evanellis/Codeforces-Python-Submissions_correct

712

B

Memory and Trident

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion: - An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down. But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.

The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.

If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.

[ "RRU\n", "UDUR\n", "RUUR\n" ]

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

In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk. In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

1,000

[ { "input": "RRU", "output": "-1" }, { "input": "UDUR", "output": "1" }, { "input": "RUUR", "output": "2" }, { "input": "DDDD", "output": "2" }, { "input": "RRRR", "output": "2" }, { "input": "RRRUUD", "output": "2" }, { "input": "UDURLRDURLRD", "output": "1" }, { "input": "RLRU", "output": "1" }, { "input": "RDDLLDLUUUDDRDRURLUUURLLDDLRLUURRLLRRLDRLLUDRLRULLDLRRLRLRLRUDUUDLULURLLDUURULURLLRRRURRRDRUUDLDRLRDRLRRDDLDLDLLUDRUDRLLLLDRDUULRUURRDLULLULDUDULRURRDDDLLUDRLUDDLDDDRRDDDULLLLDLDRLRRLRRDDRLULURRUDRDUUUULDURUDRDLDDUDUDRRURDULRRUDRLRRDLUURURDLDRLRDUDDDLDDDURURLUULRDUUULRURUDUDRRUDULLLUUUDRLLDRRDDLRUDRDRDLLRURURRRULURURRRLUUULRRRUURUUDURUDDLLDLDRLRDLDRLLDLDRDRRLRRRURUUUDRDLRLRUDRLULUUULUDDLULDLRLLRDUULLRLRURLRURULLLUDUDDLRULRDUURURLDLUURRRDURRLLDRUUDRDLLDUUDLURUDDUUUULRLLURLUDDRLRRDRURLRUDRLDDRLLL", "output": "-1" }, { "input": "LDLDLDLDLDRULD", "output": "5" }, { "input": "LULULURULLLU", "output": "5" }, { "input": "DRDRDDRR", "output": "4" }, { "input": "LR", "output": "0" }, { "input": "UL", "output": "1" }, { "input": "UD", "output": "0" }, { "input": "R", "output": "-1" }, { "input": "LU", "output": "1" }, { "input": "RR", "output": "1" }, { "input": "UDLR", "output": "0" }, { "input": "RRRRRLLUUUUDD", "output": "-1" }, { "input": "UUUUDLLLLR", "output": "3" }, { "input": "LLRUD", "output": "-1" }, { "input": "LLRDDU", "output": "1" } ]

1,473,527,350

1,450

Python 3

OK

TESTS

75

77

307,200

s = input() countr = 0 countu = 0 for i in s: if i == 'U': countu += 1 elif i == 'D': countu -= 1 elif i == 'R': countr += 1 else: countr -= 1 if (countr + countu) % 2 == 1: print(-1) else: print((abs(countr) + abs(countu)) // 2)

Title: Memory and Trident Time Limit: None seconds Memory Limit: None megabytes Problem Description: Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion: - An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down. But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string. Input Specification: The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given. Output Specification: If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1. Demo Input: ['RRU\n', 'UDUR\n', 'RUUR\n'] Demo Output: ['-1\n', '1\n', '2\n'] Note: In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk. In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

```python s = input() countr = 0 countu = 0 for i in s: if i == 'U': countu += 1 elif i == 'D': countu -= 1 elif i == 'R': countr += 1 else: countr -= 1 if (countr + countu) % 2 == 1: print(-1) else: print((abs(countr) + abs(countu)) // 2) ```

3

764

A

Taymyr is calling you

PROGRAMMING

800

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

null

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

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

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

Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

500

[ { "input": "1 1 10", "output": "10" }, { "input": "1 2 5", "output": "2" }, { "input": "2 3 9", "output": "1" }, { "input": "4 8 9", "output": "1" }, { "input": "7 9 2", "output": "0" }, { "input": "10000 10000 10000", "output": "1" }, { "input": "24 22 9235", "output": "34" }, { "input": "74 8 417", "output": "1" }, { "input": "972 1 203", "output": "0" }, { "input": "550 1 754", "output": "1" }, { "input": "860 1 884", "output": "1" }, { "input": "358 2 809", "output": "2" }, { "input": "33 27 216", "output": "0" }, { "input": "2940 1 9311", "output": "3" }, { "input": "4624 1 1953", "output": "0" }, { "input": "2696 2 7345", "output": "2" }, { "input": "3443 2 6701", "output": "0" }, { "input": "3 613 2275", "output": "1" }, { "input": "1 10000 10000", "output": "1" }, { "input": "10000 1 10000", "output": "1" }, { "input": "1 1 1", "output": "1" }, { "input": "1 1 10000", "output": "10000" }, { "input": "34 27 10000", "output": "10" }, { "input": "2 2 9999", "output": "4999" }, { "input": "2 2 1", "output": "0" }, { "input": "6 4 36", "output": "3" }, { "input": "33 6 3005", "output": "45" }, { "input": "5 1 20", "output": "4" }, { "input": "1 2 10", "output": "5" }, { "input": "2 1 100", "output": "50" }, { "input": "10 20 10000", "output": "500" }, { "input": "8 12 12", "output": "0" } ]

1,487,498,287

2,147,483,647

Python 3

OK

TESTS

32

62

4,608,000

import math if __name__ == '__main__': n, m, z = str(input()).split() n = int(n) m = int(m) z = int(z) print((z * math.gcd(n, m)) // (n * m))

Title: Taymyr is calling you Time Limit: None seconds Memory Limit: None megabytes Problem Description: Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. Input Specification: The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). Output Specification: Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. Demo Input: ['1 1 10\n', '1 2 5\n', '2 3 9\n'] Demo Output: ['10\n', '2\n', '1\n'] Note: Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

```python import math if __name__ == '__main__': n, m, z = str(input()).split() n = int(n) m = int(m) z = int(z) print((z * math.gcd(n, m)) // (n * m)) ```

3

50

A

Domino piling

PROGRAMMING

800

[ "greedy", "math" ]

A. Domino piling

2

256

You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.

In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).

Output one number — the maximal number of dominoes, which can be placed.

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

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

none

500

[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]

1,634,476,088

2,147,483,647

Python 3

OK

TESTS

35

124

0

m=input().split() m=list(map(int,m)) p=m[0]*m[1] d=p//2 print(d)

Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none

```python m=input().split() m=list(map(int,m)) p=m[0]*m[1] d=p//2 print(d) ```

3.969

734

B

Anton and Digits

PROGRAMMING

800

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

null

Recently Anton found a box with digits in his room. There are *k*2 digits 2, *k*3 digits 3, *k*5 digits 5 and *k*6 digits 6. Anton's favorite integers are 32 and 256. He decided to compose this integers from digits he has. He wants to make the sum of these integers as large as possible. Help him solve this task! Each digit can be used no more than once, i.e. the composed integers should contain no more than *k*2 digits 2, *k*3 digits 3 and so on. Of course, unused digits are not counted in the sum.

The only line of the input contains four integers *k*2, *k*3, *k*5 and *k*6 — the number of digits 2, 3, 5 and 6 respectively (0<=≤<=*k*2,<=*k*3,<=*k*5,<=*k*6<=≤<=5·106).

Print one integer — maximum possible sum of Anton's favorite integers that can be composed using digits from the box.

[ "5 1 3 4\n", "1 1 1 1\n" ]

[ "800\n", "256\n" ]

In the first sample, there are five digits 2, one digit 3, three digits 5 and four digits 6. Anton can compose three integers 256 and one integer 32 to achieve the value 256 + 256 + 256 + 32 = 800. Note, that there is one unused integer 2 and one unused integer 6. They are not counted in the answer. In the second sample, the optimal answer is to create on integer 256, thus the answer is 256.

750

[ { "input": "5 1 3 4", "output": "800" }, { "input": "1 1 1 1", "output": "256" }, { "input": "10 2 1 5", "output": "320" }, { "input": "4 2 7 2", "output": "576" }, { "input": "489 292 127 263", "output": "41856" }, { "input": "9557 5242 1190 7734", "output": "472384" }, { "input": "1480320 1969946 1158387 3940412", "output": "306848928" }, { "input": "0 0 0 0", "output": "0" }, { "input": "5000000 5000000 5000000 5000000", "output": "1280000000" }, { "input": "1048576 256 1048576 1048576", "output": "268435456" }, { "input": "2073144 2073145 0 0", "output": "66340608" }, { "input": "1000000 0 0 1", "output": "0" }, { "input": "2 1 1 1", "output": "288" }, { "input": "0 5000000 5000000 5000000", "output": "0" }, { "input": "4494839 1140434 3336818 4921605", "output": "890719296" }, { "input": "2363223 3835613 926184 3190201", "output": "283088352" }, { "input": "198044 2268164 2811743 1458798", "output": "50699264" }, { "input": "5 5 1 0", "output": "160" }, { "input": "1 1 1 4", "output": "256" }, { "input": "3 3 4 4", "output": "768" }, { "input": "1 2 0 5", "output": "32" }, { "input": "1207814 1649617 2347252 3136345", "output": "309200384" }, { "input": "78025 2308643 78025 4943733", "output": "19974400" }, { "input": "3046068 2548438 2676145 4789979", "output": "696930656" }, { "input": "4755258 2724358 2030900 4801065", "output": "607089856" }, { "input": "1359689 3792971 4451626 4497236", "output": "348080384" }, { "input": "3484483 3995744 87159 4941393", "output": "131027072" }, { "input": "1273630 1273630 980163 1711706", "output": "260312672" }, { "input": "2010798 1111442 4014004 4430228", "output": "514764288" }, { "input": "1714940 133067 3346537 3346537", "output": "439024640" }, { "input": "3731658 4548347 3731658 3731658", "output": "955304448" }, { "input": "601597 2632066 450558 450558", "output": "120176096" }, { "input": "726573 158002 568571 568571", "output": "150610240" }, { "input": "407729 4510137 3425929 3425929", "output": "104378624" }, { "input": "1781608 3826276 4384744 4384744", "output": "456091648" }, { "input": "4486284 4486284 3249460 3249460", "output": "871440128" }, { "input": "4759823 3520376 4743363 4743363", "output": "1214827648" }, { "input": "386719 4643763 2749163 1701105", "output": "99000064" }, { "input": "3173901 1042250 4102237 3173901", "output": "812518656" }, { "input": "2681845 4558270 4388852 13014", "output": "88734176" }, { "input": "1152975 910150 2776412 242825", "output": "91288000" }, { "input": "2005366 3807065 4174270 2471686", "output": "513373696" }, { "input": "2092196 2406694 3664886 85601", "output": "86124896" }, { "input": "4542228 4542228 3992410 1039690", "output": "378241856" }, { "input": "3093105 2256347 675644 570209", "output": "218176608" }, { "input": "0 1 1 1", "output": "0" }, { "input": "10 20 10 20", "output": "2560" }, { "input": "0 0 5 1", "output": "0" }, { "input": "10 0 10 11", "output": "2560" }, { "input": "2 1 32768 65536", "output": "512" }, { "input": "2 3 5 6", "output": "512" }, { "input": "5 3 2 2", "output": "608" }, { "input": "1 0 2 3", "output": "256" }, { "input": "1 0 1 1", "output": "256" }, { "input": "2 0 3 3", "output": "512" }, { "input": "2 2 2 0", "output": "64" }, { "input": "0 0 1 1", "output": "0" }, { "input": "1 0 2 2", "output": "256" }, { "input": "4 3 4 4", "output": "1024" }, { "input": "5 1 5 6100", "output": "1280" } ]

1,700,102,665

2,147,483,647

Python 3

OK

TESTS

60

46

0

a,b,c,d = map(int,input().split()) dic = {2:a,3:b,5:c,6:d} k = min([a,c,d]) res = 0 res += k*256 dic[2] -= k dic[5] -= k dic[6] -= k if dic[2] > 0 and dic[3] > 0: k = min([dic[2],dic[3]]) res += 32 *k print(res)

Title: Anton and Digits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently Anton found a box with digits in his room. There are *k*2 digits 2, *k*3 digits 3, *k*5 digits 5 and *k*6 digits 6. Anton's favorite integers are 32 and 256. He decided to compose this integers from digits he has. He wants to make the sum of these integers as large as possible. Help him solve this task! Each digit can be used no more than once, i.e. the composed integers should contain no more than *k*2 digits 2, *k*3 digits 3 and so on. Of course, unused digits are not counted in the sum. Input Specification: The only line of the input contains four integers *k*2, *k*3, *k*5 and *k*6 — the number of digits 2, 3, 5 and 6 respectively (0<=≤<=*k*2,<=*k*3,<=*k*5,<=*k*6<=≤<=5·106). Output Specification: Print one integer — maximum possible sum of Anton's favorite integers that can be composed using digits from the box. Demo Input: ['5 1 3 4\n', '1 1 1 1\n'] Demo Output: ['800\n', '256\n'] Note: In the first sample, there are five digits 2, one digit 3, three digits 5 and four digits 6. Anton can compose three integers 256 and one integer 32 to achieve the value 256 + 256 + 256 + 32 = 800. Note, that there is one unused integer 2 and one unused integer 6. They are not counted in the answer. In the second sample, the optimal answer is to create on integer 256, thus the answer is 256.

```python a,b,c,d = map(int,input().split()) dic = {2:a,3:b,5:c,6:d} k = min([a,c,d]) res = 0 res += k*256 dic[2] -= k dic[5] -= k dic[6] -= k if dic[2] > 0 and dic[3] > 0: k = min([dic[2],dic[3]]) res += 32 *k print(res) ```

3

443

A

Anton and Letters

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

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

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

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

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

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

none

500

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

1,693,201,023

2,147,483,647

Python 3

OK

TESTS

27

46

0

str = input() res = 0 arr = str[1:-1].split(', ') let = [0 for i in range(26)] for c in arr: if(c!= ''): let[ord(c)-ord('a')] = 1 for el in let: if(el==1): res += 1 print(res)

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

```python str = input() res = 0 arr = str[1:-1].split(', ') let = [0 for i in range(26)] for c in arr: if(c!= ''): let[ord(c)-ord('a')] = 1 for el in let: if(el==1): res += 1 print(res) ```

3

479

A

Expression

PROGRAMMING

1,000

[ "brute force", "math" ]

null

Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets: - 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9 Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2. It's easy to see that the maximum value that you can obtain is 9. Your task is: given *a*, *b* and *c* print the maximum value that you can get.

The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).

Print the maximum value of the expression that you can obtain.

[ "1\n2\n3\n", "2\n10\n3\n" ]

[ "9\n", "60\n" ]

none

500

[ { "input": "1\n2\n3", "output": "9" }, { "input": "2\n10\n3", "output": "60" }, { "input": "1\n1\n1", "output": "3" }, { "input": "1\n2\n1", "output": "4" }, { "input": "10\n10\n10", "output": "1000" }, { "input": "5\n1\n3", "output": "20" }, { "input": "3\n1\n5", "output": "20" }, { "input": "6\n7\n1", "output": "48" }, { "input": "1\n8\n3", "output": "27" }, { "input": "9\n7\n2", "output": "126" }, { "input": "1\n1\n10", "output": "20" }, { "input": "9\n1\n1", "output": "18" }, { "input": "10\n5\n6", "output": "300" }, { "input": "8\n9\n7", "output": "504" }, { "input": "4\n2\n10", "output": "80" }, { "input": "2\n8\n3", "output": "48" }, { "input": "3\n5\n7", "output": "105" }, { "input": "1\n10\n1", "output": "12" }, { "input": "2\n2\n2", "output": "8" }, { "input": "5\n6\n1", "output": "35" }, { "input": "10\n1\n1", "output": "20" }, { "input": "1\n6\n1", "output": "8" }, { "input": "1\n9\n1", "output": "11" }, { "input": "2\n1\n2", "output": "6" }, { "input": "2\n6\n1", "output": "14" }, { "input": "9\n2\n1", "output": "27" }, { "input": "1\n9\n2", "output": "20" }, { "input": "1\n3\n1", "output": "5" }, { "input": "2\n1\n1", "output": "4" } ]

1,693,769,151

2,147,483,647

PyPy 3

OK

TESTS

29

93

0

def solve(a: int, b: int, c: int) -> int: count=0 L=[a+b*c,a*(b+c),a*b*c,(a+b)*c,a*b+c,a+b+c] count=max(L) return (count) a=int(input()) b=int(input()) c=int(input()) print(solve(a, b, c))

Title: Expression Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets: - 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9 Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2. It's easy to see that the maximum value that you can obtain is 9. Your task is: given *a*, *b* and *c* print the maximum value that you can get. Input Specification: The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). Output Specification: Print the maximum value of the expression that you can obtain. Demo Input: ['1\n2\n3\n', '2\n10\n3\n'] Demo Output: ['9\n', '60\n'] Note: none

```python def solve(a: int, b: int, c: int) -> int: count=0 L=[a+b*c,a*(b+c),a*b*c,(a+b)*c,a*b+c,a+b+c] count=max(L) return (count) a=int(input()) b=int(input()) c=int(input()) print(solve(a, b, c)) ```

3

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

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

1,642,609,330

2,147,483,647

Python 3

OK

TESTS

102

46

0

a = input() b = input() length = len(a) i = 0 while i < length: ai = a[i] bi = b[i] if ai != bi: print("1", end='') else: print("0", end='') i += 1

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 a = input() b = input() length = len(a) i = 0 while i < length: ai = a[i] bi = b[i] if ai != bi: print("1", end='') else: print("0", end='') i += 1 ```

3.9885

926

B

Add Points

PROGRAMMING

1,800

[]

null

There are *n* points on a straight line, and the *i*-th point among them is located at *x**i*. All these coordinates are distinct. Determine the number *m* — the smallest number of points you should add on the line to make the distances between all neighboring points equal.

The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100<=000) — the number of points. The second line contains a sequence of integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the points. All these coordinates are distinct. The points can be given in an arbitrary order.

Print a single integer *m* — the smallest number of points you should add on the line to make the distances between all neighboring points equal.

[ "3\n-5 10 5\n", "6\n100 200 400 300 600 500\n", "4\n10 9 0 -1\n" ]

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

In the first example you can add one point with coordinate 0. In the second example the distances between all neighboring points are already equal, so you shouldn't add anything.

0

[ { "input": "3\n-5 10 5", "output": "1" }, { "input": "6\n100 200 400 300 600 500", "output": "0" }, { "input": "4\n10 9 0 -1", "output": "8" }, { "input": "3\n1 4 7", "output": "0" }, { "input": "3\n1 4 6", "output": "3" }, { "input": "3\n1 2 6", "output": "3" }, { "input": "3\n1 3 6", "output": "3" }, { "input": "4\n1 2 3 4", "output": "0" }, { "input": "3\n-1000000000 -999999999 1000000000", "output": "1999999998" }, { "input": "3\n-1000000000 999999999 1000000000", "output": "1999999998" }, { "input": "3\n-1000000000 -999999998 1000000000", "output": "999999998" }, { "input": "3\n-1000000000 999999998 1000000000", "output": "999999998" }, { "input": "3\n422800963 4663162 694989823", "output": "230108885" }, { "input": "5\n-268968800 -435386086 -484420288 579138544 945328473", "output": "204249819" }, { "input": "10\n711183437 845779129 -106125616 -481773790 66231250 -183390793 -711197523 -196001897 -440633306 -873649505", "output": "156311685" }, { "input": "3\n300000002 -799999998 -599999998", "output": "9" }, { "input": "5\n-166282087 234698547 -853072571 644571043 444292437", "output": "3533" }, { "input": "7\n996073710 -246878649 34663194 35526441 634003254 -704646201 -905166147", "output": "15411" }, { "input": "3\n-1000000000 1 1000000000", "output": "1999999998" }, { "input": "5\n-1000000000 1000000000 1 2 -30000", "output": "1999999996" }, { "input": "3\n422800963 0 1000000000", "output": "999999998" }, { "input": "3\n1000000000 500000001 -500000001", "output": "1499999999" } ]

1,521,305,655

4,755

Python 3

OK

TESTS

134

295

15,667,200

n = int(input()) a = [int(x) for x in input().strip().split()] a = sorted(a) d = [] for i in range(len(a) - 1): d.append(a[i + 1] - a[i]) def gcd(a, b): if(b == 0): return a else: return gcd(b, a % b) g = d[0] for e in d: g = gcd(g, e) cnt = len(a) print( int( ( a[len(a) - 1] - a[0]) / g + 1 - cnt))

Title: Add Points Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* points on a straight line, and the *i*-th point among them is located at *x**i*. All these coordinates are distinct. Determine the number *m* — the smallest number of points you should add on the line to make the distances between all neighboring points equal. Input Specification: The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100<=000) — the number of points. The second line contains a sequence of integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the points. All these coordinates are distinct. The points can be given in an arbitrary order. Output Specification: Print a single integer *m* — the smallest number of points you should add on the line to make the distances between all neighboring points equal. Demo Input: ['3\n-5 10 5\n', '6\n100 200 400 300 600 500\n', '4\n10 9 0 -1\n'] Demo Output: ['1\n', '0\n', '8\n'] Note: In the first example you can add one point with coordinate 0. In the second example the distances between all neighboring points are already equal, so you shouldn't add anything.

```python n = int(input()) a = [int(x) for x in input().strip().split()] a = sorted(a) d = [] for i in range(len(a) - 1): d.append(a[i + 1] - a[i]) def gcd(a, b): if(b == 0): return a else: return gcd(b, a % b) g = d[0] for e in d: g = gcd(g, e) cnt = len(a) print( int( ( a[len(a) - 1] - a[0]) / g + 1 - cnt)) ```

3

4

A

Watermelon

PROGRAMMING

800

[ "brute force", "math" ]

A. Watermelon

1

64

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

[ "8\n" ]

[ "YES\n" ]

For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

0

[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]

1,689,521,158

2,147,483,647

Python 3

OK

TESTS

20

92

0

w = int(input()) if w==1 or w==2: print("NO") elif w % 2 == 0: print("YES") else: print("NO")

Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

```python w = int(input()) if w==1 or w==2: print("NO") elif w % 2 == 0: print("YES") else: print("NO") ```

3.954

847

E

Packmen

PROGRAMMING

1,800

[ "binary search", "dp" ]

null

A game field is a strip of 1<=×<=*n* square cells. In some cells there are Packmen, in some cells — asterisks, other cells are empty. Packman can move to neighboring cell in 1 time unit. If there is an asterisk in the target cell then Packman eats it. Packman doesn't spend any time to eat an asterisk. In the initial moment of time all Packmen begin to move. Each Packman can change direction of its move unlimited number of times, but it is not allowed to go beyond the boundaries of the game field. Packmen do not interfere with the movement of other packmen; in one cell there can be any number of packmen moving in any directions. Your task is to determine minimum possible time after which Packmen can eat all the asterisks.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the length of the game field. The second line contains the description of the game field consisting of *n* symbols. If there is symbol '.' in position *i* — the cell *i* is empty. If there is symbol '*' in position *i* — in the cell *i* contains an asterisk. If there is symbol 'P' in position *i* — Packman is in the cell *i*. It is guaranteed that on the game field there is at least one Packman and at least one asterisk.

Print minimum possible time after which Packmen can eat all asterisks.

[ "7\n*..P*P*\n", "10\n.**PP.*P.*\n" ]

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

In the first example Packman in position 4 will move to the left and will eat asterisk in position 1. He will spend 3 time units on it. During the same 3 time units Packman in position 6 will eat both of neighboring with it asterisks. For example, it can move to the left and eat asterisk in position 5 (in 1 time unit) and then move from the position 5 to the right and eat asterisk in the position 7 (in 2 time units). So in 3 time units Packmen will eat all asterisks on the game field. In the second example Packman in the position 4 will move to the left and after 2 time units will eat asterisks in positions 3 and 2. Packmen in positions 5 and 8 will move to the right and in 2 time units will eat asterisks in positions 7 and 10, respectively. So 2 time units is enough for Packmen to eat all asterisks on the game field.

0

[ { "input": "7\n*..P*P*", "output": "3" }, { "input": "10\n.**PP.*P.*", "output": "2" }, { "input": "19\n**P.*..*..P..*.*P**", "output": "7" }, { "input": "12\nP**.*P*P*P**", "output": "3" }, { "input": "58\n..P.P*.P*.P...PPP...P*....*..*.**......*P.*P.....**P...*P*", "output": "9" }, { "input": "10\n..P*.P.*.*", "output": "4" }, { "input": "10\n***.*.*..P", "output": "9" }, { "input": "15\nP***..PPP..P*.P", "output": "3" }, { "input": "20\n.P**P**P**PP.PP**PP*", "output": "2" }, { "input": "20\n.....*.**..........P", "output": "14" }, { "input": "25\n...*..**..*.....*..*...P.", "output": "20" }, { "input": "30\n*P.*...*.**..P**...***.*...**.", "output": "15" }, { "input": "30\n.*...*.......................P", "output": "28" }, { "input": "35\n..PP.P....*PP.*.PPPP.*P.P.PPPP.*.P.", "output": "2" }, { "input": "40\n...**P*P*...P.*PP***.*..P..**.**PP**.*.*", "output": "6" }, { "input": "40\nP*....*.*....*...*..*.......*...**..***.", "output": "38" }, { "input": "45\nP.P*..P....*P.*PP*PP*.**P...PP*PP*.P.P..PP.PP", "output": "2" }, { "input": "45\n*.*.*..*.*.**.*..**..*.....**.**P....*****.**", "output": "56" }, { "input": "50\n*PP....PPPP*....*P*P..PPPPPP...***P*P.........PP..", "output": "3" }, { "input": "50\n*..***.*.****.*....P*.**...***.......**....*.***..", "output": "66" }, { "input": "55\n......P.*.....P*.*P....*..P*.P.P....**....*..........*.", "output": "22" }, { "input": "55\n*.....*.*..**..*...***..**.**.*.*.P..*.*.**...**.*..*.*", "output": "74" }, { "input": "60\n.P...P.PPP.P....P...PP.*......**...P.*.P.P*P.*...P*P..P.P..P", "output": "5" }, { "input": "60\n..*....*...***....P...........*............*....**....*...**", "output": "73" }, { "input": "65\n......PP..PP**.***.*.P.P..PP.....**PP*PP.....P..P*PP.*.*P..P*P*..", "output": "5" }, { "input": "70\n*..***.**..**....***........*.**...*...**.**..*.......**P*..*.......**", "output": "82" }, { "input": "75\n..***P*.**.P.**P.**.***.P*..**P.P*.P*.**.....*PP..P***....**PPP..**P..P..P*", "output": "6" }, { "input": "80\n*..**..*...*....*.*.**.*.*..*..P..*..**.**..*..**.*.*.*.*.***...*.*..**.*....**.", "output": "109" }, { "input": "85\n.*.....*.....**..........*P*..........*.........*...*..*...****..*..*P*..*..P.***...*", "output": "31" }, { "input": "90\n......P.*.PPP...*.P.**P..*.*.*..*P**PP**..***.PPP....P..**P*.*.*..*.P*P.*PP*.....P.*.**P**", "output": "5" }, { "input": "95\n.*..P****....****.*.***P..*.*.**P..*.***.PP.**.**....*****P**..*..*..***...*P..P****.*.P...*..*", "output": "12" }, { "input": "95\n.*.***...*...P**...*.**.*..*.*..*...****..*.*.*..*.****.*....*...*..*..**.**.********..*.*...**", "output": "105" }, { "input": "100\nPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP*PPPPPPPPPPPPP", "output": "1" }, { "input": "100\n************************************************************P***************************************", "output": "138" }, { "input": "100\n......*....*...P*P*.....**.......P.P..........*........*....PP.................P......*.P.P.......*.", "output": "11" }, { "input": "100\n**.**.**.........P*..P**.*.******.***.....***..*.*P.*.***.*.*..*.***..*********.*...***..*..*...P*..", "output": "26" }, { "input": "100\n.PPP....PPPP.*...P...PPP..*P...P.*P.PP..P.P...PPPPP..PP.P..P..P..P...P.......P..PP..P..PPPPPP.P.PPPP", "output": "2" }, { "input": "100\n...............................................*P.........*..P.........P.P....P..P.................*", "output": "18" }, { "input": "2\n*P", "output": "1" }, { "input": "2\nP*", "output": "1" } ]

1,691,834,318

2,147,483,647

PyPy 3

OK

TESTS

77

202

5,120,000

n = int(input()) string = input() player = [] food = [] for i in range(len(string)): if string[i] == "*": food.append(i) elif string[i] == "P": player.append(i) def check(t): idx = 0 for i in range(len(player)): loc = player[i] if idx == len(food):break l = food[idx] while idx!=len(food): r = food[idx] if (((abs(loc-l)+abs(l-r))<=t) or ((abs(l-r)+abs(loc-r))<=t)): idx+=1 else: break return idx>=len(food) l = 0 h = pow(10,18) while l<=h: mid = l+(h-l)//2 if check(mid): ans = mid h = mid-1 else: l= mid+1 print(ans)

Title: Packmen Time Limit: None seconds Memory Limit: None megabytes Problem Description: A game field is a strip of 1<=×<=*n* square cells. In some cells there are Packmen, in some cells — asterisks, other cells are empty. Packman can move to neighboring cell in 1 time unit. If there is an asterisk in the target cell then Packman eats it. Packman doesn't spend any time to eat an asterisk. In the initial moment of time all Packmen begin to move. Each Packman can change direction of its move unlimited number of times, but it is not allowed to go beyond the boundaries of the game field. Packmen do not interfere with the movement of other packmen; in one cell there can be any number of packmen moving in any directions. Your task is to determine minimum possible time after which Packmen can eat all the asterisks. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the length of the game field. The second line contains the description of the game field consisting of *n* symbols. If there is symbol '.' in position *i* — the cell *i* is empty. If there is symbol '*' in position *i* — in the cell *i* contains an asterisk. If there is symbol 'P' in position *i* — Packman is in the cell *i*. It is guaranteed that on the game field there is at least one Packman and at least one asterisk. Output Specification: Print minimum possible time after which Packmen can eat all asterisks. Demo Input: ['7\n*..P*P*\n', '10\n.**PP.*P.*\n'] Demo Output: ['3\n', '2\n'] Note: In the first example Packman in position 4 will move to the left and will eat asterisk in position 1. He will spend 3 time units on it. During the same 3 time units Packman in position 6 will eat both of neighboring with it asterisks. For example, it can move to the left and eat asterisk in position 5 (in 1 time unit) and then move from the position 5 to the right and eat asterisk in the position 7 (in 2 time units). So in 3 time units Packmen will eat all asterisks on the game field. In the second example Packman in the position 4 will move to the left and after 2 time units will eat asterisks in positions 3 and 2. Packmen in positions 5 and 8 will move to the right and in 2 time units will eat asterisks in positions 7 and 10, respectively. So 2 time units is enough for Packmen to eat all asterisks on the game field.

```python n = int(input()) string = input() player = [] food = [] for i in range(len(string)): if string[i] == "*": food.append(i) elif string[i] == "P": player.append(i) def check(t): idx = 0 for i in range(len(player)): loc = player[i] if idx == len(food):break l = food[idx] while idx!=len(food): r = food[idx] if (((abs(loc-l)+abs(l-r))<=t) or ((abs(l-r)+abs(loc-r))<=t)): idx+=1 else: break return idx>=len(food) l = 0 h = pow(10,18) while l<=h: mid = l+(h-l)//2 if check(mid): ans = mid h = mid-1 else: l= mid+1 print(ans) ```

3

598

A

Tricky Sum

PROGRAMMING

900

[ "math" ]

null

In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum. For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively. Calculate the answer for *t* values of *n*.

The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed. Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).

Print the requested sum for each of *t* integers *n* given in the input.

[ "2\n4\n1000000000\n" ]

[ "-4\n499999998352516354\n" ]

The answer for the first sample is explained in the statement.

0

[ { "input": "2\n4\n1000000000", "output": "-4\n499999998352516354" }, { "input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10", "output": "-1\n-3\n0\n-4\n1\n7\n14\n6\n15\n25" }, { "input": "10\n10\n9\n47\n33\n99\n83\n62\n1\n100\n53", "output": "25\n15\n1002\n435\n4696\n3232\n1827\n-1\n4796\n1305" }, { "input": "100\n901\n712\n3\n677\n652\n757\n963\n134\n205\n888\n847\n283\n591\n984\n1\n61\n540\n986\n950\n729\n104\n244\n500\n461\n251\n685\n631\n803\n526\n600\n1000\n899\n411\n219\n597\n342\n771\n348\n507\n775\n454\n102\n486\n333\n580\n431\n537\n355\n624\n23\n429\n276\n84\n704\n96\n536\n855\n653\n72\n718\n776\n658\n802\n777\n995\n285\n328\n405\n184\n555\n956\n410\n846\n853\n525\n983\n65\n549\n839\n929\n620\n725\n635\n303\n201\n878\n580\n139\n182\n69\n400\n788\n985\n792\n103\n248\n570\n839\n253\n417", "output": "404305\n251782\n0\n227457\n210832\n284857\n462120\n8535\n20605\n392670\n357082\n39164\n172890\n482574\n-1\n1765\n144024\n484545\n449679\n264039\n5206\n29380\n124228\n105469\n31116\n232909\n197350\n320760\n136555\n178254\n498454\n402504\n83644\n23580\n176457\n57631\n295560\n59704\n127756\n298654\n102263\n4999\n117319\n54589\n166444\n92074\n142407\n62168\n192954\n214\n91213\n37204\n3316\n246114\n4402\n141870\n363894\n211485\n2374\n256075\n299430\n214765\n319957\n300207\n493464\n39733\n52934\n81193\n16510\n15..." }, { "input": "1\n16", "output": "74" }, { "input": "60\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457", "output": "144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875..." }, { "input": "13\n1\n19\n31\n19\n19\n92\n74\n69\n32\n32\n91\n42\n73", "output": "-1\n128\n434\n128\n128\n4024\n2521\n2161\n402\n402\n3932\n777\n2447" }, { "input": "1\n16383", "output": "134176770" }, { "input": "16\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100", "output": "5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908" }, { "input": "1\n414234", "output": "85794061921" }, { "input": "1\n414232", "output": "85793233454" }, { "input": "3\n414231\n414231\n414231", "output": "85792819222\n85792819222\n85792819222" }, { "input": "1\n121", "output": "7127" } ]

1,678,197,871

871

PyPy 3

OK

TESTS

21

93

1,228,800

t = int(input()) for p in range(t): x = int(input()) k = 1 cnt = 0 while(k <= x): k *= 2 cnt += 1 ans = cnt print(x * (x + 1) // 2 - 2 * (2 ** ans) + 2)

Title: Tricky Sum Time Limit: None seconds Memory Limit: None megabytes Problem Description: In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum. For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively. Calculate the answer for *t* values of *n*. Input Specification: The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed. Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109). Output Specification: Print the requested sum for each of *t* integers *n* given in the input. Demo Input: ['2\n4\n1000000000\n'] Demo Output: ['-4\n499999998352516354\n'] Note: The answer for the first sample is explained in the statement.

