6 7 8 9 10\n3 2\n3 9\n2 10\n3 1\n3 10\n1 1 10\n2 10\n2 10\n3 1\n3 10\n3 9\n" ]

[ "2\n9\n11\n20\n30\n40\n39\n" ]

none

1,000

[ { "input": "10 11\n1 2 3 4 5 6 7 8 9 10\n3 2\n3 9\n2 10\n3 1\n3 10\n1 1 10\n2 10\n2 10\n3 1\n3 10\n3 9", "output": "2\n9\n11\n20\n30\n40\n39" }, { "input": "1 3\n1\n1 1 2\n2 1\n3 1", "output": "3" }, { "input": "1 1\n1\n3 1", "output": "1" }, { "input": "6 6\n202714501 613423725 367325359 862808465 182588146 745008828\n2 492\n2 1943\n1 3 173984392\n1 3 9356383\n3 4\n1 5 472683539", "output": "862810900" }, { "input": "6 5\n545129895 918519812 2334334 565540665 484519712 904102869\n3 3\n1 2 641196860\n2 779\n2 3036\n3 1", "output": "2334334\n545133710" }, { "input": "5 5\n286163265 724763447 761775891 215352371 491215007\n1 4 780579549\n1 2 638546757\n3 4\n1 3 38072627\n2 6119", "output": "780579549" }, { "input": "4 4\n529316834 995684640 949078705 317773978\n3 3\n3 1\n2 6271\n1 1 856879574", "output": "949078705\n529316834" }, { "input": "5 4\n293170637 715384768 756975575 129004053 875201151\n1 3 695181967\n3 3\n2 2288\n1 5 332270946", "output": "695181967" } ]

1,588,713,135

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

10

1,000

15,052,800

n, m = map(int, input().split()) a = [int(i) for i in input().split()] results = [] for i in range(m): op = [int(j) for j in input().split()] if (op[0] == 1): a[op[1] - 1] = op[2] elif (op[0] == 2): a = [i + op[1] for i in a] else: results.append(a[op[1] - 1]) for result in results: print (result)

Title: Sereja and Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja has got an array, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n*. Sereja is an active boy, so he is now going to complete *m* operations. Each operation will have one of the three forms: 1. Make *v**i*-th array element equal to *x**i*. In other words, perform the assignment *a**v**i*<==<=*x**i*. 1. Increase each array element by *y**i*. In other words, perform *n* assignments *a**i*<==<=*a**i*<=+<=*y**i* (1<=≤<=*i*<=≤<=*n*). 1. Take a piece of paper and write out the *q**i*-th array element. That is, the element *a**q**i*. Help Sereja, complete all his operations. Input Specification: The first line contains integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the original array. Next *m* lines describe operations, the *i*-th line describes the *i*-th operation. The first number in the *i*-th line is integer *t**i* (1<=≤<=*t**i*<=≤<=3) that represents the operation type. If *t**i*<==<=1, then it is followed by two integers *v**i* and *x**i*, (1<=≤<=*v**i*<=≤<=*n*,<=1<=≤<=*x**i*<=≤<=109). If *t**i*<==<=2, then it is followed by integer *y**i* (1<=≤<=*y**i*<=≤<=104). And if *t**i*<==<=3, then it is followed by integer *q**i* (1<=≤<=*q**i*<=≤<=*n*). Output Specification: For each third type operation print value *a**q**i*. Print the values in the order, in which the corresponding queries follow in the input. Demo Input: ['10 11\n1 2 3 4 5 6 7 8 9 10\n3 2\n3 9\n2 10\n3 1\n3 10\n1 1 10\n2 10\n2 10\n3 1\n3 10\n3 9\n'] Demo Output: ['2\n9\n11\n20\n30\n40\n39\n'] Note: none

```python n, m = map(int, input().split()) a = [int(i) for i in input().split()] results = [] for i in range(m): op = [int(j) for j in input().split()] if (op[0] == 1): a[op[1] - 1] = op[2] elif (op[0] == 2): a = [i + op[1] for i in a] else: results.append(a[op[1] - 1]) for result in results: print (result) ```

0

218

B

Airport

PROGRAMMING

1,100

[ "implementation" ]

null

Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows: - it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency). The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer? The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets. The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.

Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.

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

[ "5 5\n", "7 6\n" ]

In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum. In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane.

500

[ { "input": "4 3\n2 1 1", "output": "5 5" }, { "input": "4 3\n2 2 2", "output": "7 6" }, { "input": "10 5\n10 3 3 1 2", "output": "58 26" }, { "input": "10 1\n10", "output": "55 55" }, { "input": "10 1\n100", "output": "955 955" }, { "input": "10 2\n4 7", "output": "37 37" }, { "input": "40 10\n1 2 3 4 5 6 7 10 10 10", "output": "223 158" }, { "input": "1 1\n6", "output": "6 6" }, { "input": "1 2\n10 9", "output": "10 9" }, { "input": "2 1\n7", "output": "13 13" }, { "input": "2 2\n7 2", "output": "13 3" }, { "input": "3 2\n4 7", "output": "18 9" }, { "input": "3 3\n2 1 1", "output": "4 4" }, { "input": "3 3\n2 1 1", "output": "4 4" }, { "input": "10 10\n3 1 2 2 1 1 2 1 2 3", "output": "20 13" }, { "input": "10 2\n7 3", "output": "34 34" }, { "input": "10 1\n19", "output": "145 145" }, { "input": "100 3\n29 36 35", "output": "1731 1731" }, { "input": "100 5\n3 38 36 35 2", "output": "2019 1941" }, { "input": "510 132\n50 76 77 69 94 30 47 65 14 62 18 121 26 35 49 17 105 93 47 16 78 3 7 74 7 37 30 36 30 83 71 113 7 58 86 10 65 57 34 102 55 44 43 47 106 44 115 75 109 70 47 45 16 57 62 55 20 88 74 40 45 84 41 1 9 53 65 25 67 31 115 2 63 51 123 70 65 65 18 14 75 14 103 26 117 105 36 104 81 37 35 61 44 90 71 70 88 89 26 21 64 77 89 16 87 99 13 79 27 3 46 120 116 11 14 17 32 70 113 94 108 57 29 100 53 48 44 29 70 30 32 62", "output": "50279 5479" }, { "input": "510 123\n5 2 3 2 5 7 2 3 1 3 6 6 3 1 5 3 5 6 2 2 1 5 5 5 2 2 3 1 6 3 5 8 4 6 1 5 4 5 1 6 5 5 3 6 4 1 6 1 3 5 2 7 5 2 4 4 5 6 5 5 4 3 4 6 5 4 4 3 5 8 5 5 6 3 1 7 4 4 3 3 5 3 6 3 3 6 2 5 3 2 4 5 4 5 2 2 4 4 4 7 3 4 6 5 3 6 4 7 1 6 5 7 6 5 7 3 7 4 4 1 6 6 4", "output": "1501 1501" }, { "input": "610 33\n15 44 8 8 17 11 39 39 38 25 17 36 17 25 21 37 10 11 34 30 29 50 29 50 4 20 32 13 41 14 2 11 2", "output": "12204 8871" } ]

1,429,337,317

2,147,483,647

Python 3

OK

TESTS

33

218

0

n, m = [int(i) for i in input().split()] a = [int(i) for i in input().split()] max_ans = 0 min_ans = 0 maxli = list(a) minli = list(a) while n > 0: n -= 1 max_index = max(maxli) min_index = min(minli) max_ans += max_index min_ans += min_index max_index = maxli.index(max_index) min_index = minli.index(min_index) if maxli[max_index] == 1: del(maxli[max_index]) else: maxli[max_index] -= 1 if minli[min_index] == 1: del(minli[min_index]) else: minli[min_index] -= 1 print(max_ans, min_ans)

Title: Airport Time Limit: None seconds Memory Limit: None megabytes Problem Description: Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows: - it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency). The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer? The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets. The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total. Output Specification: Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly. Demo Input: ['4 3\n2 1 1\n', '4 3\n2 2 2\n'] Demo Output: ['5 5\n', '7 6\n'] Note: In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum. In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane.

```python n, m = [int(i) for i in input().split()] a = [int(i) for i in input().split()] max_ans = 0 min_ans = 0 maxli = list(a) minli = list(a) while n > 0: n -= 1 max_index = max(maxli) min_index = min(minli) max_ans += max_index min_ans += min_index max_index = maxli.index(max_index) min_index = minli.index(min_index) if maxli[max_index] == 1: del(maxli[max_index]) else: maxli[max_index] -= 1 if minli[min_index] == 1: del(minli[min_index]) else: minli[min_index] -= 1 print(max_ans, min_ans) ```

3

13

A

Numbers

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Numbers

1

64

Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18. Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1. Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.

Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).

Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.

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

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

In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.

0

[ { "input": "5", "output": "7/3" }, { "input": "3", "output": "2/1" }, { "input": "1000", "output": "90132/499" }, { "input": "927", "output": "155449/925" }, { "input": "260", "output": "6265/129" }, { "input": "131", "output": "3370/129" }, { "input": "386", "output": "857/12" }, { "input": "277", "output": "2864/55" }, { "input": "766", "output": "53217/382" }, { "input": "28", "output": "85/13" }, { "input": "406", "output": "7560/101" }, { "input": "757", "output": "103847/755" }, { "input": "6", "output": "9/4" }, { "input": "239", "output": "10885/237" }, { "input": "322", "output": "2399/40" }, { "input": "98", "output": "317/16" }, { "input": "208", "output": "4063/103" }, { "input": "786", "output": "55777/392" }, { "input": "879", "output": "140290/877" }, { "input": "702", "output": "89217/700" }, { "input": "948", "output": "7369/43" }, { "input": "537", "output": "52753/535" }, { "input": "984", "output": "174589/982" }, { "input": "934", "output": "157951/932" }, { "input": "726", "output": "95491/724" }, { "input": "127", "output": "3154/125" }, { "input": "504", "output": "23086/251" }, { "input": "125", "output": "3080/123" }, { "input": "604", "output": "33178/301" }, { "input": "115", "output": "2600/113" }, { "input": "27", "output": "167/25" }, { "input": "687", "output": "85854/685" }, { "input": "880", "output": "69915/439" }, { "input": "173", "output": "640/19" }, { "input": "264", "output": "6438/131" }, { "input": "785", "output": "111560/783" }, { "input": "399", "output": "29399/397" }, { "input": "514", "output": "6031/64" }, { "input": "381", "output": "26717/379" }, { "input": "592", "output": "63769/590" }, { "input": "417", "output": "32002/415" }, { "input": "588", "output": "62723/586" }, { "input": "852", "output": "131069/850" }, { "input": "959", "output": "5059/29" }, { "input": "841", "output": "127737/839" }, { "input": "733", "output": "97598/731" }, { "input": "692", "output": "87017/690" }, { "input": "69", "output": "983/67" }, { "input": "223", "output": "556/13" }, { "input": "93", "output": "246/13" }, { "input": "643", "output": "75503/641" }, { "input": "119", "output": "2833/117" }, { "input": "498", "output": "1459/16" }, { "input": "155", "output": "4637/153" }, { "input": "305", "output": "17350/303" }, { "input": "454", "output": "37893/452" }, { "input": "88", "output": "1529/86" }, { "input": "850", "output": "32645/212" }, { "input": "474", "output": "20581/236" }, { "input": "309", "output": "17731/307" }, { "input": "762", "output": "105083/760" }, { "input": "591", "output": "63761/589" }, { "input": "457", "output": "38317/455" }, { "input": "141", "output": "3832/139" }, { "input": "385", "output": "27232/383" }, { "input": "387", "output": "27628/385" }, { "input": "469", "output": "40306/467" }, { "input": "624", "output": "35285/311" }, { "input": "330", "output": "487/8" }, { "input": "31", "output": "222/29" }, { "input": "975", "output": "171679/973" }, { "input": "584", "output": "62183/582" }, { "input": "668", "output": "81127/666" }, { "input": "331", "output": "20297/329" }, { "input": "189", "output": "6789/187" }, { "input": "251", "output": "11939/249" }, { "input": "876", "output": "69196/437" }, { "input": "615", "output": "68987/613" }, { "input": "451", "output": "37258/449" }, { "input": "499", "output": "45727/497" }, { "input": "699", "output": "89117/697" }, { "input": "619", "output": "70019/617" }, { "input": "413", "output": "10515/137" }, { "input": "197", "output": "7399/195" }, { "input": "794", "output": "14281/99" }, { "input": "659", "output": "79403/657" }, { "input": "653", "output": "77695/651" }, { "input": "23", "output": "45/7" }, { "input": "430", "output": "16985/214" }, { "input": "249", "output": "11659/247" }, { "input": "837", "output": "126869/835" }, { "input": "258", "output": "12373/256" }, { "input": "995", "output": "59665/331" }, { "input": "102", "output": "504/25" }, { "input": "989", "output": "177124/987" }, { "input": "376", "output": "13008/187" }, { "input": "657", "output": "15715/131" }, { "input": "746", "output": "50509/372" }, { "input": "602", "output": "13177/120" } ]

1,591,517,097

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

278

1,228,800

a = int(input()) g = 0 for k in range(2,a): x=a l=x%k while x>1: x=x//k g+=x%k g+=l print(str(g)+'/'+str(a-2))

Title: Numbers Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18. Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1. Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10. Input Specification: Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000). Output Specification: Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator. Demo Input: ['5\n', '3\n'] Demo Output: ['7/3\n', '2/1\n'] Note: In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.

```python a = int(input()) g = 0 for k in range(2,a): x=a l=x%k while x>1: x=x//k g+=x%k g+=l print(str(g)+'/'+str(a-2)) ```

0

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

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

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,669,891,966

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

31

0

a = input() ls = 2 required = ['h','e','l','l','o'] got = [] gat = '' for i in a: got.append(i) for i in got: if i == 'l' and ls != 0: gat += i ls -= 1 elif i not in gat and i != 'l': gat += i if 'hello' in gat: print("YES") else: print("NO")

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python a = input() ls = 2 required = ['h','e','l','l','o'] got = [] gat = '' for i in a: got.append(i) for i in got: if i == 'l' and ls != 0: gat += i ls -= 1 elif i not in gat and i != 'l': gat += i if 'hello' in gat: print("YES") else: print("NO") ```

0

259

B

Little Elephant and Magic Square

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes.

The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.

Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions.

[ "0 1 1\n1 0 1\n1 1 0\n", "0 3 6\n5 0 5\n4 7 0\n" ]

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

none

1,000

[ { "input": "0 1 1\n1 0 1\n1 1 0", "output": "1 1 1\n1 1 1\n1 1 1" }, { "input": "0 3 6\n5 0 5\n4 7 0", "output": "6 3 6\n5 5 5\n4 7 4" }, { "input": "0 4 4\n4 0 4\n4 4 0", "output": "4 4 4\n4 4 4\n4 4 4" }, { "input": "0 54 48\n36 0 78\n66 60 0", "output": "69 54 48\n36 57 78\n66 60 45" }, { "input": "0 17 14\n15 0 15\n16 13 0", "output": "14 17 14\n15 15 15\n16 13 16" }, { "input": "0 97 56\n69 0 71\n84 43 0", "output": "57 97 56\n69 70 71\n84 43 83" }, { "input": "0 1099 1002\n1027 0 1049\n1074 977 0", "output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063" }, { "input": "0 98721 99776\n99575 0 99123\n98922 99977 0", "output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148" }, { "input": "0 6361 2304\n1433 0 8103\n7232 3175 0", "output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897" }, { "input": "0 99626 99582\n99766 0 99258\n99442 99398 0", "output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696" }, { "input": "0 99978 99920\n99950 0 99918\n99948 99890 0", "output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964" }, { "input": "0 840 666\n612 0 948\n894 720 0", "output": "834 840 666\n612 780 948\n894 720 726" }, { "input": "0 28 10\n12 0 24\n26 8 0", "output": "16 28 10\n12 18 24\n26 8 20" }, { "input": "0 120 83\n98 0 90\n105 68 0", "output": "79 120 83\n98 94 90\n105 68 109" }, { "input": "0 86900 85807\n85836 0 86842\n86871 85778 0", "output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368" }, { "input": "0 74 78\n78 0 74\n74 78 0", "output": "76 74 78\n78 76 74\n74 78 76" }, { "input": "0 505 681\n605 0 657\n581 757 0", "output": "707 505 681\n605 631 657\n581 757 555" }, { "input": "0 662 918\n822 0 854\n758 1014 0", "output": "934 662 918\n822 838 854\n758 1014 742" }, { "input": "0 93 95\n93 0 97\n95 97 0", "output": "97 93 95\n93 95 97\n95 97 93" }, { "input": "0 709 712\n719 0 695\n702 705 0", "output": "700 709 712\n719 707 695\n702 705 714" }, { "input": "0 7 6\n9 0 1\n4 3 0", "output": "2 7 6\n9 5 1\n4 3 8" }, { "input": "0 9 2\n3 0 7\n8 1 0", "output": "4 9 2\n3 5 7\n8 1 6" }, { "input": "0 1 43\n13 0 61\n31 73 0", "output": "67 1 43\n13 37 61\n31 73 7" }, { "input": "0 100000 100000\n100000 0 100000\n100000 100000 0", "output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000" } ]

1,630,434,281

2,147,483,647

Python 3

OK

TESTS

24

124

6,963,200

g = [] o_s = 0 f_s = 0 p_s = 0 t_s = 0 for i in range(3): l = list(map(int, input().split())) s = sum(l) if i == 0: p_s = s f_s = s else: t_s += o_s + p_s - s o_s += p_s - s p_s = s g.append(l) c_s = f_s + (f_s - t_s) // 2 for l_i in g: s = sum(l_i) a = list() for n in l_i: if n == 0: a.append(c_s - s) else: a.append(n) print(" ".join(str(x) for x in a))

Title: Little Elephant and Magic Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes. Input Specification: The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. Output Specification: Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions. Demo Input: ['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n'] Demo Output: ['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n'] Note: none

```python g = [] o_s = 0 f_s = 0 p_s = 0 t_s = 0 for i in range(3): l = list(map(int, input().split())) s = sum(l) if i == 0: p_s = s f_s = s else: t_s += o_s + p_s - s o_s += p_s - s p_s = s g.append(l) c_s = f_s + (f_s - t_s) // 2 for l_i in g: s = sum(l_i) a = list() for n in l_i: if n == 0: a.append(c_s - s) else: a.append(n) print(" ".join(str(x) for x in a)) ```

3

588

A

Duff and Meat

PROGRAMMING

900

[ "greedy" ]

null

Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat. There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future. Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.

The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days. In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.

Print the minimum money needed to keep Duff happy for *n* days, in one line.

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

[ "10\n", "8\n" ]

In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day. In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.

750

[ { "input": "3\n1 3\n2 2\n3 1", "output": "10" }, { "input": "3\n1 3\n2 1\n3 2", "output": "8" }, { "input": "1\n39 52", "output": "2028" }, { "input": "2\n25 56\n94 17", "output": "2998" }, { "input": "5\n39 21\n95 89\n73 90\n9 55\n85 32", "output": "6321" }, { "input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88", "output": "6742" }, { "input": "2\n100 68\n38 80", "output": "9384" } ]

1,618,250,706

2,147,483,647

Python 3

OK

TESTS

30

358

4,710,400

result = 0 n = int(input()) d = n amount_and_prices = [] for i in range(n): a, p = map(int, input().split()) amount_and_prices.append((a, p)) result += amount_and_prices[0][0] * amount_and_prices[0][1] d -= 1 i = 1 mini = 0 while d > 0: if(amount_and_prices[mini][1] < amount_and_prices[i][1]): result += amount_and_prices[i][0] * amount_and_prices[mini][1] else: result += amount_and_prices[i][0] * amount_and_prices[i][1] mini = i i += 1 d -= 1 print(result)

Title: Duff and Meat Time Limit: None seconds Memory Limit: None megabytes Problem Description: Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat. There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future. Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. Input Specification: The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days. In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. Output Specification: Print the minimum money needed to keep Duff happy for *n* days, in one line. Demo Input: ['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n'] Demo Output: ['10\n', '8\n'] Note: In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day. In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.

```python result = 0 n = int(input()) d = n amount_and_prices = [] for i in range(n): a, p = map(int, input().split()) amount_and_prices.append((a, p)) result += amount_and_prices[0][0] * amount_and_prices[0][1] d -= 1 i = 1 mini = 0 while d > 0: if(amount_and_prices[mini][1] < amount_and_prices[i][1]): result += amount_and_prices[i][0] * amount_and_prices[mini][1] else: result += amount_and_prices[i][0] * amount_and_prices[i][1] mini = i i += 1 d -= 1 print(result) ```

3

194

A

Exams

PROGRAMMING

900

[ "implementation", "math" ]

null

One day the Codeforces round author sat exams. He had *n* exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2. The author would need to spend too much time and effort to make the sum of his marks strictly more than *k*. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than *k*, the author's mum won't be pleased at all. The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams. Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all *n* exams equal exactly *k*.

The single input line contains space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=250) — the number of exams and the required sum of marks. It is guaranteed that there exists a way to pass *n* exams in the way that makes the sum of marks equal exactly *k*.

Print the single number — the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal *k*.

[ "4 8\n", "4 10\n", "1 3\n" ]

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

In the first sample the author has to get a 2 for all his exams. In the second sample he should get a 3 for two exams and a 2 for two more. In the third sample he should get a 3 for one exam.

500

[ { "input": "4 8", "output": "4" }, { "input": "4 10", "output": "2" }, { "input": "1 3", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "4 9", "output": "3" }, { "input": "50 234", "output": "0" }, { "input": "50 100", "output": "50" }, { "input": "50 250", "output": "0" }, { "input": "29 116", "output": "0" }, { "input": "20 69", "output": "0" }, { "input": "46 127", "output": "11" }, { "input": "3 7", "output": "2" }, { "input": "36 99", "output": "9" }, { "input": "45 104", "output": "31" }, { "input": "13 57", "output": "0" }, { "input": "25 106", "output": "0" }, { "input": "8 19", "output": "5" }, { "input": "20 69", "output": "0" }, { "input": "13 32", "output": "7" }, { "input": "47 128", "output": "13" }, { "input": "17 73", "output": "0" }, { "input": "3 7", "output": "2" }, { "input": "16 70", "output": "0" }, { "input": "1 5", "output": "0" }, { "input": "38 137", "output": "0" }, { "input": "7 20", "output": "1" }, { "input": "1 5", "output": "0" }, { "input": "36 155", "output": "0" }, { "input": "5 15", "output": "0" }, { "input": "27 75", "output": "6" }, { "input": "21 73", "output": "0" }, { "input": "2 5", "output": "1" }, { "input": "49 177", "output": "0" }, { "input": "7 20", "output": "1" }, { "input": "44 173", "output": "0" }, { "input": "49 219", "output": "0" }, { "input": "16 70", "output": "0" }, { "input": "10 28", "output": "2" } ]

1,659,064,419

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

62

0

#### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### ####################################### ####################################### ####################///# #///# #///# #///# ####################///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# ####################///# #///# #///# #///# ####################///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# ####################################################################### #///# ####################################################################### #///# ##################################### ##################################### ##################################### ##################################### a,b = list(map(int,input().split())) count = 0 if a*2 <= b: print("0") elif ((b//a)) <= a: print(a) else: print(a-((((b//a)-a)*2))) ##################################### ##################################### ##################################### ##################################### #####################################

Title: Exams Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day the Codeforces round author sat exams. He had *n* exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2. The author would need to spend too much time and effort to make the sum of his marks strictly more than *k*. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than *k*, the author's mum won't be pleased at all. The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams. Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all *n* exams equal exactly *k*. Input Specification: The single input line contains space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=250) — the number of exams and the required sum of marks. It is guaranteed that there exists a way to pass *n* exams in the way that makes the sum of marks equal exactly *k*. Output Specification: Print the single number — the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal *k*. Demo Input: ['4 8\n', '4 10\n', '1 3\n'] Demo Output: ['4\n', '2\n', '0\n'] Note: In the first sample the author has to get a 2 for all his exams. In the second sample he should get a 3 for two exams and a 2 for two more. In the third sample he should get a 3 for one exam.

```python #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### #### ####################################### ####################################### ####################///# #///# #///# #///# ####################///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# #### #///# #///# #///# #///# ####################///# #///# #///# #///# ####################///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# #///# ####################################################################### #///# ####################################################################### #///# ##################################### ##################################### ##################################### ##################################### a,b = list(map(int,input().split())) count = 0 if a*2 <= b: print("0") elif ((b//a)) <= a: print(a) else: print(a-((((b//a)-a)*2))) ##################################### ##################################### ##################################### ##################################### ##################################### ```

0

606

A

Magic Spheres

PROGRAMMING

1,200

[ "implementation" ]

null

Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)?