```python t = int(input()) for p in range(t): x = int(input()) k = 1 cnt = 0 while(k <= x): k *= 2 cnt += 1 ans = cnt print(x * (x + 1) // 2 - 2 * (2 ** ans) + 2) ```

3

255

B

Code Parsing

PROGRAMMING

1,200

[ "implementation" ]

null

Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string *s*, consisting of characters "x" and "y", and uses two following operations at runtime: 1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string. 1. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string. The input for the new algorithm is string *s*, and the algorithm works as follows: 1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string. 1. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm. Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string *s*.

The first line contains a non-empty string *s*. It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.

In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string *s*.

[ "x\n", "yxyxy\n", "xxxxxy\n" ]

[ "x\n", "y\n", "xxxx\n" ]

In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change. In the second test the transformation will be like this: 1. string "yxyxy" transforms into string "xyyxy"; 1. string "xyyxy" transforms into string "xyxyy"; 1. string "xyxyy" transforms into string "xxyyy"; 1. string "xxyyy" transforms into string "xyy"; 1. string "xyy" transforms into string "y". As a result, we've got string "y". In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".

1,000

[ { "input": "x", "output": "x" }, { "input": "yxyxy", "output": "y" }, { "input": "xxxxxy", "output": "xxxx" }, { "input": "yxyyxyyx", "output": "yy" }, { "input": "yxxyxyx", "output": "x" }, { "input": "xxx", "output": "xxx" }, { "input": "xxyxx", "output": "xxx" }, { "input": "xxxyx", "output": "xxx" }, { "input": "yxxxx", "output": "xxx" }, { "input": "xyyxyyyyyxxxxxxxyxyxyyxyyxyyxxyxyxyxxxyxxy", "output": "xx" }, { "input": "xyyxyyyyyxxxxxxxyxyxyyxyyxyyxxyxyxyxxxyxxy", "output": "xx" }, { "input": "xxxxxxxxxxxyxyyxxxxyxxxxxyxxxxxyxxxxxxxxyx", "output": "xxxxxxxxxxxxxxxxxxxxxxxxxxxx" }, { "input": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx" }, { "input": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy", "output": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy" } ]

1,699,025,954

2,147,483,647

PyPy 3

OK

TESTS

49

498

9,830,400

s=input() cx = 0 cy = 0 for c in s: if c == "x": cx+=1 else: cy+=1 if cx > cy: out = "x"*(cx-cy) else: out = "y"*(cy-cx) print(out)

Title: Code Parsing Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string *s*, consisting of characters "x" and "y", and uses two following operations at runtime: 1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string. 1. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string. The input for the new algorithm is string *s*, and the algorithm works as follows: 1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string. 1. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm. Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string *s*. Input Specification: The first line contains a non-empty string *s*. It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string. Output Specification: In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string *s*. Demo Input: ['x\n', 'yxyxy\n', 'xxxxxy\n'] Demo Output: ['x\n', 'y\n', 'xxxx\n'] Note: In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change. In the second test the transformation will be like this: 1. string "yxyxy" transforms into string "xyyxy"; 1. string "xyyxy" transforms into string "xyxyy"; 1. string "xyxyy" transforms into string "xxyyy"; 1. string "xxyyy" transforms into string "xyy"; 1. string "xyy" transforms into string "y". As a result, we've got string "y". In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".

```python s=input() cx = 0 cy = 0 for c in s: if c == "x": cx+=1 else: cy+=1 if cx > cy: out = "x"*(cx-cy) else: out = "y"*(cy-cx) print(out) ```

3

1,004

B

Sonya and Exhibition

PROGRAMMING

1,300

[ "constructive algorithms", "greedy", "implementation", "math" ]

null

Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition. There are $n$ flowers in a row in the exhibition. Sonya can put either a rose or a lily in the $i$-th position. Thus each of $n$ positions should contain exactly one flower: a rose or a lily. She knows that exactly $m$ people will visit this exhibition. The $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies. Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.

The first line contains two integers $n$ and $m$ ($1\leq n, m\leq 10^3$) — the number of flowers and visitors respectively. Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1\leq l_i\leq r_i\leq n$), meaning that $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive.

Print the string of $n$ characters. The $i$-th symbol should be «0» if you want to put a rose in the $i$-th position, otherwise «1» if you want to put a lily. If there are multiple answers, print any.

[ "5 3\n1 3\n2 4\n2 5\n", "6 3\n5 6\n1 4\n4 6\n" ]

[ "01100", "110010" ]

In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions; - in the segment $[1\ldots3]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots4]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots5]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$. The total beauty is equal to $2+2+4=8$. In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions; - in the segment $[5\ldots6]$, there are one rose and one lily, so the beauty is equal to $1\cdot 1=1$; - in the segment $[1\ldots4]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$; - in the segment $[4\ldots6]$, there are two roses and one lily, so the beauty is equal to $2\cdot 1=2$. The total beauty is equal to $1+4+2=7$.

1,000

[ { "input": "5 3\n1 3\n2 4\n2 5", "output": "01010" }, { "input": "6 3\n5 6\n1 4\n4 6", "output": "010101" }, { "input": "10 4\n3 3\n1 6\n9 9\n10 10", "output": "0101010101" }, { "input": "1 1\n1 1", "output": "0" }, { "input": "1000 10\n3 998\n2 1000\n1 999\n2 1000\n3 998\n2 1000\n3 998\n1 1000\n2 1000\n3 999", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "1000 20\n50 109\n317 370\n710 770\n440 488\n711 757\n236 278\n314 355\n131 190\n115 162\n784 834\n16 56\n677 730\n802 844\n632 689\n23 74\n647 702\n930 986\n926 983\n769 822\n508 558", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "1000 10\n138 238\n160 260\n716 816\n504 604\n98 198\n26 126\n114 214\n217 317\n121 221\n489 589", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "1000 5\n167 296\n613 753\n650 769\n298 439\n71 209", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "1000 5\n349 415\n714 773\n125 179\n1 80\n148 242", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "914 10\n587 646\n770 843\n825 875\n439 485\n465 521\n330 387\n405 480\n477 521\n336 376\n715 771", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "571 10\n13 94\n450 510\n230 293\n302 375\n304 354\n421 504\n24 87\n122 181\n221 296\n257 307", "output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..." }, { "input": "6 2\n1 6\n1 4", "output": "010101" }, { "input": "2 1\n1 2", "output": "01" } ]

1,530,841,926

2,147,483,647

Python 3

OK

TESTS

27

109

0

n,k=[int(x) for x in input().split()] for i in range(k): a=input() a='0' s='' for i in range(n): s+=a if a=='0': a='1' else: a='0' print(s)

Title: Sonya and Exhibition Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition. There are $n$ flowers in a row in the exhibition. Sonya can put either a rose or a lily in the $i$-th position. Thus each of $n$ positions should contain exactly one flower: a rose or a lily. She knows that exactly $m$ people will visit this exhibition. The $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies. Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible. Input Specification: The first line contains two integers $n$ and $m$ ($1\leq n, m\leq 10^3$) — the number of flowers and visitors respectively. Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1\leq l_i\leq r_i\leq n$), meaning that $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive. Output Specification: Print the string of $n$ characters. The $i$-th symbol should be «0» if you want to put a rose in the $i$-th position, otherwise «1» if you want to put a lily. If there are multiple answers, print any. Demo Input: ['5 3\n1 3\n2 4\n2 5\n', '6 3\n5 6\n1 4\n4 6\n'] Demo Output: ['01100', '110010'] Note: In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions; - in the segment $[1\ldots3]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots4]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots5]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$. The total beauty is equal to $2+2+4=8$. In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions; - in the segment $[5\ldots6]$, there are one rose and one lily, so the beauty is equal to $1\cdot 1=1$; - in the segment $[1\ldots4]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$; - in the segment $[4\ldots6]$, there are two roses and one lily, so the beauty is equal to $2\cdot 1=2$. The total beauty is equal to $1+4+2=7$.

```python n,k=[int(x) for x in input().split()] for i in range(k): a=input() a='0' s='' for i in range(n): s+=a if a=='0': a='1' else: a='0' print(s) ```

3

545

B

Equidistant String

PROGRAMMING

1,100

[ "greedy" ]

null

Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance: We will define the distance between two strings *s* and *t* of the same length consisting of digits zero and one as the number of positions *i*, such that *s**i* isn't equal to *t**i*. As besides everything else Susie loves symmetry, she wants to find for two strings *s* and *t* of length *n* such string *p* of length *n*, that the distance from *p* to *s* was equal to the distance from *p* to *t*. It's time for Susie to go to bed, help her find such string *p* or state that it is impossible.

The first line contains string *s* of length *n*. The second line contains string *t* of length *n*. The length of string *n* is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.

Print a string of length *n*, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes). If there are multiple possible answers, print any of them.

[ "0001\n1011\n", "000\n111\n" ]

[ "0011\n", "impossible\n" ]

In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.

1,000

[ { "input": "0001\n1011", "output": "0011" }, { "input": "000\n111", "output": "impossible" }, { "input": "1010101011111110111111001111111111111111111111101101110111111111111110110110101011111110110111111101\n0101111111000100010100001100010101000000011000000000011011000001100100001110111011111000001110011111", "output": "1111101111101100110110001110110111010101011101001001010011101011101100100110111011111100100110111111" }, { "input": "0000000001000000000000100000100001000000\n1111111011111101111111111111111111111111", "output": "0101010011010100101010110101101011010101" }, { "input": "10101000101001001101010010000101100011010011000011001001001111110010100110000001111111\n01001011110111111101111011011111110000000111111001000011010101001010000111101010000101", "output": "11101010111101101101110011001101110010010111010001001011000111011010100111001000101101" }, { "input": "1111111111111111111111111110111111111111111111111111111111111110111111101111111111111111111111111111\n1111111111111111111001111111110010111111111111111111001111111111111111111111111111111111111111111111", "output": "1111111111111111111101111110110110111111111111111111101111111110111111111111111111111111111111111111" }, { "input": "0000000000000000000000000000111111111111111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "0101010101010101010101000000101010101010101010101010101010101010101010101010101010101010101010101010" }, { "input": "00001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n11111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "01011110101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "impossible" }, { "input": "1\n1", "output": "1" }, { "input": "1\n0", "output": "impossible" }, { "input": "1111\n0000", "output": "1010" }, { "input": "1111\n1001", "output": "1101" }, { "input": "0000\n1111", "output": "0101" }, { "input": "1010\n0101", "output": "1111" } ]

1,686,001,568

2,147,483,647

PyPy 3-64

OK

TESTS

54

826

10,752,000

s = input() t = input() p = "" c = 0 for i in range(len(s)): if s[i] == t[i]: p += s[i] else: c += 1 if c % 2 == 1: p += s[i] else: p += t[i] if c % 2 == 1: print("impossible") else: print(p)

Title: Equidistant String Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance: We will define the distance between two strings *s* and *t* of the same length consisting of digits zero and one as the number of positions *i*, such that *s**i* isn't equal to *t**i*. As besides everything else Susie loves symmetry, she wants to find for two strings *s* and *t* of length *n* such string *p* of length *n*, that the distance from *p* to *s* was equal to the distance from *p* to *t*. It's time for Susie to go to bed, help her find such string *p* or state that it is impossible. Input Specification: The first line contains string *s* of length *n*. The second line contains string *t* of length *n*. The length of string *n* is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one. Output Specification: Print a string of length *n*, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes). If there are multiple possible answers, print any of them. Demo Input: ['0001\n1011\n', '000\n111\n'] Demo Output: ['0011\n', 'impossible\n'] Note: In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.

```python s = input() t = input() p = "" c = 0 for i in range(len(s)): if s[i] == t[i]: p += s[i] else: c += 1 if c % 2 == 1: p += s[i] else: p += t[i] if c % 2 == 1: print("impossible") else: print(p) ```

3

548

B

Mike and Fun

PROGRAMMING

1,400

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

null

Mike and some bears are playing a game just for fun. Mike is the judge. All bears except Mike are standing in an *n*<=×<=*m* grid, there's exactly one bear in each cell. We denote the bear standing in column number *j* of row number *i* by (*i*,<=*j*). Mike's hands are on his ears (since he's the judge) and each bear standing in the grid has hands either on his mouth or his eyes. They play for *q* rounds. In each round, Mike chooses a bear (*i*,<=*j*) and tells him to change his state i. e. if his hands are on his mouth, then he'll put his hands on his eyes or he'll put his hands on his mouth otherwise. After that, Mike wants to know the score of the bears. Score of the bears is the maximum over all rows of number of consecutive bears with hands on their eyes in that row. Since bears are lazy, Mike asked you for help. For each round, tell him the score of these bears after changing the state of a bear selected in that round.

The first line of input contains three integers *n*, *m* and *q* (1<=≤<=*n*,<=*m*<=≤<=500 and 1<=≤<=*q*<=≤<=5000). The next *n* lines contain the grid description. There are *m* integers separated by spaces in each line. Each of these numbers is either 0 (for mouth) or 1 (for eyes). The next *q* lines contain the information about the rounds. Each of them contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n* and 1<=≤<=*j*<=≤<=*m*), the row number and the column number of the bear changing his state.

After each round, print the current score of the bears.

[ "5 4 5\n0 1 1 0\n1 0 0 1\n0 1 1 0\n1 0 0 1\n0 0 0 0\n1 1\n1 4\n1 1\n4 2\n4 3\n" ]

[ "3\n4\n3\n3\n4\n" ]

none

1,000

[ { "input": "5 4 5\n0 1 1 0\n1 0 0 1\n0 1 1 0\n1 0 0 1\n0 0 0 0\n1 1\n1 4\n1 1\n4 2\n4 3", "output": "3\n4\n3\n3\n4" }, { "input": "2 2 10\n1 1\n0 1\n1 1\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 2\n1 1\n1 1", "output": "1\n2\n2\n2\n1\n1\n1\n1\n2\n1" }, { "input": "2 2 10\n1 1\n0 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n1 1\n2 1\n1 1\n2 1", "output": "2\n2\n1\n2\n1\n2\n1\n2\n2\n2" }, { "input": "5 5 30\n0 1 1 1 0\n1 1 0 1 1\n0 1 1 1 1\n0 0 1 1 0\n0 0 0 0 0\n3 2\n2 2\n2 2\n4 3\n1 4\n3 2\n4 1\n2 4\n1 4\n2 1\n5 2\n4 1\n4 1\n5 1\n2 4\n2 4\n4 4\n1 2\n3 1\n4 5\n1 2\n2 3\n1 1\n5 1\n3 4\n1 1\n5 4\n1 5\n5 4\n2 2", "output": "3\n3\n3\n3\n3\n4\n4\n4\n4\n4\n4\n4\n4\n4\n4\n4\n4\n4\n5\n5\n5\n5\n5\n5\n4\n3\n3\n4\n4\n4" }, { "input": "1 1 10\n0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "1\n0\n1\n0\n1\n0\n1\n0\n1\n0" }, { "input": "1 1 3\n1\n1 1\n1 1\n1 1", "output": "0\n1\n0" }, { "input": "1 5 5\n0 0 0 0 0\n1 2\n1 1\n1 4\n1 5\n1 3", "output": "1\n2\n2\n2\n5" }, { "input": "5 1 5\n0\n0\n0\n0\n0\n1 1\n2 1\n3 1\n4 1\n5 1", "output": "1\n1\n1\n1\n1" }, { "input": "1 1 1\n0\n1 1", "output": "1" }, { "input": "2 2 1\n1 1\n1 1\n1 1", "output": "2" } ]

1,607,518,627

2,147,483,647

PyPy 3

OK

TESTS

35

374

7,475,200

n,m,q = map(int,input().split()) grid =[]*m for i in range(n): grid.append(list(map(int,input().split()))) #print(grid) count = [] for i in range(n): c=0 tmp=0 for j in range(m): if grid[i][j]==1: c+=1 else: tmp=max(c,tmp) c=0 count.append(max(tmp,c)) #ans = max(count) #print(count) for i in range(q): x,y = map(int,input().split()) x-=1;y-=1 if grid[x][y]==0: grid[x][y]=1 else: grid[x][y]=0 c=tmp=0 for i in range(m): if grid[x][i]==1: c+=1 else: tmp=max(c,tmp) c=0 count[x]=max(tmp,c) print(max(count))

Title: Mike and Fun Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mike and some bears are playing a game just for fun. Mike is the judge. All bears except Mike are standing in an *n*<=×<=*m* grid, there's exactly one bear in each cell. We denote the bear standing in column number *j* of row number *i* by (*i*,<=*j*). Mike's hands are on his ears (since he's the judge) and each bear standing in the grid has hands either on his mouth or his eyes. They play for *q* rounds. In each round, Mike chooses a bear (*i*,<=*j*) and tells him to change his state i. e. if his hands are on his mouth, then he'll put his hands on his eyes or he'll put his hands on his mouth otherwise. After that, Mike wants to know the score of the bears. Score of the bears is the maximum over all rows of number of consecutive bears with hands on their eyes in that row. Since bears are lazy, Mike asked you for help. For each round, tell him the score of these bears after changing the state of a bear selected in that round. Input Specification: The first line of input contains three integers *n*, *m* and *q* (1<=≤<=*n*,<=*m*<=≤<=500 and 1<=≤<=*q*<=≤<=5000). The next *n* lines contain the grid description. There are *m* integers separated by spaces in each line. Each of these numbers is either 0 (for mouth) or 1 (for eyes). The next *q* lines contain the information about the rounds. Each of them contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n* and 1<=≤<=*j*<=≤<=*m*), the row number and the column number of the bear changing his state. Output Specification: After each round, print the current score of the bears. Demo Input: ['5 4 5\n0 1 1 0\n1 0 0 1\n0 1 1 0\n1 0 0 1\n0 0 0 0\n1 1\n1 4\n1 1\n4 2\n4 3\n'] Demo Output: ['3\n4\n3\n3\n4\n'] Note: none

```python n,m,q = map(int,input().split()) grid =[]*m for i in range(n): grid.append(list(map(int,input().split()))) #print(grid) count = [] for i in range(n): c=0 tmp=0 for j in range(m): if grid[i][j]==1: c+=1 else: tmp=max(c,tmp) c=0 count.append(max(tmp,c)) #ans = max(count) #print(count) for i in range(q): x,y = map(int,input().split()) x-=1;y-=1 if grid[x][y]==0: grid[x][y]=1 else: grid[x][y]=0 c=tmp=0 for i in range(m): if grid[x][i]==1: c+=1 else: tmp=max(c,tmp) c=0 count[x]=max(tmp,c) print(max(count)) ```

3

897

A

Scarborough Fair

PROGRAMMING

800

[ "implementation" ]

null

Parsley, sage, rosemary and thyme. Remember me to one who lives there. He once was the true love of mine. Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there. Willem asks his friend, Grick for directions, Grick helped them, and gave them a task. Although the girl wants to help, Willem insists on doing it by himself. Grick gave Willem a string of length *n*. Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed. Grick wants to know the final string after all the *m* operations.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains a string *s* of length *n*, consisting of lowercase English letters. Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.

Output string *s* after performing *m* operations described above.

[ "3 1\nioi\n1 1 i n\n", "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n" ]

[ "noi", "gaaak" ]

For the second example: After the first operation, the string is wxxak. After the second operation, the string is waaak. After the third operation, the string is gaaak.

500

[ { "input": "3 1\nioi\n1 1 i n", "output": "noi" }, { "input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g", "output": "gaaak" }, { "input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n4 5 a e\n3 9 f a\n1 2 c h\n4 8 a c\n3 5 e d\n3 4 g f\n2 3 d h\n2 3 d e\n1 7 d g\n2 6 e g\n2 3 d g\n5 5 h h\n2 8 g d\n8 9 a f\n5 9 c e\n1 7 f d\n1 6 e e\n5 7 c a\n8 9 b b\n2 6 e b\n6 6 g h\n1 2 b b\n1 5 a f\n5 8 f h\n1 5 e g\n3 9 f h\n6 8 g a\n4 6 h g\n1 5 f a\n5 6 a c\n4 8 e d\n1 4 d g\n7 8 b f\n5 6 h b\n3 9 c e\n1 9 b a", "output": "aahaddddh" }, { "input": "28 45\ndcbbaddjhbeefjadjchgkhgggfha\n10 25 c a\n13 19 a f\n12 28 e d\n12 27 e a\n9 20 b e\n7 17 g d\n22 26 j j\n8 16 c g\n14 16 a d\n3 10 f c\n10 26 d b\n8 17 i e\n10 19 d i\n6 21 c j\n7 22 b k\n17 19 a i\n4 18 j k\n8 25 a g\n10 27 j e\n9 18 g d\n16 23 h a\n17 26 k e\n8 16 h f\n1 15 d f\n22 28 k k\n11 20 c k\n6 11 b h\n17 17 e i\n15 22 g h\n8 18 c f\n4 16 e a\n8 25 b c\n6 24 d g\n5 9 f j\n12 19 i h\n4 25 e f\n15 25 c j\n15 27 e e\n11 20 b f\n19 27 e k\n2 21 d a\n9 27 k e\n14 24 b a\n3 6 i g\n2 26 k f", "output": "fcbbajjfjaaefefehfahfagggfha" }, { "input": "87 5\nnfinedeojadjmgafnaogekfjkjfncnliagfchjfcmellgigjjcaaoeakdolchjcecljdeblmheimkibkgdkcdml\n47 56 a k\n51 81 o d\n5 11 j h\n48 62 j d\n16 30 k m", "output": "nfinedeohadjmgafnaogemfjmjfncnliagfchjfcmellgigddckkdekkddlchdcecljdeblmheimkibkgdkcdml" }, { "input": "5 16\nacfbb\n1 2 e f\n2 5 a f\n2 3 b e\n4 4 f a\n2 3 f a\n1 2 b e\n4 5 c d\n2 4 e c\n1 4 e a\n1 3 d c\n3 5 e b\n3 5 e b\n2 2 e d\n1 3 e c\n3 3 a e\n1 5 a a", "output": "acebb" }, { "input": "94 13\nbcaaaaaaccacddcdaacbdaabbcbaddbccbccbbbddbadddcccbddadddaadbdababadaacdcdbcdadabdcdcbcbcbcbbcd\n52 77 d d\n21 92 d b\n45 48 c b\n20 25 d a\n57 88 d b\n3 91 b d\n64 73 a a\n5 83 b d\n2 69 c c\n28 89 a b\n49 67 c b\n41 62 a c\n49 87 b c", "output": "bcaaaaaaccacddcdaacddaaddcdbdddccdccddddddbdddddcdddcdddccdddcdcdcdcccdcddcdcdcddcdcdcdcdcdbcd" }, { "input": "67 39\nacbcbccccbabaabcabcaaaaaaccbcbbcbaaaacbbcccbcbabbcacccbbabbabbabaac\n4 36 a b\n25 38 a a\n3 44 b c\n35 57 b a\n4 8 a c\n20 67 c a\n30 66 b b\n27 40 a a\n2 56 a b\n10 47 c a\n22 65 c b\n29 42 a b\n1 46 c b\n57 64 b c\n20 29 b a\n14 51 c a\n12 55 b b\n20 20 a c\n2 57 c a\n22 60 c b\n16 51 c c\n31 64 a c\n17 30 c a\n23 36 c c\n28 67 a c\n37 40 a c\n37 50 b c\n29 48 c b\n2 34 b c\n21 53 b a\n26 63 a c\n23 28 c a\n51 56 c b\n32 61 b b\n64 67 b b\n21 67 b c\n8 53 c c\n40 62 b b\n32 38 c c", "output": "accccccccaaaaaaaaaaaaaaaaaaaccccccccccccccccccccccccccccccccccccccc" }, { "input": "53 33\nhhcbhfafeececbhadfbdbehdfacfchbhdbfebdfeghebfcgdhehfh\n27 41 h g\n18 35 c b\n15 46 h f\n48 53 e g\n30 41 b c\n12 30 b f\n10 37 e f\n18 43 a h\n10 52 d a\n22 48 c e\n40 53 f d\n7 12 b h\n12 51 f a\n3 53 g a\n19 41 d h\n22 29 b h\n2 30 a b\n26 28 e h\n25 35 f a\n19 31 h h\n44 44 d e\n19 22 e c\n29 44 d h\n25 33 d h\n3 53 g c\n18 44 h b\n19 28 f e\n3 22 g h\n8 17 c a\n37 51 d d\n3 28 e h\n27 50 h h\n27 46 f b", "output": "hhcbhfbfhfababbbbbbbbbbbbbbbbbeaaeaaeaaeabebdeaahahdh" }, { "input": "83 10\nfhbecdgadecabbbecedcgfdcefcbgechbedagecgdgfgdaahchdgchbeaedgafdefecdchceececfcdhcdh\n9 77 e e\n26 34 b g\n34 70 b a\n40 64 e g\n33 78 h f\n14 26 a a\n17 70 d g\n56 65 a c\n8 41 d c\n11 82 c b", "output": "fhbecdgacebabbbebegbgfgbefbggebhgegagebgggfggaafbfggbfagbgggbfggfebgbfbeebebfbdhbdh" }, { "input": "1 4\ne\n1 1 c e\n1 1 e a\n1 1 e c\n1 1 d a", "output": "a" }, { "input": "71 21\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n61 61 a a\n32 56 a a\n10 67 a a\n7 32 a a\n26 66 a a\n41 55 a a\n49 55 a a\n4 61 a a\n53 59 a a\n37 58 a a\n7 63 a a\n39 40 a a\n51 64 a a\n27 37 a a\n22 71 a a\n4 45 a a\n7 8 a a\n43 46 a a\n19 28 a a\n51 54 a a\n14 67 a a", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" }, { "input": "30 4\neaaddabedcbbcccddbabdecadcecce\n2 17 c a\n16 29 e e\n16 21 c b\n7 11 b c", "output": "eaaddacedacbaaaddbabdecadcecce" }, { "input": "48 30\naaaabaabbaababbbaabaabaababbabbbaabbbaabaaaaaaba\n3 45 a b\n1 14 a a\n15 32 a b\n37 47 a b\n9 35 a b\n36 39 b b\n6 26 a b\n36 44 a a\n28 44 b a\n29 31 b a\n20 39 a a\n45 45 a b\n21 32 b b\n7 43 a b\n14 48 a b\n14 33 a b\n39 44 a a\n9 36 b b\n4 23 b b\n9 42 b b\n41 41 b a\n30 47 a b\n8 42 b a\n14 38 b b\n3 15 a a\n35 47 b b\n14 34 a b\n38 43 a b\n1 35 b a\n16 28 b a", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbb" }, { "input": "89 29\nbabaabaaabaaaababbbbbbbabbbaaaaababbaababababbababaaabbababaaabbbbaaabaaaaaabaaabaabbabab\n39 70 b b\n3 56 b b\n5 22 b a\n4 39 a b\n41 87 b b\n34 41 a a\n10 86 a b\n29 75 a b\n2 68 a a\n27 28 b b\n42 51 b a\n18 61 a a\n6 67 b a\n47 63 a a\n8 68 a b\n4 74 b a\n19 65 a b\n8 55 a b\n5 30 a a\n3 65 a b\n16 57 a b\n34 56 b a\n1 70 a b\n59 68 b b\n29 57 b a\n47 49 b b\n49 73 a a\n32 61 b b\n29 42 a a", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaabbbbbbbbbbbbbab" }, { "input": "59 14\nfbebcfabdefbaaedcefdeecababcabebadfbccaaedaebfdaefdbbcbebbe\n5 32 e f\n8 46 e e\n31 43 e f\n3 10 e a\n53 54 f d\n55 59 d a\n39 58 e b\n54 56 f a\n9 40 b e\n28 37 d a\n7 35 e b\n7 56 c f\n23 26 e a\n15 44 e d", "output": "fbabcfabdffbaafdfffdfffababfabfbaafdffaafdabbfdabfdbbfbbbbe" }, { "input": "7 17\nbbaabab\n3 5 a b\n5 7 a a\n5 5 a a\n4 4 b a\n7 7 a a\n5 6 b b\n1 3 b a\n6 7 a b\n4 6 a b\n6 6 a a\n2 4 b a\n1 7 b a\n4 6 b b\n2 5 b b\n2 5 a b\n1 4 a a\n4 4 b a", "output": "abbabaa" }, { "input": "100 1\ndebaaagbfdgehagadabfgheegggfghghgeeeabgceffeffggcbcegfgebbdhebhfagcgadcbdbabddbcadgbgdebdfehceehcaef\n13 99 f c", "output": "debaaagbfdgehagadabcgheegggcghghgeeeabgcecceccggcbcegcgebbdhebhcagcgadcbdbabddbcadgbgdebdcehceehcaef" }, { "input": "1 1\na\n1 1 a b", "output": "b" }, { "input": "100 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n1 100 a b", "output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" }, { "input": "2 2\naa\n1 2 a b\n1 2 b c", "output": "cc" }, { "input": "3 3\naaa\n1 3 a b\n1 3 b c\n1 3 c d", "output": "ddd" }, { "input": "2 2\naa\n2 2 a b\n1 1 a b", "output": "bb" } ]

1,567,936,070

2,147,483,647

PyPy 3

OK

TESTS

47

155

0

from math import ceil def main(): n,m = [int(v) for v in input().split()] inp = input() for i in range(m): s,e,source, target = [v for v in input().split()] start = int(s)-1 end = int(e)-1 inp = inp[0:start] + inp[start:end + 1].replace(source, target) + inp[end + 1:] print(inp) if __name__ == "__main__": main()

Title: Scarborough Fair Time Limit: None seconds Memory Limit: None megabytes Problem Description: Parsley, sage, rosemary and thyme. Remember me to one who lives there. He once was the true love of mine. Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there. Willem asks his friend, Grick for directions, Grick helped them, and gave them a task. Although the girl wants to help, Willem insists on doing it by himself. Grick gave Willem a string of length *n*. Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed. Grick wants to know the final string after all the *m* operations. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains a string *s* of length *n*, consisting of lowercase English letters. Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space. Output Specification: Output string *s* after performing *m* operations described above. Demo Input: ['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n'] Demo Output: ['noi', 'gaaak'] Note: For the second example: After the first operation, the string is wxxak. After the second operation, the string is waaak. After the third operation, the string is gaaak.

```python from math import ceil def main(): n,m = [int(v) for v in input().split()] inp = input() for i in range(m): s,e,source, target = [v for v in input().split()] start = int(s)-1 end = int(e)-1 inp = inp[0:start] + inp[start:end + 1].replace(source, target) + inp[end + 1:] print(inp) if __name__ == "__main__": main() ```

3

729

E

Subordinates

PROGRAMMING

1,900

[ "constructive algorithms", "data structures", "graphs", "greedy", "sortings" ]

null

There are *n* workers in a company, each of them has a unique id from 1 to *n*. Exaclty one of them is a chief, his id is *s*. Each worker except the chief has exactly one immediate superior. There was a request to each of the workers to tell how how many superiors (not only immediate). Worker's superiors are his immediate superior, the immediate superior of the his immediate superior, and so on. For example, if there are three workers in the company, from which the first is the chief, the second worker's immediate superior is the first, the third worker's immediate superior is the second, then the third worker has two superiors, one of them is immediate and one not immediate. The chief is a superior to all the workers except himself. Some of the workers were in a hurry and made a mistake. You are to find the minimum number of workers that could make a mistake.

The first line contains two positive integers *n* and *s* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*s*<=≤<=*n*) — the number of workers and the id of the chief. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=*n*<=-<=1), where *a**i* is the number of superiors (not only immediate) the worker with id *i* reported about.

Print the minimum number of workers that could make a mistake.

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

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

In the first example it is possible that only the first worker made a mistake. Then: - the immediate superior of the first worker is the second worker, - the immediate superior of the third worker is the first worker, - the second worker is the chief.

2,000

[ { "input": "3 2\n2 0 2", "output": "1" }, { "input": "5 3\n1 0 0 4 1", "output": "2" }, { "input": "1 1\n0", "output": "0" }, { "input": "2 1\n0 0", "output": "1" }, { "input": "2 1\n0 1", "output": "0" }, { "input": "2 1\n1 0", "output": "2" }, { "input": "2 1\n1 1", "output": "1" }, { "input": "2 2\n0 0", "output": "1" }, { "input": "2 2\n0 1", "output": "2" }, { "input": "9 1\n0 1 1 1 1 1 6 7 8", "output": "3" }, { "input": "9 1\n0 1 1 1 1 5 6 7 8", "output": "3" }, { "input": "6 1\n0 1 2 2 0 0", "output": "2" }, { "input": "2 2\n1 1", "output": "1" }, { "input": "2 2\n1 0", "output": "0" }, { "input": "3 1\n0 1 2", "output": "0" }, { "input": "3 1\n2 1 1", "output": "1" }, { "input": "3 1\n0 0 2", "output": "1" }, { "input": "3 2\n2 0 1", "output": "0" }, { "input": "3 2\n2 2 1", "output": "1" }, { "input": "3 2\n2 1 1", "output": "1" }, { "input": "3 3\n1 1 0", "output": "0" }, { "input": "3 3\n2 1 2", "output": "1" }, { "input": "3 3\n2 1 0", "output": "0" }, { "input": "3 2\n2 2 2", "output": "2" }, { "input": "5 5\n0 1 1 0 0", "output": "2" }, { "input": "7 1\n4 4 6 6 6 6 5", "output": "4" }, { "input": "10 6\n3 0 0 0 0 0 0 1 0 0", "output": "7" }, { "input": "5 1\n0 0 1 3 4", "output": "1" }, { "input": "9 1\n0 0 0 2 5 5 5 5 5", "output": "3" }, { "input": "6 1\n5 2 1 3 3 1", "output": "1" }, { "input": "3 1\n1 2 2", "output": "2" } ]

1,567,565,789

2,147,483,647

PyPy 3

OK

TESTS

99

296

15,872,000

[n, s] = [int(x) for x in input().split()] a = [int(x) for x in input().split()] mistakes = 0 mistakes += (a[s-1] is not 0) a[s - 1] = 0 numSuperiors = [0]*(2*100000+100) for superiors in a: numSuperiors[superiors] += 1 cachedMistakes = 0 while numSuperiors[0] != 1: cachedMistakes += 1 numSuperiors[0] -= 1 rightIndex = len(numSuperiors) - 1 leftIndex = 0 while True: while True: if numSuperiors[leftIndex] == 0 and cachedMistakes != 0: numSuperiors[leftIndex] += 1 cachedMistakes -= 1 mistakes += 1 if numSuperiors[leftIndex] == 0: break leftIndex += 1 while numSuperiors[rightIndex] == 0: rightIndex -= 1 if leftIndex >= rightIndex: break numSuperiors[rightIndex] -= 1 cachedMistakes += 1 print(mistakes)

Title: Subordinates Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* workers in a company, each of them has a unique id from 1 to *n*. Exaclty one of them is a chief, his id is *s*. Each worker except the chief has exactly one immediate superior. There was a request to each of the workers to tell how how many superiors (not only immediate). Worker's superiors are his immediate superior, the immediate superior of the his immediate superior, and so on. For example, if there are three workers in the company, from which the first is the chief, the second worker's immediate superior is the first, the third worker's immediate superior is the second, then the third worker has two superiors, one of them is immediate and one not immediate. The chief is a superior to all the workers except himself. Some of the workers were in a hurry and made a mistake. You are to find the minimum number of workers that could make a mistake. Input Specification: The first line contains two positive integers *n* and *s* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*s*<=≤<=*n*) — the number of workers and the id of the chief. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=*n*<=-<=1), where *a**i* is the number of superiors (not only immediate) the worker with id *i* reported about. Output Specification: Print the minimum number of workers that could make a mistake. Demo Input: ['3 2\n2 0 2\n', '5 3\n1 0 0 4 1\n'] Demo Output: ['1\n', '2\n'] Note: In the first example it is possible that only the first worker made a mistake. Then: - the immediate superior of the first worker is the second worker, - the immediate superior of the third worker is the first worker, - the second worker is the chief.

```python [n, s] = [int(x) for x in input().split()] a = [int(x) for x in input().split()] mistakes = 0 mistakes += (a[s-1] is not 0) a[s - 1] = 0 numSuperiors = [0]*(2*100000+100) for superiors in a: numSuperiors[superiors] += 1 cachedMistakes = 0 while numSuperiors[0] != 1: cachedMistakes += 1 numSuperiors[0] -= 1 rightIndex = len(numSuperiors) - 1 leftIndex = 0 while True: while True: if numSuperiors[leftIndex] == 0 and cachedMistakes != 0: numSuperiors[leftIndex] += 1 cachedMistakes -= 1 mistakes += 1 if numSuperiors[leftIndex] == 0: break leftIndex += 1 while numSuperiors[rightIndex] == 0: rightIndex -= 1 if leftIndex >= rightIndex: break numSuperiors[rightIndex] -= 1 cachedMistakes += 1 print(mistakes) ```

3

895

B

XK Segments

PROGRAMMING

1,700

[ "binary search", "math", "sortings", "two pointers" ]

null

While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*. In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1).

The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.

Print one integer — the answer to the problem.

[ "4 2 1\n1 3 5 7\n", "4 2 0\n5 3 1 7\n", "5 3 1\n3 3 3 3 3\n" ]

[ "3\n", "4\n", "25\n" ]

In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4). In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4). In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25.

1,000

[ { "input": "4 2 1\n1 3 5 7", "output": "3" }, { "input": "4 2 0\n5 3 1 7", "output": "4" }, { "input": "5 3 1\n3 3 3 3 3", "output": "25" }, { "input": "5 3 4\n24 13 1 24 24", "output": "4" }, { "input": "4 2 2\n1 3 5 7", "output": "2" }, { "input": "5 1 0\n10 12 14 20 30", "output": "0" }, { "input": "7 3 1\n3 3 6 9 12 15 18", "output": "9" }, { "input": "2 5 0\n3 4", "output": "3" }, { "input": "3 5 0\n4 4 4", "output": "9" }, { "input": "3 6 0\n3 4 5", "output": "6" }, { "input": "10 2 1\n2 2 2 2 2 2 2 1 2 1", "output": "80" }, { "input": "100 2 1\n1 2 2 2 1 2 1 2 2 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 1 2 2 2 2 2 1 1 1 2 1 2 1 2 1 2 1 2 2 1 2 2 2 1 1 1 2 2 1 1 2 2 1 1 1 1 1 1 2 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 1 2 2 1 1 2 1 2 2 1 1 2 2 2 1", "output": "5200" }, { "input": "1 13 1\n13", "output": "1" }, { "input": "2 1 1000000000\n1 1000000000", "output": "1" }, { "input": "2 1000000000 1\n1 1000000000", "output": "2" }, { "input": "1 1 1\n1", "output": "1" }, { "input": "5 1000000000 1\n1000000000 1 1 1 1", "output": "5" }, { "input": "2 1000 1000000\n1 1000000000", "output": "1" }, { "input": "6 2 10\n1 1 1 20 20 20", "output": "9" }, { "input": "1 1000000000 1\n1000000000", "output": "1" }, { "input": "4 65536 65536\n1 2 3 4", "output": "0" }, { "input": "1 600000000 1\n500000000", "output": "0" }, { "input": "5 3 2\n2 3 4 5 8", "output": "2" } ]

1,511,896,089

2,147,483,647

Python 3

OK

TESTS

69

343

14,438,400

from bisect import bisect_left R=lambda:map(int,input().split()) n,x,k=R() a=sorted(R()) z=((u,((u-1)//x+k)*x) for u in a) print(sum(bisect_left(a,l+x)-bisect_left(a,max(u,l)) for u,l in z))

Title: XK Segments Time Limit: None seconds Memory Limit: None megabytes Problem Description: While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*. In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1). Input Specification: The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. Output Specification: Print one integer — the answer to the problem. Demo Input: ['4 2 1\n1 3 5 7\n', '4 2 0\n5 3 1 7\n', '5 3 1\n3 3 3 3 3\n'] Demo Output: ['3\n', '4\n', '25\n'] Note: In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4). In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4). In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25.

```python from bisect import bisect_left R=lambda:map(int,input().split()) n,x,k=R() a=sorted(R()) z=((u,((u-1)//x+k)*x) for u in a) print(sum(bisect_left(a,l+x)-bisect_left(a,max(u,l)) for u,l in z)) ```

3

950

B

Intercepted Message

PROGRAMMING

1,100

[ "greedy", "implementation" ]

null

Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information. Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive. Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages. You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.

The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message. The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message. It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106.

Print the maximum number of files the intercepted array could consist of.

[ "7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n", "3 3\n1 10 100\n1 100 10\n", "1 4\n4\n1 1 1 1\n" ]

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

In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8. In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100. In the third example the only possibility is that the archive contains a single file of size 4.

1,000

[ { "input": "7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8", "output": "3" }, { "input": "3 3\n1 10 100\n1 100 10", "output": "2" }, { "input": "1 4\n4\n1 1 1 1", "output": "1" }, { "input": "1 1\n1000000\n1000000", "output": "1" }, { "input": "3 5\n2 2 9\n2 1 4 2 4", "output": "2" }, { "input": "5 3\n1 1 4 1 2\n1 4 4", "output": "2" }, { "input": "30 50\n3 3 1 3 1 2 4 3 4 1 3 2 3 3 2 3 2 1 3 4 2 1 1 3 2 2 1 3 1 60\n4 4 1 2 2 2 3 1 3 2 1 2 4 4 2 1 2 3 1 3 4 4 3 3 4 4 4 1 2 1 3 3 1 1 3 3 4 3 2 3 2 4 1 4 2 3 2 2 3 1", "output": "12" }, { "input": "50 50\n5733 740 547 3647 5382 5109 6842 7102 5879 1502 3574 1628 7905 4357 8569 9564 8268 3542 2487 8532 425 7713 2585 925 6458 2697 2844 69 324 9030 495 4428 6724 3524 3304 4874 1303 2098 1136 1048 2464 7316 274 9586 534 2450 2368 8060 7795 70692\n1918 4122 6806 4914 6517 6278 9842 9480 6609 4221 9373 1728 9508 9778 8578 5589 2673 6618 6031 9016 4017 6671 6008 2268 5154 9614 6834 9512 9618 6424 1736 1464 6520 9812 1722 9197 2412 2699 73 968 2906 2715 6573 8675 548 7061 5455 88 5565 2544", "output": "1" }, { "input": "1 2\n2\n1 1", "output": "1" }, { "input": "1 2\n1000000\n999999 1", "output": "1" }, { "input": "2 2\n1 1\n1 1", "output": "2" }, { "input": "2 2\n500000 500000\n1 999999", "output": "1" }, { "input": "2 2\n2 3\n4 1", "output": "1" }, { "input": "2 2\n2 3\n3 2", "output": "1" }, { "input": "2 2\n2 3\n2 3", "output": "2" }, { "input": "2 3\n2 2\n1 1 2", "output": "2" }, { "input": "1 1\n1\n1", "output": "1" }, { "input": "2 3\n3 2\n2 1 2", "output": "2" }, { "input": "2 3\n2 3\n2 1 2", "output": "2" }, { "input": "50 30\n2 3 1 2 2 4 3 4 3 2 1 4 2 3 1 3 1 2 2 3 1 1 1 2 3 1 4 3 1 2 1 2 2 1 2 4 4 3 3 2 2 1 1 1 2 2 2 4 3 3\n3 3 3 4 1 4 1 4 4 1 3 4 3 1 2 4 2 1 4 2 3 1 1 2 2 1 2 4 1 41", "output": "12" }, { "input": "50 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", "output": "50" }, { "input": "31 31\n5745 258 5486 13779 20931 407 1478 49032 30787 4957 36603 1034 5011 22319 50560 34419 22036 18235 62551 89259 36093 126169 106027 1673 52983 50127 640 30714 54574 20129 45984\n5745 258 5486 13779 20931 407 1478 49032 30787 4957 36603 1034 5011 22319 50560 34419 22036 18235 62551 89259 36093 126169 106027 1673 52983 50127 640 30714 54574 20129 45984", "output": "31" }, { "input": "3 6\n8 4 1\n1 8 1 1 1 1", "output": "2" } ]

1,666,588,004

2,147,483,647

PyPy 3-64

OK

TESTS

59

77

19,865,600

n, m = list(map(int, input().split())) an = list(map(int, input().split())) am = list(map(int, input().split())) pa = 0 pb = 0 runningSuma = an[0] runningSumb = am[0] count = 0 while (pa < n and pb < m): if (runningSuma == runningSumb and runningSuma != 0): count+=1 pa+=1 pb+=1 if (pa < n and pb < m): runningSuma = an[pa] runningSumb = am[pb] elif (runningSumb > runningSuma): pa+=1 runningSuma += an[pa] else: pb+=1 runningSumb += am[pb] print(count)

Title: Intercepted Message Time Limit: None seconds Memory Limit: None megabytes Problem Description: Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information. Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive. Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages. You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct. Input Specification: The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages. The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message. The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message. It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106. Output Specification: Print the maximum number of files the intercepted array could consist of. Demo Input: ['7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n', '3 3\n1 10 100\n1 100 10\n', '1 4\n4\n1 1 1 1\n'] Demo Output: ['3\n', '2\n', '1\n'] Note: In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8. In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100. In the third example the only possibility is that the archive contains a single file of size 4.

```python n, m = list(map(int, input().split())) an = list(map(int, input().split())) am = list(map(int, input().split())) pa = 0 pb = 0 runningSuma = an[0] runningSumb = am[0] count = 0 while (pa < n and pb < m): if (runningSuma == runningSumb and runningSuma != 0): count+=1 pa+=1 pb+=1 if (pa < n and pb < m): runningSuma = an[pa] runningSumb = am[pb] elif (runningSumb > runningSuma): pa+=1 runningSuma += an[pa] else: pb+=1 runningSumb += am[pb] print(count) ```

3

39

F

Pacifist frogs

PROGRAMMING

1,300

[ "implementation" ]

F. Pacifist frogs

2

64

Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much. One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to *n* and the number of a hill is equal to the distance in meters between it and the island. The distance between the *n*-th hill and the shore is also 1 meter. Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is *d*, the frog will jump from the island on the hill *d*, then — on the hill 2*d*, then 3*d* and so on until they get to the shore (i.e. find itself beyond the hill *n*). However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.

The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=109, 1<=≤<=*m*,<=*k*<=≤<=100) — the number of hills, frogs and mosquitoes respectively. The second line contains *m* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the lengths of the frogs’ jumps. The third line contains *k* integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.

In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to *m* in the order of the jump length given in the input data.

[ "5 3 5\n2 3 4\n1 2 3 4 5\n", "1000000000 2 3\n2 5\n999999995 999999998 999999996\n" ]

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

none

0

[ { "input": "5 3 5\n2 3 4\n1 2 3 4 5", "output": "2\n2 3" }, { "input": "1000000000 2 3\n2 5\n999999995 999999998 999999996", "output": "1\n2" }, { "input": "1 1 1\n1\n1", "output": "1\n1" }, { "input": "2 2 1\n2 1\n1", "output": "1\n1" }, { "input": "3 2 2\n2 4\n3 2", "output": "1\n2" }, { "input": "10 3 6\n5 2 8\n5 6 7 8 9 10", "output": "1\n3" }, { "input": "10 10 9\n10 9 8 7 6 5 4 3 2 1\n10 9 8 7 5 4 3 2 1", "output": "1\n5" }, { "input": "20 3 5\n2 3 5\n2 5 6 10 15", "output": "1\n2" }, { "input": "20 4 8\n1 2 3 4\n2 4 6 8 10 12 14 16", "output": "1\n3" }, { "input": "10 5 5\n1 5 3 5 1\n1 6 5 7 2", "output": "3\n2 3 4" }, { "input": "20 10 5\n1 12 6 11 9 21 15 16 8 9\n11 13 15 2 1", "output": "7\n2 3 5 6 8 9 10" }, { "input": "20 10 10\n9 8 21 8 7 2 13 17 20 18\n7 16 20 3 6 1 11 18 15 17", "output": "2\n3 7" }, { "input": "20 10 10\n6 17 14 12 13 15 6 14 16 17\n1 6 16 14 7 8 9 12 10 2", "output": "4\n2 5 6 10" }, { "input": "100 30 30\n25 34 81 32 96 79 36 21 53 15 51 69 78 99 60 2 80 37 61 70 32 31 31 6 7 38 95 70 81 39\n1 50 75 8 90 69 13 57 6 4 60 19 94 52 45 42 95 88 21 22 96 2 56 61 31 78 7 62 68 72", "output": "11\n3 6 9 11 14 17 18 20 26 28 29" }, { "input": "200 35 67\n152 112 102 46 54 189 56 76 10 39 157 6 84 188 122 117 51 163 6 50 195 34 44 178 28 32 100 67 74 48 88 100 91 50 91\n126 68 138 157 92 128 183 36 175 49 168 198 116 20 31 88 61 46 12 179 137 130 185 5 171 96 184 85 37 147 50 75 93 103 160 10 120 140 59 98 131 124 121 190 169 141 165 39 47 28 90 139 148 119 73 6 51 94 21 52 89 35 97 79 3 13 142", "output": "17\n1 2 3 5 6 8 14 15 16 18 21 24 27 28 32 33 35" }, { "input": "200 72 29\n201 145 169 163 32 126 131 71 26 130 2 61 110 17 179 114 79 30 192 91 141 70 101 119 185 66 72 76 164 144 106 162 122 146 119 181 184 61 131 131 140 152 60 65 183 154 32 33 108 77 29 102 67 5 125 26 126 104 20 89 183 21 126 195 198 24 123 173 135 164 141 32\n160 65 136 22 194 110 155 138 92 118 87 40 49 191 190 99 157 3 23 17 34 123 31 81 67 86 196 45 109", "output": "59\n1 2 3 4 6 7 8 9 10 12 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 46 49 50 52 55 56 57 58 60 61 62 63 64 65 66 68 69 70 71" }, { "input": "500 46 46\n363 441 170 289 389 394 488 72 332 285 445 185 221 183 397 175 98 192 202 16 123 436 336 260 212 229 459 473 66 19 445 153 476 234 396 159 289 137 331 18 268 224 71 133 196 7\n454 64 417 129 95 162 496 300 234 359 224 354 334 155 191 82 35 319 244 126 292 108 321 93 77 311 107 487 121 431 235 100 445 68 338 467 133 307 4 220 245 84 468 141 436 363", "output": "35\n2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 21 23 24 25 26 27 28 29 32 33 35 36 37 38 39 41 43 45" }, { "input": "1000 19 27\n656 162 264 790 579 786 877 998 516 247 650 150 858 281 279 549 354 353 533\n349 411 1 248 22 649 726 382 423 832 172 864 17 658 840 572 564 287 800 919 500 575 461 40 1000 383 624", "output": "19\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19" } ]

1,582,359,797

2,147,483,647

PyPy 3

OK

TESTS

35

404

2,048,000

""" // Author : snape_here - Susanta Mukherjee """ from __future__ import division, print_function import os,sys from io import BytesIO, IOBase if sys.version_info[0] < 3: from __builtin__ import xrange as range from future_builtins import ascii, filter, hex, map, oct, zip def ii(): return int(input()) def si(): return input() def mi(): return map(int,input().split()) def li(): return list(mi()) def gcd(x, y): while y: x, y = y, x % y return x def read(): sys.stdin = open('input.txt', 'r') sys.stdout = open('output.txt', 'w') import math from functools import reduce def factorS(n): return set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))) def divisorGen(n): factors = list(factorGenerator(n)) nfactors = len(factors) f = [0] * nfactors while True: yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1) i = 0 while True: f[i] += 1 if f[i] <= factors[i][1]: break f[i] = 0 i += 1 if i >= nfactors: return mod=100000007 def main(): n,m,k=mi() d=li() l=li() a=[] for i in range(m): c=0 for j in range(k): if l[j]%d[i]==0: c+=1 a.append(c) b=min(a) e=a.count(b) f=[] for i in range(m): if a[i]==b: f.append(i+1) print(e) print(*f) # region fastio# BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") def print(*args, **kwargs): """Prints the values to a stream, or to sys.stdout by default.""" sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout) at_start = True for x in args: if not at_start: file.write(sep) file.write(str(x)) at_start = False file.write(kwargs.pop("end", "\n")) if kwargs.pop("flush", False): file.flush() if sys.version_info[0] < 3: sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # endregion if __name__ == "__main__": #read() main() #Comment read()

Title: Pacifist frogs Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much. One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to *n* and the number of a hill is equal to the distance in meters between it and the island. The distance between the *n*-th hill and the shore is also 1 meter. Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is *d*, the frog will jump from the island on the hill *d*, then — on the hill 2*d*, then 3*d* and so on until they get to the shore (i.e. find itself beyond the hill *n*). However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible. Input Specification: The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=109, 1<=≤<=*m*,<=*k*<=≤<=100) — the number of hills, frogs and mosquitoes respectively. The second line contains *m* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the lengths of the frogs’ jumps. The third line contains *k* integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces. Output Specification: In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to *m* in the order of the jump length given in the input data. Demo Input: ['5 3 5\n2 3 4\n1 2 3 4 5\n', '1000000000 2 3\n2 5\n999999995 999999998 999999996\n'] Demo Output: ['2\n2 3\n', '1\n2\n'] Note: none

```python """ // Author : snape_here - Susanta Mukherjee """ from __future__ import division, print_function import os,sys from io import BytesIO, IOBase if sys.version_info[0] < 3: from __builtin__ import xrange as range from future_builtins import ascii, filter, hex, map, oct, zip def ii(): return int(input()) def si(): return input() def mi(): return map(int,input().split()) def li(): return list(mi()) def gcd(x, y): while y: x, y = y, x % y return x def read(): sys.stdin = open('input.txt', 'r') sys.stdout = open('output.txt', 'w') import math from functools import reduce def factorS(n): return set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))) def divisorGen(n): factors = list(factorGenerator(n)) nfactors = len(factors) f = [0] * nfactors while True: yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1) i = 0 while True: f[i] += 1 if f[i] <= factors[i][1]: break f[i] = 0 i += 1 if i >= nfactors: return mod=100000007 def main(): n,m,k=mi() d=li() l=li() a=[] for i in range(m): c=0 for j in range(k): if l[j]%d[i]==0: c+=1 a.append(c) b=min(a) e=a.count(b) f=[] for i in range(m): if a[i]==b: f.append(i+1) print(e) print(*f) # region fastio# BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") def print(*args, **kwargs): """Prints the values to a stream, or to sys.stdout by default.""" sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout) at_start = True for x in args: if not at_start: file.write(sep) file.write(str(x)) at_start = False file.write(kwargs.pop("end", "\n")) if kwargs.pop("flush", False): file.flush() if sys.version_info[0] < 3: sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # endregion if __name__ == "__main__": #read() main() #Comment read() ```

3.883741

588

B

Duff in Love

PROGRAMMING

1,300

[ "math" ]

null

Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*. Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible. Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).

Print the answer in one line.

[ "10\n", "12\n" ]

[ "10\n", "6\n" ]

In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely. In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeed lovely.

1,000

[ { "input": "10", "output": "10" }, { "input": "12", "output": "6" }, { "input": "1", "output": "1" }, { "input": "2", "output": "2" }, { "input": "4", "output": "2" }, { "input": "8", "output": "2" }, { "input": "3", "output": "3" }, { "input": "31", "output": "31" }, { "input": "97", "output": "97" }, { "input": "1000000000000", "output": "10" }, { "input": "15", "output": "15" }, { "input": "894", "output": "894" }, { "input": "271", "output": "271" }, { "input": "2457", "output": "273" }, { "input": "2829", "output": "2829" }, { "input": "5000", "output": "10" }, { "input": "20", "output": "10" }, { "input": "68", "output": "34" }, { "input": "3096", "output": "258" }, { "input": "1024", "output": "2" }, { "input": "1048576", "output": "2" }, { "input": "413933789280", "output": "25870861830" }, { "input": "817634153013", "output": "817634153013" }, { "input": "56517269141", "output": "56517269141" }, { "input": "30707328551", "output": "30707328551" }, { "input": "279564127218", "output": "10354226934" }, { "input": "491159577042", "output": "18191095446" }, { "input": "734337660466", "output": "734337660466" }, { "input": "808453785117", "output": "808453785117" }, { "input": "55926835837", "output": "55926835837" }, { "input": "294809951965", "output": "294809951965" }, { "input": "537988035389", "output": "76855433627" }, { "input": "822722434952", "output": "205680608738" }, { "input": "699511759613", "output": "699511759613" }, { "input": "942689843037", "output": "104743315893" }, { "input": "663634158717", "output": "663634158717" }, { "input": "213612977250", "output": "11730" }, { "input": "999999999989", "output": "999999999989" }, { "input": "999999999988", "output": "499999999994" }, { "input": "87178291200", "output": "30030" }, { "input": "927668721948", "output": "463834360974" }, { "input": "562436815639", "output": "37927" }, { "input": "302981118597", "output": "35853" }, { "input": "5", "output": "5" }, { "input": "9", "output": "3" }, { "input": "36", "output": "6" }, { "input": "2231", "output": "2231" }, { "input": "27648", "output": "6" }, { "input": "40320", "output": "210" }, { "input": "648000", "output": "30" }, { "input": "999966000289", "output": "999983" }, { "input": "999985999949", "output": "999985999949" }, { "input": "991921850317", "output": "9973" } ]

1,552,918,974

2,147,483,647

Python 3

OK

TESTS

53

561

0

n = int(input()) member =[] i = 2 ans =1 while(1): if i*i <=n : if n%i ==0: while(n%i ==0): n //=i ans*=i member.append(i) i+=1 else:break print(ans*n if len(member)>0 else n)

Title: Duff in Love Time Limit: None seconds Memory Limit: None megabytes Problem Description: Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*. Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible. Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store. Input Specification: The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012). Output Specification: Print the answer in one line. Demo Input: ['10\n', '12\n'] Demo Output: ['10\n', '6\n'] Note: In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely. In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeed lovely.

```python n = int(input()) member =[] i = 2 ans =1 while(1): if i*i <=n : if n%i ==0: while(n%i ==0): n //=i ans*=i member.append(i) i+=1 else:break print(ans*n if len(member)>0 else n) ```

3

82

A

Double Cola

PROGRAMMING

1,100

[ "implementation", "math" ]

A. Double Cola

1

256

Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum. For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny. Write a program that will print the name of a man who will drink the *n*-th can. Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.

The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109). It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.

Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.

[ "1\n", "6\n", "1802\n" ]

[ "Sheldon\n", "Sheldon\n", "Penny\n" ]

none

500

[ { "input": "1", "output": "Sheldon" }, { "input": "6", "output": "Sheldon" }, { "input": "1802", "output": "Penny" }, { "input": "1", "output": "Sheldon" }, { "input": "2", "output": "Leonard" }, { "input": "3", "output": "Penny" }, { "input": "4", "output": "Rajesh" }, { "input": "5", "output": "Howard" }, { "input": "10", "output": "Penny" }, { "input": "534", "output": "Rajesh" }, { "input": "5033", "output": "Howard" }, { "input": "10010", "output": "Howard" }, { "input": "500000000", "output": "Penny" }, { "input": "63", "output": "Rajesh" }, { "input": "841", "output": "Leonard" }, { "input": "3667", "output": "Penny" }, { "input": "38614", "output": "Howard" }, { "input": "282798", "output": "Rajesh" }, { "input": "9266286", "output": "Rajesh" }, { "input": "27385966", "output": "Leonard" }, { "input": "121580142", "output": "Penny" }, { "input": "5", "output": "Howard" }, { "input": "300", "output": "Howard" }, { "input": "1745", "output": "Leonard" }, { "input": "8302", "output": "Rajesh" }, { "input": "184518", "output": "Sheldon" }, { "input": "1154414", "output": "Rajesh" }, { "input": "28643950", "output": "Leonard" }, { "input": "159222638", "output": "Howard" }, { "input": "24", "output": "Penny" }, { "input": "505", "output": "Penny" }, { "input": "4425", "output": "Rajesh" }, { "input": "12079", "output": "Sheldon" }, { "input": "469726", "output": "Penny" }, { "input": "3961838", "output": "Penny" }, { "input": "57710446", "output": "Leonard" }, { "input": "80719470", "output": "Howard" }, { "input": "1000000000", "output": "Penny" }, { "input": "999999999", "output": "Penny" }, { "input": "999999998", "output": "Penny" }, { "input": "5", "output": "Howard" } ]

1,590,040,232

2,147,483,647

PyPy 3

OK

TESTS

41

155

0

import math n = int(input()) names = {1:'Sheldon', 2:'Leonard', 3:'Penny', 4:'Rajesh', 5:'Howard'} r = 0 while(5*(2**r) < n): n-= 5*(2**r) r+=1 print(names[math.ceil(n/(2**r))])

Title: Double Cola Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum. For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny. Write a program that will print the name of a man who will drink the *n*-th can. Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. Input Specification: The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109). It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. Output Specification: Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. Demo Input: ['1\n', '6\n', '1802\n'] Demo Output: ['Sheldon\n', 'Sheldon\n', 'Penny\n'] Note: none

```python import math n = int(input()) names = {1:'Sheldon', 2:'Leonard', 3:'Penny', 4:'Rajesh', 5:'Howard'} r = 0 while(5*(2**r) < n): n-= 5*(2**r) r+=1 print(names[math.ceil(n/(2**r))]) ```

3.9225

173

B

Chamber of Secrets

PROGRAMMING

1,800

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

null

"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny. The Chamber of Secrets is an *n*<=×<=*m* rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below. The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position. Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.

The first line of the input contains two integer numbers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.

Print the minimum number of columns to make magic or -1 if it's impossible to do.

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

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

The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.

1,000

[ { "input": "3 3\n.#.\n...\n.#.", "output": "2" }, { "input": "4 3\n##.\n...\n.#.\n.#.", "output": "2" }, { "input": "3 3\n###\n###\n###", "output": "2" }, { "input": "3 4\n..##\n....\n..#.", "output": "2" }, { "input": "4 3\n#.#\n...\n...\n.##", "output": "2" }, { "input": "4 4\n##..\n..#.\n...#\n...#", "output": "-1" }, { "input": "5 5\n...#.\n.....\n.....\n.....\n..#.#", "output": "-1" }, { "input": "5 5\n.....\n.#...\n.....\n.....\n#.###", "output": "-1" }, { "input": "5 5\n.....\n.###.\n..##.\n..##.\n...#.", "output": "-1" }, { "input": "5 5\n.##..\n.##..\n.#.#.\n..#..\n..#..", "output": "2" }, { "input": "5 5\n...#.\n.#..#\n#.#.#\n#....\n#.#..", "output": "-1" }, { "input": "5 5\n.#.#.\n#..#.\n.#.##\n.#.##\n##.#.", "output": "2" }, { "input": "5 5\n.####\n#.###\n#####\n#.###\n#####", "output": "2" }, { "input": "5 5\n.####\n#.#.#\n#####\n#####\n#####", "output": "2" }, { "input": "5 5\n#####\n#####\n#####\n#####\n#####", "output": "2" }, { "input": "2 42\n.########.#.########.#..#############.##..\n#####.#############.#####.#.#.#####.####..", "output": "2" }, { "input": "2 2\n##\n.#", "output": "2" }, { "input": "4 3\n.#.\n...\n##.\n#.#", "output": "4" }, { "input": "4 3\n..#\n#.#\n...\n#..", "output": "4" }, { "input": "10 10\n#.........\n...#..#...\n..........\n..........\n.......#.#\n..........\n..........\n......##..\n#..#......\n.........#", "output": "10" }, { "input": "4 4\n.#..\n..##\n.#.#\n..#.", "output": "6" }, { "input": "4 3\n#.#\n.##\n...\n.#.", "output": "4" }, { "input": "4 4\n...#\n.##.\n.#.#\n..#.", "output": "6" }, { "input": "4 4\n#...\n.##.\n#.#.\n.#..", "output": "6" }, { "input": "3 3\n..#\n#.#\n#..", "output": "4" }, { "input": "4 4\n#...\n.#.#\n##..\n...#", "output": "6" }, { "input": "5 5\n#....\n.....\n..##.\n#.#..\n...#.", "output": "6" }, { "input": "4 4\n..#.\n#.#.\n#..#\n...#", "output": "6" }, { "input": "4 7\n..#....\n##.##.#\n.##....\n...##.#", "output": "6" }, { "input": "5 2\n.#\n##\n..\n#.\n#.", "output": "4" }, { "input": "4 3\n.#.\n#.#\n##.\n..#", "output": "6" } ]

1,698,962,146

2,147,483,647

Python 3

OK

TESTS

105

996

54,886,400

from collections import deque n, m = map(int, input().split()) arr = [[] for _ in range(n + m)] ret = -1 q, q2= deque([0]), deque([0]) s = set([0]) for x in range(n): row = input() for y in range(m): if row[y] == '#': arr[x].append(y + n) arr[y + n].append(x) while q: temp = q.popleft() temp2 = q2.popleft() if temp == n - 1: ret = temp2 break for x in arr[temp]: if x not in s: q.append(x) q2.append(temp2 + 1) s.add(x) print(ret)

Title: Chamber of Secrets Time Limit: None seconds Memory Limit: None megabytes Problem Description: "The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny. The Chamber of Secrets is an *n*<=×<=*m* rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below. The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position. Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber. Input Specification: The first line of the input contains two integer numbers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column. Output Specification: Print the minimum number of columns to make magic or -1 if it's impossible to do. Demo Input: ['3 3\n.#.\n...\n.#.\n', '4 3\n##.\n...\n.#.\n.#.\n'] Demo Output: ['2\n', '2\n'] Note: The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.