The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal. The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get.

If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".

[ "4 4 0\n2 1 2\n", "5 6 1\n2 7 2\n", "3 3 3\n2 2 2\n" ]

[ "Yes\n", "No\n", "Yes\n" ]

In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.

500

[ { "input": "4 4 0\n2 1 2", "output": "Yes" }, { "input": "5 6 1\n2 7 2", "output": "No" }, { "input": "3 3 3\n2 2 2", "output": "Yes" }, { "input": "0 0 0\n0 0 0", "output": "Yes" }, { "input": "0 0 0\n0 0 1", "output": "No" }, { "input": "0 1 0\n0 0 0", "output": "Yes" }, { "input": "1 0 0\n1 0 0", "output": "Yes" }, { "input": "2 2 1\n1 1 2", "output": "No" }, { "input": "1 3 1\n2 1 1", "output": "Yes" }, { "input": "1000000 1000000 1000000\n1000000 1000000 1000000", "output": "Yes" }, { "input": "1000000 500000 500000\n0 750000 750000", "output": "Yes" }, { "input": "500000 1000000 500000\n750001 0 750000", "output": "No" }, { "input": "499999 500000 1000000\n750000 750000 0", "output": "No" }, { "input": "500000 500000 0\n0 0 500000", "output": "Yes" }, { "input": "0 500001 499999\n500000 0 0", "output": "No" }, { "input": "1000000 500000 1000000\n500000 1000000 500000", "output": "Yes" }, { "input": "1000000 1000000 499999\n500000 500000 1000000", "output": "No" }, { "input": "500000 1000000 1000000\n1000000 500001 500000", "output": "No" }, { "input": "1000000 500000 500000\n0 1000000 500000", "output": "Yes" }, { "input": "500000 500000 1000000\n500001 1000000 0", "output": "No" }, { "input": "500000 999999 500000\n1000000 0 500000", "output": "No" }, { "input": "4 0 3\n2 2 1", "output": "Yes" }, { "input": "0 2 4\n2 0 2", "output": "Yes" }, { "input": "3 1 0\n1 1 1", "output": "Yes" }, { "input": "4 4 1\n1 3 2", "output": "Yes" }, { "input": "1 2 4\n2 1 3", "output": "No" }, { "input": "1 1 0\n0 0 1", "output": "No" }, { "input": "4 0 0\n0 1 1", "output": "Yes" }, { "input": "0 3 0\n1 0 1", "output": "No" }, { "input": "0 0 3\n1 0 1", "output": "Yes" }, { "input": "1 12 1\n4 0 4", "output": "Yes" }, { "input": "4 0 4\n1 2 1", "output": "Yes" }, { "input": "4 4 0\n1 1 3", "output": "No" }, { "input": "0 9 0\n2 2 2", "output": "No" }, { "input": "0 10 0\n2 2 2", "output": "Yes" }, { "input": "9 0 9\n0 8 0", "output": "Yes" }, { "input": "0 9 9\n9 0 0", "output": "No" }, { "input": "9 10 0\n0 0 9", "output": "Yes" }, { "input": "10 0 9\n0 10 0", "output": "No" }, { "input": "0 10 10\n10 0 0", "output": "Yes" }, { "input": "10 10 0\n0 0 11", "output": "No" }, { "input": "307075 152060 414033\n381653 222949 123101", "output": "No" }, { "input": "569950 228830 153718\n162186 357079 229352", "output": "No" }, { "input": "149416 303568 749016\n238307 493997 190377", "output": "No" }, { "input": "438332 298094 225324\n194220 400244 245231", "output": "No" }, { "input": "293792 300060 511272\n400687 382150 133304", "output": "No" }, { "input": "295449 518151 368838\n382897 137148 471892", "output": "No" }, { "input": "191789 291147 691092\n324321 416045 176232", "output": "Yes" }, { "input": "286845 704749 266526\n392296 104421 461239", "output": "Yes" }, { "input": "135522 188282 377041\n245719 212473 108265", "output": "Yes" }, { "input": "404239 359124 133292\n180069 184791 332544", "output": "No" }, { "input": "191906 624432 244408\n340002 367217 205432", "output": "No" }, { "input": "275980 429361 101824\n274288 302579 166062", "output": "No" }, { "input": "136092 364927 395302\n149173 343146 390922", "output": "No" }, { "input": "613852 334661 146012\n363786 326286 275233", "output": "No" }, { "input": "348369 104625 525203\n285621 215396 366411", "output": "No" }, { "input": "225307 153572 114545\n154753 153282 149967", "output": "Yes" }, { "input": "438576 124465 629784\n375118 276028 390116", "output": "Yes" }, { "input": "447521 327510 158732\n395759 178458 259139", "output": "Yes" }, { "input": "8 5 5\n5 5 5", "output": "Yes" }, { "input": "100 100 100\n1 1 1", "output": "Yes" }, { "input": "100 100 100\n0 0 0", "output": "Yes" }, { "input": "3 2 3\n2 3 2", "output": "No" }, { "input": "5 4 3\n2 2 2", "output": "Yes" }, { "input": "14 9 8\n12 5 10", "output": "Yes" }, { "input": "10 10 10\n1 1 1", "output": "Yes" }, { "input": "6 3 3\n3 3 3", "output": "Yes" }, { "input": "10 0 4\n2 4 2", "output": "Yes" }, { "input": "100 100 100\n2 2 2", "output": "Yes" }, { "input": "4 6 0\n2 1 2", "output": "Yes" }, { "input": "4 6 3\n4 2 3", "output": "Yes" }, { "input": "5 5 5\n1 1 1", "output": "Yes" }, { "input": "41 17 34\n0 19 24", "output": "Yes" }, { "input": "8 8 8\n3 3 3", "output": "Yes" }, { "input": "7 7 1\n1 1 2", "output": "Yes" }, { "input": "6 6 0\n2 2 2", "output": "Yes" }, { "input": "5 5 5\n2 2 2", "output": "Yes" }, { "input": "400 400 400\n1 1 1", "output": "Yes" }, { "input": "4 4 4\n2 2 2", "output": "Yes" } ]

1,449,697,441

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

46

0

a, b, c = [int(x) for x in input().split()] x, y, z = [int(x) for x in input().split()] a, b, c = list(sorted([a - x, b - y, c - z])) print(a,b,c) other = 0 if c < 0: print("No") exit(0) else: other += c//2 if b < 0: if b + other < 0: print("No") exit(0) else: other -= b else: other += b//2 if c < 0: if c + other < 0: print("No") exit(0) print("Yes")

Title: Magic Spheres Time Limit: None seconds Memory Limit: None megabytes Problem Description: Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)? Input Specification: The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal. The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get. Output Specification: If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No". Demo Input: ['4 4 0\n2 1 2\n', '5 6 1\n2 7 2\n', '3 3 3\n2 2 2\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n'] Note: In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.

```python a, b, c = [int(x) for x in input().split()] x, y, z = [int(x) for x in input().split()] a, b, c = list(sorted([a - x, b - y, c - z])) print(a,b,c) other = 0 if c < 0: print("No") exit(0) else: other += c//2 if b < 0: if b + other < 0: print("No") exit(0) else: other -= b else: other += b//2 if c < 0: if c + other < 0: print("No") exit(0) print("Yes") ```

0

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.

There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.

Write one line — the corresponding answer. Do not omit the leading 0s.

[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]

[ "1110001\n", "111\n", "0100\n", "00010\n" ]

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110", "output": "001100101000011111111101111011101010111001" }, { "input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001", "output": "0111010010100110110101100010000100010100000" }, { "input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100", "output": "11111110000000100101000100110111001100011001" }, { "input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011", "output": "101011011100100010100011011001101010100100010" }, { "input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001", "output": "1101001100111011010111110110101111001011110111" }, { "input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001", "output": "10010101000101000000011010011110011110011110001" }, { "input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100", "output": "011011011100000000010101110010000000101000111101" }, { "input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100", "output": "0101010111101001011011110110011101010101010100011" }, { "input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011", "output": "11001011010010111000010110011101100100001110111111" }, { "input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011", "output": "111011101010011100001111101001101011110010010110001" }, { "input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001", "output": "0100111110110011111110010010010000110111100101101101" }, { "input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100", "output": "01011001110111010111001100010011010100010000111011000" }, { "input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111", "output": "100011101001001000011011011001111000100000010100100100" }, { "input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110", "output": "1100110010000101101010111111101001001001110101110010110" }, { "input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110", "output": "01000111100111001011110010100011111111110010101100001101" }, { "input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010", "output": "110001010001000011000101110101000100001011111001011001001" }, { "input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111", "output": "1110100010111000101001001011101110011111100111000011011011" }, { "input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110", "output": "01110110101110100100110011010000001000101100101111000111011" }, { "input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011", "output": "111100101000000011101011011001110010101111000110010010000000" }, { "input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111", "output": "0100100010111110010011101010000011111110001110010110010111001" }, { "input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111", "output": "00110100000011001101101100100010110010001100000001100110011101" }, { "input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011", "output": "000000011000111011110011101000010000010100101000000011010110010" }, { "input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010", "output": "0010100110110100111100100100101101010100100111011010001001010101" }, { "input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111", "output": "11010110111100101111101001100001110100010110010110110111100110100" }, { "input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111", "output": "111111010011011100101110100110111111111001111110011010111111110000" }, { "input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110", "output": "1010101010100010001001001001100000111000010010010100010011000100000" }, { "input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000", "output": "00011111011111001000011100010011100011010100101011011000001001111110" }, { "input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111", "output": "001111000011001110100111010101111111011100110011001010010010000111011" }, { "input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101", "output": "0110001100110100010000110111000010011010011000011001010011010100010100" }, { "input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010", "output": "00010000000110110101000011001000000100100110111010011111101010001010000" }, { "input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001", "output": "000100100000000110011100100001010110101001100101110010010011111001110111" }, { "input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000", "output": "1000111100010011010110011101000000101010101100011111100001101111001010010" }, { "input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011", "output": "01000010011110111001011011110000001011000111101101101010010110001010100100" }, { "input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101", "output": "101101110110110010011100001011111100100001110000101100110000100011011100110" }, { "input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001", "output": "1110111111110010111000001100101010101011010100101010100101100011001001111111" }, { "input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111", "output": "10111011000111000101110100101000100111011011100011110110000101010001111010111" }, { "input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110", "output": "110100100110000101010010011010011001100110000111010000010100001011110111111101" }, { "input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111", "output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111" }, { "input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001", "output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001" }, { "input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110", "output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011" }, { "input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]

1,691,771,012

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

n1=input().strip() n2 =inp().strip() lst=[] for i ,x in enumerate(n1): if x==n2[i]: lst.append("0") else: lst.append("1") p ="".join(lst) print(p)

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python n1=input().strip() n2 =inp().strip() lst=[] for i ,x in enumerate(n1): if x==n2[i]: lst.append("0") else: lst.append("1") p ="".join(lst) print(p) ```

-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" } ]

1,601,708,102

2,147,483,647

Python 3

OK

TESTS

81

248

307,200

# -*- coding: utf-8 -*- """ Created on Sat Oct 3 14:33:37 2020 @author: 章斯岚 """ x=0 y=0 z=0 vector=[] n=int(input()) for i in range(n): force=input().split() vector.append(force) for i in range(n): x+=int(vector[i][0]) y+=int(vector[i][1]) z+=int(vector[i][2]) if x==0 and y==0 and 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 # -*- coding: utf-8 -*- """ Created on Sat Oct 3 14:33:37 2020 @author: 章斯岚 """ x=0 y=0 z=0 vector=[] n=int(input()) for i in range(n): force=input().split() vector.append(force) for i in range(n): x+=int(vector[i][0]) y+=int(vector[i][1]) z+=int(vector[i][2]) if x==0 and y==0 and z==0: print('YES') else: print('NO') ```

3.937428

43

B

Letter

PROGRAMMING

1,100

[ "implementation", "strings" ]

B. Letter

2

256

Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text.

The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading.

If Vasya can write the given anonymous letter, print YES, otherwise print NO

[ "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n", "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "abcdefg hijk\nk j i h g f e d c b a\n" ]

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

none

1,000

[ { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog", "output": "NO" }, { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "YES" }, { "input": "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "NO" }, { "input": "abcdefg hijk\nk j i h g f e d c b a", "output": "YES" }, { "input": "HpOKgo\neAtAVB", "output": "NO" }, { "input": "GRZGc\nLPzD", "output": "NO" }, { "input": "GtPXu\nd", "output": "NO" }, { "input": "FVF\nr ", "output": "NO" }, { "input": "HpOKgo\nogK", "output": "YES" }, { "input": "GRZGc\nZG", "output": "YES" }, { "input": "HpOKgoueAtAVBdGffvQheJDejNDHhhwyKJisugiRAH OseK yUwqPPNuThUxTfthqIUeb wS jChGOdFDarNrKRT MlwKecxWNoKEeD BbiHAruE XMlvKYVsJGPP\nAHN XvoaNwV AVBKwKjr u U K wKE D K Jy KiHsR h d W Js IHyMPK Br iSqe E fDA g H", "output": "YES" }, { "input": "GRZGcsLPzDrCSXhhNTaibJqVphhjbcPoZhCDUlzAbDnRWjHvxLKtpGiFWiGbfeDxBwCrdJmJGCGv GebAOinUsFrlqKTILOmxrFjSpEoVGoTdSSstJWVgMLKMPettxHASaQZNdOIObcTxtF qTHWBdNIKwj\nWqrxze Ji x q aT GllLrRV jMpGiMDTwwS JDsPGpAZKACmsFCOS CD Sj bCDgKF jJxa RddtLFAi VGLHH SecObzG q hPF ", "output": "YES" }, { "input": "GtPXuwdAxNhODQbjRslDDKciOALJrCifTjDQurQEBeFUUSZWwCZQPdYwZkYbrduMijFjgodAOrKIuUKwSXageZuOWMIhAMexyLRzFuzuXqBDTEaWMzVdbzhxDGSJC SsIYuYILwpiwwcObEHWpFvHeBkWYNitqYrxqgHReHcKnHbtjcWZuaxPBVPb\nTQIKyqFaewOkY lZUOOuxEw EwuKcArxRQGFYkvVWIAe SuanPeHuDjquurJu aSxwgOSw jYMwjxItNUUArQjO BIujAhSwttLWp", "output": "YES" }, { "input": "FVFSr unvtXbpKWF vPaAgNaoTqklzVqiGYcUcBIcattzBrRuNSnKUtmdGKbjcE\nUzrU K an GFGR Wc zt iBa P c T K v p V In b B c", "output": "YES" }, { "input": "lSwjnYLYtDNIZjxHiTawdh ntSzggZogcIZTuiTMWVgwyloMtEhqkrOxgIcFvwvsboXUPILPIymFAEXnhApewJXJNtFyZ\nAoxe jWZ u yImg o AZ FNI w lpj tNhT g y ZYcb rc J w Dlv", "output": "YES" }, { "input": "kvlekcdJqODUKdsJlXkRaileTmdGwUHWWgvgUokQxRzzbpFnswvNKiDnjfOFGvFcnaaiRnBGQmqoPxDHepgYasLhzjDgmvaFfVNEcSPVQCJKAbSyTGpXsAjIHr\nGjzUllNaGGKXUdYmDFpqFAKIwvTpjmqnyswWRTnxlBnavAGvavxJemrjvRJc", "output": "YES" }, { "input": "kWbvhgvvoYOhwXmgTwOSCDXrtFHhqwvMlCvsuuAUXMmWaYXiqHplFZZemhgkTuvsUtIaUxtyYauBIpjdbyYxjZ ZkaBPzwqPfqF kCqGRmXvWuabnQognnkvdNDtRUsSUvSzgBuxCMBWJifbxWegsknp\nBsH bWHJD n Ca T xq PRCv tatn Wjy sm I q s WCjFqdWe t W XUs Do eb Pfh ii hTbF O Fll", "output": "YES" }, { "input": "OTmLdkMhmDEOMQMiW ZpzEIjyElHFrNCfFQDp SZyoZaEIUIpyCHfwOUqiSkKtFHggrTBGkqfOxkChPztmPrsHoxVwAdrxbZLKxPXHlMnrkgMgiaHFopiFFiUEtKwCjpJtwdwkbJCgA bxeDIscFdmHQJLAMNhWlrZisQrHQpvbALWTwpf jnx\nDbZwrQbydCdkJMCrftiwtPFfpMiwwrfIrKidEChKECxQUBVUEfFirbGWiLkFQkdJiFtkrtkbIAEXCEDkwLpK", "output": "YES" }, { "input": "NwcGaIeSkOva\naIa", "output": "YES" }, { "input": "gSrAcVYgAdbdayzbKGhIzLDjyznLRIJH KyvilAaEddmgkBPCNzpmPNeGEbmmpAyHvUSoPvnaORrPUuafpReEGoDOQsAYnUHYfBqhdcopQfxJuGXgKnbdVMQNhJYkyjiJDKlShqBTtnnDQQzEijOMcYRGMgPGVhfIReYennKBLwDTVvcHMIHMgVpJkvzTrezxqS\nHJerIVvRyfrPgAQMTI AqGNO mQDfDwQHKgeeYmuRmozKHILvehMPOJNMRtPTAfvKvsoGKi xHEeKqDAYmQJPUXRJbIbHrgVOMGMTdvYiLui", "output": "YES" }, { "input": "ReB hksbHqQXxUgpvoNK bFqmNVCEiOyKdKcAJQRkpeohpfuqZabvrLfmpZOMcfyFBJGZwVMxiUPP pbZZtJjxhEwvrAba\nJTCpQnIViIGIdQtLnmkVzmcbBZR CoxAdTtWSYpbOglDFifqIVQ vfGKGtLpxpJHiHSWCMeRcrVOXBGBhoEnVhNTPWGTOErNtSvokcGdgZXbgTEtISUyTwaXUEIlJMmutsdCbiyrPZPJyRdOjnSuAGttLy", "output": "NO" }, { "input": "hrLzRegCuDGxTrhDgVvM KowwyYuXGzIpcXdSMgeQVfVOtJZdkhNYSegwFWWoPqcZoeapbQnyCtojgkcyezUNHGGIZrhzsKrvvcrtokIdcnqXXkCNKjrOjrnEAKBNxyDdiMVeyLvXxUYMZQRFdlcdlcxzKTeYzBlmpNiwWbNAAhWkMoGpRxkCuyqkzXdKWwGH\nJESKDOfnFdxPvUOCkrgSBEPQHJtJHzuNGstRbTCcchRWJvCcveSEAtwtOmZZiW", "output": "NO" }, { "input": "yDBxCtUygQwWqONxQCcuAvVCkMGlqgC zvkfEkwqbhMCQxnkwQIUhucCbVUyOBUcXvTNEGriTBwMDMfdsPZgWRgIUDqM\neptVnORTTyixxmWIBpSTEwOXqGZllBgSxPenYCDlFwckJlWsoVwWLAIbPOmFqcKcTcoQqahetl KLfVSyaLVebzsGwPSVbtQAeUdZAaJtfxlCEvvaRhLlVvRJhKat IaB awdqcDlrrhTbRxjEbzGwcdmdavkhcjHjzmwbxAgw", "output": "NO" }, { "input": "jlMwnnotSdlQMluKWkJwAeCetcqbIEnKeNyLWoKCGONDRBQOjbkGpUvDlmSFUJ bWhohqmmIUWTlDsvelUArAcZJBipMDwUvRfBsYzMdQnPDPAuBaeJmAxVKwUMJrwMDxNtlrtAowVWqWiwFGtmquZAcrpFsLHCrvMSMMlvQUqypAihQWrFMNoaqfs IBg\nNzeWQ bafrmDsYlpNHSGTBBgPl WIcuNhyNaNOEFvL", "output": "NO" }, { "input": "zyWvXBcUZqGqjHwZHQryBtFliLYnweXAoMKNpLaunaOlzaauWmLtywsEvWPiwxJapocAFRMjrqWJXYqfKEbBKnzLO\npsbi bsXpSeJaCkIuPWfSRADXdIClxcDCowwJzGCDTyAl", "output": "NO" }, { "input": "kKhuIwRPLCwPFfcnsyCfBdnsraGeOCcLTfXuGjqFSGPSAeDZJSS bXKFanNqWjpFnvRpWxHJspvisDlADJBioxXNbVoXeUedoPcNEpUyEeYxdJXhGzFAmpAiHotSVwbZQsuWjIVhVaEGgqbZHIoDpiEmjTtFylCwCkWWzUOoUfOHxEZvDwNpXhBWamHn\nK VpJjGhNbwCRhcfmNGVjewBFpEmPlIKeTuWiukDtEWpjgqciqglkyNfWrBLbGAKvlNWxaUelJmSlSoakSpRzePvJsshOsTYrMPXdxKpaShjyVIXGhRIAdtiGpNwtiRmGTBZhkJqIMdxMHX RMxCMYcWjcjhtCHyFnCvjjezGbkRDRiVxkbh", "output": "NO" }, { "input": "AXssNpFKyQmJcBdBdfkhhMUzfqJVgcLBddkwtnFSzSRUCjiDcdtmkzIGkCKSxWUEGhmHmciktJyGMkgCductyHx\nI nYhmJfPnvoKUiXYUBIPIcxNYTtvwPUoXERZvY ahlDpQFNMmVZqEBiYqYlHNqcpSCmhFczBlOAhsYFeqMGfqL EJsDNOgwoJfBzqijKOFcYQ", "output": "NO" }, { "input": "lkhrzDZmkdbjzYKPNMRkiwCFoZsMzBQMnxxdKKVJezSBjnLjPpUYtabcPTIaDJeDEobbWHdKOdVfMQwDXzDDcSrwVenDEYpMqfiOQ xSsqApWnAMoyhQXCKFzHvvzvUvkWwmwZrvZz\nsUzGspYpRFsHRbRgTQuCBgnFgPkisTUfFNwyEEWWRiweWWgjRkVQxgTwxOzdsOwfrGIH O gCXpzvHzfItuEHaihmugEyymSJIogYwX qAwcwIItidfnzZDhZgQHi eRjMAeVkJHceDZuJkmxGowOsmcGYYvk Ajtgi TxwihvjLViNZjvscTWvsaQUelTSivLShhEl", "output": "NO" }, { "input": "BRsVjyNhrqRHVwrJzuzRigEhdpbDmaACSPfed\nlWqKTjlrqOCUbgBBZdZDGCeQJDXawPnnDkQdZDgwrEQk", "output": "NO" }, { "input": "KRmINuyBYPwiTsdlyiNVuylToysJKmOpcLovAtwGPqrgFJQNAYvuAiyQRkeFMECVZvkDEmTauXlyjAaYRnTJXORMZRnTakBaUzSelMilejySDIZjQjzcOIrwXdvDvpeRIkoBgreyFXIyyIZutjiEBtwrmzQtPVUhvvdEtDMbXjBpoPVjGdM EXTAK JbCnw\nXZZqlJvzKKtvdNlzFPDTYxidqlsgufVzyEmO FZuLQ vVQsJESNviUCovCK NwwlbxsmPtOJNmAonCqrOZ bZ LVKAsQGmoLnYjeekvEIECFk", "output": "NO" } ]

1,668,601,980

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

92

0

# 1.x=int(input()) # ls=[] # for i in range(x): # ls.append(input()) # print(len(list(set(ls)))) m=input() n=input() lsm=list(m) lsn=list(n) index1="".join(lsm) index2="".join(lsn) ls1=list(index1.split(" ")) index3="".join(ls1) ls2=list(index2.split(" ")) index4="".join(ls2) x=0 for i in index4: if index3.count(i)>=index4.count(i): x=x+1 if x==len(index4): print("Yes") else: print("No")

Title: Letter Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text. Input Specification: The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading. Output Specification: If Vasya can write the given anonymous letter, print YES, otherwise print NO Demo Input: ['Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n', 'Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'abcdefg hijk\nk j i h g f e d c b a\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n', 'YES\n'] Note: none

```python # 1.x=int(input()) # ls=[] # for i in range(x): # ls.append(input()) # print(len(list(set(ls)))) m=input() n=input() lsm=list(m) lsn=list(n) index1="".join(lsm) index2="".join(lsn) ls1=list(index1.split(" ")) index3="".join(ls1) ls2=list(index2.split(" ")) index4="".join(ls2) x=0 for i in index4: if index3.count(i)>=index4.count(i): x=x+1 if x==len(index4): print("Yes") else: print("No") ```

0

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

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

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,651,512,284

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

46

102,400

from collections import OrderedDict s=input() s1="".join(OrderedDict.fromkeys(s)) l=s.count('l') for i in range(len(s1)): if(s1[i]=='h' and l>=2): if(s1[i+1]=='e'): if(s1[i+2]=='l'): if(s1[i+3]=='o'): print("YES") break else: print("NO") break else: print("NO") break else: print("NO") break else: continue else: print("NO")

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python from collections import OrderedDict s=input() s1="".join(OrderedDict.fromkeys(s)) l=s.count('l') for i in range(len(s1)): if(s1[i]=='h' and l>=2): if(s1[i+1]=='e'): if(s1[i+2]=='l'): if(s1[i+3]=='o'): print("YES") break else: print("NO") break else: print("NO") break else: print("NO") break else: continue else: print("NO") ```

0

630

A

Again Twenty Five!