```python from collections import deque n, m = map(int, input().split()) arr = [[] for _ in range(n + m)] ret = -1 q, q2= deque([0]), deque([0]) s = set([0]) for x in range(n): row = input() for y in range(m): if row[y] == '#': arr[x].append(y + n) arr[y + n].append(x) while q: temp = q.popleft() temp2 = q2.popleft() if temp == n - 1: ret = temp2 break for x in arr[temp]: if x not in s: q.append(x) q2.append(temp2 + 1) s.add(x) print(ret) ```

3

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": 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"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" } ]

1,596,961,863

2,147,483,647

Python 3

OK

TESTS

102

109

6,656,000

def result(a,b): r = "" for (x,y) in zip(a, b): if x != y: r += '1' else: r += '0' print(r) if __name__ == "__main__": a = input(); b = input(); result(a,b)

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 def result(a,b): r = "" for (x,y) in zip(a, b): if x != y: r += '1' else: r += '0' print(r) if __name__ == "__main__": a = input(); b = input(); result(a,b) ```

3.960352

769

A

Year of University Entrance

PROGRAMMING

800

[ "*special", "implementation", "sortings" ]

null

There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university. Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than *x* from the year of university entrance of this student, where *x* — some non-negative integer. A value *x* is not given, but it can be uniquely determined from the available data. Note that students don't join other groups. You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance.

The first line contains the positive odd integer *n* (1<=≤<=*n*<=≤<=5) — the number of groups which Igor joined. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (2010<=≤<=*a**i*<=≤<=2100) — years of student's university entrance for each group in which Igor is the member. It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly.

Print the year of Igor's university entrance.

[ "3\n2014 2016 2015\n", "1\n2050\n" ]

[ "2015\n", "2050\n" ]

In the first test the value *x* = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016. In the second test the value *x* = 0. Igor entered only the group which corresponds to the year of his university entrance.

500

[ { "input": "3\n2014 2016 2015", "output": "2015" }, { "input": "1\n2050", "output": "2050" }, { "input": "1\n2010", "output": "2010" }, { "input": "1\n2011", "output": "2011" }, { "input": "3\n2010 2011 2012", "output": "2011" }, { "input": "3\n2049 2047 2048", "output": "2048" }, { "input": "5\n2043 2042 2041 2044 2040", "output": "2042" }, { "input": "5\n2012 2013 2014 2015 2016", "output": "2014" }, { "input": "1\n2045", "output": "2045" }, { "input": "1\n2046", "output": "2046" }, { "input": "1\n2099", "output": "2099" }, { "input": "1\n2100", "output": "2100" }, { "input": "3\n2011 2010 2012", "output": "2011" }, { "input": "3\n2011 2012 2010", "output": "2011" }, { "input": "3\n2012 2011 2010", "output": "2011" }, { "input": "3\n2010 2012 2011", "output": "2011" }, { "input": "3\n2012 2010 2011", "output": "2011" }, { "input": "3\n2047 2048 2049", "output": "2048" }, { "input": "3\n2047 2049 2048", "output": "2048" }, { "input": "3\n2048 2047 2049", "output": "2048" }, { "input": "3\n2048 2049 2047", "output": "2048" }, { "input": "3\n2049 2048 2047", "output": "2048" }, { "input": "5\n2011 2014 2012 2013 2010", "output": "2012" }, { "input": "5\n2014 2013 2011 2012 2015", "output": "2013" }, { "input": "5\n2021 2023 2024 2020 2022", "output": "2022" }, { "input": "5\n2081 2079 2078 2080 2077", "output": "2079" }, { "input": "5\n2095 2099 2097 2096 2098", "output": "2097" }, { "input": "5\n2097 2099 2100 2098 2096", "output": "2098" }, { "input": "5\n2012 2010 2014 2011 2013", "output": "2012" }, { "input": "5\n2012 2011 2013 2015 2014", "output": "2013" }, { "input": "5\n2023 2024 2022 2021 2020", "output": "2022" }, { "input": "5\n2077 2078 2080 2079 2081", "output": "2079" }, { "input": "5\n2099 2096 2095 2097 2098", "output": "2097" }, { "input": "5\n2097 2100 2098 2096 2099", "output": "2098" }, { "input": "5\n2011 2014 2013 2010 2012", "output": "2012" }, { "input": "5\n2013 2011 2015 2012 2014", "output": "2013" }, { "input": "5\n2024 2020 2021 2023 2022", "output": "2022" }, { "input": "5\n2079 2080 2077 2081 2078", "output": "2079" }, { "input": "5\n2095 2097 2096 2098 2099", "output": "2097" }, { "input": "5\n2099 2096 2100 2097 2098", "output": "2098" }, { "input": "5\n2034 2033 2036 2032 2035", "output": "2034" }, { "input": "5\n2030 2031 2033 2032 2029", "output": "2031" }, { "input": "5\n2093 2092 2094 2096 2095", "output": "2094" }, { "input": "5\n2012 2015 2014 2013 2011", "output": "2013" }, { "input": "5\n2056 2057 2058 2059 2060", "output": "2058" } ]

1,646,598,680

2,147,483,647

Python 3

OK

TESTS

45

46

0

n = int(input()) print(sorted(input().split())[n//2])

Title: Year of University Entrance Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is the faculty of Computer Science in Berland. In the social net "TheContact!" for each course of this faculty there is the special group whose name equals the year of university entrance of corresponding course of students at the university. Each of students joins the group of his course and joins all groups for which the year of student's university entrance differs by no more than *x* from the year of university entrance of this student, where *x* — some non-negative integer. A value *x* is not given, but it can be uniquely determined from the available data. Note that students don't join other groups. You are given the list of groups which the student Igor joined. According to this information you need to determine the year of Igor's university entrance. Input Specification: The first line contains the positive odd integer *n* (1<=≤<=*n*<=≤<=5) — the number of groups which Igor joined. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (2010<=≤<=*a**i*<=≤<=2100) — years of student's university entrance for each group in which Igor is the member. It is guaranteed that the input data is correct and the answer always exists. Groups are given randomly. Output Specification: Print the year of Igor's university entrance. Demo Input: ['3\n2014 2016 2015\n', '1\n2050\n'] Demo Output: ['2015\n', '2050\n'] Note: In the first test the value *x* = 1. Igor entered the university in 2015. So he joined groups members of which are students who entered the university in 2014, 2015 and 2016. In the second test the value *x* = 0. Igor entered only the group which corresponds to the year of his university entrance.

```python n = int(input()) print(sorted(input().split())[n//2]) ```

3

143

A

Help Vasilisa the Wise 2

PROGRAMMING

1,000

[ "brute force", "math" ]

null

Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that. The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below. The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below. Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.

The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.

Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes). If there are several solutions, output any.

[ "3 7\n4 6\n5 5\n", "11 10\n13 8\n5 16\n", "1 2\n3 4\n5 6\n", "10 10\n10 10\n10 10\n" ]

[ "1 2\n3 4\n", "4 7\n9 1\n", "-1\n", "-1\n" ]

Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.

500

[ { "input": "3 7\n4 6\n5 5", "output": "1 2\n3 4" }, { "input": "11 10\n13 8\n5 16", "output": "4 7\n9 1" }, { "input": "1 2\n3 4\n5 6", "output": "-1" }, { "input": "10 10\n10 10\n10 10", "output": "-1" }, { "input": "5 13\n8 10\n11 7", "output": "3 2\n5 8" }, { "input": "12 17\n10 19\n13 16", "output": "-1" }, { "input": "11 11\n17 5\n12 10", "output": "9 2\n8 3" }, { "input": "12 11\n11 12\n16 7", "output": "-1" }, { "input": "5 9\n7 7\n8 6", "output": "3 2\n4 5" }, { "input": "10 7\n4 13\n11 6", "output": "-1" }, { "input": "18 10\n16 12\n12 16", "output": "-1" }, { "input": "13 6\n10 9\n6 13", "output": "-1" }, { "input": "14 16\n16 14\n18 12", "output": "-1" }, { "input": "16 10\n16 10\n12 14", "output": "-1" }, { "input": "11 9\n12 8\n11 9", "output": "-1" }, { "input": "5 14\n10 9\n10 9", "output": "-1" }, { "input": "2 4\n1 5\n3 3", "output": "-1" }, { "input": "17 16\n14 19\n18 15", "output": "-1" }, { "input": "12 12\n14 10\n16 8", "output": "9 3\n5 7" }, { "input": "15 11\n16 10\n9 17", "output": "7 8\n9 2" }, { "input": "8 10\n9 9\n13 5", "output": "6 2\n3 7" }, { "input": "13 7\n10 10\n5 15", "output": "4 9\n6 1" }, { "input": "14 11\n9 16\n16 9", "output": "-1" }, { "input": "12 8\n14 6\n8 12", "output": "-1" }, { "input": "10 6\n6 10\n4 12", "output": "-1" }, { "input": "10 8\n10 8\n4 14", "output": "-1" }, { "input": "14 13\n9 18\n14 13", "output": "-1" }, { "input": "9 14\n8 15\n8 15", "output": "-1" }, { "input": "3 8\n2 9\n6 5", "output": "-1" }, { "input": "14 17\n18 13\n15 16", "output": "-1" }, { "input": "16 14\n15 15\n17 13", "output": "9 7\n6 8" }, { "input": "14 11\n16 9\n13 12", "output": "9 5\n7 4" }, { "input": "13 10\n11 12\n7 16", "output": "4 9\n7 3" }, { "input": "14 8\n11 11\n13 9", "output": "8 6\n3 5" }, { "input": "12 11\n13 10\n10 13", "output": "-1" }, { "input": "6 5\n2 9\n5 6", "output": "-1" }, { "input": "7 8\n8 7\n12 3", "output": "-1" }, { "input": "7 11\n7 11\n6 12", "output": "-1" }, { "input": "8 5\n11 2\n8 5", "output": "-1" }, { "input": "10 16\n14 12\n14 12", "output": "-1" }, { "input": "7 9\n4 12\n5 11", "output": "-1" }, { "input": "11 13\n19 5\n12 12", "output": "-1" }, { "input": "8 12\n5 15\n11 9", "output": "2 6\n3 9" }, { "input": "16 5\n13 8\n10 11", "output": "9 7\n4 1" }, { "input": "3 14\n8 9\n10 7", "output": "2 1\n6 8" }, { "input": "12 14\n11 15\n9 17", "output": "3 9\n8 6" }, { "input": "13 7\n9 11\n14 6", "output": "8 5\n1 6" }, { "input": "13 8\n15 6\n11 10", "output": "9 4\n6 2" }, { "input": "9 6\n5 10\n3 12", "output": "1 8\n4 2" }, { "input": "12 7\n5 14\n8 11", "output": "3 9\n2 5" }, { "input": "9 12\n3 17\n10 10", "output": "-1" }, { "input": "3 8\n4 6\n5 5", "output": "-1" } ]

1,689,261,442

2,147,483,647

PyPy 3-64

OK

TESTS

52

124

1,945,600

r1, r2 = map(int, input().split()) c1, c2 = map(int, input().split()) d1, d2 = map(int, input().split()) flag = False for x in range(1, 10): for y in range(1, 10): for x1 in range(1, 10): for y1 in range(1, 10): if x + y == c1 and x1 + y1 == c2 and x + x1 == r1 and y + y1 == r2 \ and x + y1 == d1 and y + x1 == d2 and len(set(list((x, y, x1, y1)))) == 4: flag = True print(x, x1) print(y, y1) exit() if not flag: print(-1)

Title: Help Vasilisa the Wise 2 Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that. The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below. The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below. Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task. Input Specification: The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement. Output Specification: Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes). If there are several solutions, output any. Demo Input: ['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n'] Demo Output: ['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n'] Note: Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.

```python r1, r2 = map(int, input().split()) c1, c2 = map(int, input().split()) d1, d2 = map(int, input().split()) flag = False for x in range(1, 10): for y in range(1, 10): for x1 in range(1, 10): for y1 in range(1, 10): if x + y == c1 and x1 + y1 == c2 and x + x1 == r1 and y + y1 == r2 \ and x + y1 == d1 and y + x1 == d2 and len(set(list((x, y, x1, y1)))) == 4: flag = True print(x, x1) print(y, y1) exit() if not flag: print(-1) ```

3

681

A

A Good Contest

PROGRAMMING

800

[ "implementation" ]

null

Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance. Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it. Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest . The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters. It is guaranteed that all handles are distinct.

Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.

[ "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n", "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n" ]

[ "YES", "NO" ]

In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest. In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.

500

[ { "input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO" }, { "input": "1\nDb -3373 3591", "output": "NO" }, { "input": "5\nQ2bz 960 2342\nhmX 2710 -1348\ngbAe -1969 -963\nE -160 196\npsi 2665 -3155", "output": "NO" }, { "input": "9\nmwAz9lQ 1786 -1631\nnYgYFXZQfY -1849 -1775\nKU4jF -1773 -3376\nopR 3752 2931\nGl -1481 -1002\nR -1111 3778\n0i9B21DC 3650 289\nQ8L2dS0 358 -3305\ng -2662 3968", "output": "NO" }, { "input": "5\nzMSBcOUf -2883 -2238\nYN -3314 -1480\nfHpuccQn06 -1433 -589\naM1NVEPQi 399 3462\n_L 2516 -3290", "output": "NO" }, { "input": "1\na 2400 2401", "output": "YES" }, { "input": "1\nfucker 4000 4000", "output": "NO" }, { "input": "1\nJora 2400 2401", "output": "YES" }, { "input": "1\nACA 2400 2420", "output": "YES" }, { "input": "1\nAca 2400 2420", "output": "YES" }, { "input": "1\nSub_d 2401 2402", "output": "YES" }, { "input": "2\nHack 2400 2401\nDum 1243 555", "output": "YES" }, { "input": "1\nXXX 2400 2500", "output": "YES" }, { "input": "1\nfucker 2400 2401", "output": "YES" }, { "input": "1\nX 2400 2500", "output": "YES" }, { "input": "1\nvineet 2400 2401", "output": "YES" }, { "input": "1\nabc 2400 2500", "output": "YES" }, { "input": "1\naaaaa 2400 2401", "output": "YES" }, { "input": "1\nhoge 2400 2401", "output": "YES" }, { "input": "1\nInfinity 2400 2468", "output": "YES" }, { "input": "1\nBurunduk1 2400 2401", "output": "YES" }, { "input": "1\nFuck 2400 2401", "output": "YES" }, { "input": "1\nfuck 2400 2401", "output": "YES" }, { "input": "3\nApplejack 2400 2401\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "YES" }, { "input": "1\nalex 2400 2401", "output": "YES" }, { "input": "1\nA 2400 2401", "output": "YES" }, { "input": "1\na 2400 2455", "output": "YES" }, { "input": "1\nlol 2400 2401", "output": "YES" }, { "input": "2\nBurunduk1 2400 2537\nBudAlNik 2084 2214", "output": "YES" }, { "input": "1\naaaaaa 2400 2401", "output": "YES" }, { "input": "1\nBurunduk1 2400 2500", "output": "YES" }, { "input": "1\nds 2400 2410", "output": "YES" }, { "input": "1\nas 2400 2401", "output": "YES" }, { "input": "1\nabc 2400 2401", "output": "YES" }, { "input": "3\nBudAlNik 2084 2214\nsubscriber 2833 2749\nBurunduk1 2526 2537", "output": "YES" }, { "input": "1\ncaonima 2400 2401", "output": "YES" }, { "input": "1\narr 2400 2500", "output": "YES" }, { "input": "1\nx 2400 2401", "output": "YES" }, { "input": "1\narrr 2400 2500", "output": "YES" }, { "input": "1\nabc 2400 2405", "output": "YES" }, { "input": "3\nBurunduk1 2400 2420\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "1\nBurunduk1 2400 2537", "output": "YES" }, { "input": "1\nHELLO 2400 2401", "output": "YES" }, { "input": "1\neatmore 2400 2500", "output": "YES" }, { "input": "1\nb 2400 2401", "output": "YES" }, { "input": "3\nBurunduk1 2400 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "1\nApplejack 2400 2410", "output": "YES" }, { "input": "1\nabacaba 2400 2451", "output": "YES" }, { "input": "1\nrekt_n00b 2500 2600", "output": "YES" } ]

1,642,222,803

2,147,483,647

PyPy 3-64

OK

TESTS

60

109

0

n=int(input()) p=0 for i in range(n): a=input().split() if int(a[1])>=2400 and int(a[1])<int(a[2]): p=1 break if p==1: print('YES') else: print('NO')

Title: A Good Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance. Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it. Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same? Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest . The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters. It is guaranteed that all handles are distinct. Output Specification: Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. Demo Input: ['3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n', '3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n'] Demo Output: ['YES', 'NO'] Note: In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest. In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.

```python n=int(input()) p=0 for i in range(n): a=input().split() if int(a[1])>=2400 and int(a[1])<int(a[2]): p=1 break if p==1: print('YES') else: print('NO') ```

3

812

A

Sagheer and Crossroads

PROGRAMMING

1,200

[ "implementation" ]

null

Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing. An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time. Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.

The input consists of four lines with each line describing a road part given in a counter-clockwise order. Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.

On a single line, print "YES" if an accident is possible, and "NO" otherwise.

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

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

In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4. In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.

500

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"output": "NO" }, { "input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n0 0 0 1", "output": "YES" }, { "input": "1 1 0 0\n0 1 0 0\n1 1 1 0\n1 0 1 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 1", "output": "NO" }, { "input": "1 0 1 0\n1 1 0 0\n1 1 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 1 0\n1 1 0 0\n1 0 1 0\n1 0 0 0", "output": "NO" }, { "input": "0 0 1 0\n1 0 0 0\n0 0 0 1\n0 0 0 1", "output": "NO" }, { "input": "0 1 1 0\n1 1 0 1\n1 0 0 1\n1 1 1 0", "output": "YES" }, { "input": "1 0 0 0\n1 1 0 0\n1 1 0 1\n0 0 1 0", "output": "YES" }, { "input": "0 0 0 0\n1 1 0 0\n0 0 0 1\n0 0 1 0", "output": "NO" }, { "input": "0 1 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 1 0 0\n1 1 0 1\n1 0 0 1\n1 1 0 1", "output": "YES" }, { "input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 1 0 0\n0 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 1 0\n0 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 0 0\n1 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 0 0 0\n0 0 1 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n1 0 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 1 0 0", "output": "NO" }, { "input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 1 0", "output": "YES" }, { "input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 1 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 1 0\n0 0 0 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n1 0 0 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 1 0 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 1 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 1\n1 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 1\n0 0 1 0\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 1 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 1 0", "output": "NO" }, { "input": "1 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0", "output": "NO" }, { "input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0", "output": "YES" }, { "input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n1 0 0 0\n0 0 0 1\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 1 0 0\n0 0 0 1\n0 0 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 1 0\n0 0 0 1\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n1 0 0 1\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 1 0 1\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 1 1\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n1 0 0 0", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 1 0 0", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 1 0", "output": "NO" }, { "input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1", "output": "YES" }, { "input": "0 1 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 1 0\n0 0 0 0\n0 0 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 0 0\n1 0 0 0\n0 0 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 0 0\n0 1 0 0\n0 0 0 0\n0 0 0 1", "output": "YES" }, { "input": "0 0 0 0\n0 0 1 0\n0 0 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 0\n0 1 0 0\n0 0 0 1", "output": "NO" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 1 0\n0 0 0 1", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n1 0 0 1", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 1", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 1 1", "output": "YES" }, { "input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "NO" }, { "input": "1 1 1 1\n1 1 1 1\n1 1 1 1\n1 1 1 1", "output": "YES" }, { "input": "1 0 0 0\n0 1 0 0\n0 0 1 0\n0 0 0 1", "output": "YES" }, { "input": "1 1 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0", "output": "YES" }, { "input": "0 1 1 0\n1 0 1 0\n1 1 1 0\n0 0 0 1", "output": "YES" }, { "input": "1 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "1 1 1 0\n1 1 1 0\n1 1 1 0\n0 0 0 1", "output": "YES" }, { "input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1", "output": "YES" }, { "input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 0 1 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 1 1 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 1 0 1\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 0 0 1\n0 0 0 0\n0 1 0 0", "output": "YES" }, { "input": "0 0 0 1\n0 0 0 1\n1 0 0 0\n0 0 0 0", "output": "YES" } ]

1,496,328,298

1,798

PyPy 3

OK

TESTS

93

108

0

l1, s1, r1, p1 = map(int, input().split()) l2, s2, r2, p2 = map(int, input().split()) l3, s3, r3, p3 = map(int, input().split()) l4, s4, r4, p4 = map(int, input().split()) if (p1 and (r1 or s1 or l1 or r4 or l2 or s3)) or (p2 and (r2 or s2 or l2 or l3 or r1 or s4)) or \ (p3 and (r3 or s3 or l3 or l4 or r2 or s1)) or\ (p4 and (r4 or s4 or l4 or l1 or r3 or s2)): print('YES') else: print('NO')

Title: Sagheer and Crossroads Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing. An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time. Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible. Input Specification: The input consists of four lines with each line describing a road part given in a counter-clockwise order. Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light. Output Specification: On a single line, print "YES" if an accident is possible, and "NO" otherwise. Demo Input: ['1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n', '0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n', '1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4. In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.

```python l1, s1, r1, p1 = map(int, input().split()) l2, s2, r2, p2 = map(int, input().split()) l3, s3, r3, p3 = map(int, input().split()) l4, s4, r4, p4 = map(int, input().split()) if (p1 and (r1 or s1 or l1 or r4 or l2 or s3)) or (p2 and (r2 or s2 or l2 or l3 or r1 or s4)) or \ (p3 and (r3 or s3 or l3 or l4 or r2 or s1)) or\ (p4 and (r4 or s4 or l4 or l1 or r3 or s2)): print('YES') else: print('NO') ```

3

58

B

Coins

PROGRAMMING

1,300

[ "greedy" ]

B. Coins

2

256

In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly *n* Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.

The first and only line contains an integer *n* (1<=≤<=*n*<=≤<=106) which represents the denomination of the most expensive coin.

Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination *n* of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.

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

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

none

1,000

[ { "input": "10", "output": "10 5 1" }, { "input": "4", "output": "4 2 1" }, { "input": "3", "output": "3 1" }, { "input": "2", "output": "2 1" }, { "input": "5", "output": "5 1" }, { "input": "6", "output": "6 3 1" }, { "input": "7", "output": "7 1" }, { "input": "1", "output": "1" }, { "input": "8", "output": "8 4 2 1" }, { "input": "12", "output": "12 6 3 1" }, { "input": "100", "output": "100 50 25 5 1" }, { "input": "1000", "output": "1000 500 250 125 25 5 1" }, { "input": "10000", "output": "10000 5000 2500 1250 625 125 25 5 1" }, { "input": "100000", "output": "100000 50000 25000 12500 6250 3125 625 125 25 5 1" }, { "input": "1000000", "output": "1000000 500000 250000 125000 62500 31250 15625 3125 625 125 25 5 1" }, { "input": "509149", "output": "509149 1" }, { "input": "572877", "output": "572877 190959 63653 1201 1" }, { "input": "152956", "output": "152956 76478 38239 1" }, { "input": "733035", "output": "733035 244345 48869 1" }, { "input": "313114", "output": "313114 156557 3331 1" }, { "input": "893193", "output": "893193 297731 42533 1" }, { "input": "473273", "output": "473273 2243 1" }, { "input": "537000", "output": "537000 268500 134250 67125 22375 4475 895 179 1" }, { "input": "117079", "output": "117079 6887 97 1" }, { "input": "784653", "output": "784653 261551 9019 311 1" }, { "input": "627251", "output": "627251 1" }, { "input": "9", "output": "9 3 1" }, { "input": "999999", "output": "999999 333333 111111 37037 5291 481 37 1" }, { "input": "120", "output": "120 60 30 15 5 1" }, { "input": "720", "output": "720 360 180 90 45 15 5 1" }, { "input": "648", "output": "648 324 162 81 27 9 3 1" }, { "input": "2430", "output": "2430 1215 405 135 45 15 5 1" }, { "input": "119070", "output": "119070 59535 19845 6615 2205 735 245 49 7 1" }, { "input": "15", "output": "15 5 1" }, { "input": "21", "output": "21 7 1" }, { "input": "25", "output": "25 5 1" }, { "input": "100", "output": "100 50 25 5 1" }, { "input": "524287", "output": "524287 1" }, { "input": "1000000", "output": "1000000 500000 250000 125000 62500 31250 15625 3125 625 125 25 5 1" }, { "input": "600", "output": "600 300 150 75 25 5 1" }, { "input": "1000000", "output": "1000000 500000 250000 125000 62500 31250 15625 3125 625 125 25 5 1" }, { "input": "36", "output": "36 18 9 3 1" }, { "input": "1000000", "output": "1000000 500000 250000 125000 62500 31250 15625 3125 625 125 25 5 1" }, { "input": "20", "output": "20 10 5 1" }, { "input": "999983", "output": "999983 1" }, { "input": "9", "output": "9 3 1" }, { "input": "999983", "output": "999983 1" }, { "input": "20", "output": "20 10 5 1" }, { "input": "121", "output": "121 11 1" }, { "input": "1331", "output": "1331 121 11 1" } ]

1,605,891,154

2,147,483,647

Python 3

OK

TESTS

50

248

409,600

import math from collections import deque def solve(n): d = deque() d.append(n) while n % 2 == 0 : n //=2 d.append(n) for j in range(3 , int(math.sqrt(n)) + 1): while n % j == 0 : n //= j d.append(n) if n > 2 : if n not in d :d.append(n) return d n = int(input()) if n == 1 : print(1) exit(0) r = (solve(n)) if 1 not in r : r.append(1) print(*r)

Title: Coins Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly *n* Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations. Input Specification: The first and only line contains an integer *n* (1<=≤<=*n*<=≤<=106) which represents the denomination of the most expensive coin. Output Specification: Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination *n* of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them. Demo Input: ['10\n', '4\n', '3\n'] Demo Output: ['10 5 1\n', '4 2 1\n', '3 1\n'] Note: none

```python import math from collections import deque def solve(n): d = deque() d.append(n) while n % 2 == 0 : n //=2 d.append(n) for j in range(3 , int(math.sqrt(n)) + 1): while n % j == 0 : n //= j d.append(n) if n > 2 : if n not in d :d.append(n) return d n = int(input()) if n == 1 : print(1) exit(0) r = (solve(n)) if 1 not in r : r.append(1) print(*r) ```

3.937237

764

B

Timofey and cubes

PROGRAMMING

900

[ "constructive algorithms", "implementation" ]

null

Young Timofey has a birthday today! He got kit of *n* cubes as a birthday present from his parents. Every cube has a number *a**i*, which is written on it. Timofey put all the cubes in a row and went to unpack other presents. In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to *n* in their order. Dima performs several steps, on step *i* he reverses the segment of cubes from *i*-th to (*n*<=-<=*i*<=+<=1)-th. He does this while *i*<=≤<=*n*<=-<=*i*<=+<=1. After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of cubes. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109), where *a**i* is the number written on the *i*-th cube after Dima has changed their order.

Print *n* integers, separated by spaces — the numbers written on the cubes in their initial order. It can be shown that the answer is unique.

[ "7\n4 3 7 6 9 1 2\n", "8\n6 1 4 2 5 6 9 2\n" ]

[ "2 3 9 6 7 1 4", "2 1 6 2 5 4 9 6" ]

Consider the first sample. 1. At the begining row was [2, 3, 9, 6, 7, 1, 4]. 1. After first operation row was [4, 1, 7, 6, 9, 3, 2]. 1. After second operation row was [4, 3, 9, 6, 7, 1, 2]. 1. After third operation row was [4, 3, 7, 6, 9, 1, 2]. 1. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

1,000

[ { "input": "7\n4 3 7 6 9 1 2", "output": "2 3 9 6 7 1 4" }, { "input": "8\n6 1 4 2 5 6 9 2", "output": "2 1 6 2 5 4 9 6" }, { "input": "1\n1424", "output": "1424" }, { "input": "9\n-7 9 -4 9 -6 11 15 2 -10", "output": "-10 9 15 9 -6 11 -4 2 -7" }, { "input": "2\n21968 5686", "output": "5686 21968" }, { "input": "5\n241218936 -825949895 -84926813 491336344 -872198236", "output": "-872198236 -825949895 -84926813 491336344 241218936" }, { "input": "42\n-557774624 828320986 -345782722 -62979938 -681259411 -945983652 -139095040 832293378 -82572118 432027535 88438103 568183540 961782904 73543295 615958219 -5050584 322982437 -146046730 759453379 129267920 -819827396 -348156048 805080102 390723009 -771277251 -79011872 -592313207 528489973 656201270 -127795621 17284747 145139617 -565641608 83452176 -223074608 545811186 -657981923 -204657836 154779765 -476867246 180386291 202782486", "output": "202782486 828320986 -476867246 -62979938 -204657836 -945983652 545811186 832293378 83452176 432027535 145139617 568183540 -127795621 73543295 528489973 -5050584 -79011872 -146046730 390723009 129267920 -348156048 -819827396 805080102 759453379 -771277251 322982437 -592313207 615958219 656201270 961782904 17284747 88438103 -565641608 -82572118 -223074608 -139095040 -657981923 -681259411 154779765 -345782722 180386291 -557774624" }, { "input": "2\n1 2", "output": "2 1" }, { "input": "6\n1 2 3 4 5 6", "output": "6 2 4 3 5 1" }, { "input": "4\n0 1 2 3", "output": "3 1 2 0" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "10 2 8 4 6 5 7 3 9 1" }, { "input": "2\n4 5", "output": "5 4" }, { "input": "2\n1 5", "output": "5 1" }, { "input": "2\n2 5", "output": "5 2" }, { "input": "2\n3 4", "output": "4 3" }, { "input": "2\n10 0", "output": "0 10" }, { "input": "66\n8229 5299 5562 1259 2258 1126 5026 8796 3497 7111 6680 1180 1426 4821 3322 3502 400 2951 4187 417 5234 1266 7525 7648 417 9297 174 670 1790 4121 9737 938 4616 3402 2009 2461 474 3728 327 3353 9773 6818 8781 874 2209 1260 1647 6126 9287 9638 5049 1256 4459 863 1835 4822 6809 348 5053 7784 6927 2140 9173 7817 5095 4184", "output": "4184 5299 7817 1259 2140 1126 7784 8796 348 7111 4822 1180 863 4821 1256 3502 9638 2951 6126 417 1260 1266 874 7648 6818 9297 3353 670 3728 4121 2461 938 3402 4616 2009 9737 474 1790 327 174 9773 417 8781 7525 2209 5234 1647 4187 9287 400 5049 3322 4459 1426 1835 6680 6809 3497 5053 5026 6927 2258 9173 5562 5095 8229" } ]

1,656,603,266

2,147,483,647

Python 3

OK

TESTS

34

327

14,233,600

n=int(input()) ar=input().split() for i in range(1,n//2+1,2): ar[i-1],ar[-i]=ar[-i],ar[i-1] print(*ar)

Title: Timofey and cubes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Young Timofey has a birthday today! He got kit of *n* cubes as a birthday present from his parents. Every cube has a number *a**i*, which is written on it. Timofey put all the cubes in a row and went to unpack other presents. In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to *n* in their order. Dima performs several steps, on step *i* he reverses the segment of cubes from *i*-th to (*n*<=-<=*i*<=+<=1)-th. He does this while *i*<=≤<=*n*<=-<=*i*<=+<=1. After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of cubes. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109), where *a**i* is the number written on the *i*-th cube after Dima has changed their order. Output Specification: Print *n* integers, separated by spaces — the numbers written on the cubes in their initial order. It can be shown that the answer is unique. Demo Input: ['7\n4 3 7 6 9 1 2\n', '8\n6 1 4 2 5 6 9 2\n'] Demo Output: ['2 3 9 6 7 1 4', '2 1 6 2 5 4 9 6'] Note: Consider the first sample. 1. At the begining row was [2, 3, 9, 6, 7, 1, 4]. 1. After first operation row was [4, 1, 7, 6, 9, 3, 2]. 1. After second operation row was [4, 3, 9, 6, 7, 1, 2]. 1. After third operation row was [4, 3, 7, 6, 9, 1, 2]. 1. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

```python n=int(input()) ar=input().split() for i in range(1,n//2+1,2): ar[i-1],ar[-i]=ar[-i],ar[i-1] print(*ar) ```

3

931

A

Friends Meeting

PROGRAMMING

800

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

null

Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*. Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6. The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.

The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend. The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend. It is guaranteed that *a*<=≠<=*b*.

Print the minimum possible total tiredness if the friends meet in the same point.

[ "3\n4\n", "101\n99\n", "5\n10\n" ]

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

In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1. In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2. In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.

500

[ { "input": "3\n4", "output": "1" }, { "input": "101\n99", "output": "2" }, { "input": "5\n10", "output": "9" }, { "input": "1\n2", "output": "1" }, { "input": "1\n1000", "output": "250000" }, { "input": "999\n1000", "output": "1" }, { "input": "1000\n999", "output": "1" }, { "input": "1000\n1", "output": "250000" }, { "input": "2\n1", "output": "1" }, { "input": "2\n999", "output": "249001" }, { "input": "2\n998", "output": "248502" }, { "input": "999\n2", "output": "249001" }, { "input": "998\n2", "output": "248502" }, { "input": "2\n1000", "output": "249500" }, { "input": "1000\n2", "output": "249500" }, { "input": "1\n999", "output": "249500" }, { "input": "999\n1", "output": "249500" }, { "input": "188\n762", "output": "82656" }, { "input": "596\n777", "output": "8281" }, { "input": "773\n70", "output": "123904" }, { "input": "825\n729", "output": "2352" }, { "input": "944\n348", "output": "89102" }, { "input": "352\n445", "output": "2209" }, { "input": "529\n656", "output": "4096" }, { "input": "19\n315", "output": "22052" }, { "input": "138\n370", "output": "13572" }, { "input": "546\n593", "output": "576" }, { "input": "285\n242", "output": "484" }, { "input": "773\n901", "output": "4160" }, { "input": "892\n520", "output": "34782" }, { "input": "864\n179", "output": "117649" }, { "input": "479\n470", "output": "25" }, { "input": "967\n487", "output": "57840" }, { "input": "648\n106", "output": "73712" }, { "input": "58\n765", "output": "125316" }, { "input": "235\n56", "output": "8100" }, { "input": "285\n153", "output": "4422" }, { "input": "943\n13", "output": "216690" }, { "input": "675\n541", "output": "4556" }, { "input": "4\n912", "output": "206570" } ]

1,659,896,004

2,147,483,647

PyPy 3-64

OK

TESTS

40

77

0

a = int(input()) b = int(input()) def tiredness(point_one, point_two): difference = abs(point_one - point_two) steps_for_each = int(difference / 2) result = steps_for_each * (steps_for_each + 1) if (difference % 2 != 0): result += (steps_for_each + 1) return result print(tiredness(a, b))

Title: Friends Meeting Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*. Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6. The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point. Input Specification: The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend. The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend. It is guaranteed that *a*<=≠<=*b*. Output Specification: Print the minimum possible total tiredness if the friends meet in the same point. Demo Input: ['3\n4\n', '101\n99\n', '5\n10\n'] Demo Output: ['1\n', '2\n', '9\n'] Note: In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1. In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2. In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.

```python a = int(input()) b = int(input()) def tiredness(point_one, point_two): difference = abs(point_one - point_two) steps_for_each = int(difference / 2) result = steps_for_each * (steps_for_each + 1) if (difference % 2 != 0): result += (steps_for_each + 1) return result print(tiredness(a, b)) ```

3

670

D2

Magic Powder - 2

PROGRAMMING

1,500

[ "binary search", "implementation" ]

null

The term of this problem is the same as the previous one, the only exception — increased restrictions.

The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*k*<=≤<=109) — the number of ingredients and the number of grams of the magic powder. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, needed to bake one cookie. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=109), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, which Apollinaria has.

Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.

[ "1 1000000000\n1\n1000000000\n", "10 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n1 1 1 1 1 1 1 1 1 1\n", "3 1\n2 1 4\n11 3 16\n", "4 3\n4 3 5 6\n11 12 14 20\n" ]

[ "2000000000\n", "0\n", "4\n", "3\n" ]

none

500

[ { "input": "1 1000000000\n1\n1000000000", "output": "2000000000" }, { "input": "10 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "3 1\n2 1 4\n11 3 16", "output": "4" }, { "input": "4 3\n4 3 5 6\n11 12 14 20", "output": "3" }, { "input": "10 926\n5 6 8 1 2 5 1 8 4 4\n351 739 998 725 953 970 906 691 707 1000", "output": "137" }, { "input": "20 925\n7 3 1 2 1 3 1 3 1 2 3 1 5 8 1 3 7 3 4 2\n837 898 965 807 786 670 626 873 968 745 878 359 760 781 829 882 777 740 907 779", "output": "150" }, { "input": "30 300\n1 4 2 1 2 5 6 4 1 3 2 1 1 1 1 1 2 3 1 3 4 2 2 3 2 2 2 1 1 1\n997 817 767 860 835 809 817 565 630 804 586 953 977 356 905 890 958 916 740 583 902 945 313 956 871 729 976 707 516 788", "output": "164" }, { "input": "40 538\n1 3 3 1 4 1 1 1 1 5 3 3 4 1 4 2 7 1 4 1 1 2 2 1 1 1 1 4 1 4 2 3 3 3 1 3 4 1 3 5\n975 635 795 835 982 965 639 787 688 796 988 779 839 942 491 696 396 995 718 810 796 879 957 783 844 765 968 783 647 214 995 868 318 453 989 889 504 962 945 925", "output": "104" }, { "input": "1 1\n1000000000\n1000000000", "output": "1" }, { "input": "50 530\n2 3 3 1 1 1 3 4 4 2 4 2 5 1 3 1 2 6 1 1 2 5 3 2 1 5 1 3 3 2 1 1 1 1 2 1 1 2 2 1 4 2 1 3 1 2 1 1 4 2\n959 972 201 990 675 679 972 268 976 886 488 924 795 959 647 994 969 862 898 646 763 797 978 763 995 641 923 856 829 921 934 883 904 986 728 980 1000 775 716 745 833 832 999 651 571 626 827 456 636 795", "output": "133" }, { "input": "60 735\n3 1 4 7 1 7 3 1 1 5 4 7 3 3 3 2 5 3 1 2 3 6 1 1 1 1 1 2 5 3 2 1 3 5 2 1 2 2 2 2 1 3 3 3 6 4 3 5 1 3 2 2 1 3 1 1 1 7 1 2\n596 968 975 493 665 571 598 834 948 941 737 649 923 848 950 907 929 865 227 836 956 796 861 801 746 667 539 807 405 355 501 879 994 890 573 848 597 873 130 985 924 426 999 550 586 924 601 807 994 878 410 817 922 898 982 525 611 685 806 847", "output": "103" }, { "input": "70 130\n2 1 2 2 3 3 2 5 2 2 3 3 3 1 1 4 3 5 3 2 1 3 7 1 2 7 5 2 1 6 3 4 1 2 1 1 1 1 3 6 4 2 2 8 2 2 4 1 4 2 1 4 4 3 5 1 1 1 1 1 2 3 1 5 1 3 3 4 2 2\n473 311 758 768 797 572 656 898 991 534 989 702 934 767 777 799 1000 655 806 727 718 948 834 965 832 778 706 861 799 874 745 970 772 967 984 886 835 795 832 837 950 952 475 891 947 952 903 929 689 478 725 945 585 943 771 631 729 887 557 738 824 758 999 786 669 992 918 762 964 941", "output": "119" }, { "input": "80 979\n2 1 1 1 2 1 1 1 3 1 4 4 2 1 1 3 1 1 2 1 4 1 1 2 5 4 8 1 3 6 5 7 2 3 4 1 2 2 6 1 2 2 4 1 1 2 3 2 8 1 1 3 3 4 1 1 2 1 4 4 1 4 3 2 6 5 2 1 4 1 2 3 2 1 3 3 1 2 1 3\n498 976 513 869 917 914 664 656 957 893 981 947 985 693 576 958 987 822 981 718 884 729 295 683 485 998 730 894 731 975 739 854 906 740 987 976 606 689 990 775 522 994 920 893 529 651 989 799 643 215 946 987 297 868 425 810 694 908 736 903 970 751 625 904 955 945 839 777 977 974 905 900 666 680 799 873 565 919 536 686", "output": "128" }, { "input": "1 1000\n1000\n1000", "output": "2" }, { "input": "1 1000\n1\n1000", "output": "2000" }, { "input": "1 1\n4\n6", "output": "1" }, { "input": "1 1\n1\n1", "output": "2" }, { "input": "1 1\n1000000000\n1", "output": "0" }, { "input": "1 1000000000\n1000000000\n1000000000", "output": "2" }, { "input": "1 1000000000\n100000000\n1000000000", "output": "20" }, { "input": "1 1000000000\n654321\n1000000000", "output": "3056" }, { "input": "1 786543211\n12345678\n987654321", "output": "143" }, { "input": "20 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "1" }, { "input": "13 961166382\n312 781 329 6 813 68 872 146 33 382 134 744 353\n904116983 692808881 382512127 462746212 130261069 450407093 112139570 386577765 867425952 354457553 248439687 817592835 682387533", "output": "714283" }, { "input": "1 1000000000\n1\n2", "output": "1000000002" }, { "input": "6 1000000000\n1 1 1 1 1 1\n1000000000 1 1000000000 1000000000 1000000000 1000000000", "output": "1000000000" }, { "input": "2 1000000000\n1 1\n1 1", "output": "500000001" }, { "input": "1 1\n10\n2", "output": "0" }, { "input": "2 1\n2 2\n1 1", "output": "0" }, { "input": "2 1000000000\n1 1\n1000000000 1000000000", "output": "1500000000" }, { "input": "2 6\n1 3\n6 2", "output": "2" } ]

1,673,029,284

2,147,483,647

PyPy 3-64

OK

TESTS

150

140

19,456,000

n,k = map(int,input().split()) a = list((map(int,input().split()))) b = list((map(int,input().split()))) low = 0 high = 1e18 while (low < high): mid = (low + high + 1)//2 req = 0 for i in range(0,n): req = req + max(0,(mid*a[i])-b[i]) if req > k: high = mid - 1 else: low = mid print(int(low))

Title: Magic Powder - 2 Time Limit: None seconds Memory Limit: None megabytes Problem Description: The term of this problem is the same as the previous one, the only exception — increased restrictions. Input Specification: The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*k*<=≤<=109) — the number of ingredients and the number of grams of the magic powder. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, needed to bake one cookie. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=109), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, which Apollinaria has. Output Specification: Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder. Demo Input: ['1 1000000000\n1\n1000000000\n', '10 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n1 1 1 1 1 1 1 1 1 1\n', '3 1\n2 1 4\n11 3 16\n', '4 3\n4 3 5 6\n11 12 14 20\n'] Demo Output: ['2000000000\n', '0\n', '4\n', '3\n'] Note: none

```python n,k = map(int,input().split()) a = list((map(int,input().split()))) b = list((map(int,input().split()))) low = 0 high = 1e18 while (low < high): mid = (low + high + 1)//2 req = 0 for i in range(0,n): req = req + max(0,(mid*a[i])-b[i]) if req > k: high = mid - 1 else: low = mid print(int(low)) ```

3

10

A

Power Consumption Calculation

PROGRAMMING

900

[ "implementation" ]

A. Power Consumption Calculation

1

256

Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].

The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.

Output the answer to the problem.

[ "1 3 2 1 5 10\n0 10\n", "2 8 4 2 5 10\n20 30\n50 100\n" ]

[ "30", "570" ]

none

0

[ { "input": "1 3 2 1 5 10\n0 10", "output": "30" }, { "input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570" }, { "input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436", "output": "8445" }, { "input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945", "output": "52870" }, { "input": "5 41 20 33 43 4\n46 465\n598 875\n967 980\n1135 1151\n1194 1245", "output": "46995" }, { "input": "6 88 28 100 53 36\n440 445\n525 614\n644 844\n1238 1261\n1305 1307\n1425 1434", "output": "85540" }, { "input": "7 46 61 55 28 59\n24 26\n31 61\n66 133\n161 612\n741 746\n771 849\n1345 1357", "output": "67147" }, { "input": "8 83 18 30 28 5\n196 249\n313 544\n585 630\n718 843\n1040 1194\n1207 1246\n1268 1370\n1414 1422", "output": "85876" }, { "input": "9 31 65 27 53 54\n164 176\n194 210\n485 538\n617 690\n875 886\n888 902\n955 957\n1020 1200\n1205 1282", "output": "38570" }, { "input": "30 3 1 58 44 7\n11 13\n14 32\n37 50\n70 74\n101 106\n113 129\n184 195\n197 205\n213 228\n370 394\n443 446\n457 460\n461 492\n499 585\n602 627\n709 776\n812 818\n859 864\n910 913\n918 964\n1000 1010\n1051 1056\n1063 1075\n1106 1145\n1152 1189\n1211 1212\n1251 1259\n1272 1375\n1412 1417\n1430 1431", "output": "11134" }, { "input": "30 42 3 76 28 26\n38 44\n55 66\n80 81\n84 283\n298 314\n331 345\n491 531\n569 579\n597 606\n612 617\n623 701\n723 740\n747 752\n766 791\n801 827\n842 846\n853 891\n915 934\n945 949\n955 964\n991 1026\n1051 1059\n1067 1179\n1181 1191\n1214 1226\n1228 1233\n1294 1306\n1321 1340\n1371 1374\n1375 1424", "output": "59043" }, { "input": "30 46 5 93 20 46\n12 34\n40 41\n54 58\n100 121\n162 182\n220 349\n358 383\n390 398\n401 403\n408 409\n431 444\n466 470\n471 535\n556 568\n641 671\n699 709\n767 777\n786 859\n862 885\n912 978\n985 997\n1013 1017\n1032 1038\n1047 1048\n1062 1080\n1094 1097\n1102 1113\n1122 1181\n1239 1280\n1320 1369", "output": "53608" }, { "input": "30 50 74 77 4 57\n17 23\n24 61\n67 68\n79 87\n93 101\n104 123\n150 192\n375 377\n398 414\n461 566\n600 633\n642 646\n657 701\n771 808\n812 819\n823 826\n827 833\n862 875\n880 891\n919 920\n928 959\n970 1038\n1057 1072\n1074 1130\n1165 1169\n1171 1230\n1265 1276\n1279 1302\n1313 1353\n1354 1438", "output": "84067" }, { "input": "30 54 76 95 48 16\n9 11\n23 97\n112 116\n126 185\n214 223\n224 271\n278 282\n283 348\n359 368\n373 376\n452 463\n488 512\n532 552\n646 665\n681 685\n699 718\n735 736\n750 777\n791 810\n828 838\n841 858\n874 1079\n1136 1171\n1197 1203\n1210 1219\n1230 1248\n1280 1292\n1324 1374\n1397 1435\n1438 1439", "output": "79844" }, { "input": "30 58 78 12 41 28\n20 26\n27 31\n35 36\n38 99\n103 104\n106 112\n133 143\n181 246\n248 251\n265 323\n350 357\n378 426\n430 443\n466 476\n510 515\n517 540\n542 554\n562 603\n664 810\n819 823\n826 845\n869 895\n921 973\n1002 1023\n1102 1136\n1143 1148\n1155 1288\n1316 1388\n1394 1403\n1434 1437", "output": "82686" }, { "input": "30 62 80 97 25 47\n19 20\n43 75\n185 188\n199 242\n252 258\n277 310\n316 322\n336 357\n398 399\n404 436\n443 520\n549 617\n637 649\n679 694\n705 715\n725 730\n731 756\n768 793\n806 833\n834 967\n1003 1079\n1088 1097\n1100 1104\n1108 1121\n1127 1164\n1240 1263\n1274 1307\n1367 1407\n1419 1425\n1433 1437", "output": "92356" }, { "input": "30 100 48 14 9 7\n26 55\n75 107\n145 146\n158 167\n197 199\n201 213\n222 238\n257 265\n271 338\n357 439\n454 462\n463 562\n633 660\n670 691\n696 829\n844 864\n880 888\n891 894\n895 900\n950 973\n1004 1007\n1018 1021\n1045 1049\n1062 1073\n1138 1184\n1240 1272\n1278 1315\n1403 1410\n1412 1418\n1426 1434", "output": "107490" }, { "input": "30 3 50 32 1 18\n46 60\n66 94\n99 141\n162 239\n309 334\n400 550\n594 630\n659 663\n664 701\n735 736\n741 755\n756 801\n823 854\n865 915\n937 964\n984 986\n990 1007\n1008 1031\n1052 1064\n1073 1080\n1190 1191\n1192 1254\n1265 1266\n1272 1273\n1274 1277\n1284 1302\n1308 1312\n1358 1383\n1388 1389\n1390 1401", "output": "25258" }, { "input": "30 7 52 49 46 37\n2 73\n129 130\n131 140\n143 150\n187 190\n217 238\n248 271\n375 426\n465 470\n475 503\n506 541\n569 571\n578 585\n680 683\n690 738\n742 761\n868 877\n890 892\n904 915\n935 961\n1076 1100\n1103 1114\n1232 1244\n1246 1247\n1257 1261\n1271 1276\n1280 1298\n1346 1360\n1373 1382\n1407 1419", "output": "23992" } ]

1,542,623,981

2,147,483,647

Python 3

OK

TESTS

30

278

204,800

from collections import deque n, p1, p2, p3, t1, t2 = map(int, input().split()) d = {1: p1, 2: p2, 3: p3} LR = [tuple(map(int, input().split())) for _ in range(n)] q = deque(sorted( [(LR[_][0], 'left') for _ in range(n)] + [(LR[_][1], 'right') for _ in range(n)] )) p, m, t, T = 0, 1, LR[0][0], LR[-1][1] while q: tau, tp = q.popleft() if tau > T: break p += (tau - t) * d[m] # print(f'time = {tau}, type = {tp}, mode = {m}, power used = {p}') t = tau if tp == 'left': m = 1 q = deque(list(filter(lambda _: _[1] in {'left', 'right'}, q))) if tp == 'right': q.append((tau + t1, 'mode2')) q.append((tau + t1 + t2, 'mode3')) q = deque(sorted(list(q))) if tp == 'mode2': m = 2 if tp == 'mode3': m = 3 print(p)

Title: Power Consumption Calculation Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*]. Input Specification: The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work. Output Specification: Output the answer to the problem. Demo Input: ['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n'] Demo Output: ['30', '570'] Note: none

```python from collections import deque n, p1, p2, p3, t1, t2 = map(int, input().split()) d = {1: p1, 2: p2, 3: p3} LR = [tuple(map(int, input().split())) for _ in range(n)] q = deque(sorted( [(LR[_][0], 'left') for _ in range(n)] + [(LR[_][1], 'right') for _ in range(n)] )) p, m, t, T = 0, 1, LR[0][0], LR[-1][1] while q: tau, tp = q.popleft() if tau > T: break p += (tau - t) * d[m] # print(f'time = {tau}, type = {tp}, mode = {m}, power used = {p}') t = tau if tp == 'left': m = 1 q = deque(list(filter(lambda _: _[1] in {'left', 'right'}, q))) if tp == 'right': q.append((tau + t1, 'mode2')) q.append((tau + t1 + t2, 'mode3')) q = deque(sorted(list(q))) if tp == 'mode2': m = 2 if tp == 'mode3': m = 3 print(p) ```

3.860619

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" } ]

1,563,456,351

2,147,483,647

Python 3

OK

TESTS

81

248

0

m=[] s=[] l=[] for _ in range(int(input())): l.append(list(map(int,input().split()))) for j in range(len(l[0])): for i in range(len(l)): m.append(l[i][j]) s.append(m) m=[] if all(list(map(lambda x:sum(x)==0,s))): 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 m=[] s=[] l=[] for _ in range(int(input())): l.append(list(map(int,input().split()))) for j in range(len(l[0])): for i in range(len(l)): m.append(l[i][j]) s.append(m) m=[] if all(list(map(lambda x:sum(x)==0,s))): print('YES') else: print('NO') ```

3.938

828

A

Restaurant Tables

PROGRAMMING

1,200

[ "implementation" ]

null

In a small restaurant there are *a* tables for one person and *b* tables for two persons. It it known that *n* groups of people come today, each consisting of one or two people. If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group. If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group. You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.

The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables. The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people.

Print the total number of people the restaurant denies service to.

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

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

In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served. In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients.

500

[ { "input": "4 1 2\n1 2 1 1", "output": "0" }, { "input": "4 1 1\n1 1 2 1", "output": "2" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "2 1 2\n2 2", "output": "0" }, { "input": "5 1 3\n1 2 2 2 1", "output": "1" }, { "input": "7 6 1\n1 1 1 1 1 1 1", "output": "0" }, { "input": "10 2 1\n2 1 2 2 2 2 1 2 1 2", "output": "13" }, { "input": "20 4 3\n2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 1 2", "output": "25" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "1 1 1\n2", "output": "0" }, { "input": "1 200000 200000\n2", "output": "0" }, { "input": "30 10 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2", "output": "20" }, { "input": "4 1 2\n1 1 1 2", "output": "2" }, { "input": "6 2 3\n1 2 1 1 1 2", "output": "2" }, { "input": "6 1 4\n1 1 1 1 1 2", "output": "2" }, { "input": "6 1 3\n1 1 1 1 2 2", "output": "4" }, { "input": "6 1 3\n1 1 1 1 1 2", "output": "2" }, { "input": "6 4 2\n2 1 2 2 1 1", "output": "2" }, { "input": "3 10 1\n2 2 2", "output": "4" }, { "input": "5 1 3\n1 1 1 1 2", "output": "2" }, { "input": "5 2 2\n1 1 1 1 2", "output": "2" }, { "input": "15 5 5\n1 1 1 1 1 1 1 1 1 1 2 2 2 2 2", "output": "10" }, { "input": "5 1 2\n1 1 1 1 1", "output": "0" }, { "input": "3 6 1\n2 2 2", "output": "4" }, { "input": "5 3 3\n2 2 2 2 2", "output": "4" }, { "input": "8 3 3\n1 1 1 1 1 1 2 2", "output": "4" }, { "input": "5 1 2\n1 1 1 2 1", "output": "2" }, { "input": "6 1 4\n1 2 2 1 2 2", "output": "2" }, { "input": "2 1 1\n2 2", "output": "2" }, { "input": "2 2 1\n2 2", "output": "2" }, { "input": "5 8 1\n2 2 2 2 2", "output": "8" }, { "input": "3 1 4\n1 1 2", "output": "0" }, { "input": "7 1 5\n1 1 1 1 1 1 2", "output": "2" }, { "input": "6 1 3\n1 1 1 2 1 1", "output": "0" }, { "input": "6 1 2\n1 1 1 2 2 2", "output": "6" }, { "input": "8 1 4\n2 1 1 1 2 2 2 2", "output": "6" }, { "input": "4 2 3\n2 2 2 2", "output": "2" }, { "input": "3 1 1\n1 1 2", "output": "2" }, { "input": "5 1 1\n2 2 2 2 2", "output": "8" }, { "input": "10 1 5\n1 1 1 1 1 2 2 2 2 2", "output": "8" }, { "input": "5 1 2\n1 1 1 2 2", "output": "4" }, { "input": "4 1 1\n1 1 2 2", "output": "4" }, { "input": "7 1 2\n1 1 1 1 1 1 1", "output": "2" }, { "input": "5 1 4\n2 2 2 2 2", "output": "2" }, { "input": "6 2 3\n1 1 1 1 2 2", "output": "2" }, { "input": "5 2 2\n2 1 2 1 2", "output": "2" }, { "input": "4 6 1\n2 2 2 2", "output": "6" }, { "input": "6 1 4\n1 1 2 1 1 2", "output": "2" }, { "input": "7 1 3\n1 1 1 1 2 2 2", "output": "6" }, { "input": "4 1 2\n1 1 2 2", "output": "2" }, { "input": "3 1 2\n1 1 2", "output": "0" }, { "input": "6 1 3\n1 2 1 1 2 1", "output": "2" }, { "input": "6 1 3\n1 1 1 2 2 2", "output": "4" }, { "input": "10 2 2\n1 1 1 1 2 2 2 2 2 2", "output": "12" }, { "input": "10 1 4\n1 1 1 1 1 2 2 2 2 2", "output": "10" }, { "input": "3 10 2\n2 2 2", "output": "2" }, { "input": "4 3 1\n1 2 2 2", "output": "4" }, { "input": "7 1 4\n1 1 1 1 1 2 2", "output": "4" }, { "input": "3 4 1\n2 2 2", "output": "4" }, { "input": "4 1 2\n2 1 1 2", "output": "2" }, { "input": "10 1 2\n1 1 1 1 1 1 1 1 1 2", "output": "6" }, { "input": "5 1 3\n1 1 2 1 2", "output": "2" }, { "input": "6 1 3\n1 1 1 1 2 1", "output": "2" }, { "input": "6 1 4\n1 1 1 2 2 2", "output": "2" }, { "input": "7 1 2\n1 2 1 1 1 1 1", "output": "3" }, { "input": "6 2 2\n1 1 1 1 1 1", "output": "0" }, { "input": "6 1 2\n1 1 2 1 1 1", "output": "2" }, { "input": "3 3 1\n2 2 1", "output": "2" }, { "input": "8 4 2\n1 1 1 1 1 1 1 2", "output": "2" }, { "input": "9 1 4\n1 1 1 1 1 2 2 2 2", "output": "8" }, { "input": "5 10 1\n2 2 2 2 2", "output": "8" }, { "input": "3 5 1\n2 2 2", "output": "4" }, { "input": "5 100 1\n2 2 2 2 2", "output": "8" }, { "input": "4 1 2\n1 1 1 1", "output": "0" }, { "input": "4 1 1\n1 1 1 1", "output": "1" }, { "input": "7 2 2\n1 1 1 1 1 1 1", "output": "1" } ]

1,613,939,161

2,147,483,647

PyPy 3

OK

TESTS

97

155

12,492,800

n, a, b = map(int, input().split()) groups = list(map(int, input().split())) # b = b * 2 halfFilled = 0 ans = 0 for i in range(n): if groups[i] == 1: if a > 0: a -= 1 elif b > 0: b -= 1 halfFilled += 1 elif halfFilled > 0: halfFilled -= 1 else: ans += 1 else: if b > 0: b -= 1 else: ans += 2 print(ans)

Title: Restaurant Tables Time Limit: None seconds Memory Limit: None megabytes Problem Description: In a small restaurant there are *a* tables for one person and *b* tables for two persons. It it known that *n* groups of people come today, each consisting of one or two people. If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group. If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group. You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to. Input Specification: The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables. The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people. Output Specification: Print the total number of people the restaurant denies service to. Demo Input: ['4 1 2\n1 2 1 1\n', '4 1 1\n1 1 2 1\n'] Demo Output: ['0\n', '2\n'] Note: In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served. In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients.

```python n, a, b = map(int, input().split()) groups = list(map(int, input().split())) # b = b * 2 halfFilled = 0 ans = 0 for i in range(n): if groups[i] == 1: if a > 0: a -= 1 elif b > 0: b -= 1 halfFilled += 1 elif halfFilled > 0: halfFilled -= 1 else: ans += 1 else: if b > 0: b -= 1 else: ans += 2 print(ans) ```

3

710

A

King Moves

PROGRAMMING

800

[ "implementation" ]

null

The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king. Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).

The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.

Print the only integer *x* — the number of moves permitted for the king.