PROGRAMMING

800

[ "number theory" ]

null

The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions." Could you pass the interview in the machine vision company in IT City?

The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.

Output the last two digits of 5*n* without spaces between them.

[ "2\n" ]

[ "25" ]

none

0

[ { "input": "2", "output": "25" }, { "input": "7", "output": "25" }, { "input": "1000000000000000000", "output": "25" }, { "input": "2000000000000000000", "output": "25" }, { "input": "987654321012345678", "output": "25" } ]

1,689,592,955

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

2

500

19,046,400

k=int(input()) mm="" l=5**k l=str(l) m=0 for i in range(len(l)-1,-1,-1): #i=str(i) if m!=2: mm+=l[i] m+=1 print(mm[::-1])

Title: Again Twenty Five! Time Limit: None seconds Memory Limit: None megabytes Problem Description: The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions." Could you pass the interview in the machine vision company in IT City? Input Specification: The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5. Output Specification: Output the last two digits of 5*n* without spaces between them. Demo Input: ['2\n'] Demo Output: ['25'] Note: none

```python k=int(input()) mm="" l=5**k l=str(l) m=0 for i in range(len(l)-1,-1,-1): #i=str(i) if m!=2: mm+=l[i] m+=1 print(mm[::-1]) ```

0

620

A

Professor GukiZ's Robot

PROGRAMMING

800

[ "implementation", "math" ]

null

Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.

The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot. The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.

Print the only integer *d* — the minimal number of steps to get the finish position.

[ "0 0\n4 5\n", "3 4\n6 1\n" ]

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

In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position. In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times.

0

[ { "input": "0 0\n4 5", "output": "5" }, { "input": "3 4\n6 1", "output": "3" }, { "input": "0 0\n4 6", "output": "6" }, { "input": "1 1\n-3 -5", "output": "6" }, { "input": "-1 -1\n-10 100", "output": "101" }, { "input": "1 -1\n100 -100", "output": "99" }, { "input": "-1000000000 -1000000000\n1000000000 1000000000", "output": "2000000000" }, { "input": "-1000000000 -1000000000\n0 999999999", "output": "1999999999" }, { "input": "0 0\n2 1", "output": "2" }, { "input": "10 0\n100 0", "output": "90" }, { "input": "1 5\n6 4", "output": "5" }, { "input": "0 0\n5 4", "output": "5" }, { "input": "10 1\n20 1", "output": "10" }, { "input": "1 1\n-3 4", "output": "4" }, { "input": "-863407280 504312726\n786535210 -661703810", "output": "1649942490" }, { "input": "-588306085 -741137832\n341385643 152943311", "output": "929691728" }, { "input": "0 0\n4 0", "output": "4" }, { "input": "93097194 -48405232\n-716984003 -428596062", "output": "810081197" }, { "input": "9 1\n1 1", "output": "8" }, { "input": "4 6\n0 4", "output": "4" }, { "input": "2 4\n5 2", "output": "3" }, { "input": "-100000000 -100000000\n100000000 100000123", "output": "200000123" }, { "input": "5 6\n5 7", "output": "1" }, { "input": "12 16\n12 1", "output": "15" }, { "input": "0 0\n5 1", "output": "5" }, { "input": "0 1\n1 1", "output": "1" }, { "input": "-44602634 913365223\n-572368780 933284951", "output": "527766146" }, { "input": "-2 0\n2 -2", "output": "4" }, { "input": "0 0\n3 1", "output": "3" }, { "input": "-458 2\n1255 4548", "output": "4546" }, { "input": "-5 -4\n-3 -3", "output": "2" }, { "input": "4 5\n7 3", "output": "3" }, { "input": "-1000000000 -999999999\n1000000000 999999998", "output": "2000000000" }, { "input": "-1000000000 -1000000000\n1000000000 -1000000000", "output": "2000000000" }, { "input": "-464122675 -898521847\n656107323 -625340409", "output": "1120229998" }, { "input": "-463154699 -654742385\n-699179052 -789004997", "output": "236024353" }, { "input": "982747270 -593488945\n342286841 -593604186", "output": "640460429" }, { "input": "-80625246 708958515\n468950878 574646184", "output": "549576124" }, { "input": "0 0\n1 0", "output": "1" }, { "input": "109810 1\n2 3", "output": "109808" }, { "input": "-9 0\n9 9", "output": "18" }, { "input": "9 9\n9 9", "output": "0" }, { "input": "1 1\n4 3", "output": "3" }, { "input": "1 2\n45 1", "output": "44" }, { "input": "207558188 -313753260\n-211535387 -721675423", "output": "419093575" }, { "input": "-11 0\n0 0", "output": "11" }, { "input": "-1000000000 1000000000\n1000000000 -1000000000", "output": "2000000000" }, { "input": "0 0\n1 1", "output": "1" }, { "input": "0 0\n0 1", "output": "1" }, { "input": "0 0\n-1 1", "output": "1" }, { "input": "0 0\n-1 0", "output": "1" }, { "input": "0 0\n-1 -1", "output": "1" }, { "input": "0 0\n0 -1", "output": "1" }, { "input": "0 0\n1 -1", "output": "1" }, { "input": "10 90\n90 10", "output": "80" }, { "input": "851016864 573579544\n-761410925 -380746263", "output": "1612427789" }, { "input": "1 9\n9 9", "output": "8" }, { "input": "1000 1000\n1000 1000", "output": "0" }, { "input": "1 9\n9 1", "output": "8" }, { "input": "1 90\n90 90", "output": "89" }, { "input": "100 100\n1000 1000", "output": "900" }, { "input": "-1 0\n0 0", "output": "1" }, { "input": "-750595959 -2984043\n649569876 -749608783", "output": "1400165835" }, { "input": "958048496 712083589\n423286949 810566863", "output": "534761547" }, { "input": "146316710 53945094\n-523054748 147499505", "output": "669371458" }, { "input": "50383856 -596516251\n-802950224 -557916272", "output": "853334080" }, { "input": "-637204864 -280290367\n-119020929 153679771", "output": "518183935" }, { "input": "-100 -100\n-60 -91", "output": "40" }, { "input": "337537326 74909428\n-765558776 167951547", "output": "1103096102" }, { "input": "0 81\n18 90", "output": "18" }, { "input": "283722202 -902633305\n-831696497 -160868946", "output": "1115418699" }, { "input": "1000 1000\n-1000 1000", "output": "2000" }, { "input": "5 6\n4 8", "output": "2" }, { "input": "40572000 597493595\n-935051731 368493185", "output": "975623731" }, { "input": "-5 5\n5 5", "output": "10" } ]

1,559,229,183

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

93

0

import sys line1 = sys.stdin.readline().rstrip("\n").split(" ") line2 = sys.stdin.readline().rstrip("\n").split(" ") x1 = int(line1[0]) y1 = int(line1[1]) x2 = int(line2[0]) y2 = int(line2[1]) diagonalSteps = min(abs(y2)-abs(y1), abs(x2)-abs(x1)) remainingSteps = max(abs(y2)-abs(y1), abs(x2)-abs(x1)) - diagonalSteps print(diagonalSteps+remainingSteps)

Title: Professor GukiZ's Robot Time Limit: None seconds Memory Limit: None megabytes Problem Description: Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position. Input Specification: The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot. The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot. Output Specification: Print the only integer *d* — the minimal number of steps to get the finish position. Demo Input: ['0 0\n4 5\n', '3 4\n6 1\n'] Demo Output: ['5\n', '3\n'] Note: In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position. In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times.

```python import sys line1 = sys.stdin.readline().rstrip("\n").split(" ") line2 = sys.stdin.readline().rstrip("\n").split(" ") x1 = int(line1[0]) y1 = int(line1[1]) x2 = int(line2[0]) y2 = int(line2[1]) diagonalSteps = min(abs(y2)-abs(y1), abs(x2)-abs(x1)) remainingSteps = max(abs(y2)-abs(y1), abs(x2)-abs(x1)) - diagonalSteps print(diagonalSteps+remainingSteps) ```

0

5

A

Chat Servers Outgoing Traffic

PROGRAMMING

1,000

[ "implementation" ]

A. Chat Server's Outgoing Traffic

1

64

Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands: - Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command). Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands. Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message. As Polycarp has no time, he is asking for your help in solving this problem.

Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following: - +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command. <name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line. It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc. All names are case-sensitive.

Print a single number — answer to the problem.

[ "+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n", "+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n" ]

[ "9\n", "14\n" ]

none

0

[ { "input": "+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate", "output": "9" }, { "input": "+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate", "output": "14" }, { "input": "+Dmitry\n+Mike\nDmitry:All letters will be used\nDmitry:qwertyuiopasdfghjklzxcvbnm QWERTYUIOPASDFGHJKLZXCVBNM\nDmitry:And digits too\nDmitry:1234567890 0987654321\n-Dmitry", "output": "224" }, { "input": "+Dmitry\n+Mike\n+Kate\nDmitry:", "output": "0" }, { "input": "+Dmitry\nDmitry:No phrases with spaces at the beginning and at the end\n+Mike\nDmitry:spaces spaces\n-Dmitry", "output": "86" }, { "input": "+XqD\n+aT537\nXqD:x6ZPjMR1DDKG2\nXqD:lLCriywPnB\n-XqD", "output": "46" }, { "input": "+8UjgAJ\n8UjgAJ:02hR7UBc1tqqfL\n-8UjgAJ\n+zdi\n-zdi", "output": "14" }, { "input": "+6JPKkgXDrA\n+j6JHjv70An\n+QGtsceK0zJ\n6JPKkgXDrA:o4\n+CSmwi9zDra\nQGtsceK0zJ:Zl\nQGtsceK0zJ:0\nj6JHjv70An:7\nj6JHjv70An:B\nQGtsceK0zJ:OO", "output": "34" }, { "input": "+1aLNq9S7uLV\n-1aLNq9S7uLV\n+O9ykq3xDJv\n-O9ykq3xDJv\n+54Yq1xJq14F\n+0zJ5Vo0RDZ\n-54Yq1xJq14F\n-0zJ5Vo0RDZ\n+lxlH7sdolyL\n-lxlH7sdolyL", "output": "0" }, { "input": "+qlHEc2AuYy\nqlHEc2AuYy:YYRwD0 edNZgpE nGfOguRWnMYpTpGUVM aXDKGXo1Gv1tHL9\nqlHEc2AuYy:yvh3GsPcImqrvoUcBNQcP6ezwpU0 xAVltaKZp94VKiNao\nqlHEc2AuYy:zuCO6Opey L eu7lTwysaSk00zjpv zrDfbt8l hpHfu\n+pErDMxgVgh\nqlHEc2AuYy:I1FLis mmQbZtd8Ui7y 1vcax6yZBMhVRdD6Ahlq7MNCw\nqlHEc2AuYy:lz MFUNJZhlqBYckHUDlNhLiEkmecRh1o0t7alXBvCRVEFVx\npErDMxgVgh:jCyMbu1dkuEj5TzbBOjyUhpfC50cL8R900Je3R KxRgAI dT\nqlHEc2AuYy:62b47eabo2hf vSUD7KioN ZHki6WB6gh3u GKv5rgwyfF\npErDMxgVgh:zD5 9 ympl4wR gy7a7eAGAn5xVdGP9FbL6hRCZAR6O4pT6zb", "output": "615" }, { "input": "+adabacaba0", "output": "0" }, { "input": "+acabadab\n+caba0aba", "output": "0" }, { "input": "+dabaca\n-dabaca\n+aba0ab", "output": "0" }, { "input": "+cab\n+abac\n-abac\n+baca", "output": "0" }, { "input": "+cabadabac\n-cabadabac\n+abacaba1ab\n-abacaba1ab\n+ba0abaca", "output": "0" }, { "input": "+adabacaba\n-adabacaba\n+aca\naca:caba\n-aca\n+bacaba\n-bacaba\n+aba\n-aba\n+bad", "output": "4" }, { "input": "+acabadab\n-acabadab\n+aba0abacab\n+baca\n+abacaba0ab\n-baca\n-abacaba0ab\n-aba0abacab\n+cab\n-cab\n+abacabada\n-abacabada\n+badabaca\n-badabaca\n+badaba", "output": "0" }, { "input": "+badabac\nbadabac:abacabad\n-badabac\n+0ab\n-0ab\n+dabacab\n-dabacab\n+a0ab\n-a0ab\n+0abaca\n-0abaca\n+dabac\n-dabac\n+abaca\n-abaca\n+bacabada\n-bacabada\n+aca\n-aca\n+abadabaca\n-abadabaca\n+acaba\n-acaba\n+abacabadab\n-abacabadab", "output": "8" } ]

1,688,916,794

2,147,483,647

PyPy 3-64

OK

TESTS

30

280

0

sum0 = 0 set0 = set() while True: try: s = input() if s[0] == '+': set0.add(s[1:]) elif s[0] == '-': set0.remove(s[1:]) elif s[0] != '+' and s[0] != '-': lst = s.split(':') len0 = len(lst[1]) sum0 += len(set0) * len0 except: print(sum0) break

Title: Chat Servers Outgoing Traffic Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands: - Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command). Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands. Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message. As Polycarp has no time, he is asking for your help in solving this problem. Input Specification: Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following: - +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command. <name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line. It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc. All names are case-sensitive. Output Specification: Print a single number — answer to the problem. Demo Input: ['+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n', '+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n'] Demo Output: ['9\n', '14\n'] Note: none

```python sum0 = 0 set0 = set() while True: try: s = input() if s[0] == '+': set0.add(s[1:]) elif s[0] == '-': set0.remove(s[1:]) elif s[0] != '+' and s[0] != '-': lst = s.split(':') len0 = len(lst[1]) sum0 += len(set0) * len0 except: print(sum0) break ```

3.86

56

A

Bar

PROGRAMMING

1,000

[ "implementation" ]

A. Bar

2

256

According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks? The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators. Only the drinks from the list given above should be considered alcohol.

Print a single number which is the number of people Vasya should check to guarantee the law enforcement.

[ "5\n18\nVODKA\nCOKE\n19\n17\n" ]

[ "2\n" ]

In the sample test the second and fifth clients should be checked.

500

[ { "input": "5\n18\nVODKA\nCOKE\n19\n17", "output": "2" }, { "input": "2\n2\nGIN", "output": "2" }, { "input": "3\nWHISKEY\n3\nGIN", "output": "3" }, { "input": "4\n813\nIORBQITQXMPTFAEMEQDQIKFGKGOTNKTOSZCBRPXJLUKVLVHJYNRUJXK\nRUM\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWINWJW", "output": "1" }, { "input": "4\nSAKE\nSAKE\n13\n2", "output": "4" }, { "input": "2\n0\n17", "output": "2" }, { "input": "1\n0", "output": "1" } ]

1,612,298,213

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

154

0

import sys def get_string(): return sys.stdin.readline().strip() def probA (*argv): a = argv b = 0 while x<a: x+= 1 c = get_string() if c == "ABSINTH" or c == "BEER" or c == "BRANDY" or c == "CHAMPAGNE" or c == "GIN" or c == "RUM" or c == "SAKE" or c == "TEQUILA" or c == "VODKA" or c == "WHISKEY" or c == "WINE": b+=1 if c.isdigit(): if(int(c)<18): b+= 1 print (b)

Title: Bar Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks? The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators. Only the drinks from the list given above should be considered alcohol. Output Specification: Print a single number which is the number of people Vasya should check to guarantee the law enforcement. Demo Input: ['5\n18\nVODKA\nCOKE\n19\n17\n'] Demo Output: ['2\n'] Note: In the sample test the second and fifth clients should be checked.

```python import sys def get_string(): return sys.stdin.readline().strip() def probA (*argv): a = argv b = 0 while x<a: x+= 1 c = get_string() if c == "ABSINTH" or c == "BEER" or c == "BRANDY" or c == "CHAMPAGNE" or c == "GIN" or c == "RUM" or c == "SAKE" or c == "TEQUILA" or c == "VODKA" or c == "WHISKEY" or c == "WINE": b+=1 if c.isdigit(): if(int(c)<18): b+= 1 print (b) ```

0

852

F

Product transformation

PROGRAMMING

2,200

[ "combinatorics", "math", "number theory" ]

null

Consider an array *A* with *N* elements, all being the same integer *a*. Define the product transformation as a simultaneous update *A**i*<==<=*A**i*·*A**i*<=+<=1, that is multiplying each element to the element right to it for , with the last number *A**N* remaining the same. For example, if we start with an array *A* with *a*<==<=2 and *N*<==<=4, then after one product transformation *A*<==<=[4,<= 4,<= 4,<= 2], and after two product transformations *A*<==<=[16,<= 16,<= 8,<= 2]. Your simple task is to calculate the array *A* after *M* product transformations. Since the numbers can get quite big you should output them modulo *Q*.

The first and only line of input contains four integers *N*, *M*, *a*, *Q* (7<=≤<=*Q*<=≤<=109<=+<=123, 2<=≤<=*a*<=≤<=106<=+<=123, , is prime), where is the multiplicative order of the integer *a* modulo *Q*, see notes for definition.

You should output the array *A* from left to right.

[ "2 2 2 7\n" ]

[ "1 2 " ]

The multiplicative order of a number *a* modulo *Q* <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/38b13c1f6db75ae72784f8602e8230429b26cf2a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, is the smallest natural number *x* such that *a**x* *mod* *Q* = 1. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/64b53b8160eddc004a5b65223bf29dd636bb1832.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

0

[]

1,608,318,754

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

140

40,243,200

# #include <cstdio> # using namespace std; # #define N 2000050 # int n,m,a,q,i,pw[N],ni[N],c[N]; # int main() # { # scanf("%d%d%d%d",&n,&m,&a,&q); # int y=1;pw[0]=1;pw[1]=a; # while (pw[y]!=1) y++,pw[y]=1LL*pw[y-1]*a%q; # ni[1]=1; # for (i=2;i<y;i++) ni[i]=-1LL*(y/i)*ni[y%i]%y; # int x=1;c[0]=1; # for (i=1;i<n;i++) x=1LL*x*(m-i+1)%y*ni[i]%y,c[i]=(c[i-1]+x)%y; # for (i=1;i<=n;i++) printf("%d ",pw[(c[n-i]+y)%y]);puts(""); # return 0; # } N = 2000050 n,m,a,q = map(int,input().split()) pw = [0]*N ni = [0]*N c = [0]*N y = 1 pw[0] = 1 pw[1] = a while(pw[y]!=1): y+=1 pw[y] = pw[y-1]*a % q ni[1] = 1 for i in range(2,y): ni[i] = -(y/i)*ni[y%i]%y x = 1 c[0] = 1 for i in range(1,n): x *= (m-i+1) % y * ni[i] % y for i in range(1,n+1): print(pw[(c[n-i]+y)%y])

Title: Product transformation Time Limit: None seconds Memory Limit: None megabytes Problem Description: Consider an array *A* with *N* elements, all being the same integer *a*. Define the product transformation as a simultaneous update *A**i*<==<=*A**i*·*A**i*<=+<=1, that is multiplying each element to the element right to it for , with the last number *A**N* remaining the same. For example, if we start with an array *A* with *a*<==<=2 and *N*<==<=4, then after one product transformation *A*<==<=[4,<= 4,<= 4,<= 2], and after two product transformations *A*<==<=[16,<= 16,<= 8,<= 2]. Your simple task is to calculate the array *A* after *M* product transformations. Since the numbers can get quite big you should output them modulo *Q*. Input Specification: The first and only line of input contains four integers *N*, *M*, *a*, *Q* (7<=≤<=*Q*<=≤<=109<=+<=123, 2<=≤<=*a*<=≤<=106<=+<=123, , is prime), where is the multiplicative order of the integer *a* modulo *Q*, see notes for definition. Output Specification: You should output the array *A* from left to right. Demo Input: ['2 2 2 7\n'] Demo Output: ['1 2 '] Note: The multiplicative order of a number *a* modulo *Q* <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/38b13c1f6db75ae72784f8602e8230429b26cf2a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, is the smallest natural number *x* such that *a**x* *mod* *Q* = 1. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/64b53b8160eddc004a5b65223bf29dd636bb1832.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

```python # #include <cstdio> # using namespace std; # #define N 2000050 # int n,m,a,q,i,pw[N],ni[N],c[N]; # int main() # { # scanf("%d%d%d%d",&n,&m,&a,&q); # int y=1;pw[0]=1;pw[1]=a; # while (pw[y]!=1) y++,pw[y]=1LL*pw[y-1]*a%q; # ni[1]=1; # for (i=2;i<y;i++) ni[i]=-1LL*(y/i)*ni[y%i]%y; # int x=1;c[0]=1; # for (i=1;i<n;i++) x=1LL*x*(m-i+1)%y*ni[i]%y,c[i]=(c[i-1]+x)%y; # for (i=1;i<=n;i++) printf("%d ",pw[(c[n-i]+y)%y]);puts(""); # return 0; # } N = 2000050 n,m,a,q = map(int,input().split()) pw = [0]*N ni = [0]*N c = [0]*N y = 1 pw[0] = 1 pw[1] = a while(pw[y]!=1): y+=1 pw[y] = pw[y-1]*a % q ni[1] = 1 for i in range(2,y): ni[i] = -(y/i)*ni[y%i]%y x = 1 c[0] = 1 for i in range(1,n): x *= (m-i+1) % y * ni[i] % y for i in range(1,n+1): print(pw[(c[n-i]+y)%y]) ```

0

845

B

Luba And The Ticket

PROGRAMMING

1,600

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

null

Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky. The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.

You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.

Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.

[ "000000\n", "123456\n", "111000\n" ]

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

In the first example the ticket is already lucky, so the answer is 0. In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required. In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.

0

[ { "input": "000000", "output": "0" }, { "input": "123456", "output": "2" }, { "input": "111000", "output": "1" }, { "input": "120111", "output": "0" }, { "input": "999999", "output": "0" }, { "input": "199880", "output": "1" }, { "input": "899889", "output": "1" }, { "input": "899888", "output": "1" }, { "input": "505777", "output": "2" }, { "input": "999000", "output": "3" }, { "input": "989010", "output": "3" }, { "input": "651894", "output": "1" }, { "input": "858022", "output": "2" }, { "input": "103452", "output": "1" }, { "input": "999801", "output": "2" }, { "input": "999990", "output": "1" }, { "input": "697742", "output": "1" }, { "input": "242367", "output": "2" }, { "input": "099999", "output": "1" }, { "input": "198999", "output": "1" }, { "input": "023680", "output": "1" }, { "input": "999911", "output": "2" }, { "input": "000990", "output": "2" }, { "input": "117099", "output": "1" }, { "input": "990999", "output": "1" }, { "input": "000111", "output": "1" }, { "input": "000444", "output": "2" }, { "input": "202597", "output": "2" }, { "input": "000333", "output": "1" }, { "input": "030039", "output": "1" }, { "input": "000009", "output": "1" }, { "input": "006456", "output": "1" }, { "input": "022995", "output": "3" }, { "input": "999198", "output": "1" }, { "input": "223456", "output": "2" }, { "input": "333665", "output": "2" }, { "input": "123986", "output": "2" }, { "input": "599257", "output": "1" }, { "input": "101488", "output": "3" }, { "input": "111399", "output": "2" }, { "input": "369009", "output": "1" }, { "input": "024887", "output": "2" }, { "input": "314347", "output": "1" }, { "input": "145892", "output": "1" }, { "input": "321933", "output": "1" }, { "input": "100172", "output": "1" }, { "input": "222455", "output": "2" }, { "input": "317596", "output": "1" }, { "input": "979245", "output": "2" }, { "input": "000018", "output": "1" }, { "input": "101389", "output": "2" }, { "input": "123985", "output": "2" }, { "input": "900000", "output": "1" }, { "input": "132069", "output": "1" }, { "input": "949256", "output": "1" }, { "input": "123996", "output": "2" }, { "input": "034988", "output": "2" }, { "input": "320869", "output": "2" }, { "input": "089753", "output": "1" }, { "input": "335667", "output": "2" }, { "input": "868580", "output": "1" }, { "input": "958031", "output": "2" }, { "input": "117999", "output": "2" }, { "input": "000001", "output": "1" }, { "input": "213986", "output": "2" }, { "input": "123987", "output": "3" }, { "input": "111993", "output": "2" }, { "input": "642479", "output": "1" }, { "input": "033788", "output": "2" }, { "input": "766100", "output": "2" }, { "input": "012561", "output": "1" }, { "input": "111695", "output": "2" }, { "input": "123689", "output": "2" }, { "input": "944234", "output": "1" }, { "input": "154999", "output": "2" }, { "input": "333945", "output": "1" }, { "input": "371130", "output": "1" }, { "input": "977330", "output": "2" }, { "input": "777544", "output": "2" }, { "input": "111965", "output": "2" }, { "input": "988430", "output": "2" }, { "input": "123789", "output": "3" }, { "input": "111956", "output": "2" }, { "input": "444776", "output": "2" }, { "input": "001019", "output": "1" }, { "input": "011299", "output": "2" }, { "input": "011389", "output": "2" }, { "input": "999333", "output": "2" }, { "input": "126999", "output": "2" }, { "input": "744438", "output": "0" }, { "input": "588121", "output": "3" }, { "input": "698213", "output": "2" }, { "input": "652858", "output": "1" }, { "input": "989304", "output": "3" }, { "input": "888213", "output": "3" }, { "input": "969503", "output": "2" }, { "input": "988034", "output": "2" }, { "input": "889444", "output": "2" }, { "input": "990900", "output": "1" }, { "input": "301679", "output": "2" }, { "input": "434946", "output": "1" }, { "input": "191578", "output": "2" }, { "input": "118000", "output": "2" }, { "input": "636915", "output": "0" }, { "input": "811010", "output": "1" }, { "input": "822569", "output": "1" }, { "input": "122669", "output": "2" }, { "input": "010339", "output": "2" }, { "input": "213698", "output": "2" }, { "input": "895130", "output": "2" }, { "input": "000900", "output": "1" }, { "input": "191000", "output": "2" }, { "input": "001000", "output": "1" }, { "input": "080189", "output": "2" }, { "input": "990000", "output": "2" }, { "input": "201984", "output": "2" }, { "input": "002667", "output": "2" }, { "input": "877542", "output": "2" }, { "input": "301697", "output": "2" }, { "input": "211597", "output": "2" }, { "input": "420337", "output": "1" }, { "input": "024768", "output": "2" }, { "input": "878033", "output": "2" }, { "input": "788024", "output": "2" }, { "input": "023869", "output": "2" }, { "input": "466341", "output": "1" }, { "input": "696327", "output": "1" }, { "input": "779114", "output": "2" }, { "input": "858643", "output": "1" }, { "input": "011488", "output": "3" }, { "input": "003669", "output": "2" }, { "input": "202877", "output": "3" }, { "input": "738000", "output": "2" }, { "input": "567235", "output": "2" }, { "input": "887321", "output": "3" }, { "input": "401779", "output": "2" }, { "input": "989473", "output": "2" }, { "input": "004977", "output": "3" }, { "input": "023778", "output": "2" }, { "input": "809116", "output": "1" }, { "input": "042762", "output": "1" }, { "input": "777445", "output": "2" }, { "input": "769302", "output": "2" }, { "input": "023977", "output": "2" }, { "input": "990131", "output": "2" } ]

1,686,758,850

2,147,483,647

PyPy 3-64

OK

TESTS

145

124

5,939,200

import sys input = lambda: sys.stdin.readline().rstrip() import itertools S = [int(c) for c in input()] ans = 6 for a in range(10): for b in range(10): for c in range(10): for d in range(10): for e in range(10): for f in range(10): if a+b+c==d+e+f: cnt = 0 A = [a,b,c,d,e,f] for i in range(6): if A[i]!=S[i]: cnt+=1 ans = min(ans, cnt) print(ans)

Title: Luba And The Ticket Time Limit: None seconds Memory Limit: None megabytes Problem Description: Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky. The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits. Input Specification: You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0. Output Specification: Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky. Demo Input: ['000000\n', '123456\n', '111000\n'] Demo Output: ['0\n', '2\n', '1\n'] Note: In the first example the ticket is already lucky, so the answer is 0. In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required. In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.

```python import sys input = lambda: sys.stdin.readline().rstrip() import itertools S = [int(c) for c in input()] ans = 6 for a in range(10): for b in range(10): for c in range(10): for d in range(10): for e in range(10): for f in range(10): if a+b+c==d+e+f: cnt = 0 A = [a,b,c,d,e,f] for i in range(6): if A[i]!=S[i]: cnt+=1 ans = min(ans, cnt) print(ans) ```

3

337

A

Puzzles

PROGRAMMING

900

[ "greedy" ]

null

The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.

The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.

Print a single integer — the least possible difference the teacher can obtain.

[ "4 6\n10 12 10 7 5 22\n" ]

[ "5\n" ]

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

500

[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]

1,669,039,823

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

8

1,000

9,932,800

from itertools import combinations n,m=map(int,input().split(" ")) l=list(map(int,input().split(" "))) c=combinations(l,n) a=[] for i in c: minn=min(i) maxx=max(i) a.append(maxx-minn) print(min(a))

Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

```python from itertools import combinations n,m=map(int,input().split(" ")) l=list(map(int,input().split(" "))) c=combinations(l,n) a=[] for i in c: minn=min(i) maxx=max(i) a.append(maxx-minn) print(min(a)) ```

0

48

B

Land Lot

PROGRAMMING

1,200

[ "brute force", "implementation" ]

B. Land Lot

2

256

Vasya has a beautiful garden where wonderful fruit trees grow and yield fantastic harvest every year. But lately thieves started to sneak into the garden at nights and steal the fruit too often. Vasya can’t spend the nights in the garden and guard the fruit because there’s no house in the garden! Vasya had been saving in for some time and finally he decided to build the house. The rest is simple: he should choose in which part of the garden to build the house. In the evening he sat at his table and drew the garden’s plan. On the plan the garden is represented as a rectangular checkered field *n*<=×<=*m* in size divided into squares whose side length is 1. In some squares Vasya marked the trees growing there (one shouldn’t plant the trees too close to each other that’s why one square contains no more than one tree). Vasya wants to find a rectangular land lot *a*<=×<=*b* squares in size to build a house on, at that the land lot border should go along the lines of the grid that separates the squares. All the trees that grow on the building lot will have to be chopped off. Vasya loves his garden very much, so help him choose the building land lot location so that the number of chopped trees would be as little as possible.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) which represent the garden location. The next *n* lines contain *m* numbers 0 or 1, which describe the garden on the scheme. The zero means that a tree doesn’t grow on this square and the 1 means that there is a growing tree. The last line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=50). Note that Vasya can choose for building an *a*<=×<=*b* rectangle as well a *b*<=×<=*a* one, i.e. the side of the lot with the length of *a* can be located as parallel to the garden side with the length of *n*, as well as parallel to the garden side with the length of *m*.

Print the minimum number of trees that needs to be chopped off to select a land lot *a*<=×<=*b* in size to build a house on. It is guaranteed that at least one lot location can always be found, i. e. either *a*<=≤<=*n* and *b*<=≤<=*m*, or *a*<=≤<=*m* и *b*<=≤<=*n*.

[ "2 2\n1 0\n1 1\n1 1\n", "4 5\n0 0 1 0 1\n0 1 1 1 0\n1 0 1 0 1\n1 1 1 1 1\n2 3\n" ]

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

In the second example the upper left square is (1,1) and the lower right is (3,2).

0

[ { "input": "2 2\n1 0\n1 1\n1 1", "output": "0" }, { "input": "4 5\n0 0 1 0 1\n0 1 1 1 0\n1 0 1 0 1\n1 1 1 1 1\n2 3", "output": "2" }, { "input": "3 3\n0 0 0\n0 0 0\n0 0 0\n1 2", "output": "0" }, { "input": "3 3\n1 1 1\n1 1 1\n1 1 1\n2 1", "output": "2" }, { "input": "3 2\n1 1\n1 1\n1 0\n2 1", "output": "1" }, { "input": "2 3\n1 0 1\n0 1 0\n3 2", "output": "3" }, { "input": "1 1\n0\n1 1", "output": "0" }, { "input": "1 1\n1\n1 1", "output": "1" }, { "input": "3 4\n1 0 1 0\n0 1 0 1\n1 0 1 0\n2 2", "output": "2" }, { "input": "3 4\n1 1 1 1\n1 0 0 1\n1 1 1 1\n3 1", "output": "1" }, { "input": "10 10\n1 1 1 0 0 0 0 1 1 0\n1 1 1 0 1 1 0 1 1 1\n1 0 1 1 0 1 1 1 1 0\n0 1 1 1 1 1 1 1 1 1\n1 1 1 1 0 1 1 1 1 1\n1 1 1 1 0 0 1 1 1 1\n1 1 1 1 0 1 1 1 0 1\n0 1 1 1 1 1 1 0 1 0\n1 1 1 1 1 0 0 1 0 1\n1 1 0 1 0 1 1 1 1 0\n5 4", "output": "12" }, { "input": "10 10\n0 1 1 1 1 1 1 0 1 1\n0 1 1 1 1 1 0 0 1 1\n1 1 0 0 1 1 0 0 0 0\n0 0 0 0 1 0 1 1 1 0\n1 0 1 0 1 0 1 1 1 1\n1 0 0 1 1 1 1 1 0 1\n0 0 0 1 1 0 1 1 1 0\n1 0 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1\n0 0 0 1 1 0 0 1 1 1\n1 10", "output": "4" }, { "input": "10 10\n1 0 1 1 1 1 0 0 1 1\n1 1 1 1 1 1 1 1 0 1\n1 0 0 1 1 1 1 1 1 1\n1 0 1 1 1 1 0 1 1 1\n0 0 1 0 1 1 1 1 1 1\n1 1 1 0 0 1 1 1 1 1\n0 1 1 0 1 1 0 1 1 0\n1 0 1 1 1 0 1 1 1 1\n1 0 1 1 1 0 1 1 0 1\n1 1 0 1 1 1 0 0 1 0\n10 1", "output": "4" }, { "input": "10 7\n0 1 1 0 0 1 1\n1 1 0 0 0 0 1\n0 1 0 0 0 1 0\n0 1 0 1 1 1 1\n1 1 0 1 0 0 1\n0 1 0 0 0 0 0\n0 1 0 0 1 0 1\n0 1 0 1 1 0 0\n1 1 0 1 1 1 0\n1 1 0 0 0 1 0\n1 8", "output": "0" }, { "input": "10 8\n1 1 0 1 1 1 0 0\n0 1 0 1 1 1 1 1\n1 1 0 0 1 0 0 1\n0 1 1 1 1 0 1 0\n0 1 1 0 1 1 0 1\n0 1 1 0 0 1 0 1\n1 0 0 0 1 1 0 1\n0 1 1 0 1 1 1 1\n0 1 1 1 0 1 0 1\n1 1 0 1 1 0 1 1\n4 9", "output": "20" }, { "input": "10 10\n1 0 1 1 1 1 1 1 1 1\n1 1 1 0 1 1 0 1 1 1\n1 1 1 0 1 1 1 1 0 1\n1 1 0 1 1 1 0 0 0 1\n0 1 0 1 1 1 0 1 1 1\n1 0 1 0 1 0 1 1 1 1\n1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 0 1 1\n1 1 1 1 0 1 1 1 1 1\n0 1 1 1 1 0 1 1 0 1\n10 10", "output": "80" }, { "input": "10 10\n0 1 1 0 0 0 1 0 0 0\n0 0 1 1 1 1 0 1 0 0\n1 1 0 1 1 0 0 1 0 0\n1 0 0 0 0 0 0 0 1 0\n0 0 0 1 0 0 0 1 0 0\n0 1 0 0 1 0 0 0 1 0\n0 1 0 1 1 1 1 0 0 0\n1 0 0 1 0 1 0 0 0 0\n0 0 0 0 1 0 0 0 0 0\n1 1 0 0 0 0 0 0 1 0\n3 7", "output": "4" }, { "input": "10 10\n1 1 1 0 1 1 1 1 0 0\n1 1 1 1 1 0 0 0 0 1\n0 1 1 0 0 1 1 1 0 0\n1 1 1 1 0 1 1 1 1 1\n1 0 0 1 0 1 1 1 1 1\n1 1 1 1 1 1 0 1 0 1\n1 1 1 1 1 1 1 1 0 0\n0 1 0 0 1 1 1 1 1 1\n0 1 1 1 0 1 0 1 0 0\n1 1 0 1 0 1 1 1 1 0\n6 7", "output": "27" }, { "input": "10 8\n0 1 1 1 1 1 1 0\n0 0 1 1 1 1 1 1\n0 1 0 1 1 1 1 0\n0 0 1 0 1 0 1 1\n0 1 1 1 1 1 1 1\n0 1 0 1 0 0 1 1\n0 0 0 0 0 0 0 1\n1 1 1 1 1 0 1 1\n1 1 1 0 1 1 1 0\n1 1 0 0 1 1 0 1\n8 10", "output": "51" }, { "input": "10 1\n0\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1 5", "output": "4" } ]

1,646,487,215

2,147,483,647

Python 3

OK

TESTS

30

186

0

def calcTrees(cont, startRow, startCol, a, b): ans = 0 for i in range(startRow, startRow + a): for j in range(startCol, startCol + b): ans += cont[i][j] return ans n, m = [int(item) for item in input().split(' ')] cont = list() for i in range(n): cont.append([int(item) for item in input().split(' ')]) a, b = [int(item) for item in input().split(' ')] ans = a*b for i in range(n): for j in range(m): if i + a <= n and j + b <= m: helper = calcTrees(cont,i,j,a,b) if ans > helper: ans = helper if i + b <= n and j + a <= m: helper = calcTrees(cont,i,j,b,a) if ans > helper: ans = helper print(ans)

Title: Land Lot Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya has a beautiful garden where wonderful fruit trees grow and yield fantastic harvest every year. But lately thieves started to sneak into the garden at nights and steal the fruit too often. Vasya can’t spend the nights in the garden and guard the fruit because there’s no house in the garden! Vasya had been saving in for some time and finally he decided to build the house. The rest is simple: he should choose in which part of the garden to build the house. In the evening he sat at his table and drew the garden’s plan. On the plan the garden is represented as a rectangular checkered field *n*<=×<=*m* in size divided into squares whose side length is 1. In some squares Vasya marked the trees growing there (one shouldn’t plant the trees too close to each other that’s why one square contains no more than one tree). Vasya wants to find a rectangular land lot *a*<=×<=*b* squares in size to build a house on, at that the land lot border should go along the lines of the grid that separates the squares. All the trees that grow on the building lot will have to be chopped off. Vasya loves his garden very much, so help him choose the building land lot location so that the number of chopped trees would be as little as possible. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) which represent the garden location. The next *n* lines contain *m* numbers 0 or 1, which describe the garden on the scheme. The zero means that a tree doesn’t grow on this square and the 1 means that there is a growing tree. The last line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=50). Note that Vasya can choose for building an *a*<=×<=*b* rectangle as well a *b*<=×<=*a* one, i.e. the side of the lot with the length of *a* can be located as parallel to the garden side with the length of *n*, as well as parallel to the garden side with the length of *m*. Output Specification: Print the minimum number of trees that needs to be chopped off to select a land lot *a*<=×<=*b* in size to build a house on. It is guaranteed that at least one lot location can always be found, i. e. either *a*<=≤<=*n* and *b*<=≤<=*m*, or *a*<=≤<=*m* и *b*<=≤<=*n*. Demo Input: ['2 2\n1 0\n1 1\n1 1\n', '4 5\n0 0 1 0 1\n0 1 1 1 0\n1 0 1 0 1\n1 1 1 1 1\n2 3\n'] Demo Output: ['0\n', '2\n'] Note: In the second example the upper left square is (1,1) and the lower right is (3,2).

```python def calcTrees(cont, startRow, startCol, a, b): ans = 0 for i in range(startRow, startRow + a): for j in range(startCol, startCol + b): ans += cont[i][j] return ans n, m = [int(item) for item in input().split(' ')] cont = list() for i in range(n): cont.append([int(item) for item in input().split(' ')]) a, b = [int(item) for item in input().split(' ')] ans = a*b for i in range(n): for j in range(m): if i + a <= n and j + b <= m: helper = calcTrees(cont,i,j,a,b) if ans > helper: ans = helper if i + b <= n and j + a <= m: helper = calcTrees(cont,i,j,b,a) if ans > helper: ans = helper print(ans) ```

3.9535

186

A

Comparing Strings

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters. Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.

The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.

Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".

[ "ab\nba\n", "aa\nab\n" ]

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

- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".

500

[ { "input": "ab\nba", "output": "YES" }, { "input": "aa\nab", "output": "NO" }, { "input": "a\nza", "output": "NO" }, { "input": "vvea\nvvae", "output": "YES" }, { "input": "rtfabanpc\natfabrnpc", "output": "YES" }, { "input": "mt\ntm", "output": "YES" }, { "input": "qxolmbkkt\naovlajmlf", "output": "NO" }, { "input": "b\ng", "output": "NO" }, { "input": "ab\naba", "output": "NO" }, { "input": "ba\na", "output": "NO" }, { "input": "a\nab", "output": "NO" }, { "input": "a\naa", "output": "NO" }, { "input": "a\nz", "output": "NO" }, { "input": "aabb\nbbaa", "output": "NO" }, { "input": "ab\nbd", "output": "NO" }, { "input": "bac\ndae", "output": "NO" }, { "input": "abc\nakl", "output": "NO" }, { "input": "cb\naa", "output": "NO" }, { "input": "abaab\naabba", "output": "NO" }, { "input": "aab\naaa", "output": "NO" }, { "input": "abcde\nedcba", "output": "NO" }, { "input": "abab\nbaba", "output": "NO" }, { "input": "ab\nbac", "output": "NO" }, { "input": "abcd\naqcb", "output": "NO" }, { "input": "abc\nbad", "output": "NO" }, { "input": "ab\nca", "output": "NO" }, { "input": "abc\nab", "output": "NO" }, { "input": "ab\nbae", "output": "NO" }, { "input": "aaaabcccca\naaaadccccb", "output": "NO" }, { "input": "acaa\nabca", "output": "NO" }, { "input": "aba\naab", "output": "YES" }, { "input": "ah\nla", "output": "NO" }, { "input": "aab\naa", "output": "NO" }, { "input": "abab\nabcd", "output": "NO" }, { "input": "ed\nab", "output": "NO" } ]

1,609,740,068

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

30

434

6,144,000

from sys import stdin,stdout,exit import math import bisect from math import gcd,floor,sqrt,log from bisect import bisect_left as bl,bisect_right as br MOD:int=1000000007 g1 = list(stdin.readline().strip()) g2 = list(stdin.readline().strip()) g1.sort() g2.sort() if g1 == g2: stdout.write('YES\n') else: stdout.write('NO\n')

Title: Comparing Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters. Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not. Input Specification: The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length. Output Specification: Print "YES", if the dwarves belong to the same race. Otherwise, print "NO". Demo Input: ['ab\nba\n', 'aa\nab\n'] Demo Output: ['YES\n', 'NO\n'] Note: - First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".

```python from sys import stdin,stdout,exit import math import bisect from math import gcd,floor,sqrt,log from bisect import bisect_left as bl,bisect_right as br MOD:int=1000000007 g1 = list(stdin.readline().strip()) g2 = list(stdin.readline().strip()) g1.sort() g2.sort() if g1 == g2: stdout.write('YES\n') else: stdout.write('NO\n') ```

0

920

A

Water The Garden

PROGRAMMING

1,000

[ "implementation" ]

null

It is winter now, and Max decided it's about time he watered the garden. The garden can be represented as *n* consecutive garden beds, numbered from 1 to *n*. *k* beds contain water taps (*i*-th tap is located in the bed *x**i*), which, if turned on, start delivering water to neighbouring beds. If the tap on the bed *x**i* is turned on, then after one second has passed, the bed *x**i* will be watered; after two seconds have passed, the beds from the segment [*x**i*<=-<=1,<=*x**i*<=+<=1] will be watered (if they exist); after *j* seconds have passed (*j* is an integer number), the beds from the segment [*x**i*<=-<=(*j*<=-<=1),<=*x**i*<=+<=(*j*<=-<=1)] will be watered (if they exist). Nothing changes during the seconds, so, for example, we can't say that the segment [*x**i*<=-<=2.5,<=*x**i*<=+<=2.5] will be watered after 2.5 seconds have passed; only the segment [*x**i*<=-<=2,<=*x**i*<=+<=2] will be watered at that moment. Max wants to turn on all the water taps at the same moment, and now he wonders, what is the minimum number of seconds that have to pass after he turns on some taps until the whole garden is watered. Help him to find the answer!

The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=200). Then *t* test cases follow. The first line of each test case contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200, 1<=≤<=*k*<=≤<=*n*) — the number of garden beds and water taps, respectively. Next line contains *k* integers *x**i* (1<=≤<=*x**i*<=≤<=*n*) — the location of *i*-th water tap. It is guaranteed that for each condition *x**i*<=-<=1<=<<=*x**i* holds. It is guaranteed that the sum of *n* over all test cases doesn't exceed 200. Note that in hacks you have to set *t*<==<=1.

For each test case print one integer — the minimum number of seconds that have to pass after Max turns on some of the water taps, until the whole garden is watered.

[ "3\n5 1\n3\n3 3\n1 2 3\n4 1\n1\n" ]

[ "3\n1\n4\n" ]

The first example consists of 3 tests: 1. There are 5 garden beds, and a water tap in the bed 3. If we turn it on, then after 1 second passes, only bed 3 will be watered; after 2 seconds pass, beds [1, 3] will be watered, and after 3 seconds pass, everything will be watered. 1. There are 3 garden beds, and there is a water tap in each one. If we turn all of them on, then everything will be watered after 1 second passes. 1. There are 4 garden beds, and only one tap in the bed 1. It will take 4 seconds to water, for example, bed 4.

0

[ { "input": "3\n5 1\n3\n3 3\n1 2 3\n4 1\n1", "output": "3\n1\n4" }, { "input": "26\n1 1\n1\n2 1\n2\n2 1\n1\n2 2\n1 2\n3 1\n3\n3 1\n2\n3 2\n2 3\n3 1\n1\n3 2\n1 3\n3 2\n1 2\n3 3\n1 2 3\n4 1\n4\n4 1\n3\n4 2\n3 4\n4 1\n2\n4 2\n2 4\n4 2\n2 3\n4 3\n2 3 4\n4 1\n1\n4 2\n1 4\n4 2\n1 3\n4 3\n1 3 4\n4 2\n1 2\n4 3\n1 2 4\n4 3\n1 2 3\n4 4\n1 2 3 4", "output": "1\n2\n2\n1\n3\n2\n2\n3\n2\n2\n1\n4\n3\n3\n3\n2\n2\n2\n4\n2\n2\n2\n3\n2\n2\n1" }, { "input": "31\n5 1\n5\n5 1\n4\n5 2\n4 5\n5 1\n3\n5 2\n3 5\n5 2\n3 4\n5 3\n3 4 5\n5 1\n2\n5 2\n2 5\n5 2\n2 4\n5 3\n2 4 5\n5 2\n2 3\n5 3\n2 3 5\n5 3\n2 3 4\n5 4\n2 3 4 5\n5 1\n1\n5 2\n1 5\n5 2\n1 4\n5 3\n1 4 5\n5 2\n1 3\n5 3\n1 3 5\n5 3\n1 3 4\n5 4\n1 3 4 5\n5 2\n1 2\n5 3\n1 2 5\n5 3\n1 2 4\n5 4\n1 2 4 5\n5 3\n1 2 3\n5 4\n1 2 3 5\n5 4\n1 2 3 4\n5 5\n1 2 3 4 5", "output": "5\n4\n4\n3\n3\n3\n3\n4\n2\n2\n2\n3\n2\n2\n2\n5\n3\n2\n2\n3\n2\n2\n2\n4\n2\n2\n2\n3\n2\n2\n1" }, { "input": "1\n200 1\n200", "output": "200" }, { "input": "1\n5 1\n5", "output": "5" }, { "input": "1\n177 99\n1 4 7 10 11 13 14 15 16 17 19 21 22 24 25 26 27 28 32 34 35 38 39 40 42 45 46 52 54 55 57 58 59 60 62 64 65 67 70 71 74 77 78 79 80 81 83 84 88 92 93 94 95 100 101 102 104 106 107 108 109 110 112 113 114 115 116 118 122 123 124 125 127 128 129 130 134 135 137 138 139 140 142 146 148 149 154 158 160 161 162 165 166 167 169 171 172 173 176", "output": "4" }, { "input": "1\n69 12\n5 7 10 11 12 18 20 27 28 31 47 67", "output": "11" }, { "input": "1\n74 7\n19 39 40 47 55 57 61", "output": "19" }, { "input": "1\n170 11\n14 18 37 39 80 83 103 112 124 127 131", "output": "40" }, { "input": "1\n200 1\n8", "output": "193" }, { "input": "1\n155 53\n2 3 7 9 10 11 12 20 24 26 28 31 38 39 40 51 53 56 58 63 65 66 69 70 72 74 79 81 83 88 90 92 100 103 104 106 111 113 114 115 116 121 124 126 127 128 133 136 142 143 145 148 150", "output": "6" }, { "input": "1\n161 69\n2 5 8 11 12 13 17 18 23 25 28 29 30 33 34 35 36 38 39 44 45 49 52 53 56 57 58 60 62 70 71 74 76 77 82 83 86 90 94 95 97 104 105 108 109 112 113 118 120 123 126 127 132 135 137 139 140 141 142 143 144 146 147 148 151 152 153 154 161", "output": "5" }, { "input": "1\n8 4\n1 2 3 7", "output": "3" }, { "input": "1\n12 2\n5 12", "output": "5" }, { "input": "1\n13 2\n6 12", "output": "6" }, { "input": "1\n13 2\n8 13", "output": "8" }, { "input": "1\n10 4\n1 2 3 5", "output": "6" }, { "input": "1\n200 2\n50 150", "output": "51" }, { "input": "1\n110 2\n1 110", "output": "55" } ]

1,517,996,730

2,147,483,647

PyPy 3

OK

TESTS

21

108

19,968,000

t = int(input()) for _ in range(t): n, k = map(int, input().split()) pos = list(map(int, input().split())) i = pos[0] max_dist = 0 for j in pos[1:]: max_dist = max(max_dist , j - i - 1) i = j print(max(max_dist // 2 + max_dist % 2, pos[0] - 1, n - pos[-1]) + 1) # print("\n")

Title: Water The Garden Time Limit: None seconds Memory Limit: None megabytes Problem Description: It is winter now, and Max decided it's about time he watered the garden. The garden can be represented as *n* consecutive garden beds, numbered from 1 to *n*. *k* beds contain water taps (*i*-th tap is located in the bed *x**i*), which, if turned on, start delivering water to neighbouring beds. If the tap on the bed *x**i* is turned on, then after one second has passed, the bed *x**i* will be watered; after two seconds have passed, the beds from the segment [*x**i*<=-<=1,<=*x**i*<=+<=1] will be watered (if they exist); after *j* seconds have passed (*j* is an integer number), the beds from the segment [*x**i*<=-<=(*j*<=-<=1),<=*x**i*<=+<=(*j*<=-<=1)] will be watered (if they exist). Nothing changes during the seconds, so, for example, we can't say that the segment [*x**i*<=-<=2.5,<=*x**i*<=+<=2.5] will be watered after 2.5 seconds have passed; only the segment [*x**i*<=-<=2,<=*x**i*<=+<=2] will be watered at that moment. Max wants to turn on all the water taps at the same moment, and now he wonders, what is the minimum number of seconds that have to pass after he turns on some taps until the whole garden is watered. Help him to find the answer! Input Specification: The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=200). Then *t* test cases follow. The first line of each test case contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200, 1<=≤<=*k*<=≤<=*n*) — the number of garden beds and water taps, respectively. Next line contains *k* integers *x**i* (1<=≤<=*x**i*<=≤<=*n*) — the location of *i*-th water tap. It is guaranteed that for each condition *x**i*<=-<=1<=<<=*x**i* holds. It is guaranteed that the sum of *n* over all test cases doesn't exceed 200. Note that in hacks you have to set *t*<==<=1. Output Specification: For each test case print one integer — the minimum number of seconds that have to pass after Max turns on some of the water taps, until the whole garden is watered. Demo Input: ['3\n5 1\n3\n3 3\n1 2 3\n4 1\n1\n'] Demo Output: ['3\n1\n4\n'] Note: The first example consists of 3 tests: 1. There are 5 garden beds, and a water tap in the bed 3. If we turn it on, then after 1 second passes, only bed 3 will be watered; after 2 seconds pass, beds [1, 3] will be watered, and after 3 seconds pass, everything will be watered. 1. There are 3 garden beds, and there is a water tap in each one. If we turn all of them on, then everything will be watered after 1 second passes. 1. There are 4 garden beds, and only one tap in the bed 1. It will take 4 seconds to water, for example, bed 4.