[ "e4\n" ]

[ "8\n" ]

none

0

[ { "input": "e4", "output": "8" }, { "input": "a1", "output": "3" }, { "input": "h8", "output": "3" }, { "input": "a4", "output": "5" }, { "input": "g7", "output": "8" }, { "input": "e1", "output": "5" }, { "input": "b2", "output": "8" }, { "input": "c7", "output": "8" }, { "input": "h6", "output": "5" }, { "input": "c8", "output": "5" }, { "input": "h2", "output": "5" }, { "input": "h5", "output": "5" }, { "input": "a8", "output": "3" }, { "input": "f8", "output": "5" }, { "input": "h1", "output": "3" }, { "input": "f2", "output": "8" }, { "input": "e8", "output": "5" }, { "input": "h3", "output": "5" }, { "input": "b8", "output": "5" }, { "input": "g8", "output": "5" }, { "input": "d8", "output": "5" }, { "input": "h4", "output": "5" }, { "input": "b1", "output": "5" }, { "input": "a2", "output": "5" } ]

1,600,501,570

2,147,483,647

Python 3

OK

TESTS

24

108

307,200

s = input() if(s == 'a1'): print(3) elif(s == 'a8'): print(3) elif(s == 'h1'): print(3) elif(s == 'h8'): print(3) elif(s[0] == 'a'): print(5) elif(s[0] == 'h'): print(5) elif(s[1] == '1'): print(5) elif(s[1] == '8'): print(5) else: print(8)

Title: King Moves Time Limit: None seconds Memory Limit: None megabytes Problem Description: The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king. Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)). Input Specification: The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'. Output Specification: Print the only integer *x* — the number of moves permitted for the king. Demo Input: ['e4\n'] Demo Output: ['8\n'] Note: none

```python s = input() if(s == 'a1'): print(3) elif(s == 'a8'): print(3) elif(s == 'h1'): print(3) elif(s == 'h8'): print(3) elif(s[0] == 'a'): print(5) elif(s[0] == 'h'): print(5) elif(s[1] == '1'): print(5) elif(s[1] == '8'): print(5) else: print(8) ```

3

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" } ]

1,574,281,960

2,147,483,647

PyPy 3

OK

TESTS

32

280

0

n = int(input()) lists = list(map(int, input().rstrip().split())) for i in range(len(lists)): for j in range(len(lists)): if i != j: if abs(lists[i]-lists[j])%2 == 1: c = 0 else: c = 1 break if c == 0: index = i + 1 print(index)

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()) lists = list(map(int, input().rstrip().split())) for i in range(len(lists)): for j in range(len(lists)): if i != j: if abs(lists[i]-lists[j])%2 == 1: c = 0 else: c = 1 break if c == 0: index = i + 1 print(index) ```

3.93

115

A

Party

PROGRAMMING

900

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

null

A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed?

The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.

Print a single integer denoting the minimum number of groups that will be formed in the party.

[ "5\n-1\n1\n2\n1\n-1\n" ]

[ "3\n" ]

For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5

500

[ { "input": "5\n-1\n1\n2\n1\n-1", "output": "3" }, { "input": "4\n-1\n1\n2\n3", "output": "4" }, { "input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11", "output": "4" }, { "input": "6\n-1\n-1\n2\n3\n1\n1", "output": "3" }, { "input": "3\n-1\n1\n1", "output": "2" }, { "input": "1\n-1", "output": "1" }, { "input": "2\n2\n-1", "output": "2" }, { "input": "2\n-1\n-1", "output": "1" }, { "input": "3\n2\n-1\n1", "output": "3" }, { "input": "3\n-1\n-1\n-1", "output": "1" }, { "input": "5\n4\n5\n1\n-1\n4", "output": "3" }, { "input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7", "output": "4" }, { "input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4", "output": "5" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1", "output": "2" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1", "output": "1" }, { "input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11", "output": "12" }, { "input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2", "output": "11" }, { "input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6", "output": "6" }, { "input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1", "output": "3" }, { "input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1", "output": "2" }, { "input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1", "output": "3" }, { "input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8", "output": "5" }, { "input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10", "output": "6" }, { "input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1", "output": "4" }, { "input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9", "output": "4" }, { "input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8", "output": "3" }, { "input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11", "output": "7" }, { "input": "7\n7\n6\n-1\n-1\n4\n4\n-1", "output": "3" }, { "input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34", "output": "6" }, { "input": "5\n2\n3\n4\n5\n-1", "output": "5" } ]

1,660,048,677

2,147,483,647

Python 3

OK

TESTS

106

124

409,600

from collections import deque n = int(input()) company = {} maximum = 0 for i in range(1, n+1): e = int(input()) if e not in company: company[e] = [i] else: company[e].append(i) def bfs(root): q = deque() q.append((root, 0)) level = 0 while q: curr, l = q.popleft() level = l if curr in company: for element in company[curr]: q.append((element, l+1)) return level print(bfs(-1))

Title: Party Time Limit: None seconds Memory Limit: None megabytes Problem Description: A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. Output Specification: Print a single integer denoting the minimum number of groups that will be formed in the party. Demo Input: ['5\n-1\n1\n2\n1\n-1\n'] Demo Output: ['3\n'] Note: For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5

```python from collections import deque n = int(input()) company = {} maximum = 0 for i in range(1, n+1): e = int(input()) if e not in company: company[e] = [i] else: company[e].append(i) def bfs(root): q = deque() q.append((root, 0)) level = 0 while q: curr, l = q.popleft() level = l if curr in company: for element in company[curr]: q.append((element, l+1)) return level print(bfs(-1)) ```

3

215

B

Olympic Medal

PROGRAMMING

1,300

[ "greedy", "math" ]

null

The World Programming Olympics Medal is a metal disk, consisting of two parts: the first part is a ring with outer radius of *r*1 cm, inner radius of *r*2 cm, (0<=<<=*r*2<=<<=*r*1) made of metal with density *p*1 g/cm3. The second part is an inner disk with radius *r*2 cm, it is made of metal with density *p*2 g/cm3. The disk is nested inside the ring. The Olympic jury decided that *r*1 will take one of possible values of *x*1,<=*x*2,<=...,<=*x**n*. It is up to jury to decide which particular value *r*1 will take. Similarly, the Olympic jury decided that *p*1 will take one of possible value of *y*1,<=*y*2,<=...,<=*y**m*, and *p*2 will take a value from list *z*1,<=*z*2,<=...,<=*z**k*. According to most ancient traditions the ratio between the outer ring mass *m**out* and the inner disk mass *m**in* must equal , where *A*,<=*B* are constants taken from ancient books. Now, to start making medals, the jury needs to take values for *r*1, *p*1, *p*2 and calculate the suitable value of *r*2. The jury wants to choose the value that would maximize radius *r*2. Help the jury find the sought value of *r*2. Value *r*2 doesn't have to be an integer. Medal has a uniform thickness throughout the area, the thickness of the inner disk is the same as the thickness of the outer ring.

The first input line contains an integer *n* and a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*. The second input line contains an integer *m* and a sequence of integers *y*1,<=*y*2,<=...,<=*y**m*. The third input line contains an integer *k* and a sequence of integers *z*1,<=*z*2,<=...,<=*z**k*. The last line contains two integers *A* and *B*. All numbers given in the input are positive and do not exceed 5000. Each of the three sequences contains distinct numbers. The numbers in the lines are separated by spaces.

Print a single real number — the sought value *r*2 with absolute or relative error of at most 10<=-<=6. It is guaranteed that the solution that meets the problem requirements exists.

[ "3 1 2 3\n1 2\n3 3 2 1\n1 2\n", "4 2 3 6 4\n2 1 2\n3 10 6 8\n2 1\n" ]

[ "2.683281573000\n", "2.267786838055\n" ]

In the first sample the jury should choose the following values: *r*1 = 3, *p*1 = 2, *p*2 = 1.

500

[ { "input": "3 1 2 3\n1 2\n3 3 2 1\n1 2", "output": "2.683281573000" }, { "input": "4 2 3 6 4\n2 1 2\n3 10 6 8\n2 1", "output": "2.267786838055" }, { "input": "1 5\n1 3\n1 7\n515 892", "output": "3.263613058533" }, { "input": "2 3 2\n3 2 3 1\n2 2 1\n733 883", "output": "2.655066678191" }, { "input": "2 4 2\n3 1 2 3\n2 2 3\n676 769", "output": "3.176161549164" }, { "input": "2 4 2\n3 2 3 1\n2 3 1\n772 833", "output": "3.496252962144" }, { "input": "2 1 2\n3 2 3 1\n2 1 3\n452 219", "output": "1.539383784060" }, { "input": "2 3 2\n3 3 2 1\n2 3 2\n417 202", "output": "1.946150045603" }, { "input": "2 1 2\n3 1 2 3\n2 3 2\n596 206", "output": "1.168651298016" }, { "input": "2 1 2\n3 3 1 2\n2 2 3\n306 406", "output": "1.631654093847" }, { "input": "2 3 2\n3 3 1 2\n2 2 1\n881 165", "output": "1.799345811354" }, { "input": "2 2 4\n3 1 2 3\n2 2 1\n618 401", "output": "3.251156175034" }, { "input": "10 24 2621 2533 3148 3544 4273 4921 2950 3780 4483\n10 1687 4906 4246 2814 1874 3020 3039 3971 102 492\n10 3458 2699 2463 4395 3607 550 1608 958 3970 3077\n4 891", "output": "4919.762124668494" }, { "input": "1 5000\n1 5000\n1 1\n1 5000", "output": "4999.999900000003" }, { "input": "1 1\n1 1\n1 5000\n5000 1", "output": "0.000199999996" }, { "input": "3 5000 4999 4998\n3 5000 4999 4998\n4 1 2 3 4\n1 5000", "output": "4999.999900000003" }, { "input": "3 1 2 3\n3 1 2 3\n3 5000 4999 4998\n5000 1", "output": "0.001039438331" }, { "input": "3 1 2 3\n1 2\n3 3 2 1\n54 58", "output": "2.478139719747" }, { "input": "3 1 2 3\n1 2\n3 3 2 1\n52 56", "output": "2.479181611624" }, { "input": "3 1 2 3\n1 2\n3 3 2 1\n51 55", "output": "2.479731502196" }, { "input": "3 1 2 3\n1 2\n3 3 2 1\n55 59", "output": "2.477645721991" }, { "input": "3 1 2 3\n1 2\n3 1 2 3\n53 57", "output": "2.478651362102" }, { "input": "13 1 2 3 4 5 6 7 8 9 10 11 12 13\n1 14\n2 15 16\n17 18", "output": "9.165151389912" } ]

1,657,442,307

2,147,483,647

Python 3

OK

TESTS

31

92

512,000

from math import sqrt max_ans = 0 r1s = list(map(int,input().split()))[1:] p1s = list(map(int,input().split()))[1:] p2s = list(map(int,input().split()))[1:] A,B = map(int,input().split()) p2 = min(p2s) r1 = max(r1s) for p1 in p1s : r2 = sqrt((r1**2*p1*B)/((A*p2)+(B*p1))) if r2 > max_ans : max_ans = r2 print(max_ans)

Title: Olympic Medal Time Limit: None seconds Memory Limit: None megabytes Problem Description: The World Programming Olympics Medal is a metal disk, consisting of two parts: the first part is a ring with outer radius of *r*1 cm, inner radius of *r*2 cm, (0<=<<=*r*2<=<<=*r*1) made of metal with density *p*1 g/cm3. The second part is an inner disk with radius *r*2 cm, it is made of metal with density *p*2 g/cm3. The disk is nested inside the ring. The Olympic jury decided that *r*1 will take one of possible values of *x*1,<=*x*2,<=...,<=*x**n*. It is up to jury to decide which particular value *r*1 will take. Similarly, the Olympic jury decided that *p*1 will take one of possible value of *y*1,<=*y*2,<=...,<=*y**m*, and *p*2 will take a value from list *z*1,<=*z*2,<=...,<=*z**k*. According to most ancient traditions the ratio between the outer ring mass *m**out* and the inner disk mass *m**in* must equal , where *A*,<=*B* are constants taken from ancient books. Now, to start making medals, the jury needs to take values for *r*1, *p*1, *p*2 and calculate the suitable value of *r*2. The jury wants to choose the value that would maximize radius *r*2. Help the jury find the sought value of *r*2. Value *r*2 doesn't have to be an integer. Medal has a uniform thickness throughout the area, the thickness of the inner disk is the same as the thickness of the outer ring. Input Specification: The first input line contains an integer *n* and a sequence of integers *x*1,<=*x*2,<=...,<=*x**n*. The second input line contains an integer *m* and a sequence of integers *y*1,<=*y*2,<=...,<=*y**m*. The third input line contains an integer *k* and a sequence of integers *z*1,<=*z*2,<=...,<=*z**k*. The last line contains two integers *A* and *B*. All numbers given in the input are positive and do not exceed 5000. Each of the three sequences contains distinct numbers. The numbers in the lines are separated by spaces. Output Specification: Print a single real number — the sought value *r*2 with absolute or relative error of at most 10<=-<=6. It is guaranteed that the solution that meets the problem requirements exists. Demo Input: ['3 1 2 3\n1 2\n3 3 2 1\n1 2\n', '4 2 3 6 4\n2 1 2\n3 10 6 8\n2 1\n'] Demo Output: ['2.683281573000\n', '2.267786838055\n'] Note: In the first sample the jury should choose the following values: *r*1 = 3, *p*1 = 2, *p*2 = 1.

```python from math import sqrt max_ans = 0 r1s = list(map(int,input().split()))[1:] p1s = list(map(int,input().split()))[1:] p2s = list(map(int,input().split()))[1:] A,B = map(int,input().split()) p2 = min(p2s) r1 = max(r1s) for p1 in p1s : r2 = sqrt((r1**2*p1*B)/((A*p2)+(B*p1))) if r2 > max_ans : max_ans = r2 print(max_ans) ```

3

670

B

Game of Robots

PROGRAMMING

1,000

[ "implementation" ]

null

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

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

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

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

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

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

750

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

1,649,154,766

2,147,483,647

PyPy 3-64

OK

TESTS

143

155

17,612,800

n,k=list(map(int,input().split())) arr=list(map(int,input().split())) c=1 s=0 while True: s=s+c if s>(k-1): break else: c=c+1 continue v=k-(s-c) print(arr[v-1])

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

```python n,k=list(map(int,input().split())) arr=list(map(int,input().split())) c=1 s=0 while True: s=s+c if s>(k-1): break else: c=c+1 continue v=k-(s-c) print(arr[v-1]) ```

3

237

A

Free Cash

PROGRAMMING

1,000

[ "implementation" ]

null

Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately. Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe. Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors. Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe. Note that the time is given in the chronological order. All time is given within one 24-hour period.

Print a single integer — the minimum number of cashes, needed to serve all clients next day.

[ "4\n8 0\n8 10\n8 10\n8 45\n", "3\n0 12\n10 11\n22 22\n" ]

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

In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away. In the second sample all visitors will come in different times, so it will be enough one cash.

500

[ { "input": "4\n8 0\n8 10\n8 10\n8 45", "output": "2" }, { "input": "3\n0 12\n10 11\n22 22", "output": "1" }, { "input": "5\n12 8\n15 27\n15 27\n16 2\n19 52", "output": "2" }, { "input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23", "output": "3" }, { "input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6", "output": "3" }, { "input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32", "output": "6" }, { "input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30", "output": "10" }, { "input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20", "output": "8" }, { "input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49", "output": "1" }, { "input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16", "output": "5" }, { "input": "2\n0 24\n1 0", "output": "1" }, { "input": "1\n0 0", "output": "1" }, { "input": "1\n1 5", "output": "1" }, { "input": "1\n1 1", "output": "1" }, { "input": "3\n0 0\n0 0\n0 0", "output": "3" }, { "input": "1\n5 0", "output": "1" }, { "input": "5\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "5" }, { "input": "1\n10 10", "output": "1" }, { "input": "1\n8 0", "output": "1" }, { "input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "10" }, { "input": "2\n0 0\n0 1", "output": "1" }, { "input": "2\n8 5\n8 5", "output": "2" } ]

1,623,203,415

2,147,483,647

PyPy 3

OK

TESTS

67

1,746

12,800,000

def solve(n, ll): cnt = 1 temp = [1] for i in range(1, n): if ll[i] == ll[i-1]: temp.append(temp[i-1] + 1) cnt = max(cnt, temp[i]) else: temp.append(1) cnt = max(cnt, 1) return cnt n = int(input()) ll = [] for i in range(n): l = list(map(int, input().split())) ll.append(l) print(solve(n, ll))

Title: Free Cash Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately. Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe. Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors. Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe. Note that the time is given in the chronological order. All time is given within one 24-hour period. Output Specification: Print a single integer — the minimum number of cashes, needed to serve all clients next day. Demo Input: ['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n'] Demo Output: ['2\n', '1\n'] Note: In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away. In the second sample all visitors will come in different times, so it will be enough one cash.

```python def solve(n, ll): cnt = 1 temp = [1] for i in range(1, n): if ll[i] == ll[i-1]: temp.append(temp[i-1] + 1) cnt = max(cnt, temp[i]) else: temp.append(1) cnt = max(cnt, 1) return cnt n = int(input()) ll = [] for i in range(n): l = list(map(int, input().split())) ll.append(l) print(solve(n, ll)) ```

3

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

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

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

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

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

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

none

500

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

1,646,325,965

2,147,483,647

PyPy 3-64

OK

TESTS

40

62

0

def solve(): s = input() prev = -1 for i in 'hello': if i in s[prev + 1:]: prev = s[prev + 1:].find(i) + prev + 1 else: return 'NO' return 'YES' print(solve())

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 def solve(): s = input() prev = -1 for i in 'hello': if i in s[prev + 1:]: prev = s[prev + 1:].find(i) + prev + 1 else: return 'NO' return 'YES' print(solve()) ```

3.969

374

B

Inna and Nine

PROGRAMMING

1,500

[ "combinatorics", "greedy" ]

null

Inna loves digit 9 very much. That's why she asked Dima to write a small number consisting of nines. But Dima must have misunderstood her and he wrote a very large number *a*, consisting of digits from 1 to 9. Inna wants to slightly alter the number Dima wrote so that in the end the number contained as many digits nine as possible. In one move, Inna can choose two adjacent digits in a number which sum equals 9 and replace them by a single digit 9. For instance, Inna can alter number 14545181 like this: 14545181<=→<=1945181<=→<=194519<=→<=19919. Also, she can use this method to transform number 14545181 into number 19991. Inna will not transform it into 149591 as she can get numbers 19919 and 19991 which contain more digits nine. Dima is a programmer so he wants to find out how many distinct numbers containing as many digits nine as possible Inna can get from the written number. Help him with this challenging task.

The first line of the input contains integer *a* (1<=≤<=*a*<=≤<=10100000). Number *a* doesn't have any zeroes.

In a single line print a single integer — the answer to the problem. It is guaranteed that the answer to the problem doesn't exceed 263<=-<=1. 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.

[ "369727\n", "123456789987654321\n", "1\n" ]

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

Notes to the samples In the first sample Inna can get the following numbers: 369727 → 99727 → 9997, 369727 → 99727 → 9979. In the second sample, Inna can act like this: 123456789987654321 → 12396789987654321 → 1239678998769321.

1,000

[ { "input": "369727", "output": "2" }, { "input": "123456789987654321", "output": "1" }, { "input": "1", "output": "1" }, { "input": "3636363636363454545454543636363636454545452727272727218181818181999111777", "output": "1512" }, { "input": "1188", "output": "1" }, { "input": "121212912121291299129191219", "output": "1" }, { "input": "181818918181891918918181918189181818181891818191818191819189", "output": "54" }, { "input": "12191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219", "output": "1" }, { "input": "444444444444445444444444454444444444444444444444445544444444444444444444444444444444444444444554444444444444444444444444444444444444445454444444444444444444444444444444444444454444444444444444444444444444444444445444444444444444444444444444444444444444444445444444444444444444444444444444444456666666666666666666663555555555555555888888888882333333312567312389542179415242164512341234213443123412341293412341234123412746129342154796124123459123491238471234213451692341278451234125934", "output": "96" }, { "input": "123456789", "output": "1" }, { "input": "3639272918194549", "output": "16" }, { "input": "1121314151617181921222324252627282931323334353637383941424344454647484951525354556575859616263646566768697172737475767787981828384858687888991929394959696979899", "output": "256" }, { "input": "14545181", "output": "2" }, { "input": "272727272", "output": "5" }, { "input": "1212121217272727121", "output": "4" }, { "input": "1212172727", "output": "3" }, { "input": "181817272727", "output": "12" } ]

1,557,907,859

2,147,483,647

Python 3

OK

TESTS

25

202

204,800

s = input() n = len(s) res = 1 cnt = 1 for i in range(n-1): if int(s[i]) + int(s[i+1]) == 9: cnt+=1 else: if cnt % 2 == 1: res = res * (cnt // 2 + 1) cnt = 1 if cnt % 2 == 1: res = res * (cnt // 2 + 1) print(res)

Title: Inna and Nine Time Limit: None seconds Memory Limit: None megabytes Problem Description: Inna loves digit 9 very much. That's why she asked Dima to write a small number consisting of nines. But Dima must have misunderstood her and he wrote a very large number *a*, consisting of digits from 1 to 9. Inna wants to slightly alter the number Dima wrote so that in the end the number contained as many digits nine as possible. In one move, Inna can choose two adjacent digits in a number which sum equals 9 and replace them by a single digit 9. For instance, Inna can alter number 14545181 like this: 14545181<=→<=1945181<=→<=194519<=→<=19919. Also, she can use this method to transform number 14545181 into number 19991. Inna will not transform it into 149591 as she can get numbers 19919 and 19991 which contain more digits nine. Dima is a programmer so he wants to find out how many distinct numbers containing as many digits nine as possible Inna can get from the written number. Help him with this challenging task. Input Specification: The first line of the input contains integer *a* (1<=≤<=*a*<=≤<=10100000). Number *a* doesn't have any zeroes. Output Specification: In a single line print a single integer — the answer to the problem. It is guaranteed that the answer to the problem doesn't exceed 263<=-<=1. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['369727\n', '123456789987654321\n', '1\n'] Demo Output: ['2\n', '1\n', '1\n'] Note: Notes to the samples In the first sample Inna can get the following numbers: 369727 → 99727 → 9997, 369727 → 99727 → 9979. In the second sample, Inna can act like this: 123456789987654321 → 12396789987654321 → 1239678998769321.

```python s = input() n = len(s) res = 1 cnt = 1 for i in range(n-1): if int(s[i]) + int(s[i+1]) == 9: cnt+=1 else: if cnt % 2 == 1: res = res * (cnt // 2 + 1) cnt = 1 if cnt % 2 == 1: res = res * (cnt // 2 + 1) print(res) ```

3

379

A

New Year Candles

PROGRAMMING

1,000

[ "implementation" ]

null

Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.

The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).

Print a single integer — the number of hours Vasily can light up the room for.

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

[ "7\n", "8\n" ]

Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.

500

[ { "input": "4 2", "output": "7" }, { "input": "6 3", "output": "8" }, { "input": "1000 1000", "output": "1001" }, { "input": "123 5", "output": "153" }, { "input": "1000 2", "output": "1999" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "1 4", "output": "1" }, { "input": "2 2", "output": "3" }, { "input": "3 2", "output": "5" }, { "input": "3 3", "output": "4" }, { "input": "999 2", "output": "1997" }, { "input": "1000 3", "output": "1499" }, { "input": "1000 4", "output": "1333" }, { "input": "1 1000", "output": "1" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "10 4", "output": "13" }, { "input": "4 3", "output": "5" }, { "input": "91 5", "output": "113" }, { "input": "777 17", "output": "825" }, { "input": "100 4", "output": "133" }, { "input": "5 3", "output": "7" }, { "input": "6 4", "output": "7" }, { "input": "26 8", "output": "29" }, { "input": "9 4", "output": "11" }, { "input": "20 3", "output": "29" }, { "input": "17 3", "output": "25" } ]

1,654,104,074

2,147,483,647

PyPy 3-64

OK

TESTS

32

46

0

a,b = list(map(int, input().split())) ans = a while a >= b: c = a // b d = a % b ans += c a = c a += d print(ans)

Title: New Year Candles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. Input Specification: The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). Output Specification: Print a single integer — the number of hours Vasily can light up the room for. Demo Input: ['4 2\n', '6 3\n'] Demo Output: ['7\n', '8\n'] Note: Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.

```python a,b = list(map(int, input().split())) ans = a while a >= b: c = a // b d = a % b ans += c a = c a += d print(ans) ```

3

119

A

Epic Game

PROGRAMMING

800

[ "implementation" ]

null

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

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

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

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

[ "0", "1" ]

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

500

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

1,654,494,438

2,147,483,647

Python 3

OK

TESTS

45

62

0

#!/usr/bin/env/python # -*- coding: utf-8 -*- import math a, b, n = list(map(int, input().split())) ab = [a, b] turn = 0 while True: g = math.gcd(ab[turn], n) if g <= n: n -= g else: print(1 - turn) break turn = 1 - turn

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

```python #!/usr/bin/env/python # -*- coding: utf-8 -*- import math a, b, n = list(map(int, input().split())) ab = [a, b] turn = 0 while True: g = math.gcd(ab[turn], n) if g <= n: n -= g else: print(1 - turn) break turn = 1 - turn ```

3

758

A

Holiday Of Equality

PROGRAMMING

800

[ "implementation", "math" ]

null

In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.

The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.

In the only line print the integer *S* — the minimum number of burles which are had to spend.

[ "5\n0 1 2 3 4\n", "5\n1 1 0 1 1\n", "3\n1 3 1\n", "1\n12\n" ]

[ "10", "1", "4", "0" ]

In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.

500

[ { "input": "5\n0 1 2 3 4", "output": "10" }, { "input": "5\n1 1 0 1 1", "output": "1" }, { "input": "3\n1 3 1", "output": "4" }, { "input": "1\n12", "output": "0" }, { "input": "3\n1 2 3", "output": "3" }, { "input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281", "output": "5464380" }, { "input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353", "output": "9535765" }, { "input": "3\n1 3 2", "output": "3" }, { "input": "3\n2 1 3", "output": "3" }, { "input": "3\n2 3 1", "output": "3" }, { "input": "3\n3 1 2", "output": "3" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n228503", "output": "0" }, { "input": "2\n32576 550340", "output": "517764" }, { "input": "3\n910648 542843 537125", "output": "741328" }, { "input": "4\n751720 572344 569387 893618", "output": "787403" }, { "input": "6\n433864 631347 597596 794426 713555 231193", "output": "1364575" }, { "input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107", "output": "4647430" }, { "input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025", "output": "13488674" }, { "input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763", "output": "21993384" }, { "input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351", "output": "26984185" }, { "input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642", "output": "32816391" }, { "input": "1\n0", "output": "0" }, { "input": "1\n1000000", "output": "0" }, { "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": "0" }, { "input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99000000" }, { "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 1000000", "output": "99000000" }, { "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 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "99000000" }, { "input": "3\n0 0 0", "output": "0" }, { "input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "3\n5 0 0", "output": "10" }, { "input": "5\n2 10 0 0 0", "output": "38" } ]

1,649,950,113

2,147,483,647

Python 3

OK

TESTS

41

46

0

n = int(input()) w = list(map(int,input().split())) b = w[:] for i in range(len(w)): if w[i] != max(w): w[i] = w[i] + (max(w)-w[i]) print(sum(w)-sum(b))

Title: Holiday Of Equality Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. Input Specification: The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. Output Specification: In the only line print the integer *S* — the minimum number of burles which are had to spend. Demo Input: ['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n'] Demo Output: ['10', '1', '4', '0'] Note: In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.

```python n = int(input()) w = list(map(int,input().split())) b = w[:] for i in range(len(w)): if w[i] != max(w): w[i] = w[i] + (max(w)-w[i]) print(sum(w)-sum(b)) ```

3

599

A

Patrick and Shopping

PROGRAMMING

800

[ "implementation" ]

null

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house. Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.

The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths. - *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

[ "10 20 30\n", "1 1 5\n" ]

[ "60\n", "4\n" ]

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.

500

[ { "input": "10 20 30", "output": "60" }, { "input": "1 1 5", "output": "4" }, { "input": "100 33 34", "output": "134" }, { "input": "777 777 777", "output": "2331" }, { "input": "2 2 8", "output": "8" }, { "input": "12 34 56", "output": "92" }, { "input": "789 101112 131415", "output": "203802" }, { "input": "27485716 99999999 35182", "output": "55041796" }, { "input": "1 293548 5", "output": "12" }, { "input": "12059 259855 5874875", "output": "543828" }, { "input": "46981 105809 585858", "output": "305580" }, { "input": "9889 1221 2442", "output": "7326" }, { "input": "100500 200600 300700", "output": "601800" }, { "input": "318476 318476 318476", "output": "955428" }, { "input": "23985 3353 75633", "output": "54676" }, { "input": "120 1298 2222", "output": "2836" }, { "input": "98437 23487 666672", "output": "243848" }, { "input": "100000000 100000000 100000000", "output": "300000000" }, { "input": "2 5 2", "output": "8" }, { "input": "1 1000 1", "output": "4" }, { "input": "1 100000000 1", "output": "4" } ]

1,623,515,805

2,147,483,647

Python 3

OK

TESTS

21

62

0

a,b,c=map(int,input().split()) print(min(2*(a+b+c-max(a,b,c)),a+b+c))

Title: Patrick and Shopping Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house. Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. Input Specification: The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths. - *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. Output Specification: Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. Demo Input: ['10 20 30\n', '1 1 5\n'] Demo Output: ['60\n', '4\n'] Note: The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.

```python a,b,c=map(int,input().split()) print(min(2*(a+b+c-max(a,b,c)),a+b+c)) ```

3

978

A

Remove Duplicates

PROGRAMMING

800

[ "implementation" ]

null

Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.

The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.

In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.

[ "6\n1 5 5 1 6 1\n", "5\n2 4 2 4 4\n", "5\n6 6 6 6 6\n" ]

[ "3\n5 6 1 \n", "2\n2 4 \n", "1\n6 \n" ]

In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.

0

[ { "input": "6\n1 5 5 1 6 1", "output": "3\n5 6 1 " }, { "input": "5\n2 4 2 4 4", "output": "2\n2 4 " }, { "input": "5\n6 6 6 6 6", "output": "1\n6 " }, { "input": "7\n1 2 3 4 2 2 3", "output": "4\n1 4 2 3 " }, { "input": "9\n100 100 100 99 99 99 100 100 100", "output": "2\n99 100 " }, { "input": "27\n489 489 487 488 750 230 43 645 42 42 489 42 973 42 973 750 645 355 868 112 868 489 750 489 887 489 868", "output": "13\n487 488 230 43 42 973 645 355 112 750 887 489 868 " }, { "input": "40\n151 421 421 909 117 222 909 954 227 421 227 954 954 222 421 227 421 421 421 151 421 227 222 222 222 222 421 183 421 227 421 954 222 421 954 421 222 421 909 421", "output": "8\n117 151 183 227 954 222 909 421 " }, { "input": "48\n2 2 2 903 903 2 726 2 2 2 2 2 2 2 2 2 2 726 2 2 2 2 2 2 2 726 2 2 2 2 62 2 2 2 2 2 2 2 2 726 62 726 2 2 2 903 903 2", "output": "4\n62 726 903 2 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "13\n5 37 375 5 37 33 37 375 37 2 3 3 2", "output": "6\n5 33 375 37 3 2 " }, { "input": "50\n1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 4 3 2 5 1 6 6 6 6 6", "output": "6\n4 3 2 5 1 6 " }, { "input": "47\n233 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "2\n233 1 " }, { "input": "47\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", "output": "1\n1 " }, { "input": "2\n964 964", "output": "1\n964 " }, { "input": "2\n1000 1000", "output": "1\n1000 " }, { "input": "1\n1000", "output": "1\n1000 " }, { "input": "45\n991 991 996 996 992 992 999 1000 998 1000 992 999 996 999 991 991 999 993 992 999 1000 997 992 999 996 991 994 996 991 999 1000 993 999 997 999 992 991 997 991 998 998 995 998 994 993", "output": "10\n996 1000 999 992 997 991 995 998 994 993 " }, { "input": "6\n994 993 1000 998 991 994", "output": "5\n993 1000 998 991 994 " }, { "input": "48\n992 995 992 991 994 992 995 999 996 993 999 995 993 992 1000 992 997 996 991 993 992 998 998 998 999 995 992 992 993 992 992 995 996 995 997 991 997 991 999 994 994 997 1000 998 1000 992 1000 999", "output": "10\n993 996 995 991 994 997 998 992 1000 999 " }, { "input": "3\n6 6 3", "output": "2\n6 3 " }, { "input": "36\n999 1000 993 993 1000 999 996 997 998 995 995 997 999 995 1000 998 998 994 993 998 994 999 1000 995 996 994 991 991 999 996 993 999 996 998 991 997", "output": "9\n1000 995 994 993 999 996 998 991 997 " }, { "input": "49\n991 997 995 991 995 998 994 996 997 991 998 993 992 999 996 995 991 991 999 997 995 993 997 1000 997 993 993 994 999 994 992 991 1000 996 994 996 996 999 992 992 1000 991 997 993 991 994 997 991 996", "output": "10\n998 995 999 992 1000 993 994 997 991 996 " }, { "input": "2\n1000 999", "output": "2\n1000 999 " }, { "input": "28\n997 994 991 994 994 995 1000 992 995 994 994 995 991 996 991 996 991 999 999 993 994 997 995 992 991 992 998 1000", "output": "10\n996 999 993 994 997 995 991 992 998 1000 " }, { "input": "15\n991 995 995 1000 991 993 997 998 1000 994 1000 992 991 995 992", "output": "8\n993 997 998 994 1000 991 995 992 " }, { "input": "6\n1000 1 1000 2 1000 1", "output": "3\n2 1000 1 " }, { "input": "6\n1000 1 1000 2 1 1000", "output": "3\n2 1 1000 " }, { "input": "7\n1 1000 1 999 1000 1 2", "output": "4\n999 1000 1 2 " } ]

1,658,244,961

361

Python 3

OK

TESTS

28

46

102,400

from collections import defaultdict n = int(input()) a = list(map(int, input().split())) d = defaultdict(lambda : 0) for i in a: d[i] = 1 ans = [] for i in a[::-1]: if d[i] == 1: ans.append(i) d[i] = 0 print(len(ans)) for i in ans[::-1]: print(i, end = " ") print()

Title: Remove Duplicates Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. Input Specification: The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. Output Specification: In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. Demo Input: ['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n'] Demo Output: ['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n'] Note: In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.

```python from collections import defaultdict n = int(input()) a = list(map(int, input().split())) d = defaultdict(lambda : 0) for i in a: d[i] = 1 ans = [] for i in a[::-1]: if d[i] == 1: ans.append(i) d[i] = 0 print(len(ans)) for i in ans[::-1]: print(i, end = " ") print() ```

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" } ]

1,678,889,395

2,147,483,647

Python 3

OK

TESTS

30

92

0

word = input() count_uc = 0 count_lc = 0 for c in word: if c.isupper(): count_uc += 1 else: count_lc += 1 if count_uc > count_lc: print(word.upper()) else: print(word.lower())

Title: Word Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. Input Specification: The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. Output Specification: Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. Demo Input: ['HoUse\n', 'ViP\n', 'maTRIx\n'] Demo Output: ['house\n', 'VIP\n', 'matrix\n'] Note: none

```python word = input() count_uc = 0 count_lc = 0 for c in word: if c.isupper(): count_uc += 1 else: count_lc += 1 if count_uc > count_lc: print(word.upper()) else: print(word.lower()) ```

3.977

385

A

Bear and Raspberry

PROGRAMMING

1,000

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

null

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

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

Print a single integer — the answer to the problem.