```python t = int(input()) for _ in range(t): n, k = map(int, input().split()) pos = list(map(int, input().split())) i = pos[0] max_dist = 0 for j in pos[1:]: max_dist = max(max_dist , j - i - 1) i = j print(max(max_dist // 2 + max_dist % 2, pos[0] - 1, n - pos[-1]) + 1) # print("\n") ```

3

71

A

Way Too Long Words

PROGRAMMING

800

[ "strings" ]

A. Way Too Long Words

1

256

Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.

Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.

[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]

[ "word\nl10n\ni18n\np43s\n" ]

none

500

[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]

1,673,434,457

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

l=int(input()) for i in range(l): l==str(input()) if (len(l)>10): l = l[0]+str(len(l)-2)+l[-1] print(l)

Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none

```python l=int(input()) for i in range(l): l==str(input()) if (len(l)>10): l = l[0]+str(len(l)-2)+l[-1] print(l) ```

-1

841

A

Generous Kefa

PROGRAMMING

900

[ "brute force", "implementation" ]

null

One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons.

Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary.

[ "4 2\naabb\n", "6 3\naacaab\n" ]

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

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

500

[ { "input": "4 2\naabb", "output": "YES" }, { "input": "6 3\naacaab", "output": "NO" }, { "input": "2 2\nlu", "output": "YES" }, { "input": "5 3\novvoo", "output": "YES" }, { "input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf", "output": "YES" }, { "input": "81 3\nooycgmvvrophvcvpoupepqllqttwcocuilvyxbyumdmmfapvpnxhjhxfuagpnntonibicaqjvwfhwxhbv", "output": "NO" }, { "input": "100 100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "YES" }, { "input": "100 1\nnubcvvjvbjgnjsdkajimdcxvewbcytvfkihunycdrlconddlwgzjasjlsrttlrzsumzpyumpveglfqzmaofbshbojmwuwoxxvrod", "output": "NO" }, { "input": "100 13\nvyldolgryldqrvoldvzvrdrgorlorszddtgqvrlisxxrxdxlqtvtgsrqlzixoyrozxzogqxlsgzdddzqrgitxxritoolzolgrtvl", "output": "YES" }, { "input": "18 6\njzwtnkvmscqhmdlsxy", "output": "YES" }, { "input": "21 2\nfscegcqgzesefghhwcexs", "output": "NO" }, { "input": "32 22\ncduamsptaklqtxlyoutlzepxgyfkvngc", "output": "YES" }, { "input": "49 27\noxyorfnkzwsfllnyvdhdanppuzrnbxehugvmlkgeymqjlmfxd", "output": "YES" }, { "input": "50 24\nxxutzjwbggcwvxztttkmzovtmuwttzcbwoztttohzzxghuuthv", "output": "YES" }, { "input": "57 35\nglxshztrqqfyxthqamagvtmrdparhelnzrqvcwqxjytkbuitovkdxueul", "output": "YES" }, { "input": "75 23\nittttiiuitutuiiuuututiuttiuiuutuuuiuiuuuuttuuttuutuiiuiuiiuiitttuututuiuuii", "output": "NO" }, { "input": "81 66\nfeqevfqfebhvubhuuvfuqheuqhbeeuebehuvhffvbqvqvfbqqvvhevqffbqqhvvqhfeehuhqeqhueuqqq", "output": "YES" }, { "input": "93 42\npqeiafraiavfcteumflpcbpozcomlvpovlzdbldvoopnhdoeqaopzthiuzbzmeieiatthdeqovaqfipqlddllmfcrrnhb", "output": "YES" }, { "input": "100 53\nizszyqyndzwzyzgsdagdwdazadiawizinagqqgczaqqnawgijziziawzszdjdcqjdjqiwgadydcnqisaayjiqqsscwwzjzaycwwc", "output": "YES" }, { "input": "100 14\nvkrdcqbvkwuckpmnbydmczdxoagdsgtqxvhaxntdcxhjcrjyvukhugoglbmyoaqexgtcfdgemmizoniwtmisqqwcwfusmygollab", "output": "YES" }, { "input": "100 42\naaaaaiiiiaiiiaaiaiiaaiiiiiaaaaaiaiiiaiiiiaiiiaaaaaiiiaaaiiaaiiiaiiiaiaaaiaiiiiaaiiiaiiaiaiiaiiiaaaia", "output": "NO" }, { "input": "100 89\ntjbkmydejporbqhcbztkcumxjjgsrvxpuulbhzeeckkbchpbxwhedrlhjsabcexcohgdzouvsgphjdthpuqrlkgzxvqbuhqxdsmf", "output": "YES" }, { "input": "100 100\njhpyiuuzizhubhhpxbbhpyxzhbpjphzppuhiahihiappbhuypyauhizpbibzixjbzxzpbphuiaypyujappuxiyuyaajaxjupbahb", "output": "YES" }, { "input": "100 3\nsszoovvzysavsvzsozzvoozvysozsaszayaszasaysszzzysosyayyvzozovavzoyavsooaoyvoozvvozsaosvayyovazzszzssa", "output": "NO" }, { "input": "100 44\ndluthkxwnorabqsukgnxnvhmsmzilyulpursnxkdsavgemiuizbyzebhyjejgqrvuckhaqtuvdmpziesmpmewpvozdanjyvwcdgo", "output": "YES" }, { "input": "100 90\ntljonbnwnqounictqqctgonktiqoqlocgoblngijqokuquoolciqwnctgoggcbojtwjlculoikbggquqncittwnjbkgkgubnioib", "output": "YES" }, { "input": "100 79\nykxptzgvbqxlregvkvucewtydvnhqhuggdsyqlvcfiuaiddnrrnstityyehiamrggftsqyduwxpuldztyzgmfkehprrneyvtknmf", "output": "YES" }, { "input": "100 79\naagwekyovbviiqeuakbqbqifwavkfkutoriovgfmittulhwojaptacekdirgqoovlleeoqkkdukpadygfwavppohgdrmymmulgci", "output": "YES" }, { "input": "100 93\nearrehrehenaddhdnrdddhdahnadndheeennrearrhraharddreaeraddhehhhrdnredanndneheddrraaneerreedhnadnerhdn", "output": "YES" }, { "input": "100 48\nbmmaebaebmmmbbmxvmammbvvebvaemvbbaxvbvmaxvvmveaxmbbxaaemxmxvxxxvxbmmxaaaevvaxmvamvvmaxaxavexbmmbmmev", "output": "YES" }, { "input": "100 55\nhsavbkehaaesffaeeffakhkhfehbbvbeasahbbbvkesbfvkefeesesevbsvfkbffakvshsbkahfkfakebsvafkbvsskfhfvaasss", "output": "YES" }, { "input": "100 2\ncscffcffsccffsfsfffccssfsscfsfsssffcffsscfccssfffcfscfsscsccccfsssffffcfcfsfffcsfsccffscffcfccccfffs", "output": "NO" }, { "input": "100 3\nzrgznxgdpgfoiifrrrsjfuhvtqxjlgochhyemismjnanfvvpzzvsgajcbsulxyeoepjfwvhkqogiiwqxjkrpsyaqdlwffoockxnc", "output": "NO" }, { "input": "100 5\njbltyyfjakrjeodqepxpkjideulofbhqzxjwlarufwzwsoxhaexpydpqjvhybmvjvntuvhvflokhshpicbnfgsqsmrkrfzcrswwi", "output": "NO" }, { "input": "100 1\nfnslnqktlbmxqpvcvnemxcutebdwepoxikifkzaaixzzydffpdxodmsxjribmxuqhueifdlwzytxkklwhljswqvlejedyrgguvah", "output": "NO" }, { "input": "100 21\nddjenetwgwmdtjbpzssyoqrtirvoygkjlqhhdcjgeurqpunxpupwaepcqkbjjfhnvgpyqnozhhrmhfwararmlcvpgtnopvjqsrka", "output": "YES" }, { "input": "100 100\nnjrhiauqlgkkpkuvciwzivjbbplipvhslqgdkfnmqrxuxnycmpheenmnrglotzuyxycosfediqcuadklsnzjqzfxnbjwvfljnlvq", "output": "YES" }, { "input": "100 100\nbbbbbbbtbbttbtbbbttbttbtbbttttbbbtbttbbbtbttbtbbttttbbbbbtbbttbtbbtbttbbbtbtbtbtbtbtbbbttbbtbtbtbbtb", "output": "YES" }, { "input": "14 5\nfssmmsfffmfmmm", "output": "NO" }, { "input": "2 1\nff", "output": "NO" }, { "input": "2 1\nhw", "output": "YES" }, { "input": "2 2\nss", "output": "YES" }, { "input": "1 1\nl", "output": "YES" }, { "input": "100 50\nfffffttttttjjjuuuvvvvvdddxxxxwwwwgggbsssncccczzyyyyyhhhhhkrreeeeeeaaaaaiiillllllllooooqqqqqqmmpppppp", "output": "YES" }, { "input": "100 50\nbbbbbbbbgggggggggggaaaaaaaahhhhhhhhhhpppppppppsssssssrrrrrrrrllzzzzzzzeeeeeeekkkkkkkwwwwwwwwjjjjjjjj", "output": "YES" }, { "input": "100 50\nwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxzzzzzzzzzzzzzzzzzzbbbbbbbbbbbbbbbbbbbbjjjjjjjjjjjjjjjjjjjjjjjj", "output": "YES" }, { "input": "100 80\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm", "output": "YES" }, { "input": "100 10\nbbttthhhhiiiiiiijjjjjvvvvpppssssseeeeeeewwwwgggkkkkkkkkmmmddddduuuzzzzllllnnnnnxxyyyffffccraaaaooooq", "output": "YES" }, { "input": "100 20\nssssssssssbbbbbbbhhhhhhhyyyyyyyzzzzzzzzzzzzcccccxxxxxxxxxxddddmmmmmmmeeeeeeejjjjjjjjjwwwwwwwtttttttt", "output": "YES" }, { "input": "1 2\na", "output": "YES" }, { "input": "3 1\nabb", "output": "NO" }, { "input": "2 1\naa", "output": "NO" }, { "input": "2 1\nab", "output": "YES" }, { "input": "6 2\naaaaaa", "output": "NO" }, { "input": "8 4\naaaaaaaa", "output": "NO" }, { "input": "4 2\naaaa", "output": "NO" }, { "input": "4 3\naaaa", "output": "NO" }, { "input": "1 3\na", "output": "YES" }, { "input": "4 3\nzzzz", "output": "NO" }, { "input": "4 1\naaaa", "output": "NO" }, { "input": "3 4\nabc", "output": "YES" }, { "input": "2 5\nab", "output": "YES" }, { "input": "2 4\nab", "output": "YES" }, { "input": "1 10\na", "output": "YES" }, { "input": "5 2\nzzzzz", "output": "NO" }, { "input": "53 26\naaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "NO" }, { "input": "4 1\nabab", "output": "NO" }, { "input": "4 1\nabcb", "output": "NO" }, { "input": "4 2\nabbb", "output": "NO" }, { "input": "5 2\nabccc", "output": "NO" }, { "input": "2 3\nab", "output": "YES" }, { "input": "4 3\nbbbs", "output": "YES" }, { "input": "10 2\nazzzzzzzzz", "output": "NO" }, { "input": "1 2\nb", "output": "YES" }, { "input": "1 3\nb", "output": "YES" }, { "input": "4 5\nabcd", "output": "YES" }, { "input": "4 6\naabb", "output": "YES" }, { "input": "5 2\naaaab", "output": "NO" }, { "input": "3 5\naaa", "output": "YES" }, { "input": "5 3\nazzzz", "output": "NO" }, { "input": "4 100\naabb", "output": "YES" }, { "input": "3 10\naaa", "output": "YES" }, { "input": "3 4\naaa", "output": "YES" }, { "input": "12 5\naaaaabbbbbbb", "output": "NO" }, { "input": "5 2\naabbb", "output": "NO" }, { "input": "10 5\nzzzzzzzzzz", "output": "NO" }, { "input": "2 4\naa", "output": "YES" }, { "input": "1 5\na", "output": "YES" }, { "input": "10 5\naaaaaaaaaa", "output": "NO" }, { "input": "6 3\naaaaaa", "output": "NO" }, { "input": "7 1\nabcdeee", "output": "NO" }, { "input": "18 3\naaaaaabbbbbbcccccc", "output": "NO" }, { "input": "8 2\naabbccdd", "output": "YES" }, { "input": "4 2\nzzzz", "output": "NO" }, { "input": "4 2\nabaa", "output": "NO" }, { "input": "3 2\naaa", "output": "NO" }, { "input": "3 1\nzzz", "output": "NO" }, { "input": "5 4\nzzzzz", "output": "NO" }, { "input": "6 2\naabbbc", "output": "NO" }, { "input": "3 6\naaa", "output": "YES" }, { "input": "2 1\nzz", "output": "NO" }, { "input": "10 3\naaaeeeeeee", "output": "NO" }, { "input": "4 5\naabb", "output": "YES" }, { "input": "3 1\naaa", "output": "NO" }, { "input": "5 2\naazzz", "output": "NO" }, { "input": "6 2\nabbbbc", "output": "NO" }, { "input": "4 2\nxxxx", "output": "NO" }, { "input": "6 3\nzzzzzz", "output": "NO" }, { "input": "3 2\nabb", "output": "YES" }, { "input": "3 2\nzzz", "output": "NO" }, { "input": "6 5\nzzzzzz", "output": "NO" }, { "input": "6 3\nbcaaaa", "output": "NO" }, { "input": "100 100\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "YES" }, { "input": "3 6\nabc", "output": "YES" } ]

1,503,572,506

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

a, b = map(int, input().split()) s = [0]*26 for i in input(): s[ord(i)-97] += 1 print 'NO' if max(s) > b else 'YES'

Title: Generous Kefa Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons. Output Specification: Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary. Demo Input: ['4 2\naabb\n', '6 3\naacaab\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

```python a, b = map(int, input().split()) s = [0]*26 for i in input(): s[ord(i)-97] += 1 print 'NO' if max(s) > b else 'YES' ```

-1

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" } ]

1,615,836,528

2,147,483,647

Python 3

OK

TESTS

40

124

0

#https://codeforces.com/problemset/problem/41/A a=input() b=input() b=list(b) b.reverse() if a==''.join(b): 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 #https://codeforces.com/problemset/problem/41/A a=input() b=input() b=list(b) b.reverse() if a==''.join(b): print("YES") else: print("NO") ```

3.969

141

A

Amusing Joke

PROGRAMMING

800

[ "implementation", "sortings", "strings" ]

null

So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.

The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.

Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.

[ "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n", "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n", "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n" ]

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

In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".

500

[ { "input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS", "output": "YES" }, { "input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI", "output": "NO" }, { "input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER", "output": "NO" }, { "input": "B\nA\nAB", "output": "YES" }, { "input": "ONDOL\nJNPB\nONLNJBODP", "output": "YES" }, { "input": "Y\nW\nYW", "output": "YES" }, { "input": "OI\nM\nIMO", "output": "YES" }, { "input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF", "output": "YES" }, { "input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB", "output": "NO" }, { "input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH", "output": "NO" }, { "input": "IQ\nOQ\nQOQIGGKFNHJSGCGM", "output": "NO" }, { "input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR", "output": "YES" }, { "input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY", "output": "YES" }, { "input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX", "output": "YES" }, { "input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW", "output": "YES" }, { "input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU", "output": "YES" }, { "input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK", "output": "YES" }, { "input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE", "output": "NO" }, { "input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ", "output": "NO" }, { "input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI", "output": "NO" }, { "input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT", "output": "NO" }, { "input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY", "output": "NO" }, { "input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT", "output": "NO" }, { "input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY", "output": "NO" }, { "input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI", "output": "NO" }, { "input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE", "output": "NO" }, { "input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH", "output": "NO" }, { "input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU", "output": "NO" }, { "input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG", "output": "NO" }, { "input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM", "output": "NO" }, { "input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW", "output": "NO" }, { "input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ", "output": "YES" }, { "input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS", "output": "YES" }, { "input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH", "output": "YES" }, { "input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB", "output": "YES" }, { "input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP", "output": "YES" }, { "input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR", "output": "YES" }, { "input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT", "output": "YES" }, { "input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH", "output": "YES" }, { "input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG", "output": "YES" }, { "input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF", "output": "YES" }, { "input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA", "output": "NO" }, { "input": "IXFDY\nJRMOU\nDF", "output": "NO" }, { "input": "JPSPZ\nUGCUB\nJMZZZZZZZZ", "output": "NO" }, { "input": "AC\nA\nBBA", "output": "NO" }, { "input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB", "output": "NO" }, { "input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW", "output": "NO" }, { "input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE", "output": "NO" }, { "input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH", "output": "NO" }, { "input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ", "output": "NO" }, { "input": "GIO\nPRL\nPRL", "output": "NO" }, { "input": "A\nB\nABC", "output": "NO" }, { "input": "KKK\nKKK\nZZZZZ", "output": "NO" }, { "input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA", "output": "NO" } ]

1,688,805,461

2,147,483,647

Python 3

OK

TESTS

54

92

0

a1=input() a=[] for i in range(len(a1)): a.append(a1[i]) b1=input() b=[] for i in range(len(b1)): b.append(b1[i]) c1=input() c=[] for i in range(len(c1)): c.append(c1[i]) ok='YES' for i in range(len(a)): if a[i] not in c: ok='NO' else: c.pop(c.index(a[i])) for i in range(len(b)): if b[i] not in c: ok='NO' else: c.pop(c.index(b[i])) if len(c)!=0: ok='NO' print(ok) # Sat Jul 08 2023 11:37:39 GMT+0300 (Moscow Standard Time)

Title: Amusing Joke Time Limit: None seconds Memory Limit: None megabytes Problem Description: So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door. The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters. Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. Input Specification: The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. Output Specification: Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. Demo Input: ['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left. In the second sample letter "P" is missing from the pile and there's an extra letter "L". In the third sample there's an extra letter "L".

```python a1=input() a=[] for i in range(len(a1)): a.append(a1[i]) b1=input() b=[] for i in range(len(b1)): b.append(b1[i]) c1=input() c=[] for i in range(len(c1)): c.append(c1[i]) ok='YES' for i in range(len(a)): if a[i] not in c: ok='NO' else: c.pop(c.index(a[i])) for i in range(len(b)): if b[i] not in c: ok='NO' else: c.pop(c.index(b[i])) if len(c)!=0: ok='NO' print(ok) # Sat Jul 08 2023 11:37:39 GMT+0300 (Moscow Standard Time) ```

3

520

A

Pangram

PROGRAMMING

800

[ "implementation", "strings" ]

null

A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices. You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string. The second line contains the string. The string consists only of uppercase and lowercase Latin letters.

Output "YES", if the string is a pangram and "NO" otherwise.

[ "12\ntoosmallword\n", "35\nTheQuickBrownFoxJumpsOverTheLazyDog\n" ]

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

none

500

[ { "input": "12\ntoosmallword", "output": "NO" }, { "input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog", "output": "YES" }, { "input": "1\na", "output": "NO" }, { "input": "26\nqwertyuiopasdfghjklzxcvbnm", "output": "YES" }, { "input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "YES" }, { "input": "48\nthereisasyetinsufficientdataforameaningfulanswer", "output": "NO" }, { "input": "30\nToBeOrNotToBeThatIsTheQuestion", "output": "NO" }, { "input": "30\njackdawslovemybigsphinxofquarz", "output": "NO" }, { "input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY", "output": "YES" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "NO" }, { "input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx", "output": "YES" }, { "input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ", "output": "YES" }, { "input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ", "output": "YES" }, { "input": "25\nnxYTzLFwzNolAumjgcAboyxAj", "output": "NO" }, { "input": "26\npRWdodGdxUESvcScPGbUoooZsC", "output": "NO" }, { "input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj", "output": "NO" }, { "input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa", "output": "YES" }, { "input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK", "output": "NO" }, { "input": "26\nvCUFRKElZOnjmXGylWQaHDiPst", "output": "NO" }, { "input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY", "output": "NO" }, { "input": "26\npGiFluRteQwkaVoPszJyNBChxM", "output": "NO" }, { "input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY", "output": "NO" }, { "input": "26\nLndjgvAEuICHKxPwqYztosrmBN", "output": "NO" }, { "input": "26\nMdaXJrCipnOZLykfqHWEStevbU", "output": "NO" }, { "input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba", "output": "NO" }, { "input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo", "output": "NO" }, { "input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa", "output": "NO" }, { "input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO", "output": "NO" }, { "input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe", "output": "NO" }, { "input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG", "output": "NO" }, { "input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF", "output": "NO" }, { "input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu", "output": "NO" }, { "input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf", "output": "NO" }, { "input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE", "output": "NO" }, { "input": "26\nDqspXZJTuONYieKgaHLMBwfVSC", "output": "NO" }, { "input": "26\necOyUkqNljFHRVXtIpWabGMLDz", "output": "NO" }, { "input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI", "output": "NO" }, { "input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA", "output": "NO" }, { "input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz", "output": "NO" }, { "input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ", "output": "NO" }, { "input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF", "output": "NO" }, { "input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB", "output": "NO" }, { "input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg", "output": "NO" }, { "input": "26\nmnbvcxxlkjhgfdsapoiuytrewq", "output": "NO" }, { "input": "26\naAbcdefghijklmnopqrstuvwxy", "output": "NO" }, { "input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl", "output": "NO" }, { "input": "25\nabcdefghijklmnopqrstuvwxy", "output": "NO" }, { "input": "26\nabcdefhijklmnopqrstVxyzABC", "output": "NO" }, { "input": "25\nqwertyuiopasdfghjklxcvbnm", "output": "NO" }, { "input": "34\nTheQuickBrownFoxJumpsOverTheLayDog", "output": "NO" }, { "input": "26\nabcdefghigklmnopqrstuvwxyz", "output": "NO" }, { "input": "26\nabcdefghijklmnopqrstuvwxyA", "output": "NO" }, { "input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM", "output": "NO" }, { "input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog", "output": "NO" }, { "input": "25\nbcdefghijklmnopqrstuvwxyz", "output": "NO" }, { "input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa", "output": "NO" }, { "input": "26\nabcdefghiklmnopqrstvxyzABC", "output": "NO" }, { "input": "26\nabcdefghijklmnopqrstuvwxzZ", "output": "NO" }, { "input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY", "output": "NO" } ]

1,696,479,881

2,147,483,647

Python 3

OK

TESTS

59

46

0

l=int(input()) s=input() d="abcdefghijklmnopqrstuvwxyz" for i in d: if i not in s.lower(): print("NO") break else: print("YES")

Title: Pangram Time Limit: None seconds Memory Limit: None megabytes Problem Description: A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices. You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string. The second line contains the string. The string consists only of uppercase and lowercase Latin letters. Output Specification: Output "YES", if the string is a pangram and "NO" otherwise. Demo Input: ['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n'] Demo Output: ['NO\n', 'YES\n'] Note: none

```python l=int(input()) s=input() d="abcdefghijklmnopqrstuvwxyz" for i in d: if i not in s.lower(): print("NO") break else: print("YES") ```

3

915

G

Coprime Arrays

PROGRAMMING

2,300

[ "math", "number theory" ]

null

Let's call an array *a* of size *n* coprime iff *gcd*(*a*1,<=*a*2,<=...,<=*a**n*)<==<=1, where *gcd* is the greatest common divisor of the arguments. You are given two numbers *n* and *k*. For each *i* (1<=≤<=*i*<=≤<=*k*) you have to determine the number of coprime arrays *a* of size *n* such that for every *j* (1<=≤<=*j*<=≤<=*n*) 1<=≤<=*a**j*<=≤<=*i*. Since the answers can be very large, you have to calculate them modulo 109<=+<=7.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=2·106) — the size of the desired arrays and the maximum upper bound on elements, respectively.

Since printing 2·106 numbers may take a lot of time, you have to output the answer in such a way: Let *b**i* be the number of coprime arrays with elements in range [1,<=*i*], taken modulo 109<=+<=7. You have to print , taken modulo 109<=+<=7. Here denotes bitwise xor operation (^ in C++ or Java, xor in Pascal).

[ "3 4\n", "2000000 8\n" ]

[ "82\n", "339310063\n" ]

Explanation of the example: Since the number of coprime arrays is large, we will list the arrays that are non-coprime, but contain only elements in range [1, *i*]: For *i* = 1, the only array is coprime. *b*1 = 1. For *i* = 2, array [2, 2, 2] is not coprime. *b*2 = 7. For *i* = 3, arrays [2, 2, 2] and [3, 3, 3] are not coprime. *b*3 = 25. For *i* = 4, arrays [2, 2, 2], [3, 3, 3], [2, 2, 4], [2, 4, 2], [2, 4, 4], [4, 2, 2], [4, 2, 4], [4, 4, 2] and [4, 4, 4] are not coprime. *b*4 = 55.

0

[ { "input": "3 4", "output": "82" }, { "input": "2000000 8", "output": "339310063" }, { "input": "1000 1000", "output": "293255159" }, { "input": "400000 400000", "output": "641589365" }, { "input": "1000 2000", "output": "946090030" }, { "input": "400000 800000", "output": "700177418" }, { "input": "1000 3000", "output": "599681537" }, { "input": "400000 1200000", "output": "599314521" }, { "input": "1000 4000", "output": "369962559" }, { "input": "400000 1600000", "output": "383930351" }, { "input": "1000 5000", "output": "792761486" }, { "input": "400000 2000000", "output": "581254555" }, { "input": "2000 1000", "output": "895983599" }, { "input": "800000 400000", "output": "194918098" }, { "input": "2000 2000", "output": "993145929" }, { "input": "800000 800000", "output": "959396582" }, { "input": "2000 3000", "output": "801808235" }, { "input": "800000 1200000", "output": "456936609" }, { "input": "2000 4000", "output": "531261925" }, { "input": "800000 1600000", "output": "58127171" }, { "input": "2000 5000", "output": "460721768" }, { "input": "800000 2000000", "output": "607152577" }, { "input": "3000 1000", "output": "547188062" }, { "input": "1200000 400000", "output": "535771854" }, { "input": "3000 2000", "output": "612556768" }, { "input": "1200000 800000", "output": "925258449" }, { "input": "3000 3000", "output": "113732600" }, { "input": "1200000 1200000", "output": "468573037" }, { "input": "3000 4000", "output": "81501584" }, { "input": "1200000 1600000", "output": "549284678" }, { "input": "3000 5000", "output": "674745410" }, { "input": "1200000 2000000", "output": "340808789" }, { "input": "4000 1000", "output": "312642124" }, { "input": "1600000 400000", "output": "180485439" }, { "input": "4000 2000", "output": "467116350" }, { "input": "1600000 800000", "output": "990942172" }, { "input": "4000 3000", "output": "441359443" }, { "input": "1600000 1200000", "output": "186789429" }, { "input": "4000 4000", "output": "927028696" }, { "input": "1600000 1600000", "output": "373782836" }, { "input": "4000 5000", "output": "107116959" }, { "input": "1600000 2000000", "output": "160805957" }, { "input": "5000 1000", "output": "834732252" }, { "input": "2000000 400000", "output": "313670498" }, { "input": "5000 2000", "output": "273324715" }, { "input": "2000000 800000", "output": "204745061" }, { "input": "5000 3000", "output": "649058184" }, { "input": "2000000 1200000", "output": "749062788" }, { "input": "5000 4000", "output": "332100759" }, { "input": "2000000 1600000", "output": "267687597" }, { "input": "5000 5000", "output": "785334012" }, { "input": "2000000 2000000", "output": "219072617" }, { "input": "1 1", "output": "0" }, { "input": "1 2000000", "output": "986000" }, { "input": "2000000 1", "output": "0" } ]

1,638,042,191

2,147,483,647

PyPy 3-64

OK

TESTS

55

2,948

51,916,800

import sys input = sys.stdin.readline n, k = map(int, input().split()) mod = pow(10, 9) + 7 ans = 0 p2 = [pow(i, n, mod) for i in range(k + 1)] dp0 = [1] * (k + 1) dp = [0] * (k + 1) cnt = 0 for i in range(2, k + 1): dp0i = dp0[i] dp[i] += dp0i dp[i] %= mod c = 2 for j in range(2 * i, k + 1, i): dp0[j] -= dp0i dp0[j] %= mod dp[j] += (p2[c] - p2[c - 1]) % mod * dp0i % mod dp[j] %= mod c += 1 cnt += dp[i] cnt %= mod ans += ((p2[i] - cnt) % mod) ^ i ans %= mod print(ans)

Title: Coprime Arrays Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's call an array *a* of size *n* coprime iff *gcd*(*a*1,<=*a*2,<=...,<=*a**n*)<==<=1, where *gcd* is the greatest common divisor of the arguments. You are given two numbers *n* and *k*. For each *i* (1<=≤<=*i*<=≤<=*k*) you have to determine the number of coprime arrays *a* of size *n* such that for every *j* (1<=≤<=*j*<=≤<=*n*) 1<=≤<=*a**j*<=≤<=*i*. Since the answers can be very large, you have to calculate them modulo 109<=+<=7. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=2·106) — the size of the desired arrays and the maximum upper bound on elements, respectively. Output Specification: Since printing 2·106 numbers may take a lot of time, you have to output the answer in such a way: Let *b**i* be the number of coprime arrays with elements in range [1,<=*i*], taken modulo 109<=+<=7. You have to print , taken modulo 109<=+<=7. Here denotes bitwise xor operation (^ in C++ or Java, xor in Pascal). Demo Input: ['3 4\n', '2000000 8\n'] Demo Output: ['82\n', '339310063\n'] Note: Explanation of the example: Since the number of coprime arrays is large, we will list the arrays that are non-coprime, but contain only elements in range [1, *i*]: For *i* = 1, the only array is coprime. *b*1 = 1. For *i* = 2, array [2, 2, 2] is not coprime. *b*2 = 7. For *i* = 3, arrays [2, 2, 2] and [3, 3, 3] are not coprime. *b*3 = 25. For *i* = 4, arrays [2, 2, 2], [3, 3, 3], [2, 2, 4], [2, 4, 2], [2, 4, 4], [4, 2, 2], [4, 2, 4], [4, 4, 2] and [4, 4, 4] are not coprime. *b*4 = 55.

```python import sys input = sys.stdin.readline n, k = map(int, input().split()) mod = pow(10, 9) + 7 ans = 0 p2 = [pow(i, n, mod) for i in range(k + 1)] dp0 = [1] * (k + 1) dp = [0] * (k + 1) cnt = 0 for i in range(2, k + 1): dp0i = dp0[i] dp[i] += dp0i dp[i] %= mod c = 2 for j in range(2 * i, k + 1, i): dp0[j] -= dp0i dp0[j] %= mod dp[j] += (p2[c] - p2[c - 1]) % mod * dp0i % mod dp[j] %= mod c += 1 cnt += dp[i] cnt %= mod ans += ((p2[i] - cnt) % mod) ^ i ans %= mod print(ans) ```

3

922

A

Cloning Toys

PROGRAMMING

1,300

[ "implementation" ]

null

Imp likes his plush toy a lot. Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies. Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.

The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one).

Print "Yes", if the desired configuration is possible, and "No" otherwise. You can print each letter in arbitrary case (upper or lower).

[ "6 3\n", "4 2\n", "1000 1001\n" ]

[ "Yes\n", "No\n", "Yes\n" ]

In the first example, Imp has to apply the machine twice to original toys and then twice to copies.

500

[ { "input": "6 3", "output": "Yes" }, { "input": "4 2", "output": "No" }, { "input": "1000 1001", "output": "Yes" }, { "input": "1000000000 999999999", "output": "Yes" }, { "input": "81452244 81452247", "output": "No" }, { "input": "188032448 86524683", "output": "Yes" }, { "input": "365289629 223844571", "output": "No" }, { "input": "247579518 361164458", "output": "No" }, { "input": "424836699 793451637", "output": "No" }, { "input": "602093880 930771525", "output": "No" }, { "input": "779351061 773124120", "output": "Yes" }, { "input": "661640950 836815080", "output": "No" }, { "input": "543930839 974134967", "output": "No" }, { "input": "16155311 406422145", "output": "No" }, { "input": "81601559 445618240", "output": "No" }, { "input": "963891449 582938127", "output": "No" }, { "input": "141148629 351661795", "output": "No" }, { "input": "318405810 783948974", "output": "No" }, { "input": "495662991 921268861", "output": "No" }, { "input": "1 0", "output": "No" }, { "input": "0 1", "output": "Yes" }, { "input": "0 0", "output": "No" }, { "input": "453462237 167520068", "output": "Yes" }, { "input": "630719418 9872663", "output": "Yes" }, { "input": "807976599 442159843", "output": "No" }, { "input": "690266488 579479730", "output": "No" }, { "input": "771581370 589752968", "output": "No" }, { "input": "948838551 727072855", "output": "No" }, { "input": "831128440 790763814", "output": "No" }, { "input": "303352912 928083702", "output": "No" }, { "input": "185642801 65403588", "output": "Yes" }, { "input": "67932690 202723476", "output": "No" }, { "input": "540157163 340043363", "output": "No" }, { "input": "422447052 772330542", "output": "No" }, { "input": "599704233 541054210", "output": "Yes" }, { "input": "481994122 678374097", "output": "No" }, { "input": "48564714 743566477", "output": "No" }, { "input": "225821895 880886365", "output": "No" }, { "input": "403079076 313173543", "output": "Yes" }, { "input": "1000000000 1000000000", "output": "No" }, { "input": "1 1", "output": "No" }, { "input": "1 2", "output": "Yes" }, { "input": "2 1", "output": "No" }, { "input": "2 2", "output": "No" }, { "input": "2 0", "output": "No" }, { "input": "0 2", "output": "No" }, { "input": "1000000000 1", "output": "No" }, { "input": "777777 0", "output": "No" }, { "input": "10 1", "output": "No" }, { "input": "7 0", "output": "No" }, { "input": "3 0", "output": "No" }, { "input": "3 2", "output": "Yes" } ]

1,651,308,172

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

42

31

0

x,y=map(int,input().split()) z=y-1 if(x==1 and y==0): print("No") elif(x>=z): if((x-z)%2==0): print("Yes") else: print("No") else: print("No")

Title: Cloning Toys Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imp likes his plush toy a lot. Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies. Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies. Input Specification: The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one). Output Specification: Print "Yes", if the desired configuration is possible, and "No" otherwise. You can print each letter in arbitrary case (upper or lower). Demo Input: ['6 3\n', '4 2\n', '1000 1001\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n'] Note: In the first example, Imp has to apply the machine twice to original toys and then twice to copies.

```python x,y=map(int,input().split()) z=y-1 if(x==1 and y==0): print("No") elif(x>=z): if((x-z)%2==0): print("Yes") else: print("No") else: print("No") ```

0

47

A

Triangular numbers

PROGRAMMING

800

[ "brute force", "math" ]

A. Triangular numbers

2

256

A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number). Your task is to find out if a given integer is a triangular number.

The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.

If the given integer is a triangular number output YES, otherwise output NO.

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

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

none

500

[ { "input": "1", "output": "YES" }, { "input": "2", "output": "NO" }, { "input": "3", "output": "YES" }, { "input": "4", "output": "NO" }, { "input": "5", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "7", "output": "NO" }, { "input": "8", "output": "NO" }, { "input": "12", "output": "NO" }, { "input": "10", "output": "YES" }, { "input": "11", "output": "NO" }, { "input": "9", "output": "NO" }, { "input": "14", "output": "NO" }, { "input": "15", "output": "YES" }, { "input": "16", "output": "NO" }, { "input": "20", "output": "NO" }, { "input": "21", "output": "YES" }, { "input": "22", "output": "NO" }, { "input": "121", "output": "NO" }, { "input": "135", "output": "NO" }, { "input": "136", "output": "YES" }, { "input": "137", "output": "NO" }, { "input": "152", "output": "NO" }, { "input": "153", "output": "YES" }, { "input": "154", "output": "NO" }, { "input": "171", "output": "YES" }, { "input": "189", "output": "NO" }, { "input": "190", "output": "YES" }, { "input": "191", "output": "NO" }, { "input": "210", "output": "YES" }, { "input": "211", "output": "NO" }, { "input": "231", "output": "YES" }, { "input": "232", "output": "NO" }, { "input": "252", "output": "NO" }, { "input": "253", "output": "YES" }, { "input": "254", "output": "NO" }, { "input": "275", "output": "NO" }, { "input": "276", "output": "YES" }, { "input": "277", "output": "NO" }, { "input": "299", "output": "NO" }, { "input": "300", "output": "YES" }, { "input": "301", "output": "NO" }, { "input": "324", "output": "NO" }, { "input": "325", "output": "YES" }, { "input": "326", "output": "NO" }, { "input": "350", "output": "NO" }, { "input": "351", "output": "YES" }, { "input": "352", "output": "NO" }, { "input": "377", "output": "NO" }, { "input": "378", "output": "YES" }, { "input": "406", "output": "YES" }, { "input": "434", "output": "NO" }, { "input": "435", "output": "YES" }, { "input": "464", "output": "NO" }, { "input": "465", "output": "YES" }, { "input": "495", "output": "NO" }, { "input": "496", "output": "YES" }, { "input": "497", "output": "NO" }, { "input": "331", "output": "NO" }, { "input": "413", "output": "NO" }, { "input": "177", "output": "NO" }, { "input": "122", "output": "NO" }, { "input": "478", "output": "NO" }, { "input": "188", "output": "NO" }, { "input": "21", "output": "YES" }, { "input": "474", "output": "NO" }, { "input": "21", "output": "YES" }, { "input": "48", "output": "NO" }, { "input": "498", "output": "NO" }, { "input": "499", "output": "NO" }, { "input": "500", "output": "NO" } ]

1,627,491,163

2,147,483,647

Python 3

OK

TESTS

71

124

6,758,400

n = int(input()) i = 1 while i*(i+1)/2 < n: i += 1 # at this point, we have i*(i+1)/2 >= n if i*(i+1)/2 == n: print("YES") else: print("NO")

Title: Triangular numbers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number). Your task is to find out if a given integer is a triangular number. Input Specification: The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer. Output Specification: If the given integer is a triangular number output YES, otherwise output NO. Demo Input: ['1\n', '2\n', '3\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: none

```python n = int(input()) i = 1 while i*(i+1)/2 < n: i += 1 # at this point, we have i*(i+1)/2 >= n if i*(i+1)/2 == n: print("YES") else: print("NO") ```

3.956411

262

A

Roma and Lucky Numbers

PROGRAMMING

800

[ "implementation" ]

null

Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers. Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.

The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has. The numbers in the lines are separated by single spaces.

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

[ "3 4\n1 2 4\n", "3 2\n447 44 77\n" ]

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

In the first sample all numbers contain at most four lucky digits, so the answer is 3. In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.

500

[ { "input": "3 4\n1 2 4", "output": "3" }, { "input": "3 2\n447 44 77", "output": "2" }, { "input": "2 2\n507978501 180480073", "output": "2" }, { "input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089", "output": "9" }, { "input": "6 100\n170427799 37215529 675016434 168544291 683447134 950090227", "output": "6" }, { "input": "4 2\n194041605 706221269 69909135 257655784", "output": "3" }, { "input": "4 2\n9581849 67346651 530497 272158241", "output": "4" }, { "input": "3 47\n378261451 163985731 230342101", "output": "3" }, { "input": "2 3\n247776868 480572137", "output": "1" }, { "input": "7 77\n366496749 549646417 278840199 119255907 33557677 379268590 150378796", "output": "7" }, { "input": "40 31\n32230963 709031779 144328646 513494529 36547831 416998222 84161665 318773941 170724397 553666286 368402971 48581613 31452501 368026285 47903381 939151438 204145360 189920160 288159400 133145006 314295423 450219949 160203213 358403181 478734385 29331901 31051111 110710191 567314089 139695685 111511396 87708701 317333277 103301481 110400517 634446253 481551313 39202255 105948 738066085", "output": "40" }, { "input": "1 8\n55521105", "output": "1" }, { "input": "49 3\n34644511 150953622 136135827 144208961 359490601 86708232 719413689 188605873 64330753 488776302 104482891 63360106 437791390 46521319 70778345 339141601 136198441 292941209 299339510 582531183 555958105 437904637 74219097 439816011 236010407 122674666 438442529 186501223 63932449 407678041 596993853 92223251 849265278 480265849 30983497 330283357 186901672 20271344 794252593 123774176 27851201 52717531 479907210 196833889 149331196 82147847 255966471 278600081 899317843", "output": "44" }, { "input": "26 2\n330381357 185218042 850474297 483015466 296129476 1205865 538807493 103205601 160403321 694220263 416255901 7245756 507755361 88187633 91426751 1917161 58276681 59540376 576539745 595950717 390256887 105690055 607818885 28976353 488947089 50643601", "output": "22" }, { "input": "38 1\n194481717 126247087 815196361 106258801 381703249 283859137 15290101 40086151 213688513 577996947 513899717 371428417 107799271 11136651 5615081 323386401 381128815 34217126 17709913 520702093 201694245 570931849 169037023 417019726 282437316 7417126 271667553 11375851 185087449 410130883 383045677 5764771 905017051 328584026 215330671 299553233 15838255 234532105", "output": "20" }, { "input": "44 9\n683216389 250581469 130029957 467020047 188395565 206237982 63257361 68314981 732878407 563579660 199133851 53045209 665723851 16273169 10806790 556633156 350593410 474645249 478790761 708234243 71841230 18090541 19836685 146373571 17947452 534010506 46933264 377035021 311636557 75193963 54321761 12759959 71120181 548816939 23608621 31876417 107672995 72575155 369667956 20574379 210596751 532163173 75726739 853719629", "output": "44" }, { "input": "8 6\n204157376 10514197 65483881 347219841 263304577 296402721 11739011 229776191", "output": "8" }, { "input": "38 29\n333702889 680931737 61137217 203030505 68728281 11414209 642645708 590904616 3042901 607198177 189041074 700764043 813035201 198341461 126403544 401436841 420826465 45046581 20249976 46978855 46397957 706610773 24701041 57954481 51603266 593109701 385569073 178982291 582152863 287317968 1474090 34825141 432421977 130257781 151516903 540852403 548392 117246529", "output": "38" }, { "input": "19 3\n562569697 549131571 50676718 84501863 74567295 702372009 365895280 451459937 40378543 167666701 158635641 53639293 442332661 825055617 100109161 326616021 862332843 533271196 4791547", "output": "18" }, { "input": "1 1\n44", "output": "0" }, { "input": "1 1\n4", "output": "1" }, { "input": "10 3\n444 447 774 777 7777 4447 4 7 7 4", "output": "8" } ]

1,616,610,305

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

124

0

n,k=map(int,input().split()) l=[] c=0 l.extend(map(int,input().split())) for i in l: if not i//10**k : c+=1 print(c)

Title: Roma and Lucky Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers. Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem. Input Specification: The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has. The numbers in the lines are separated by single spaces. Output Specification: In a single line print a single integer — the answer to the problem. Demo Input: ['3 4\n1 2 4\n', '3 2\n447 44 77\n'] Demo Output: ['3\n', '2\n'] Note: In the first sample all numbers contain at most four lucky digits, so the answer is 3. In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.

```python n,k=map(int,input().split()) l=[] c=0 l.extend(map(int,input().split())) for i in l: if not i//10**k : c+=1 print(c) ```

0

108

A

Palindromic Times

PROGRAMMING

1,000

[ "implementation", "strings" ]

A. Palindromic Times

2

256

Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues. On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome. In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment. However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.

The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.

Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.