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

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

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

500

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

1,683,474,046

2,147,483,647

Python 3

OK

TESTS

32

46

0

n,c=map(int, input().split()) l=list(map(int, input().split())) l1=[] for i in range(n-1): a=l[i]-l[i+1]-c l1.append(a) if (max(l1)>0): print(max(l1)) else: print(0)

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

```python n,c=map(int, input().split()) l=list(map(int, input().split())) l1=[] for i in range(n-1): a=l[i]-l[i+1]-c l1.append(a) if (max(l1)>0): print(max(l1)) else: print(0) ```

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" } ]

1,624,115,612

2,147,483,647

Python 3

OK

TESTS

30

186

512,000

from sys import stdin, stdout from string import ascii_letters, ascii_lowercase, ascii_uppercase from collections import Counter # with open('testcase.txt', 'r') as stdin: # _ = stdin.readline() string = stdin.readline().rstrip() lower = tuple(i for i in string if i in ascii_lowercase) if len(lower) >= len(string)-len(lower): string = string.lower() else: string = string.upper() stdout.write(string)

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 from sys import stdin, stdout from string import ascii_letters, ascii_lowercase, ascii_uppercase from collections import Counter # with open('testcase.txt', 'r') as stdin: # _ = stdin.readline() string = stdin.readline().rstrip() lower = tuple(i for i in string if i in ascii_lowercase) if len(lower) >= len(string)-len(lower): string = string.lower() else: string = string.upper() stdout.write(string) ```

3.952546

408

B

Garland

PROGRAMMING

1,200

[ "implementation" ]

null

Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter. The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland. Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.

The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color. The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.

Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.

[ "aaabbac\naabbccac\n", "a\nz\n" ]

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

In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6. In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*.

1,000

[ { "input": "aaabbac\naabbccac", "output": "6" }, { "input": "a\nz", "output": "-1" }, { "input": "r\nr", "output": "1" }, { "input": "stnsdn\nndnndsn", "output": "4" }, { "input": "yqfqfp\ntttwtqq", "output": "-1" }, { "input": "zzbbrrtrtzr\ntbbtrrrzr", "output": "9" }, { "input": "ivvfisvsvii\npaihjinno", "output": "-1" }, { "input": "zbvwnlgkshqerxptyod\nz", "output": "1" }, { "input": "xlktwjymocqrahnbesf\nfoo", "output": "2" }, { "input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae", "output": "77" }, { "input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl", "output": "-1" }, { "input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh", "output": "199" }, { "input": "aaccddff\nabcdf", "output": "-1" } ]

1,492,535,611

2,147,483,647

Python 3

OK

TESTS

21

140

9,420,800

#!/usr/bin/python import re import inspect from sys import argv, exit def rstr(): return input() def rstrs(splitchar=' '): return [i for i in input().split(splitchar)] def rint(): return int(input()) def rints(splitchar=' '): return [int(i) for i in rstrs(splitchar)] def varnames(obj, namespace=globals()): return [name for name in namespace if namespace[name] is obj] def pvar(var, override=False): prnt(varnames(var), var) def prnt(*args, override=False): if '-v' in argv or override: print(*args) # Faster IO pq = [] def penq(s): if not isinstance(s, str): s = str(s) pq.append(s) def pdump(): s = ('\n'.join(pq)).encode() os.write(1, s) if __name__ == '__main__': s1 = [c for c in rstr()] s2 = [c for c in rstr()] counts1 = [len([c for c in s1 if ord(c)==ch]) for ch in range(ord('a'), ord('z')+1)] counts2 = [len([c for c in s2 if ord(c)==ch]) for ch in range(ord('a'), ord('z')+1)] total = 0 for c1, c2 in zip(counts1, counts2): if c2 != 0 and c1 == 0: print('-1') exit(0) if c2 > c1: ans = c1 else: ans = c2 total += ans print(total)

Title: Garland Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter. The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland. Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. Input Specification: The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color. The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. Output Specification: Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. Demo Input: ['aaabbac\naabbccac\n', 'a\nz\n'] Demo Output: ['6\n', '-1'] Note: In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6. In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*.

```python #!/usr/bin/python import re import inspect from sys import argv, exit def rstr(): return input() def rstrs(splitchar=' '): return [i for i in input().split(splitchar)] def rint(): return int(input()) def rints(splitchar=' '): return [int(i) for i in rstrs(splitchar)] def varnames(obj, namespace=globals()): return [name for name in namespace if namespace[name] is obj] def pvar(var, override=False): prnt(varnames(var), var) def prnt(*args, override=False): if '-v' in argv or override: print(*args) # Faster IO pq = [] def penq(s): if not isinstance(s, str): s = str(s) pq.append(s) def pdump(): s = ('\n'.join(pq)).encode() os.write(1, s) if __name__ == '__main__': s1 = [c for c in rstr()] s2 = [c for c in rstr()] counts1 = [len([c for c in s1 if ord(c)==ch]) for ch in range(ord('a'), ord('z')+1)] counts2 = [len([c for c in s2 if ord(c)==ch]) for ch in range(ord('a'), ord('z')+1)] total = 0 for c1, c2 in zip(counts1, counts2): if c2 != 0 and c1 == 0: print('-1') exit(0) if c2 > c1: ans = c1 else: ans = c2 total += ans print(total) ```

3

166

A

Rank List

PROGRAMMING

1,100

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

null

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

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

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

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

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

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

500

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

1,696,606,826

2,147,483,647

Python 3

OK

TESTS

46

124

307,200

from collections import Counter from itertools import accumulate testcases = [] n, k = [int(x) for x in input().split(' ')] for _ in range(n): points = [int(x) for x in input().split(' ')] testcases.append(tuple(points)) # # print(testcases) # points_approx = [] cnt = Counter(testcases) sorted_scores = sorted(set(testcases), key =lambda x: (x[0], -x[1]), reverse=True) # print(sorted_scores) positions_ord = [cnt[sorted_score] for sorted_score in sorted_scores] acc = list(accumulate(positions_ord)) ans_i = 0 while k > acc[ans_i]: ans_i += 1 print(positions_ord[ans_i])

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

```python from collections import Counter from itertools import accumulate testcases = [] n, k = [int(x) for x in input().split(' ')] for _ in range(n): points = [int(x) for x in input().split(' ')] testcases.append(tuple(points)) # # print(testcases) # points_approx = [] cnt = Counter(testcases) sorted_scores = sorted(set(testcases), key =lambda x: (x[0], -x[1]), reverse=True) # print(sorted_scores) positions_ord = [cnt[sorted_score] for sorted_score in sorted_scores] acc = list(accumulate(positions_ord)) ans_i = 0 while k > acc[ans_i]: ans_i += 1 print(positions_ord[ans_i]) ```

3

911

B

Two Cakes

PROGRAMMING

1,200

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

null

It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces. Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met: 1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes. To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake. Help Ivan to calculate this number *x*!

The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.

Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake.

[ "5 2 3\n", "4 7 10\n" ]

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

In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it. In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.

0

[ { "input": "5 2 3", "output": "1" }, { "input": "4 7 10", "output": "3" }, { "input": "100 100 100", "output": "2" }, { "input": "10 100 3", "output": "3" }, { "input": "2 9 29", "output": "9" }, { "input": "4 6 10", "output": "3" }, { "input": "3 70 58", "output": "35" }, { "input": "5 7 10", "output": "3" }, { "input": "5 30 22", "output": "10" }, { "input": "5 5 6", "output": "2" }, { "input": "2 4 3", "output": "3" }, { "input": "10 10 31", "output": "3" }, { "input": "2 1 1", "output": "1" }, { "input": "10 98 99", "output": "19" }, { "input": "4 10 16", "output": "5" }, { "input": "11 4 8", "output": "1" }, { "input": "5 10 14", "output": "4" }, { "input": "6 7 35", "output": "7" }, { "input": "5 6 7", "output": "2" }, { "input": "4 15 3", "output": "3" }, { "input": "7 48 77", "output": "16" }, { "input": "4 4 10", "output": "3" }, { "input": "4 7 20", "output": "6" }, { "input": "5 2 8", "output": "2" }, { "input": "3 2 3", "output": "1" }, { "input": "14 95 1", "output": "1" }, { "input": "99 82 53", "output": "1" }, { "input": "10 71 27", "output": "9" }, { "input": "5 7 8", "output": "2" }, { "input": "11 77 77", "output": "12" }, { "input": "10 5 28", "output": "3" }, { "input": "7 3 12", "output": "2" }, { "input": "10 15 17", "output": "3" }, { "input": "7 7 7", "output": "1" }, { "input": "4 11 18", "output": "6" }, { "input": "3 3 4", "output": "2" }, { "input": "9 2 10", "output": "1" }, { "input": "100 90 20", "output": "1" }, { "input": "3 2 2", "output": "1" }, { "input": "12 45 60", "output": "8" }, { "input": "3 94 79", "output": "47" }, { "input": "41 67 34", "output": "2" }, { "input": "9 3 23", "output": "2" }, { "input": "10 20 57", "output": "7" }, { "input": "55 27 30", "output": "1" }, { "input": "100 100 10", "output": "1" }, { "input": "20 8 70", "output": "3" }, { "input": "3 3 3", "output": "1" }, { "input": "4 9 15", "output": "5" }, { "input": "3 1 3", "output": "1" }, { "input": "2 94 94", "output": "94" }, { "input": "5 3 11", "output": "2" }, { "input": "4 3 2", "output": "1" }, { "input": "12 12 100", "output": "9" }, { "input": "6 75 91", "output": "25" }, { "input": "3 4 3", "output": "2" }, { "input": "3 2 5", "output": "2" }, { "input": "6 5 15", "output": "3" }, { "input": "4 3 6", "output": "2" }, { "input": "3 9 9", "output": "4" }, { "input": "26 93 76", "output": "6" }, { "input": "41 34 67", "output": "2" }, { "input": "6 12 6", "output": "3" }, { "input": "5 20 8", "output": "5" }, { "input": "2 1 3", "output": "1" }, { "input": "35 66 99", "output": "4" }, { "input": "30 7 91", "output": "3" }, { "input": "5 22 30", "output": "10" }, { "input": "8 19 71", "output": "10" }, { "input": "3 5 6", "output": "3" }, { "input": "5 3 8", "output": "2" }, { "input": "2 4 2", "output": "2" }, { "input": "4 3 7", "output": "2" }, { "input": "5 20 10", "output": "5" }, { "input": "5 100 50", "output": "25" }, { "input": "6 3 10", "output": "2" }, { "input": "2 90 95", "output": "90" }, { "input": "4 8 6", "output": "3" }, { "input": "6 10 3", "output": "2" }, { "input": "3 3 5", "output": "2" }, { "input": "5 33 33", "output": "11" }, { "input": "5 5 8", "output": "2" }, { "input": "19 24 34", "output": "3" }, { "input": "5 5 12", "output": "3" }, { "input": "8 7 10", "output": "2" }, { "input": "5 56 35", "output": "17" }, { "input": "4 3 5", "output": "1" }, { "input": "18 100 50", "output": "8" }, { "input": "5 6 8", "output": "2" }, { "input": "5 98 100", "output": "33" }, { "input": "6 5 8", "output": "2" }, { "input": "3 40 80", "output": "40" }, { "input": "4 8 11", "output": "4" }, { "input": "66 100 99", "output": "3" }, { "input": "17 100 79", "output": "10" }, { "input": "3 2 10", "output": "2" }, { "input": "99 100 99", "output": "2" }, { "input": "21 100 5", "output": "5" }, { "input": "3 10 2", "output": "2" }, { "input": "4 100 63", "output": "33" }, { "input": "2 2 10", "output": "2" }, { "input": "5 94 79", "output": "31" }, { "input": "4 12 5", "output": "4" }, { "input": "5 5 40", "output": "5" }, { "input": "99 99 99", "output": "1" }, { "input": "8 97 44", "output": "16" }, { "input": "11 4 10", "output": "1" }, { "input": "6 3 3", "output": "1" }, { "input": "7 3 4", "output": "1" }, { "input": "8 4 4", "output": "1" }, { "input": "9 4 5", "output": "1" }, { "input": "12 6 6", "output": "1" }, { "input": "4 48 89", "output": "29" }, { "input": "8 3 6", "output": "1" }, { "input": "4 6 3", "output": "2" }, { "input": "5 5 1", "output": "1" }, { "input": "11 6 5", "output": "1" }, { "input": "4 5 4", "output": "2" }, { "input": "6 6 4", "output": "1" }, { "input": "2 1 2", "output": "1" }, { "input": "4 1 3", "output": "1" }, { "input": "3 3 1", "output": "1" }, { "input": "9 4 6", "output": "1" }, { "input": "6 5 6", "output": "1" }, { "input": "2 2 3", "output": "2" }, { "input": "4 5 1", "output": "1" }, { "input": "13 6 7", "output": "1" }, { "input": "14 7 7", "output": "1" }, { "input": "12 97 13", "output": "8" }, { "input": "4 2 9", "output": "2" }, { "input": "10 20 59", "output": "7" }, { "input": "12 34 56", "output": "7" }, { "input": "4 5 9", "output": "3" }, { "input": "2 2 2", "output": "2" }, { "input": "4 66 41", "output": "22" } ]

1,630,435,668

2,147,483,647

Python 3

OK

TESTS

135

78

6,963,200

n,a,b=map(int,input().split()) if a+b==n: print(1) else: p=0 a,b=min(a,b),max(a,b) for i in range(1,n//2+1): r=min(a//i,b//(n-i)) p=max(p,r) print(p)

Title: Two Cakes Time Limit: None seconds Memory Limit: None megabytes Problem Description: It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces. Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met: 1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes. To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake. Help Ivan to calculate this number *x*! Input Specification: The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively. Output Specification: Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake. Demo Input: ['5 2 3\n', '4 7 10\n'] Demo Output: ['1\n', '3\n'] Note: In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it. In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.

```python n,a,b=map(int,input().split()) if a+b==n: print(1) else: p=0 a,b=min(a,b),max(a,b) for i in range(1,n//2+1): r=min(a//i,b//(n-i)) p=max(p,r) print(p) ```

3

888

B

Buggy Robot

PROGRAMMING

1,000

[ "greedy" ]

null

Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform: - U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*). Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!

The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100). The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.

Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.

[ "4\nLDUR\n", "5\nRRRUU\n", "6\nLLRRRR\n" ]

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

none

0

[ { "input": "4\nLDUR", "output": "4" }, { "input": "5\nRRRUU", "output": "0" }, { "input": "6\nLLRRRR", "output": "4" }, { "input": "88\nLLUUULRDRRURDDLURRLRDRLLRULRUUDDLLLLRRDDURDURRLDURRLDRRRUULDDLRRRDDRRLUULLURDURUDDDDDLDR", "output": "76" }, { "input": "89\nLDLLLDRDUDURRRRRUDULDDDLLUDLRLRLRLDLDUULRDUDLRRDLUDLURRDDRRDLDUDUUURUUUDRLUDUDLURDLDLLDDU", "output": "80" }, { "input": "90\nRRRDUULLLRDUUDDRLDLRLUDURDRDUUURUURDDRRRURLDDDUUDRLLLULURDRDRURLDRRRRUULDULDDLLLRRLRDLLLLR", "output": "84" }, { "input": "91\nRLDRLRRLLDLULULLURULLRRULUDUULLUDULDUULURUDRUDUURDULDUDDUUUDRRUUDLLRULRULURLDRDLDRURLLLRDDD", "output": "76" }, { "input": "92\nRLRDDLULRLLUURRDDDLDDDLDDUURRRULLRDULDULLLUUULDUDLRLRRDRDRDDULDRLUDRDULDRURUDUULLRDRRLLDRLRR", "output": "86" }, { "input": "93\nRLLURLULRURDDLUURLUDDRDLUURLRDLRRRDUULLRDRRLRLDURRDLLRDDLLLDDDLDRRURLLDRUDULDDRRULRRULRLDRDLR", "output": "84" }, { "input": "94\nRDULDDDLULRDRUDRUUDUUDRRRULDRRUDURUULRDUUDLULLLUDURRDRDLUDRULRRRULUURUDDDDDUDLLRDLDRLLRUUURLUL", "output": "86" }, { "input": "95\nRDLUUULLUURDDRLDLLRRRULRLRDULULRULRUDURLULDDDRLURLDRULDUDUUULLRDDURUULULLDDLDRDRLLLURLRDLLDDDDU", "output": "86" }, { "input": "96\nRDDRLRLLDDULRLRURUDLRLDUDRURLLUUDLLURDLRRUURDRRUDRURLLDLLRDURDURLRLUDURULLLRDUURULUUULRRURRDLURL", "output": "84" }, { "input": "97\nRURDDLRLLRULUDURDLRLLUUDURRLLUDLLLDUDRUULDRUUURURULRDLDRRLLUUUDLLLDDLLLLRLLDUDRRDLLUDLRURUDULRLUR", "output": "82" }, { "input": "98\nRUDURLULLDDLLRDLLRDDLLLLRLDDDDRRRDDRRURLDRLLRUUUDLUUUDDDUDRUURLURUUDUUDRULRRULLRRLRULLULDLUURLULRD", "output": "92" }, { "input": "99\nRRULLDULRRDRULLDUDRUDDDRLLUUDRDDUDURLDDRUUDRRUUURRRURDDLDUURDLRLURRDDLUDDLUDURDRUDDURLURURLRUDRURLD", "output": "86" }, { "input": "100\nUDRLRRLLRRLRRRDDLLDDDLULLDDLURUURUULUDDDRDDLLRDLLUURLRDRLRRLRLLLULDUDDUURRLRDULDRDURRRRRRULDRRDLDRRL", "output": "88" }, { "input": "1\nU", "output": "0" }, { "input": "5\nUUULD", "output": "2" }, { "input": "1\nD", "output": "0" }, { "input": "5\nURLUL", "output": "2" }, { "input": "5\nDDDRU", "output": "2" }, { "input": "2\nLR", "output": "2" }, { "input": "8\nDDRDLDUR", "output": "4" }, { "input": "6\nLLLLUD", "output": "2" }, { "input": "13\nRRRLLLLLLLLLL", "output": "6" } ]

1,598,463,530

2,147,483,647

Python 3

OK

TESTS

25

109

0

n = int(input()) s = input() l_count = 0 r_count = 0 u_count = 0 d_count = 0 for i in s: if i == "L": l_count+=1 elif i == "R": r_count+=1 elif i == "U": u_count+=1 else: d_count+=1 count = 0 count+=abs(l_count - r_count) count+=abs(u_count - d_count) print(n - count)

Title: Buggy Robot Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform: - U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*). Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations! Input Specification: The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100). The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R. Output Specification: Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell. Demo Input: ['4\nLDUR\n', '5\nRRRUU\n', '6\nLLRRRR\n'] Demo Output: ['4\n', '0\n', '4\n'] Note: none

```python n = int(input()) s = input() l_count = 0 r_count = 0 u_count = 0 d_count = 0 for i in s: if i == "L": l_count+=1 elif i == "R": r_count+=1 elif i == "U": u_count+=1 else: d_count+=1 count = 0 count+=abs(l_count - r_count) count+=abs(u_count - d_count) print(n - count) ```

3

A

Shortest path of the king

PROGRAMMING

1,000

[ "greedy", "shortest paths" ]

A. Shortest path of the king

1

64

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this. In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

The first line contains the chessboard coordinates of square *s*, the second line — of square *t*. Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.

In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD. L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

[ "a8\nh1\n" ]

[ "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n" ]

none

0

[ { "input": "a8\nh1", "output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD" }, { "input": "b2\nb4", "output": "2\nU\nU" }, { "input": "a5\na5", "output": "0" }, { "input": "h1\nb2", "output": "6\nLU\nL\nL\nL\nL\nL" }, { "input": "c5\nh2", "output": "5\nRD\nRD\nRD\nR\nR" }, { "input": "e1\nf2", "output": "1\nRU" }, { "input": "g4\nd2", "output": "3\nLD\nLD\nL" }, { "input": "a8\nb2", "output": "6\nRD\nD\nD\nD\nD\nD" }, { "input": "d4\nh2", "output": "4\nRD\nRD\nR\nR" }, { "input": "c5\na2", "output": "3\nLD\nLD\nD" }, { "input": "h5\nf8", "output": "3\nLU\nLU\nU" }, { "input": "e6\nb6", "output": "3\nL\nL\nL" }, { "input": "a6\ng4", "output": "6\nRD\nRD\nR\nR\nR\nR" }, { "input": "f7\nc2", "output": "5\nLD\nLD\nLD\nD\nD" }, { "input": "b7\nh8", "output": "6\nRU\nR\nR\nR\nR\nR" }, { "input": "g7\nd6", "output": "3\nLD\nL\nL" }, { "input": "c8\na3", "output": "5\nLD\nLD\nD\nD\nD" }, { "input": "h8\nf1", "output": "7\nLD\nLD\nD\nD\nD\nD\nD" }, { "input": "d1\nb7", "output": "6\nLU\nLU\nU\nU\nU\nU" }, { "input": "a7\ne5", "output": "4\nRD\nRD\nR\nR" }, { "input": "d6\nb1", "output": "5\nLD\nLD\nD\nD\nD" }, { "input": "f5\ng5", "output": "1\nR" }, { "input": "h4\nd1", "output": "4\nLD\nLD\nLD\nL" }, { "input": "b3\na5", "output": "2\nLU\nU" }, { "input": "d2\nf1", "output": "2\nRD\nR" }, { "input": "f1\nc5", "output": "4\nLU\nLU\nLU\nU" }, { "input": "a8\nh1", "output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD" }, { "input": "c7\ne5", "output": "2\nRD\nRD" }, { "input": "e7\nb1", "output": "6\nLD\nLD\nLD\nD\nD\nD" }, { "input": "g8\na8", "output": "6\nL\nL\nL\nL\nL\nL" }, { "input": "g6\nf2", "output": "4\nLD\nD\nD\nD" }, { "input": "g4\nc4", "output": "4\nL\nL\nL\nL" }, { "input": "g2\na6", "output": "6\nLU\nLU\nLU\nLU\nL\nL" }, { "input": "f8\nf8", "output": "0" }, { "input": "f5\nd2", "output": "3\nLD\nLD\nD" } ]

1,575,311,277

2,147,483,647

Python 3

OK

TESTS

35

218

307,200

d = {"a":1, "b":2, "c":3, "d":4, "e":5, "f":6, "g":7, "h":8} ne = {"a":"b", "b":"c", "c":"d", "d":"e", "e":"f", "f":"g", "g":"h"} pr = {"h":"g", "g":"f", "f":"e", "e":"d", "d":"c", "c":"b", "b":"a"} s = list(input()) st = list(input()) num1 = int(s[1]) num2 = int(st[1]) ans = 0 res = "" while True: if s[0] < st[0]: if num1 < num2: res += "RU\n" num1 += 1 s[0] = ne[s[0]] ans += 1 elif num1 > num2: res += "RD\n" num1 -= 1 s[0] = ne[s[0]] ans += 1 else: res += "R\n" s[0] = ne[s[0]] ans += 1 elif s[0] > st[0]: if num1 < num2: res += "LU\n" num1 += 1 s[0] = pr[s[0]] ans += 1 elif num1 > num2: res += "LD\n" num1 -= 1 s[0] = pr[s[0]] ans += 1 else: res += "L\n" s[0] = pr[s[0]] ans += 1 else: if num1 < num2: res += "U\n" num1 += 1 ans += 1 elif num1 > num2: res += "D\n" num1 -= 1 ans += 1 else: break print(ans) print(res)

Title: Shortest path of the king Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this. In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to). Input Specification: The first line contains the chessboard coordinates of square *s*, the second line — of square *t*. Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8. Output Specification: In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD. L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them. Demo Input: ['a8\nh1\n'] Demo Output: ['7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n'] Note: none

```python d = {"a":1, "b":2, "c":3, "d":4, "e":5, "f":6, "g":7, "h":8} ne = {"a":"b", "b":"c", "c":"d", "d":"e", "e":"f", "f":"g", "g":"h"} pr = {"h":"g", "g":"f", "f":"e", "e":"d", "d":"c", "c":"b", "b":"a"} s = list(input()) st = list(input()) num1 = int(s[1]) num2 = int(st[1]) ans = 0 res = "" while True: if s[0] < st[0]: if num1 < num2: res += "RU\n" num1 += 1 s[0] = ne[s[0]] ans += 1 elif num1 > num2: res += "RD\n" num1 -= 1 s[0] = ne[s[0]] ans += 1 else: res += "R\n" s[0] = ne[s[0]] ans += 1 elif s[0] > st[0]: if num1 < num2: res += "LU\n" num1 += 1 s[0] = pr[s[0]] ans += 1 elif num1 > num2: res += "LD\n" num1 -= 1 s[0] = pr[s[0]] ans += 1 else: res += "L\n" s[0] = pr[s[0]] ans += 1 else: if num1 < num2: res += "U\n" num1 += 1 ans += 1 elif num1 > num2: res += "D\n" num1 -= 1 ans += 1 else: break print(ans) print(res) ```

3.888711

231

A

Team

PROGRAMMING

800

[ "brute force", "greedy" ]

null

One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.

The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.

Print a single integer — the number of problems the friends will implement on the contest.

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

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

In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

500

[ { "input": "3\n1 1 0\n1 1 1\n1 0 0", "output": "2" }, { "input": "2\n1 0 0\n0 1 1", "output": "1" }, { "input": "1\n1 0 0", "output": "0" }, { "input": "2\n1 0 0\n1 1 1", "output": "1" }, { "input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0", "output": "1" }, { "input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0", "output": "4" }, { "input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1", "output": "10" }, { "input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1", "output": "29" }, { "input": "1\n1 1 1", "output": "1" }, { "input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1", "output": "4" }, { "input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1", "output": "16" } ]

1,698,332,610

2,147,483,647

Python 3

OK

TESTS

21

92

0

def dadd(n): total = 0 for i in range(n): l = input().split() l = [int(x) for x in l] if sum(l) >= 2: total += 1 return total n = int(input()) print(dadd(n))

Title: Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input Specification: The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Specification: Print a single integer — the number of problems the friends will implement on the contest. Demo Input: ['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n'] Demo Output: ['2\n', '1\n'] Note: In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

```python def dadd(n): total = 0 for i in range(n): l = input().split() l = [int(x) for x in l] if sum(l) >= 2: total += 1 return total n = int(input()) print(dadd(n)) ```

3

877

A

Alex and broken contest

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems. But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name. It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita". Names are case sensitive.

The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.

Print "YES", if problem is from this contest, and "NO" otherwise.

[ "Alex_and_broken_contest\n", "NikitaAndString\n", "Danil_and_Olya\n" ]

[ "NO", "YES", "NO" ]

none

500

[ { "input": "Alex_and_broken_contest", "output": "NO" }, { "input": "NikitaAndString", "output": "YES" }, { "input": "Danil_and_Olya", "output": "NO" }, { "input": "Slava____and_the_game", "output": "YES" }, { "input": "Olya_and_energy_drinks", "output": "YES" }, { "input": "Danil_and_part_time_job", "output": "YES" }, { "input": "Ann_and_books", "output": "YES" }, { "input": "Olya", "output": "YES" }, { "input": "Nikita", "output": "YES" }, { "input": "Slava", "output": "YES" }, { "input": "Vanya", "output": "NO" }, { "input": "I_dont_know_what_to_write_here", "output": "NO" }, { "input": "danil_and_work", "output": "NO" }, { "input": "Ann", "output": "YES" }, { "input": "Batman_Nananananananan_Batman", "output": "NO" }, { "input": "Olya_Nikita_Ann_Slava_Danil", "output": "NO" }, { "input": "its_me_Mario", "output": "NO" }, { "input": "A", "output": "NO" }, { "input": "Wake_up_Neo", "output": "NO" }, { "input": "Hardest_problem_ever", "output": "NO" }, { "input": "Nikita_Nikita", "output": "NO" }, { "input": "____________________________________________________________________________________________________", "output": "NO" }, { "input": "Nikitb", "output": "NO" }, { "input": "Unn", "output": "NO" }, { "input": "oLya_adn_smth", "output": "NO" }, { "input": "FloorISLava", "output": "NO" }, { "input": "ann", "output": "NO" }, { "input": "aa", "output": "NO" }, { "input": "AAnnnnn", "output": "YES" }, { "input": "AnnAnn", "output": "NO" }, { "input": "Annn", "output": "YES" }, { "input": "Dilzhan", "output": "NO" }, { "input": "Danilaaa", "output": "YES" }, { "input": "AndAnn", "output": "YES" }, { "input": "OlyaAnnAnn", "output": "NO" }, { "input": "DanilDanilOlya", "output": "NO" }, { "input": "DDanil", "output": "YES" }, { "input": "AnnAnnDanil", "output": "NO" }, { "input": "And_Danil", "output": "YES" }, { "input": "abcddddDanil", "output": "YES" }, { "input": "DanilOlyaOlya", "output": "NO" }, { "input": "Nikitaaa", "output": "YES" }, { "input": "aaabbba", "output": "NO" }, { "input": "Ann_Ann_Danil", "output": "NO" }, { "input": "Danil_Danil_Nikita", "output": "NO" }, { "input": "AlexaaaaaaBBBBBOlyaDDDDD", "output": "YES" }, { "input": "IloveDaniland", "output": "YES" }, { "input": "AnAnn", "output": "YES" }, { "input": "Danil_Danil_Olya", "output": "NO" }, { "input": "DanilDanilSlava", "output": "NO" }, { "input": "DanilDanil", "output": "NO" }, { "input": "OlyOlya", "output": "YES" }, { "input": "NikitaNikitb", "output": "YES" }, { "input": "ababaca", "output": "NO" }, { "input": "AnnNikitaNikitaNikitaNikita__good_luck", "output": "NO" } ]

1,606,407,061

2,147,483,647

Python 3

OK

TESTS

55

109

0

s=input() c=s.count("Danil")+s.count("Olya")+s.count("Slava")+s.count("Ann")+s.count("Nikita") if c==1: print("YES") else: print("NO")

Title: Alex and broken contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems. But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name. It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita". Names are case sensitive. Input Specification: The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem. Output Specification: Print "YES", if problem is from this contest, and "NO" otherwise. Demo Input: ['Alex_and_broken_contest\n', 'NikitaAndString\n', 'Danil_and_Olya\n'] Demo Output: ['NO', 'YES', 'NO'] Note: none

```python s=input() c=s.count("Danil")+s.count("Olya")+s.count("Slava")+s.count("Ann")+s.count("Nikita") if c==1: print("YES") else: print("NO") ```

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" } ]

1,694,687,162

2,147,483,647

Python 3

OK

TESTS

20

31

0

raj = int(input()) for _ in range(raj): r= input() b = len(r) if b > 10: print(r[0] + str(b - 2) + r[b - 1]) else: print(r)

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 raj = int(input()) for _ in range(raj): r= input() b = len(r) if b > 10: print(r[0] + str(b - 2) + r[b - 1]) else: print(r) ```

3.9845

270

A

Fancy Fence

PROGRAMMING

1,100

[ "geometry", "implementation", "math" ]

null

Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?

The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.

For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.

[ "3\n30\n60\n90\n" ]

[ "NO\nYES\nYES\n" ]

In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.