[ "12:21\n", "23:59\n" ]

[ "13:31\n", "00:00\n" ]

none

500

[ { "input": "12:21", "output": "13:31" }, { "input": "23:59", "output": "00:00" }, { "input": "15:51", "output": "20:02" }, { "input": "10:44", "output": "11:11" }, { "input": "04:02", "output": "04:40" }, { "input": "02:11", "output": "02:20" }, { "input": "12:15", "output": "12:21" }, { "input": "07:07", "output": "10:01" }, { "input": "00:17", "output": "01:10" }, { "input": "04:55", "output": "05:50" }, { "input": "02:17", "output": "02:20" }, { "input": "07:56", "output": "10:01" }, { "input": "00:29", "output": "01:10" }, { "input": "23:31", "output": "23:32" }, { "input": "19:30", "output": "20:02" }, { "input": "12:14", "output": "12:21" }, { "input": "17:32", "output": "20:02" }, { "input": "03:44", "output": "04:40" }, { "input": "07:15", "output": "10:01" }, { "input": "18:42", "output": "20:02" }, { "input": "08:56", "output": "10:01" }, { "input": "04:50", "output": "05:50" }, { "input": "14:32", "output": "14:41" }, { "input": "23:23", "output": "23:32" }, { "input": "08:35", "output": "10:01" }, { "input": "03:32", "output": "04:40" }, { "input": "07:59", "output": "10:01" }, { "input": "14:12", "output": "14:41" }, { "input": "23:52", "output": "00:00" }, { "input": "16:36", "output": "20:02" }, { "input": "17:50", "output": "20:02" }, { "input": "06:59", "output": "10:01" }, { "input": "16:50", "output": "20:02" }, { "input": "00:00", "output": "01:10" }, { "input": "23:59", "output": "00:00" }, { "input": "23:33", "output": "00:00" } ]

1,611,335,435

2,147,483,647

Python 3

OK

TESTS

36

154

307,200

s=input() if s[0]=="0": h=s[1] else: h=s[:2] if s[3]=="0": m=s[3] else: m=s[3:] def pallin(x,y): if len(x)==1: x="0"+x if len(y)==1: y="0"+y if x[0]==y[1] and x[1]==y[0]: return True else: return False while(1): if int(m)+1==60: h=str((int(h)+1)%24) m=str(0) else: m=str(int(m)+1) if pallin(h,m): if len(h)==1: h="0"+h if len(m)==1: m="0"+m print(h+":"+m) break

Title: Palindromic Times Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues. On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome. In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment. However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. Input Specification: The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. Output Specification: Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. Demo Input: ['12:21\n', '23:59\n'] Demo Output: ['13:31\n', '00:00\n'] Note: none

```python s=input() if s[0]=="0": h=s[1] else: h=s[:2] if s[3]=="0": m=s[3] else: m=s[3:] def pallin(x,y): if len(x)==1: x="0"+x if len(y)==1: y="0"+y if x[0]==y[1] and x[1]==y[0]: return True else: return False while(1): if int(m)+1==60: h=str((int(h)+1)%24) m=str(0) else: m=str(int(m)+1) if pallin(h,m): if len(h)==1: h="0"+h if len(m)==1: m="0"+m print(h+":"+m) break ```

3.960928

17

A

Noldbach problem

PROGRAMMING

1,000

[ "brute force", "math", "number theory" ]

A. Noldbach problem

2

64

Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. Two prime numbers are called neighboring if there are no other prime numbers between them. You are to help Nick, and find out if he is right or wrong.

The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).

Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.

[ "27 2\n", "45 7\n" ]

[ "YES", "NO" ]

In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.

0

[ { "input": "27 2", "output": "YES" }, { "input": "45 7", "output": "NO" }, { "input": "2 0", "output": "YES" }, { "input": "15 1", "output": "YES" }, { "input": "17 1", "output": "YES" }, { "input": "34 5", "output": "NO" }, { "input": "37 4", "output": "YES" }, { "input": "43 5", "output": "YES" }, { "input": "47 7", "output": "NO" }, { "input": "50 5", "output": "YES" }, { "input": "57 6", "output": "YES" }, { "input": "60 8", "output": "NO" }, { "input": "62 7", "output": "YES" }, { "input": "76 9", "output": "NO" }, { "input": "69 7", "output": "YES" }, { "input": "113 10", "output": "YES" }, { "input": "141 11", "output": "YES" }, { "input": "207 16", "output": "NO" }, { "input": "231 18", "output": "NO" }, { "input": "296 19", "output": "YES" }, { "input": "332 20", "output": "YES" }, { "input": "378 24", "output": "NO" }, { "input": "428 23", "output": "YES" }, { "input": "497 27", "output": "NO" }, { "input": "640 32", "output": "YES" }, { "input": "798 35", "output": "YES" }, { "input": "802 35", "output": "YES" }, { "input": "864 40", "output": "NO" }, { "input": "953 44", "output": "NO" }, { "input": "995 44", "output": "YES" }, { "input": "1000 44", "output": "YES" }, { "input": "1000 0", "output": "YES" }, { "input": "1000 1000", "output": "NO" }, { "input": "2 1000", "output": "NO" }, { "input": "2 0", "output": "YES" } ]

1,693,883,481

2,147,483,647

PyPy 3-64

OK

TESTS

35

124

1,331,200

n,k=map(int,input().split()) primes=[1]*(n+1) primes[0]=primes[1]=0 p=2 while(p*p<=n): if primes[p]: for i in range(p*p,n+1,p): primes[i]=0 p+=1 prime=[] for i in range(n+1): if primes[i]: prime.append(i) count=set() for i in range(len(prime)-1): if prime[i]+prime[i+1]+1<=n and primes[prime[i]+prime[i+1]+1]==1: count.add(prime[i]+prime[i+1]+1) if len(count)>=k: print("YES") else: print("NO")

Title: Noldbach problem Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. Two prime numbers are called neighboring if there are no other prime numbers between them. You are to help Nick, and find out if he is right or wrong. Input Specification: The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). Output Specification: Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. Demo Input: ['27 2\n', '45 7\n'] Demo Output: ['YES', 'NO'] Note: In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.

```python n,k=map(int,input().split()) primes=[1]*(n+1) primes[0]=primes[1]=0 p=2 while(p*p<=n): if primes[p]: for i in range(p*p,n+1,p): primes[i]=0 p+=1 prime=[] for i in range(n+1): if primes[i]: prime.append(i) count=set() for i in range(len(prime)-1): if prime[i]+prime[i+1]+1<=n and primes[prime[i]+prime[i+1]+1]==1: count.add(prime[i]+prime[i+1]+1) if len(count)>=k: print("YES") else: print("NO") ```

3.959082

932

B

Recursive Queries

PROGRAMMING

1,300

[ "binary search", "data structures", "dfs and similar" ]

null

Let us define two functions *f* and *g* on positive integer numbers. You need to process *Q* queries. In each query, you will be given three integers *l*, *r* and *k*. You need to print the number of integers *x* between *l* and *r* inclusive, such that *g*(*x*)<==<=*k*.

The first line of the input contains an integer *Q* (1<=≤<=*Q*<=≤<=2<=×<=105) representing the number of queries. *Q* lines follow, each of which contains 3 integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=106,<=1<=≤<=*k*<=≤<=9).

For each query, print a single line containing the answer for that query.

[ "4\n22 73 9\n45 64 6\n47 55 7\n2 62 4\n", "4\n82 94 6\n56 67 4\n28 59 9\n39 74 4\n" ]

[ "1\n4\n0\n8\n", "3\n1\n1\n5\n" ]

In the first example: - *g*(33) = 9 as *g*(33) = *g*(3 × 3) = *g*(9) = 9 - *g*(47) = *g*(48) = *g*(60) = *g*(61) = 6 - There are no such integers between 47 and 55. - *g*(4) = *g*(14) = *g*(22) = *g*(27) = *g*(39) = *g*(40) = *g*(41) = *g*(58) = 4

1,000

[ { "input": "4\n22 73 9\n45 64 6\n47 55 7\n2 62 4", "output": "1\n4\n0\n8" }, { "input": "4\n82 94 6\n56 67 4\n28 59 9\n39 74 4", "output": "3\n1\n1\n5" } ]

1,685,159,550

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

1,372

186,572,800

# from math import gcd # from queue import Queue # from collections import deque # from collections import Counter q=[[0 for a in range(10)] for b in range(10**6+1)] # print(q) def prod(w): ans=1 l=list(str(w)) for i in range(len(l)): ans*=int(l[i]) return(ans) def rec(n): if n<10: return(n) else: return(rec(prod(n))) for i in range(1,10**6+1): q[i][rec(i)]=q[i-1][rec(i)]+1 qu=int(input()) temp=0 for _ in range(qu): l,r,k=map(int,input().split()) temp=q[r][k]-q[l-1][k] print(temp)

Title: Recursive Queries Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let us define two functions *f* and *g* on positive integer numbers. You need to process *Q* queries. In each query, you will be given three integers *l*, *r* and *k*. You need to print the number of integers *x* between *l* and *r* inclusive, such that *g*(*x*)<==<=*k*. Input Specification: The first line of the input contains an integer *Q* (1<=≤<=*Q*<=≤<=2<=×<=105) representing the number of queries. *Q* lines follow, each of which contains 3 integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=106,<=1<=≤<=*k*<=≤<=9). Output Specification: For each query, print a single line containing the answer for that query. Demo Input: ['4\n22 73 9\n45 64 6\n47 55 7\n2 62 4\n', '4\n82 94 6\n56 67 4\n28 59 9\n39 74 4\n'] Demo Output: ['1\n4\n0\n8\n', '3\n1\n1\n5\n'] Note: In the first example: - *g*(33) = 9 as *g*(33) = *g*(3 × 3) = *g*(9) = 9 - *g*(47) = *g*(48) = *g*(60) = *g*(61) = 6 - There are no such integers between 47 and 55. - *g*(4) = *g*(14) = *g*(22) = *g*(27) = *g*(39) = *g*(40) = *g*(41) = *g*(58) = 4

```python # from math import gcd # from queue import Queue # from collections import deque # from collections import Counter q=[[0 for a in range(10)] for b in range(10**6+1)] # print(q) def prod(w): ans=1 l=list(str(w)) for i in range(len(l)): ans*=int(l[i]) return(ans) def rec(n): if n<10: return(n) else: return(rec(prod(n))) for i in range(1,10**6+1): q[i][rec(i)]=q[i-1][rec(i)]+1 qu=int(input()) temp=0 for _ in range(qu): l,r,k=map(int,input().split()) temp=q[r][k]-q[l-1][k] print(temp) ```

0

734

A

Anton and Danik

PROGRAMMING

800

[ "implementation", "strings" ]

null

Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.

If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes).

[ "6\nADAAAA\n", "7\nDDDAADA\n", "6\nDADADA\n" ]

[ "Anton\n", "Danik\n", "Friendship\n" ]

In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

500

[ { "input": "6\nADAAAA", "output": "Anton" }, { "input": "7\nDDDAADA", "output": "Danik" }, { "input": "6\nDADADA", "output": "Friendship" }, { "input": "10\nDDDDADDADD", "output": "Danik" }, { "input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA", "output": "Anton" }, { "input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD", "output": "Friendship" }, { "input": "1\nA", "output": "Anton" }, { "input": "1\nD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "4\nDADA", "output": "Friendship" }, { "input": "4\nDAAD", "output": "Friendship" }, { "input": "3\nADD", "output": "Danik" }, { "input": "3\nDAD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA", "output": "Anton" } ]

1,697,793,810

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

31

0

n = int(input()) w = str(input()) if w.count('A') > w.count('D'): print("Anton") else: print("Danik")

Title: Anton and Danik Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game. Output Specification: If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes). Demo Input: ['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n'] Demo Output: ['Anton\n', 'Danik\n', 'Friendship\n'] Note: In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

```python n = int(input()) w = str(input()) if w.count('A') > w.count('D'): print("Anton") else: print("Danik") ```

0

658

A

Bear and Reverse Radewoosh

PROGRAMMING

800

[ "implementation" ]

null

Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order. There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1. A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points. Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie. You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.

The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points. The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores. The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.

Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.

[ "3 2\n50 85 250\n10 15 25\n", "3 6\n50 85 250\n10 15 25\n", "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n" ]

[ "Limak\n", "Radewoosh\n", "Tie\n" ]

In the first sample, there are 3 problems. Limak solves them as follows: 1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points. So, Limak got 30 + 35 + 150 = 215 points. Radewoosh solves problem in the reversed order: 1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points. Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins. In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway. In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.

500

[ { "input": "3 2\n50 85 250\n10 15 25", "output": "Limak" }, { "input": "3 6\n50 85 250\n10 15 25", "output": "Radewoosh" }, { "input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76", "output": "Tie" }, { "input": "4 1\n3 5 6 9\n1 2 4 8", "output": "Limak" }, { "input": "4 1\n1 3 6 10\n1 5 7 8", "output": "Radewoosh" }, { "input": "4 1\n2 4 5 10\n2 3 9 10", "output": "Tie" }, { "input": "18 4\n68 97 121 132 146 277 312 395 407 431 458 461 595 634 751 855 871 994\n1 2 3 4 9 10 13 21 22 29 31 34 37 38 39 41 48 49", "output": "Radewoosh" }, { "input": "50 1\n5 14 18 73 137 187 195 197 212 226 235 251 262 278 287 304 310 322 342 379 393 420 442 444 448 472 483 485 508 515 517 523 559 585 618 627 636 646 666 682 703 707 780 853 937 951 959 989 991 992\n30 84 113 173 199 220 235 261 266 277 300 306 310 312 347 356 394 396 397 409 414 424 446 462 468 487 507 517 537 566 594 643 656 660 662 668 706 708 773 774 779 805 820 827 868 896 929 942 961 995", "output": "Tie" }, { "input": "4 1\n4 6 9 10\n2 3 4 5", "output": "Radewoosh" }, { "input": "4 1\n4 6 9 10\n3 4 5 7", "output": "Radewoosh" }, { "input": "4 1\n1 6 7 10\n2 7 8 10", "output": "Tie" }, { "input": "4 1\n4 5 7 9\n1 4 5 8", "output": "Limak" }, { "input": "50 1\n6 17 44 82 94 127 134 156 187 211 212 252 256 292 294 303 352 355 379 380 398 409 424 434 480 524 584 594 631 714 745 756 777 778 789 793 799 821 841 849 859 878 879 895 925 932 944 952 958 990\n15 16 40 42 45 71 99 100 117 120 174 181 186 204 221 268 289 332 376 394 403 409 411 444 471 487 499 539 541 551 567 589 619 623 639 669 689 722 735 776 794 822 830 840 847 907 917 927 936 988", "output": "Radewoosh" }, { "input": "50 10\n25 49 52 73 104 117 127 136 149 164 171 184 226 251 257 258 286 324 337 341 386 390 428 453 464 470 492 517 543 565 609 634 636 660 678 693 710 714 729 736 739 749 781 836 866 875 956 960 977 979\n2 4 7 10 11 22 24 26 27 28 31 35 37 38 42 44 45 46 52 53 55 56 57 59 60 61 64 66 67 68 69 71 75 76 77 78 79 81 83 85 86 87 89 90 92 93 94 98 99 100", "output": "Limak" }, { "input": "50 10\n11 15 25 71 77 83 95 108 143 150 182 183 198 203 213 223 279 280 346 348 350 355 375 376 412 413 415 432 470 545 553 562 589 595 607 633 635 637 688 719 747 767 771 799 842 883 905 924 942 944\n1 3 5 6 7 10 11 12 13 14 15 16 19 20 21 23 25 32 35 36 37 38 40 41 42 43 47 50 51 54 55 56 57 58 59 60 62 63 64 65 66 68 69 70 71 72 73 75 78 80", "output": "Radewoosh" }, { "input": "32 6\n25 77 141 148 157 159 192 196 198 244 245 255 332 392 414 457 466 524 575 603 629 700 738 782 838 841 845 847 870 945 984 985\n1 2 4 5 8 9 10 12 13 14 15 16 17 18 20 21 22 23 24 26 28 31 38 39 40 41 42 43 45 47 48 49", "output": "Radewoosh" }, { "input": "5 1\n256 275 469 671 842\n7 9 14 17 26", "output": "Limak" }, { "input": "2 1000\n1 2\n1 2", "output": "Tie" }, { "input": "3 1\n1 50 809\n2 8 800", "output": "Limak" }, { "input": "1 13\n866\n10", "output": "Tie" }, { "input": "15 1\n9 11 66 128 199 323 376 386 393 555 585 718 935 960 971\n3 11 14 19 20 21 24 26 32 38 40 42 44 47 50", "output": "Limak" }, { "input": "1 10\n546\n45", "output": "Tie" }, { "input": "50 20\n21 43 51 99 117 119 158 167 175 190 196 244 250 316 335 375 391 403 423 428 451 457 460 480 487 522 539 559 566 584 598 602 604 616 626 666 675 730 771 787 828 841 861 867 886 889 898 970 986 991\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", "output": "Limak" }, { "input": "50 21\n13 20 22 38 62 84 118 135 141 152 170 175 194 218 227 229 232 253 260 263 278 313 329 357 396 402 422 452 454 533 575 576 580 594 624 644 653 671 676 759 789 811 816 823 831 833 856 924 933 987\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", "output": "Tie" }, { "input": "1 36\n312\n42", "output": "Tie" }, { "input": "1 1000\n1\n1000", "output": "Tie" }, { "input": "1 1\n1000\n1", "output": "Tie" }, { "input": "50 35\n9 17 28 107 136 152 169 174 186 188 201 262 291 312 324 330 341 358 385 386 393 397 425 431 479 498 502 523 530 540 542 554 578 588 622 623 684 696 709 722 784 819 836 845 850 932 945 969 983 984\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", "output": "Tie" }, { "input": "50 20\n12 113 116 120 138 156 167 183 185 194 211 228 234 261 278 287 310 317 346 361 364 397 424 470 496 522 527 536 611 648 668 704 707 712 717 752 761 766 815 828 832 864 872 885 889 901 904 929 982 993\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", "output": "Limak" } ]

1,582,806,917

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

19

109

307,200

n,c=map(int,input().split()) l1=list(map(int,input().split())) l2=list(map(int,input().split())) s1,a=0,0 d,e=[],[] for i in range(n): s1+=l2[i] a=l1[i]-c*s1 if a>0: d.append(a) else: d.append(0) s1,b=0,0 for i in range(n-1,0,-1): s1+=l2[i] b=l1[i]-c*s1 if b>0: e.append(b) else: e.append(0) sr1=sum(d) sr2=sum(e) if sr1>sr2: print('Limak') elif sr2>sr1: print('Radewoosh') else: print('Tie')

Title: Bear and Reverse Radewoosh Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order. There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1. A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points. Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie. You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems. Input Specification: The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points. The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores. The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem. Output Specification: Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points. Demo Input: ['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n'] Demo Output: ['Limak\n', 'Radewoosh\n', 'Tie\n'] Note: In the first sample, there are 3 problems. Limak solves them as follows: 1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points. So, Limak got 30 + 35 + 150 = 215 points. Radewoosh solves problem in the reversed order: 1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points. Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins. In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway. In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.

```python n,c=map(int,input().split()) l1=list(map(int,input().split())) l2=list(map(int,input().split())) s1,a=0,0 d,e=[],[] for i in range(n): s1+=l2[i] a=l1[i]-c*s1 if a>0: d.append(a) else: d.append(0) s1,b=0,0 for i in range(n-1,0,-1): s1+=l2[i] b=l1[i]-c*s1 if b>0: e.append(b) else: e.append(0) sr1=sum(d) sr2=sum(e) if sr1>sr2: print('Limak') elif sr2>sr1: print('Radewoosh') else: print('Tie') ```

0

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" } ]

1,684,162,422

2,147,483,647

PyPy 3

OK

TESTS

47

93

0

l, k = map(int, input().split()) start = ord('a') res = "" for i in range(l): res += chr(start) start = (start+1) if start - ord('a') >= k: start = ord('a') print(res)

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 l, k = map(int, input().split()) start = ord('a') res = "" for i in range(l): res += chr(start) start = (start+1) if start - ord('a') >= k: start = ord('a') print(res) ```

3

59

A

Word

PROGRAMMING

800

[ "implementation", "strings" ]

A. Word

2

256

Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.

The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.

Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.

[ "HoUse\n", "ViP\n", "maTRIx\n" ]

[ "house\n", "VIP\n", "matrix\n" ]

none

500

[ { "input": "HoUse", "output": "house" }, { "input": "ViP", "output": "VIP" }, { "input": "maTRIx", "output": "matrix" }, { "input": "BNHWpnpawg", "output": "bnhwpnpawg" }, { "input": "VTYGP", "output": "VTYGP" }, { "input": "CHNenu", "output": "chnenu" }, { "input": "ERPZGrodyu", "output": "erpzgrodyu" }, { "input": "KSXBXWpebh", "output": "KSXBXWPEBH" }, { "input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv", "output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv" }, { "input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd", "output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd" }, { "input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv", "output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv" }, { "input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg", "output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg" }, { "input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak", "output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK" }, { "input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw", "output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW" }, { "input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB", "output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB" }, { "input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge", "output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge" }, { "input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw", "output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw" }, { "input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks", "output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks" }, { "input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz", "output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ" }, { "input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg", "output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG" }, { "input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO", "output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO" }, { "input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd", "output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD" }, { "input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas", "output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas" }, { "input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm", "output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM" }, { "input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn", "output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn" }, { "input": "t", "output": "t" }, { "input": "N", "output": "N" }, { "input": "kv", "output": "kv" }, { "input": "Ur", "output": "ur" }, { "input": "CN", "output": "CN" } ]

1,690,201,591

2,147,483,647

Python 3

OK

TESTS

30

92

0

s = input() k1 = 0 k2 = 0 l = len(s) for i in range(0,l): if s[i] >= 'a' and s[i] <= 'z': k1 += 1 else: k2 += 1 if k1 >= k2: s = s.lower() else: s = s.upper() print(s)

Title: Word Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. Input Specification: The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. Output Specification: Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. Demo Input: ['HoUse\n', 'ViP\n', 'maTRIx\n'] Demo Output: ['house\n', 'VIP\n', 'matrix\n'] Note: none

```python s = input() k1 = 0 k2 = 0 l = len(s) for i in range(0,l): if s[i] >= 'a' and s[i] <= 'z': k1 += 1 else: k2 += 1 if k1 >= k2: s = s.lower() else: s = s.upper() print(s) ```

3.977

483

A

Counterexample

PROGRAMMING

1,100

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

null

Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one. Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime. You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*. More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.

The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).

Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order. If the counterexample does not exist, print the single number -1.

[ "2 4\n", "10 11\n", "900000000000000009 900000000000000029\n" ]

[ "2 3 4\n", "-1\n", "900000000000000009 900000000000000010 900000000000000021\n" ]

In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are. In the second sample you cannot form a group of three distinct integers, so the answer is -1. In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.

500

[ { "input": "2 4", "output": "2 3 4" }, { "input": "10 11", "output": "-1" }, { "input": "900000000000000009 900000000000000029", "output": "900000000000000009 900000000000000010 900000000000000021" }, { "input": "640097987171091791 640097987171091835", "output": "640097987171091792 640097987171091793 640097987171091794" }, { "input": "19534350415104721 19534350415104725", "output": "19534350415104722 19534350415104723 19534350415104724" }, { "input": "933700505788726243 933700505788726280", "output": "933700505788726244 933700505788726245 933700505788726246" }, { "input": "1 3", "output": "-1" }, { "input": "1 4", "output": "2 3 4" }, { "input": "1 1", "output": "-1" }, { "input": "266540997167959130 266540997167959164", "output": "266540997167959130 266540997167959131 266540997167959132" }, { "input": "267367244641009850 267367244641009899", "output": "267367244641009850 267367244641009851 267367244641009852" }, { "input": "268193483524125978 268193483524125993", "output": "268193483524125978 268193483524125979 268193483524125980" }, { "input": "269019726702209402 269019726702209432", "output": "269019726702209402 269019726702209403 269019726702209404" }, { "input": "269845965585325530 269845965585325576", "output": "269845965585325530 269845965585325531 269845965585325532" }, { "input": "270672213058376250 270672213058376260", "output": "270672213058376250 270672213058376251 270672213058376252" }, { "input": "271498451941492378 271498451941492378", "output": "-1" }, { "input": "272324690824608506 272324690824608523", "output": "272324690824608506 272324690824608507 272324690824608508" }, { "input": "273150934002691930 273150934002691962", "output": "273150934002691930 273150934002691931 273150934002691932" }, { "input": "996517375802030516 996517375802030524", "output": "996517375802030516 996517375802030517 996517375802030518" }, { "input": "997343614685146644 997343614685146694", "output": "997343614685146644 997343614685146645 997343614685146646" }, { "input": "998169857863230068 998169857863230083", "output": "998169857863230068 998169857863230069 998169857863230070" }, { "input": "998996101041313492 998996101041313522", "output": "998996101041313492 998996101041313493 998996101041313494" }, { "input": "999822344219396916 999822344219396961", "output": "999822344219396916 999822344219396917 999822344219396918" }, { "input": "648583102513043 648583102513053", "output": "648583102513044 648583102513045 648583102513046" }, { "input": "266540997167959130 266540997167959131", "output": "-1" }, { "input": "267367244641009850 267367244641009850", "output": "-1" }, { "input": "268193483524125978 268193483524125979", "output": "-1" }, { "input": "269019726702209402 269019726702209402", "output": "-1" }, { "input": "269845965585325530 269845965585325530", "output": "-1" }, { "input": "270672213058376250 270672213058376254", "output": "270672213058376250 270672213058376251 270672213058376252" }, { "input": "271498451941492378 271498451941492379", "output": "-1" }, { "input": "272324690824608506 272324690824608508", "output": "272324690824608506 272324690824608507 272324690824608508" }, { "input": "273150934002691930 273150934002691931", "output": "-1" }, { "input": "996517375802030516 996517375802030518", "output": "996517375802030516 996517375802030517 996517375802030518" }, { "input": "997343614685146644 997343614685146644", "output": "-1" }, { "input": "2147483647 2147483649", "output": "-1" }, { "input": "3 5", "output": "-1" }, { "input": "1 7", "output": "2 3 4" }, { "input": "9 12", "output": "9 11 12" }, { "input": "4 4", "output": "-1" }, { "input": "11 13", "output": "-1" }, { "input": "2 2", "output": "-1" } ]

1,645,427,268

2,147,483,647

Python 3

OK

TESTS

42

46

0

l,r = map(int,input().split()) res='' if abs(l-r) < 2 : res='-1' elif l%2 : l+=1 if abs(l-r) < 2 : res='-1' else : res=str(l)+' '+str(l+1)+' '+str(l+2) else : res=str(l)+' '+str(l+1)+' '+str(l+2) print(res)

Title: Counterexample Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one. Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime. You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*. More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime. Input Specification: The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50). Output Specification: Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order. If the counterexample does not exist, print the single number -1. Demo Input: ['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n'] Demo Output: ['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n'] Note: In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are. In the second sample you cannot form a group of three distinct integers, so the answer is -1. In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.

```python l,r = map(int,input().split()) res='' if abs(l-r) < 2 : res='-1' elif l%2 : l+=1 if abs(l-r) < 2 : res='-1' else : res=str(l)+' '+str(l+1)+' '+str(l+2) else : res=str(l)+' '+str(l+1)+' '+str(l+2) print(res) ```

3

448

A

Rewards

PROGRAMMING

800

[ "implementation" ]

null

Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces.

Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

[ "1 1 1\n1 1 1\n4\n", "1 1 3\n2 3 4\n2\n", "1 0 0\n1 0 0\n1\n" ]

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

none

500

[ { "input": "1 1 1\n1 1 1\n4", "output": "YES" }, { "input": "1 1 3\n2 3 4\n2", "output": "YES" }, { "input": "1 0 0\n1 0 0\n1", "output": "NO" }, { "input": "0 0 0\n0 0 0\n1", "output": "YES" }, { "input": "100 100 100\n100 100 100\n100", "output": "YES" }, { "input": "100 100 100\n100 100 100\n1", "output": "NO" }, { "input": "1 10 100\n100 10 1\n20", "output": "NO" }, { "input": "1 1 1\n0 0 0\n1", "output": "YES" }, { "input": "0 0 0\n1 1 1\n1", "output": "YES" }, { "input": "5 5 5\n0 0 0\n2", "output": "NO" }, { "input": "0 0 0\n10 10 10\n2", "output": "NO" }, { "input": "21 61 39\n63 58 69\n44", "output": "YES" }, { "input": "18 95 4\n7 1 75\n46", "output": "YES" }, { "input": "64 27 81\n72 35 23\n48", "output": "YES" }, { "input": "6 6 6\n11 11 11\n7", "output": "NO" }, { "input": "1 2 3\n2 4 6\n3", "output": "NO" }, { "input": "1 2 3\n2 4 6\n4", "output": "YES" }, { "input": "99 99 99\n99 99 99\n89", "output": "NO" }, { "input": "5 0 0\n15 0 0\n2", "output": "NO" }, { "input": "10 10 10\n0 0 0\n1", "output": "NO" }, { "input": "1 1 1\n1 1 1\n15", "output": "YES" }, { "input": "2 3 5\n2 3 5\n2", "output": "NO" }, { "input": "2 2 2\n3 3 5\n3", "output": "NO" }, { "input": "1 2 2\n2 4 4\n1", "output": "NO" }, { "input": "1 2 3\n1 5 5\n2", "output": "NO" } ]

1,681,237,423

2,147,483,647

Python 3

OK

TESTS

25

46

0

def main(): a1, a2, a3 = map(int, input().split()) b1, b2, b3 = map(int, input().split()) n = int(input()) if (a1 + a2 + a3 + 4) // 5 + (b1 + b2 + b3 + 9) // 10 <= n: print('YES') else: print('NO') if __name__ == '__main__': main()

Title: Rewards Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled. Input Specification: The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces. Output Specification: Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes). Demo Input: ['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: none

```python def main(): a1, a2, a3 = map(int, input().split()) b1, b2, b3 = map(int, input().split()) n = int(input()) if (a1 + a2 + a3 + 4) // 5 + (b1 + b2 + b3 + 9) // 10 <= n: print('YES') else: print('NO') if __name__ == '__main__': main() ```

3

621

A

Wet Shark and Odd and Even

PROGRAMMING

900

[ "implementation" ]

null

Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.

The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

Print the maximum possible even sum that can be obtained if we use some of the given integers.

[ "3\n1 2 3\n", "5\n999999999 999999999 999999999 999999999 999999999\n" ]

[ "6", "3999999996" ]

In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

500

[ { "input": "3\n1 2 3", "output": "6" }, { "input": "5\n999999999 999999999 999999999 999999999 999999999", "output": "3999999996" }, { "input": "1\n1", "output": "0" }, { "input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98", "output": "870" }, { "input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25", "output": "840" }, { "input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44", "output": "922" }, { "input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59", "output": "674" }, { "input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82", "output": "848" }, { "input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93", "output": "902" }, { "input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20", "output": "798" }, { "input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31", "output": "632" }, { "input": "1\n4", "output": "4" }, { "input": "2\n1 4", "output": "4" }, { "input": "3\n1 2 4", "output": "6" }, { "input": "2\n9 3", "output": "12" }, { "input": "2\n1000000000 1001", "output": "1000000000" }, { "input": "3\n1 8 4", "output": "12" }, { "input": "3\n7 4 4", "output": "8" }, { "input": "5\n2 3 4 5 3", "output": "14" }, { "input": "2\n4 5", "output": "4" }, { "input": "3\n2 4 5", "output": "6" }, { "input": "3\n2 2 3", "output": "4" }, { "input": "2\n2 3", "output": "2" }, { "input": "4\n2 3 7 7", "output": "16" }, { "input": "2\n999999999 2", "output": "2" }, { "input": "2\n2 5", "output": "2" }, { "input": "3\n5 3 1", "output": "8" }, { "input": "4\n3 2 5 7", "output": "14" } ]

1,521,479,366

2,147,483,647

Python 3

OK

TESTS

87

155

14,438,400

count = int(input()) total, smallest = 0, 10**9 numbers = map(int, input().split()) for x in numbers: if x < smallest and x % 2: smallest = x total += x if total % 2 == 0: print (total) else: print (total - smallest)

Title: Wet Shark and Odd and Even Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. Input Specification: The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. Output Specification: Print the maximum possible even sum that can be obtained if we use some of the given integers. Demo Input: ['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n'] Demo Output: ['6', '3999999996'] Note: In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

```python count = int(input()) total, smallest = 0, 10**9 numbers = map(int, input().split()) for x in numbers: if x < smallest and x % 2: smallest = x total += x if total % 2 == 0: print (total) else: print (total - smallest) ```

3

88

B

Keyboard

PROGRAMMING

1,500

[ "implementation" ]

B. Keyboard

1

256

Vasya learns to type. He has an unusual keyboard at his disposal: it is rectangular and it has *n* rows of keys containing *m* keys in each row. Besides, the keys are of two types. Some of the keys have lowercase Latin letters on them and some of the keys work like the "Shift" key on standard keyboards, that is, they make lowercase letters uppercase. Vasya can press one or two keys with one hand. However, he can only press two keys if the Euclidean distance between the centers of the keys does not exceed *x*. The keys are considered as squares with a side equal to 1. There are no empty spaces between neighbouring keys. Vasya is a very lazy boy, that's why he tries to type with one hand as he eats chips with his other one. However, it is possible that some symbol can't be typed with one hand only, because the distance between it and the closest "Shift" key is strictly larger than *x*. In this case he will have to use his other hand. Having typed the symbol, Vasya returns other hand back to the chips. You are given Vasya's keyboard and the text. Count the minimum number of times Vasya will have to use the other hand.

The first line contains three integers *n*, *m*, *x* (1<=≤<=*n*,<=*m*<=≤<=30,<=1<=≤<=*x*<=≤<=50). Next *n* lines contain descriptions of all the keyboard keys. Each line contains the descriptions of exactly *m* keys, without spaces. The letter keys are marked with the corresponding lowercase letters. The "Shift" keys are marked with the "S" symbol. Then follow the length of the text *q* (1<=≤<=*q*<=≤<=5·105). The last line contains the text *T*, which consists of *q* symbols, which are uppercase and lowercase Latin letters.

If Vasya can type the text, then print the minimum number of times he will have to use his other hand. Otherwise, print "-1" (without the quotes).

[ "2 2 1\nab\ncd\n1\nA\n", "2 2 1\nab\ncd\n1\ne\n", "2 2 1\nab\ncS\n5\nabcBA\n", "3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG\n" ]

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

In the first sample the symbol "A" is impossible to print as there's no "Shift" key on the keyboard. In the second sample the symbol "e" is impossible to print as there's no such key on the keyboard. In the fourth sample the symbols "T", "G" are impossible to print with one hand. The other letters that are on the keyboard can be printed. Those symbols come up in the text twice, thus, the answer is 2.

1,000

[ { "input": "2 2 1\nab\ncd\n1\nA", "output": "-1" }, { "input": "2 2 1\nab\ncd\n1\ne", "output": "-1" }, { "input": "2 2 1\nab\ncS\n5\nabcBA", "output": "1" }, { "input": "3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG", "output": "2" }, { "input": "10 9 3\noboxlgpey\nyxcuwkkmp\njuqeflhwq\nsfnxqtjqS\nkkudcnyjl\nhgjlcrkjq\njnofqksxn\nqbhsnuguv\nlvahnifao\nebwnnlrwe\n35\nCodeforcesBetaRoundproblemAtestfive", "output": "4" }, { "input": "2 7 4\niuqtieo\nysxcgmS\n2\nsQ", "output": "1" }, { "input": "1 2 4\nbS\n8\nbBbbbBbb", "output": "0" }, { "input": "7 8 5\nfqiubjpm\nqbshcsyk\ncjbxpbef\nptwpmapx\nryazscbm\nqnvsgzrf\nhtardzkz\n9\nuxrmwkayy", "output": "0" }, { "input": "8 6 4\nefvmov\nkeofnw\npwajpe\nknptky\nSibruu\nrgdukk\nbsxosd\nhovgSe\n10\nECreruXmsC", "output": "-1" }, { "input": "10 3 2\nukk\neqt\nfex\nqSh\ntvz\nfjn\niol\nehd\nnte\ngyx\n5\ncgQxI", "output": "-1" }, { "input": "10 10 19\nowqjcaSpqn\nvgrhboqahn\nbzziocjmbu\npurqsmiSop\nxcsifctjhy\nycyytwoamk\nrnjfxsxowl\nnkgcywcdff\nbazljrisqv\nkcakigSekq\n100\nzewpATtssQVicNrlRrcoifTutTAfFMUEfDFKoNyQbSrSYxTGMadNkRpmJvoEqUsqPYgAdQreaUrwDKMNFWiwdRRCcJBPorfMVMoK", "output": "0" }, { "input": "10 10 26\nwxmssptheb\nzpxbxsyxsy\nqbjkpaywqp\nfwhnuzjcgq\nycgaanzedz\njrycrbzqfs\ngswwakybus\nfhtxhljedz\noSepmyjosv\ndwviycevdn\n100\nyapwUfnyPzgZyFvAHGKWVbXQHkuhJDoUTvCAtdMMCQmKchxKkilUTECOqYJFUSHPqKiRKhDXZgHxwApDWlShdwakmVCgaeKCLOMX", "output": "0" }, { "input": "10 10 3\nrvouufmnqu\nbyukrnmnhr\nzjggwxgvkz\ntcagkSitiw\nhryajgtpwc\njragfhqoks\nkgroxxkuvp\nbpgrkqiyns\njbuhjjkziw\nomjmbaggsw\n100\nCpRzrPqPngYvrVJFCWRPMRwrpXcbtiwfoFcAkRaNjzpMMKOQAzBxSrxGbIHaYgmSqhhxhZTmhFttKnhFzRfKxYXshUZRvtKJIzZq", "output": "12" }, { "input": "10 10 2\nfriuxvShvg\nerslojqtgu\nzeqsmdewry\nwvhbeeyeSu\ngkofbjaavr\ntwkcdxugps\nnzlylSmafu\nstamkpxnzt\nuwxwximkrm\nmzxyboazbl\n100\nmRIfAtrLKmztpVkAmojDCiIgseBwlUilBIixDQhqNhNAqVLLIobuCIretLdSvixNNdCiouFMXtwHZFlObCeaygmIiFBfaCirbmCa", "output": "19" }, { "input": "10 10 2\nbddahSqkmk\npxbocxayjs\nottvdazstk\nlaxuidqlqb\nkfjwdpdfat\nxlipuubkgv\niqyomzfktm\niwbgidmwyu\nrngqkeupsf\nbqndtekryw\n100\nMNQgWFLhHycqwjSsbTkbgMYAIHFYARRmOsinYMFjOxxnLjiKfeiBbMpoeTdzUMORPaAxRNfvdAPFaKkPdxdAjjJgGCxkDzmSasqq", "output": "37" }, { "input": "10 10 2\nnxcwdrsmrv\nSyjahsosvp\nvkrqbxhgbv\nwkxywavtnn\nepkyoviqbi\nsfmpvhuwwq\nnlsostrotx\ntcdguorhny\nimixrqzSdu\nxzhdhdwibt\n100\nUzzaWiRFYbAqxIDMrRBBDoGQhSzSqSLEddAiJsZcxbemdeuddamNYdWOvzlYSCuHIRpnuxdNxAsnZMiLXBYwnrMcrbNeLrUYhZOB", "output": "17" }, { "input": "10 10 23\nhtyvouoiqi\nvySvsfqadv\nxvqyqjyutq\npjcrrphzbk\nhlqfyoqfmo\nezcSwleoew\nxkwqrajxyg\nngSiftgoso\njyndgicccr\nlgjvokydhp\n100\nJzVVfotldIRcyjhTNRcFlTxFeZKRwavZxYcvdDOQyUvTmryFRuRBcRvmscegtspkPuchqlFEKbrfpTOSlSFOARsbbvSenMwNmaRj", "output": "0" }, { "input": "10 10 7\nifcwalsdbj\njpykymrbei\nrylzgkyefh\noilvvexpjp\niptgodpfim\ndSrqejaixu\npksxlsniwa\nmoSenxtfbc\noqssptcenz\nqdhmouvyas\n100\nqtMDVUXJpSEFgPsLKyRJVRbfVoYaCKJDnQDLFVngVjSPzzVyMnMyuyahMRiBJuNhKtgpVqvukUolLvYEmidvXotgQUJukYwIweUW", "output": "0" }, { "input": "10 10 1\nmdxafehbkr\nyuhenybjps\ntvfwmiwcoh\njmzrepzjvx\nnqyorkSnuk\ntSmztmwidv\ncmmajnlqrw\nfiqewpdwax\nuesmkdcplt\nlgkomdcqbo\n100\nmcEQmAvFqKYMXLHQUDeIulkmAMRkIUtbKihTFJwJYQfcAelNrZWSAwHunwZTrdHaRWokgCyLqbubOpEHuZiDVoFHjvkMSoBPyGOI", "output": "39" }, { "input": "10 10 2\nnhfafdwqhh\neyvitpcthk\nrpiotuoqzh\nnxxnhuaxee\nyevrtirzwf\nkbtSsamyel\nfeenjvxsmo\nkqpenxjmde\nlqsamthlwp\njdyyqsbtbk\n100\nUHucxPWDaKonVpXEctuqYUAQnrFEZaTYxhoacNbHIMevlbDejXjitEzyVrTfcfBHWRMdJvaTkbkqccyHjtzpTbKmRAXwlXCtFKNX", "output": "29" }, { "input": "10 10 1\nsufnxxpdnx\nvttibpllhv\nlvbrjmfdjx\ngmtexvrnfh\nygsqrsSwxd\nkxbbjxgbzs\nedutwocmzd\nfebjgknyai\nvcvquagvrs\ndrdoarhgoc\n100\nZoZJXhUWyaLgBTpgbznABKHuyFcKzJmGaMhoKkKfyOGacLwBspaKtAEdwMZJFYiZUFNDxdDIDgKSCRvsbGUOXRqalbpuEqkduYpW", "output": "44" }, { "input": "10 10 2\ncstcrltzsl\nblotmquzvj\nuiitiytlgx\nwumpfdaprd\ntfxohqpztn\nvfrpsccddo\nneegusrkxw\niijfjozqjq\nioegbvuhew\npjjpqdxvqu\n100\nkPCBONfZLkeXzWVuSgvinPENazcnRoBcUHXwRzPyvNIiDlDSeKOYmiUmjooXuzTCtIRxKDAYeTLgjsenxHoymVazMALUADQpjVjV", "output": "-1" }, { "input": "10 10 1\nqztnjglyrc\nnukswgzajl\nqnpbdwjvbb\nliiakzcrlz\nnolwfzzvxd\nmqvhiySttx\nqwuizSjuto\nqbgwiwjukx\nkomyvblgkc\ntkzlxzgsru\n100\nYUzTZDzLFkMUhjQWbwljJCRyZGFzgJcozvROiwPktRGxkMKiPyiTvhDrtusPYhMgVAOFIjAvlpzcrUvrMrMbhkpUiyAytKfYOGTF", "output": "37" }, { "input": "10 10 1\nmgziiihbkq\niobjknuogh\nvntwahSopu\nsjsxjpaqvm\nwqgrodizst\nselzugktoi\nvbhfzvgjfn\nliqlfdcjhf\nbpbtpmimxb\npksfiydpfw\n100\nwAVZXEhKTuajdCauVTIwgnfbxWuUSmtXkjHZtNVcfTsiSAPLdpdEFdTJLZRjptUcRhAmrNjKMXmuDGatAQoaIpbddnzRGHsJrhoq", "output": "39" }, { "input": "10 10 2\nshbqxycvfm\notydudkttw\nqhatsxsngz\nixvyujtyjc\nsbvqhnjbak\neggcguuuka\nxcydfgjzeb\nytpdkcdrsq\nefqlpywggu\nfcnfrhnouo\n100\nHPqtuVckdUOhsnuhnbpekWvWKUnAEaOCihpeEvmaOKOPcIZiMixGJGEuXAaOxuUNyrIesmldLEqGnvyDKPDvFkCbRebCORHmUgeV", "output": "-1" }, { "input": "1 1 50\nS\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn", "output": "-1" }, { "input": "1 1 50\nS\n1\nS", "output": "-1" }, { "input": "1 1 50\na\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn", "output": "-1" }, { "input": "1 1 50\nz\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn", "output": "-1" }, { "input": "2 1 2\nS\nc\n4\nCSSA", "output": "-1" } ]

1,600,879,876

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

12

139

307,200

def sqrt(a): return a**0.5 n , m , x = map(int,input().split()) keyboard = [] for i in range(n): keyboard.append(input()) #print(keyboard) q = int(input()) text = input() shifts = [] cost = dict() for i in range(n): for j in range(m): if keyboard[i][j] == 'S': shifts.append((i , j)) else: cost[keyboard[i][j]] = 0 if (len(shifts) > 0 ): for i in range(n): for j in range(m): if keyboard[i][j] != 'S': distance= min((i - si)**2 + (j - sj)**2 for si , sj in shifts ) letter = keyboard[i][j].upper() #print(letter) if distance <= x ** 2: cost[letter] = 0 elif distance > x **2 : cost[letter] = 1 res = 0 for letter in text : if letter in cost: res += cost[letter] else: res = -1 break print(res)

Title: Keyboard Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya learns to type. He has an unusual keyboard at his disposal: it is rectangular and it has *n* rows of keys containing *m* keys in each row. Besides, the keys are of two types. Some of the keys have lowercase Latin letters on them and some of the keys work like the "Shift" key on standard keyboards, that is, they make lowercase letters uppercase. Vasya can press one or two keys with one hand. However, he can only press two keys if the Euclidean distance between the centers of the keys does not exceed *x*. The keys are considered as squares with a side equal to 1. There are no empty spaces between neighbouring keys. Vasya is a very lazy boy, that's why he tries to type with one hand as he eats chips with his other one. However, it is possible that some symbol can't be typed with one hand only, because the distance between it and the closest "Shift" key is strictly larger than *x*. In this case he will have to use his other hand. Having typed the symbol, Vasya returns other hand back to the chips. You are given Vasya's keyboard and the text. Count the minimum number of times Vasya will have to use the other hand. Input Specification: The first line contains three integers *n*, *m*, *x* (1<=≤<=*n*,<=*m*<=≤<=30,<=1<=≤<=*x*<=≤<=50). Next *n* lines contain descriptions of all the keyboard keys. Each line contains the descriptions of exactly *m* keys, without spaces. The letter keys are marked with the corresponding lowercase letters. The "Shift" keys are marked with the "S" symbol. Then follow the length of the text *q* (1<=≤<=*q*<=≤<=5·105). The last line contains the text *T*, which consists of *q* symbols, which are uppercase and lowercase Latin letters. Output Specification: If Vasya can type the text, then print the minimum number of times he will have to use his other hand. Otherwise, print "-1" (without the quotes). Demo Input: ['2 2 1\nab\ncd\n1\nA\n', '2 2 1\nab\ncd\n1\ne\n', '2 2 1\nab\ncS\n5\nabcBA\n', '3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG\n'] Demo Output: ['-1\n', '-1\n', '1\n', '2\n'] Note: In the first sample the symbol "A" is impossible to print as there's no "Shift" key on the keyboard. In the second sample the symbol "e" is impossible to print as there's no such key on the keyboard. In the fourth sample the symbols "T", "G" are impossible to print with one hand. The other letters that are on the keyboard can be printed. Those symbols come up in the text twice, thus, the answer is 2.

```python def sqrt(a): return a**0.5 n , m , x = map(int,input().split()) keyboard = [] for i in range(n): keyboard.append(input()) #print(keyboard) q = int(input()) text = input() shifts = [] cost = dict() for i in range(n): for j in range(m): if keyboard[i][j] == 'S': shifts.append((i , j)) else: cost[keyboard[i][j]] = 0 if (len(shifts) > 0 ): for i in range(n): for j in range(m): if keyboard[i][j] != 'S': distance= min((i - si)**2 + (j - sj)**2 for si , sj in shifts ) letter = keyboard[i][j].upper() #print(letter) if distance <= x ** 2: cost[letter] = 0 elif distance > x **2 : cost[letter] = 1 res = 0 for letter in text : if letter in cost: res += cost[letter] else: res = -1 break print(res) ```

0

285

C

Building Permutation

PROGRAMMING

1,200

[ "greedy", "implementation", "sortings" ]

null

Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*. You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.

The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).

Print a single number — the minimum number of moves. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

[ "2\n3 0\n", "3\n-1 -1 2\n" ]

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

In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1). In the second sample you need 6 moves to build permutation (1, 3, 2).

1,500

[ { "input": "2\n3 0", "output": "2" }, { "input": "3\n-1 -1 2", "output": "6" }, { "input": "5\n-3 5 -3 3 3", "output": "10" }, { "input": "10\n9 6 -2 4 1 1 1 9 6 2", "output": "18" }, { "input": "9\n2 0 0 6 5 4 1 9 3", "output": "15" }, { "input": "100\n-77 57 -95 -23 53 -28 82 -83 38 -73 85 28 25 6 -43 4 -10 -30 -9 -92 14 34 -93 61 36 -100 90 -68 28 16 100 -3 97 30 36 -55 62 -62 53 74 -50 -23 67 11 22 -30 -19 83 7 84 43 90 -65 -75 -15 97 90 15 66 2 13 -91 91 -44 46 51 51 -58 95 77 20 30 76 79 91 60 76 2 82 42 -93 94 -57 88 65 -95 -66 100 -9 33 -67 54 -99 97 53 13 54 66 60 -48", "output": "3459" } ]

1,676,534,611

2,147,483,647

Python 3

OK

TESTS

33

374

24,268,800

n = int(input()) a = [int(i) for i in input().split()] a.sort() print(sum(abs(a[i] - (i + 1)) for i in range(n)))

Title: Building Permutation Time Limit: None seconds Memory Limit: None megabytes Problem Description: Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*. You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). Output Specification: Print a single number — the minimum number of moves. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['2\n3 0\n', '3\n-1 -1 2\n'] Demo Output: ['2\n', '6\n'] Note: In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1). In the second sample you need 6 moves to build permutation (1, 3, 2).

```python n = int(input()) a = [int(i) for i in input().split()] a.sort() print(sum(abs(a[i] - (i + 1)) for i in range(n))) ```

3