500

[ { "input": "3\n30\n60\n90", "output": "NO\nYES\nYES" }, { "input": "6\n1\n2\n3\n170\n179\n25", "output": "NO\nNO\nNO\nYES\nYES\nNO" } ]

1,665,491,968

2,147,483,647

Python 3

OK

TESTS

3

92

0

turn = int(input()) example = [180*(n-2)/n for n in range(3, 361)] for i in range(turn): a = int(input()) if a in example: print('YES') else: print('NO')

Title: Fancy Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? Input Specification: The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. Output Specification: For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. Demo Input: ['3\n30\n60\n90\n'] Demo Output: ['NO\nYES\nYES\n'] Note: In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.

```python turn = int(input()) example = [180*(n-2)/n for n in range(3, 361)] for i in range(turn): a = int(input()) if a in example: print('YES') else: print('NO') ```

3

208

A

Dubstep

PROGRAMMING

900

[ "strings" ]

null

Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.

The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.

Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.

[ "WUBWUBABCWUB\n", "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n" ]

[ "ABC ", "WE ARE THE CHAMPIONS MY FRIEND " ]

In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

500

[ { "input": "WUBWUBABCWUB", "output": "ABC " }, { "input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB", "output": "WE ARE THE CHAMPIONS MY FRIEND " }, { "input": "WUBWUBWUBSR", "output": "SR " }, { "input": "RWUBWUBWUBLWUB", "output": "R L " }, { "input": "ZJWUBWUBWUBJWUBWUBWUBL", "output": "ZJ J L " }, { "input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB", "output": "C B E Q " }, { "input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB", "output": "JKD WBIRAQKF YE WV " }, { "input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB", "output": "KSDHEMIXUJ R S H " }, { "input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB", "output": "OG X I KO " }, { "input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH", "output": "Q QQ I WW JOPJPBRH " }, { "input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB", "output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C " }, { "input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV", "output": "E IQMJNIQ GZZBQZAUHYP PMR DCV " }, { "input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB", "output": "FV BPS RXNETCJ JDMBH B V B " }, { "input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL", "output": "FBQ IDFSY CTWDM SXO QI L " }, { "input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL", "output": "I QLHD YIIKZDFQ CX U K NL " }, { "input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE", "output": "K UPDYXGOKU AGOAH IZD IY V P E " }, { "input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB", "output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ " }, { "input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB", "output": "PAMJGY XGPQM TKGSXUY E N H E " }, { "input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB", "output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB " }, { "input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM", "output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M " }, { "input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW", "output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W " }, { "input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG", "output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G " }, { "input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN", "output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N " }, { "input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG", "output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG " }, { "input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB", "output": "MP OR DLGK 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KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R " }, { "input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB", "output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM " }, { "input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB", "output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H " }, { "input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB", "output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S " }, { "input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB", "output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J " }, { "input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF", "output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F " }, { "input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY", "output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y " }, { "input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB", "output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO " }, { "input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW", "output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W " }, { "input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD", "output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D " }, { "input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB", "output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A " }, { "input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB", "output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y " }, { "input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB", "output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R " }, { "input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY", "output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y " }, { "input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE", "output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E " }, { "input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS", "output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S " }, { "input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP", "output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP " }, { "input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB", "output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y " }, { "input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL", "output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL " }, { "input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO", "output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O " }, { "input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU", "output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU " }, { "input": "A", "output": "A " }, { "input": "WUBA", "output": "A " }, { "input": "WUBWUBA", "output": "A " }, { "input": "AWUBWUBWUB", "output": "A " }, { "input": "AWUBBWUBCWUBD", "output": "A B C D " }, { "input": "WUBWWUBWUBWUBUWUBWUBBWUB", "output": "W U B " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "WUWUBBWWUBUB", "output": "WU BW UB " }, { "input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB", "output": "WUAB " }, { "input": "ZWUB", "output": "Z " }, { "input": "WU", "output": "WU " }, { "input": "UB", "output": "UB " }, { "input": "U", "output": "U " }, { "input": "WUBW", "output": "W " }, { "input": "WUBWU", "output": "WU " }, { "input": "WUWUB", "output": "WU " }, { "input": "UBWUB", "output": "UB " }, { "input": "WUWUBUBWUBUWUB", "output": "WU UB U " }, { "input": "WUBWWUBAWUB", "output": "W A " }, { "input": "WUUUUU", "output": "WUUUUU " } ]

1,684,430,548

2,147,483,647

PyPy 3

OK

TESTS

71

186

0

a=input().split("WUB") for char in a: if char!='': print(char,end=" ")

Title: Dubstep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. Input Specification: The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. Output Specification: Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. Demo Input: ['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n'] Demo Output: ['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND '] Note: In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

```python a=input().split("WUB") for char in a: if char!='': print(char,end=" ") ```

3

166

A

Rank List

PROGRAMMING

1,100

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

null

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

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

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

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

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

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

500

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1,634,649,902

2,147,483,647

PyPy 3

OK

TESTS

46

154

0

n,k=map(int,input().split()) a=[] for i in range(n): l,m=map(int,input().split()) a.append([-l,m]) a.sort() print(a.count(a[k-1]))

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

```python n,k=map(int,input().split()) a=[] for i in range(n): l,m=map(int,input().split()) a.append([-l,m]) a.sort() print(a.count(a[k-1])) ```

3

198

A

About Bacteria

PROGRAMMING

1,700

[ "implementation", "math" ]

null

Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high. At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into *k* bacteria. After that some abnormal effects create *b* more bacteria in the test tube. Thus, if at the beginning of some second the test tube had *x* bacteria, then at the end of the second it will have *kx*<=+<=*b* bacteria. The experiment showed that after *n* seconds there were exactly *z* bacteria and the experiment ended at this point. For the second experiment Qwerty is going to sterilize the test tube and put there *t* bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least *z* bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment. Help Qwerty and find the minimum number of seconds needed to get a tube with at least *z* bacteria in the second experiment.

The first line contains four space-separated integers *k*, *b*, *n* and *t* (1<=≤<=*k*,<=*b*,<=*n*,<=*t*<=≤<=106) — the parameters of bacterial growth, the time Qwerty needed to grow *z* bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.

Print a single number — the minimum number of seconds Qwerty needs to grow at least *z* bacteria in the tube.

[ "3 1 3 5\n", "1 4 4 7\n", "2 2 4 100\n" ]

[ "2", "3", "0" ]

none

500

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"1 1 2 2", "output": "1" }, { "input": "100000 100000 10 1000000", "output": "9" }, { "input": "2 2 5 4", "output": "4" }, { "input": "999997 1 100000 1000000", "output": "99999" }, { "input": "5 2 100 38", "output": "98" }, { "input": "1 4 1 5", "output": "0" }, { "input": "1 2149 1000000 1000000", "output": "999535" }, { "input": "99999 99999 10 1000000", "output": "9" }, { "input": "999998 1 1000000 1000000", "output": "999999" }, { "input": "1 1 10 2", "output": "9" }, { "input": "1 1 100 1000", "output": "0" }, { "input": "1 1 1000000 553211", "output": "446790" }, { "input": "1 1 10 1", "output": "10" }, { "input": "3 1 3 1", "output": "3" }, { "input": "888888 2 4 999999", "output": "3" }, { "input": "3 5 10 29", "output": "8" }, { "input": "1 1 100 2", "output": "99" }, { "input": "5 5 2 1", "output": "2" }, { "input": "50000 42 1337 999999", "output": "1336" }, { "input": "2 345678 908765 987654", "output": "908764" }, { "input": "1 7 15 7", "output": "15" }, { "input": "842717 8581 19342 851297", "output": "19342" }, { "input": "5 4 1 4", "output": "1" }, { "input": "2 2 5 94", "output": "0" }, { "input": "2 100000 5 10", "output": "5" }, { "input": "722229 410423 118215 838505", "output": "118215" }, { "input": "3 1 3 13", "output": "1" }, { "input": "900000 1 100 1000000", "output": "99" }, { "input": "2 4 4 36", "output": "1" }, { "input": "999990 1 1000000 1000000", "output": "999999" }, { "input": "100000 100000 1000000 1000000", "output": "999999" }, { "input": "999998 1 5 1000000", "output": "4" }, { "input": "1 1 10 4", "output": "7" }, { "input": "2 3 4 5", "output": "3" }, { "input": "3 1 3 40", "output": "0" }, { "input": "1 10 10 100", "output": "1" }, { "input": "999987 123456 1000000 1", "output": "1000000" } ]

1,389,008,324

2,147,483,647

Python 3

OK

TESTS

114

124

307,200

from math import log k, b, n, t = map(int, input().split()) if k == 1: print(max((n * b + b - t) // b, 0)) else: print(max(0, n - int(log((k * t - t + b) / (k - 1 + b)) / log(k))))

Title: About Bacteria Time Limit: None seconds Memory Limit: None megabytes Problem Description: Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high. At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into *k* bacteria. After that some abnormal effects create *b* more bacteria in the test tube. Thus, if at the beginning of some second the test tube had *x* bacteria, then at the end of the second it will have *kx*<=+<=*b* bacteria. The experiment showed that after *n* seconds there were exactly *z* bacteria and the experiment ended at this point. For the second experiment Qwerty is going to sterilize the test tube and put there *t* bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least *z* bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment. Help Qwerty and find the minimum number of seconds needed to get a tube with at least *z* bacteria in the second experiment. Input Specification: The first line contains four space-separated integers *k*, *b*, *n* and *t* (1<=≤<=*k*,<=*b*,<=*n*,<=*t*<=≤<=106) — the parameters of bacterial growth, the time Qwerty needed to grow *z* bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly. Output Specification: Print a single number — the minimum number of seconds Qwerty needs to grow at least *z* bacteria in the tube. Demo Input: ['3 1 3 5\n', '1 4 4 7\n', '2 2 4 100\n'] Demo Output: ['2', '3', '0'] Note: none

```python from math import log k, b, n, t = map(int, input().split()) if k == 1: print(max((n * b + b - t) // b, 0)) else: print(max(0, n - int(log((k * t - t + b) / (k - 1 + b)) / log(k)))) ```

3

996

A

Hit the Lottery

PROGRAMMING

800

[ "dp", "greedy" ]

null

Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?

The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).

Output the minimum number of bills that Allen could receive.

[ "125\n", "43\n", "1000000000\n" ]

[ "3\n", "5\n", "10000000\n" ]

In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.

500

[ { "input": "125", "output": "3" }, { "input": "43", "output": "5" }, { "input": "1000000000", "output": "10000000" }, { "input": "4", "output": "4" }, { "input": "5", "output": "1" }, { "input": "1", "output": "1" }, { "input": "74", "output": "8" }, { "input": "31", "output": "3" }, { "input": "59", "output": "8" }, { "input": "79", "output": "9" }, { "input": "7", "output": "3" }, { "input": "55", "output": "4" }, { "input": "40", "output": "2" }, { "input": "719", "output": "13" }, { "input": "847", "output": "13" }, { "input": "225", "output": "4" }, { "input": "4704", "output": "51" }, { "input": "1132", "output": "15" }, { "input": "7811", "output": "80" }, { "input": "7981", "output": "84" }, { "input": "82655", "output": "830" }, { "input": "6364", "output": "70" }, { "input": "74611", "output": "748" }, { "input": "45391", "output": "459" }, { "input": "620448", "output": "6210" }, { "input": "265145", "output": "2654" }, { "input": "671704", "output": "6721" }, { "input": "365173", "output": "3658" }, { "input": "7130872", "output": "71314" }, { "input": "9628747", "output": "96292" }, { "input": "8898399", "output": "88993" }, { "input": "9497953", "output": "94985" }, { "input": "19070947", "output": "190714" }, { "input": "20185520", "output": "201856" }, { "input": "91402248", "output": "914028" }, { "input": "27035533", "output": "270360" }, { "input": "22717528", "output": "227180" }, { "input": "24403439", "output": "244041" }, { "input": "163565555", "output": "1635659" }, { "input": "152541908", "output": "1525423" }, { "input": "668439837", "output": "6684403" }, { "input": "20", "output": "1" }, { "input": "999999999", "output": "10000009" }, { "input": "10", "output": "1" }, { "input": "7", "output": "3" }, { "input": "6", "output": "2" }, { "input": "99999999", "output": "1000009" }, { "input": "1000001", "output": "10001" }, { "input": "2521", "output": "27" } ]

1,696,850,263

2,147,483,647

PyPy 3-64

OK

TESTS

49

124

0

num= int(input()) ans= 0 arr= [100, 20, 10, 5, 1] for i in arr: count = num//i ans += count num-= count *i print(ans)

Title: Hit the Lottery Time Limit: None seconds Memory Limit: None megabytes Problem Description: Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? Input Specification: The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). Output Specification: Output the minimum number of bills that Allen could receive. Demo Input: ['125\n', '43\n', '1000000000\n'] Demo Output: ['3\n', '5\n', '10000000\n'] Note: In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.

```python num= int(input()) ans= 0 arr= [100, 20, 10, 5, 1] for i in arr: count = num//i ans += count num-= count *i print(ans) ```

3

854

A

Fraction

PROGRAMMING

800

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

null

Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1). During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation. Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.

In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.

Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.

[ "3\n", "4\n", "12\n" ]

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

none

500

[ { "input": "3", "output": "1 2" }, { "input": "4", "output": "1 3" }, { "input": "12", "output": "5 7" }, { "input": "34", "output": "15 19" }, { "input": "13", "output": "6 7" }, { "input": "11", "output": "5 6" }, { "input": "24", "output": "11 13" }, { "input": "17", "output": "8 9" }, { "input": "10", "output": "3 7" }, { "input": "69", "output": "34 35" }, { "input": "100", "output": "49 51" }, { "input": "57", "output": "28 29" }, { "input": "1000", "output": "499 501" }, { "input": "999", "output": "499 500" }, { "input": "998", "output": "497 501" }, { "input": "997", "output": "498 499" }, { "input": "996", "output": "497 499" }, { "input": "995", "output": "497 498" }, { "input": "994", "output": "495 499" }, { "input": "5", "output": "2 3" }, { "input": "6", "output": "1 5" }, { "input": "8", "output": "3 5" }, { "input": "9", "output": "4 5" }, { "input": "423", "output": "211 212" }, { "input": "876", "output": "437 439" }, { "input": "29", "output": "14 15" } ]

1,606,538,297

2,147,483,647

PyPy 3

OK

TESTS

26

140

1,331,200

import math n=int(input()) answer=0 numerator=0 denominator=0 for a in range(1, 1001): for b in range(a+1, 1001): if a+b==n: if math.gcd(a, b)==1: if a/b>answer: answer=a/b numerator=a denominator=b print(numerator, denominator)

Title: Fraction Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1). During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation. Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. Input Specification: In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. Output Specification: Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. Demo Input: ['3\n', '4\n', '12\n'] Demo Output: ['1 2\n', '1 3\n', '5 7\n'] Note: none

```python import math n=int(input()) answer=0 numerator=0 denominator=0 for a in range(1, 1001): for b in range(a+1, 1001): if a+b==n: if math.gcd(a, b)==1: if a/b>answer: answer=a/b numerator=a denominator=b print(numerator, denominator) ```

3

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" } ]

1,553,863,159

2,147,483,647

Python 3

OK

TESTS

71

218

0

n = int(input()) n = (8 * n + 1) ** .5 print('NYOE S'[not n % 1 and int(n) & 1::2])

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()) n = (8 * n + 1) ** .5 print('NYOE S'[not n % 1 and int(n) & 1::2]) ```

3.9455

1,003

A

Polycarp's Pockets

PROGRAMMING

800

[ "implementation" ]

null

Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.

The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.

Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.

[ "6\n1 2 4 3 3 2\n", "1\n100\n" ]

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

none

0

[ { "input": "6\n1 2 4 3 3 2", "output": "2" }, { "input": "1\n100", "output": "1" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" }, { "input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71", "output": "51" }, { "input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29", "output": "1" }, { "input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24", "output": "2" }, { "input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6", "output": "10" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "99" }, { "input": "7\n1 2 3 3 3 1 2", "output": "3" }, { "input": "5\n1 2 3 4 5", "output": "1" }, { "input": "7\n1 2 3 4 5 6 7", "output": "1" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "1" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "1" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "1" }, { "input": "3\n2 1 1", "output": "2" }, { "input": "11\n1 2 3 4 5 6 7 8 9 1 1", "output": "3" }, { "input": "12\n1 2 1 1 1 1 1 1 1 1 1 1", "output": "11" }, { "input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "14" }, { "input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15" }, { "input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "16" }, { "input": "3\n1 1 1", "output": "3" }, { "input": "3\n1 2 3", "output": "1" }, { "input": "10\n1 1 1 1 2 2 1 1 9 10", "output": "6" }, { "input": "2\n1 1", "output": "2" }, { "input": "56\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "56" }, { "input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92", "output": "6" }, { "input": "10\n1 2 2 3 3 3 4 4 4 4", "output": "4" }, { "input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21", "output": "3" }, { "input": "5\n5 5 5 5 1", "output": "4" }, { "input": "6\n2 3 2 5 2 6", "output": "3" }, { "input": "3\n58 59 58", "output": "2" }, { "input": "9\n1 2 3 4 5 6 7 8 8", "output": "2" }, { "input": "97\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "97" }, { "input": "3\n95 95 4", "output": "2" }, { "input": "3\n2 2 5", "output": "2" } ]

1,647,089,262

2,147,483,647

Python 3

OK

TESTS

37

46

0

n=int(input()) a=list(map(int,input().split())) b=[] for i in range(n): b.append(a.count(a[i])) print(max(b))

Title: Polycarp's Pockets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. Input Specification: The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. Output Specification: Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. Demo Input: ['6\n1 2 4 3 3 2\n', '1\n100\n'] Demo Output: ['2\n', '1\n'] Note: none

```python n=int(input()) a=list(map(int,input().split())) b=[] for i in range(n): b.append(a.count(a[i])) print(max(b)) ```

3

15

A

Cottage Village

PROGRAMMING

1,200

[ "implementation", "sortings" ]

A. Cottage Village

2

64

A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other. The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village. Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?

The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000).

Output the amount of possible positions of the new house.

[ "2 2\n0 4\n6 2\n", "2 2\n0 4\n5 2\n", "2 3\n0 4\n5 2\n" ]

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

It is possible for the *x*-coordinate of the new house to have non-integer value.

0

[ { "input": "2 2\n0 4\n6 2", "output": "4" }, { "input": "2 2\n0 4\n5 2", "output": "3" }, { "input": "2 3\n0 4\n5 2", "output": "2" }, { "input": "1 1\n1 1", "output": "2" }, { "input": "1 2\n2 1", "output": "2" }, { "input": "2 1\n2 1\n1 1", "output": "2" }, { "input": "2 2\n0 4\n7 4", "output": "4" }, { "input": "4 1\n-12 1\n-14 1\n4 1\n-11 1", "output": "5" }, { "input": "6 15\n19 1\n2 3\n6 2\n-21 2\n-15 2\n23 1", "output": "2" }, { "input": "10 21\n-61 6\n55 2\n-97 1\n37 1\n-39 1\n26 2\n21 1\n64 3\n-68 1\n-28 6", "output": "6" }, { "input": "26 51\n783 54\n-850 6\n-997 59\n573 31\n-125 20\n472 52\n101 5\n-561 4\n625 35\n911 14\n-47 33\n677 55\n-410 54\n13 53\n173 31\n968 30\n-497 7\n832 42\n271 59\n-638 52\n-301 51\n378 36\n-813 7\n-206 22\n-737 37\n-911 9", "output": "35" }, { "input": "14 101\n121 88\n-452 91\n635 28\n-162 59\n-872 26\n-996 8\n468 86\n742 63\n892 89\n-249 107\n300 51\n-753 17\n-620 31\n-13 34", "output": "16" }, { "input": "3 501\n827 327\n-85 480\n-999 343", "output": "6" }, { "input": "2 999\n-999 471\n530 588", "output": "4" }, { "input": "22 54\n600 43\n806 19\n-269 43\n-384 78\n222 34\n392 10\n318 30\n488 73\n-756 49\n-662 22\n-568 50\n-486 16\n-470 2\n96 66\n864 16\n934 15\n697 43\n-154 30\n775 5\n-876 71\n-33 78\n-991 31", "output": "30" }, { "input": "17 109\n52 7\n216 24\n-553 101\n543 39\n391 92\n-904 67\n95 34\n132 14\n730 103\n952 118\n-389 41\n-324 36\n-74 2\n-147 99\n-740 33\n233 1\n-995 3", "output": "16" }, { "input": "4 512\n-997 354\n-568 216\n-234 221\n603 403", "output": "4" }, { "input": "3 966\n988 5\n15 2\n-992 79", "output": "6" }, { "input": "2 1000\n-995 201\n206 194", "output": "4" }, { "input": "50 21\n-178 1\n49 1\n-98 1\n-220 1\n152 1\n-160 3\n17 2\n77 1\n-24 1\n214 2\n-154 2\n-141 1\n79 1\n206 1\n8 1\n-208 1\n36 1\n231 3\n-2 2\n-130 2\n-14 2\n34 1\n-187 2\n14 1\n-83 2\n-241 1\n149 2\n73 1\n-233 3\n-45 1\n197 1\n145 2\n-127 2\n-229 4\n-85 1\n-66 1\n-76 2\n104 1\n175 1\n70 1\n131 3\n-108 1\n-5 4\n140 1\n33 1\n248 3\n-36 3\n134 1\n-183 1\n56 2", "output": "9" }, { "input": "50 1\n37 1\n-38 1\n7 1\n47 1\n-4 1\n24 1\n-32 1\n-23 1\n-3 1\n-19 1\n5 1\n-50 1\n11 1\n-11 1\n49 1\n-39 1\n0 1\n43 1\n-10 1\n6 1\n19 1\n1 1\n27 1\n29 1\n-47 1\n-40 1\n-46 1\n-26 1\n-42 1\n-37 1\n13 1\n-29 1\n-30 1\n3 1\n44 1\n10 1\n4 1\n-14 1\n-2 1\n34 1\n18 1\n-33 1\n-44 1\n9 1\n-36 1\n-7 1\n25 1\n22 1\n-20 1\n-41 1", "output": "43" }, { "input": "50 1\n-967 7\n696 7\n-366 4\n557 1\n978 2\n800 4\n-161 2\n-773 2\n-248 2\n134 3\n869 6\n-932 2\n-262 14\n191 3\n669 2\n72 5\n0 1\n757 8\n859 2\n-131 8\n-169 3\n543 10\n-120 2\n-87 8\n-936 6\n-620 3\n-281 11\n684 3\n886 10\n497 4\n380 4\n833 1\n-727 6\n470 11\n584 9\n66 6\n-609 12\n-661 4\n-57 8\n628 7\n635 4\n-924 3\n-982 4\n-201 7\n-9 8\n-560 9\n712 7\n-330 8\n-191 1\n-892 7", "output": "96" }, { "input": "1 1000\n0 1000", "output": "2" } ]

1,651,216,262

2,147,483,647

PyPy 3-64

OK

TESTS

35

342

8,192,000

import abc import itertools import math from math import gcd as gcd import sys import queue import itertools from heapq import heappop, heappush import random def solve(): n, t = map(int, input().split()) t = 2 * t res = 2 h = [] for i in range(n): x, a = map(int, input().split()) l, r = 2 * x - a, 2 * x + a h.append([l, r]) h.sort(key=lambda x: (x[0], x[1])) for i in range(1, n): r1, l2 = h[i - 1][1], h[i][0] if l2 - r1 == t: res += 1 elif l2 - r1 > t: res += 2 print(res) if __name__ == '__main__': multi_test = 0 if multi_test == 1: t = int(sys.stdin.readline()) for _ in range(t): solve() else: solve()

Title: Cottage Village Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other. The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village. Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house? Input Specification: The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000). Output Specification: Output the amount of possible positions of the new house. Demo Input: ['2 2\n0 4\n6 2\n', '2 2\n0 4\n5 2\n', '2 3\n0 4\n5 2\n'] Demo Output: ['4\n', '3\n', '2\n'] Note: It is possible for the *x*-coordinate of the new house to have non-integer value.

```python import abc import itertools import math from math import gcd as gcd import sys import queue import itertools from heapq import heappop, heappush import random def solve(): n, t = map(int, input().split()) t = 2 * t res = 2 h = [] for i in range(n): x, a = map(int, input().split()) l, r = 2 * x - a, 2 * x + a h.append([l, r]) h.sort(key=lambda x: (x[0], x[1])) for i in range(1, n): r1, l2 = h[i - 1][1], h[i][0] if l2 - r1 == t: res += 1 elif l2 - r1 > t: res += 2 print(res) if __name__ == '__main__': multi_test = 0 if multi_test == 1: t = int(sys.stdin.readline()) for _ in range(t): solve() else: solve() ```

3.853465

104

A

Blackjack

PROGRAMMING

800

[ "implementation" ]

A. Blackjack

2

256

One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.

The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.

Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.

[ "12\n", "20\n", "10\n" ]

[ "4", "15", "0" ]

In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

500

[ { "input": "12", "output": "4" }, { "input": "20", "output": "15" }, { "input": "10", "output": "0" }, { "input": "11", "output": "4" }, { "input": "15", "output": "4" }, { "input": "18", "output": "4" }, { "input": "25", "output": "0" }, { "input": "22", "output": "0" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "3", "output": "0" }, { "input": "4", "output": "0" }, { "input": "5", "output": "0" }, { "input": "6", "output": "0" }, { "input": "7", "output": "0" }, { "input": "8", "output": "0" }, { "input": "9", "output": "0" }, { "input": "13", "output": "4" }, { "input": "14", "output": "4" }, { "input": "16", "output": "4" }, { "input": "17", "output": "4" }, { "input": "19", "output": "4" }, { "input": "21", "output": "4" }, { "input": "23", "output": "0" }, { "input": "24", "output": "0" } ]

1,696,065,870

2,147,483,647

Python 3

OK

TESTS

25

92

0

cards = [(i + j) for i in (('S', 'H', 'D', 'C')) for j in (tuple('23456789jqka') + ('10',))] cards.remove('Sq') cards_points = {} for card in cards: if card[-1] in ('j', 'q', 'k', '0'): cards_points[card] = 10 elif card[-1] == 'a': cards_points[card] = [1, 11] else: cards_points[card] = int(str(card[-1])) n = int(input()) if n <= 10 or n > 21: print(0) else: count = 0 for card in cards_points: if isinstance(cards_points[card], list): if (n - 10) in cards_points[card]: count += 1 elif (n - 10) == cards_points[card]: count += 1 print(count)

Title: Blackjack Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. Input Specification: The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. Output Specification: Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. Demo Input: ['12\n', '20\n', '10\n'] Demo Output: ['4', '15', '0'] Note: In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

```python cards = [(i + j) for i in (('S', 'H', 'D', 'C')) for j in (tuple('23456789jqka') + ('10',))] cards.remove('Sq') cards_points = {} for card in cards: if card[-1] in ('j', 'q', 'k', '0'): cards_points[card] = 10 elif card[-1] == 'a': cards_points[card] = [1, 11] else: cards_points[card] = int(str(card[-1])) n = int(input()) if n <= 10 or n > 21: print(0) else: count = 0 for card in cards_points: if isinstance(cards_points[card], list): if (n - 10) in cards_points[card]: count += 1 elif (n - 10) == cards_points[card]: count += 1 print(count) ```

3.977

268

B

Buttons

PROGRAMMING

1,000

[ "implementation", "math" ]

null

Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens. Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock. Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.

A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has.

In a single line print the number of times Manao has to push a button in the worst-case scenario.

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

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

Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.

1,000

[ { "input": "2", "output": "3" }, { "input": "3", "output": "7" }, { "input": "4", "output": "14" }, { "input": "1", "output": "1" }, { "input": "10", "output": "175" }, { "input": "2000", "output": "1333335000" }, { "input": "1747", "output": "888644743" }, { "input": "889", "output": "117099969" }, { "input": "1999", "output": "1331335999" }, { "input": "914", "output": "127259419" }, { "input": "996", "output": "164675486" }, { "input": "17", "output": "833" }, { "input": "50", "output": "20875" }, { "input": "91", "output": "125671" }, { "input": "92", "output": "129858" }, { "input": "256", "output": "2796416" }, { "input": "512", "output": "22370048" }, { "input": "666", "output": "49235271" }, { "input": "667", "output": "49457383" }, { "input": "314", "output": "5160119" }, { "input": "1241", "output": "318541121" }, { "input": "1500", "output": "562501250" }, { "input": "1837", "output": "1033182073" }, { "input": "1000", "output": "166667500" } ]

1,691,731,356

2,147,483,647

PyPy 3-64

OK

TESTS

24

122

1,331,200

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

Title: Buttons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens. Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock. Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario. Input Specification: A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has. Output Specification: In a single line print the number of times Manao has to push a button in the worst-case scenario. Demo Input: ['2\n', '3\n'] Demo Output: ['3\n', '7\n'] Note: Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.

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

3

629

B

Far Relative’s Problem

PROGRAMMING

1,100

[ "brute force" ]

null

Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible. Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party. Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends. Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive.

Print the maximum number of people that may come to Famil Door's party.

[ "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n", "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n" ]

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

In the first sample, friends 3 and 4 can come on any day in range [117, 128]. In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

1,000

[ { "input": "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128", "output": "2" }, { "input": "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200", "output": "4" }, { "input": "1\nF 68 307", "output": "0" }, { "input": "40\nM 55 363\nF 117 252\nM 157 282\nF 322 345\nM 330 363\nF 154 231\nF 216 352\nF 357 365\nM 279 292\nF 353 359\nF 82 183\nM 78 297\nM 231 314\nM 107 264\nF 34 318\nM 44 244\nF 42 339\nM 253 307\nM 128 192\nF 119 328\nM 135 249\nF 303 358\nF 348 352\nF 8 364\nF 126 303\nM 226 346\nF 110 300\nF 47 303\nF 201 311\nF 287 288\nM 270 352\nM 227 351\nF 8 111\nF 39 229\nM 163 315\nF 269 335\nF 147 351\nF 96 143\nM 97 99\nM 177 295", "output": "22" }, { "input": "2\nF 1 1\nM 1 1", "output": "2" }, { "input": "4\nM 1 2\nM 2 3\nF 3 4\nF 4 5", "output": "2" }, { "input": "1\nF 1 2", "output": "0" }, { "input": "18\nF 3 4\nF 3 4\nF 3 4\nF 3 4\nF 5 6\nF 5 6\nM 3 4\nM 3 4\nM 5 6\nM 5 6\nM 5 6\nM 5 6\nF 7 8\nF 7 8\nF 7 8\nM 7 8\nM 7 8\nM 7 8", "output": "6" }, { "input": "2\nM 1 1\nF 1 1", "output": "2" }, { "input": "3\nM 1 1\nF 1 1\nM 1 1", "output": "2" } ]

1,459,066,479

2,147,483,647

Python 3

OK

TESTS

76

78

4,608,000

n = int(input()) a = [0 for i in range(368)] b = [0 for i in range(368)] for i in range(n): s = input().split() if s[0] == 'M': a[int(s[1])] += 1 a[int(s[2]) + 1] -= 1 else: b[int(s[1])] += 1 b[int(s[2]) + 1] -= 1 ans = 0 for i in range(1, 367): a[i] += a[i - 1] b[i] += b[i - 1] ans = max(ans, min(a[i], b[i])) print(ans << 1)

Title: Far Relative’s Problem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible. Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party. Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends. Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive. Output Specification: Print the maximum number of people that may come to Famil Door's party. Demo Input: ['4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n', '6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n'] Demo Output: ['2\n', '4\n'] Note: In the first sample, friends 3 and 4 can come on any day in range [117, 128]. In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

```python n = int(input()) a = [0 for i in range(368)] b = [0 for i in range(368)] for i in range(n): s = input().split() if s[0] == 'M': a[int(s[1])] += 1 a[int(s[2]) + 1] -= 1 else: b[int(s[1])] += 1 b[int(s[2]) + 1] -= 1 ans = 0 for i in range(1, 367): a[i] += a[i - 1] b[i] += b[i - 1] ans = max(ans, min(a[i], b[i])) print(ans << 1) ```

3

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" } ]

1,426,998,090

2,147,483,647

Python 3

OK

TESTS

33

278

10,240,000

I=lambda:list(map(int,input().split())) n,m=I();a=sorted(I());b=[] for i in a:b+=list(range(1,i+1)) b=sorted(b) i=mini=p=0 while((i<m)and(p+a[i]<=n)): p+=a[i] mini+=(1+a[i])*a[i]//2 i+=1 if p<n: mini+=(2*a[i]-n+p+1)*(n-p)//2 print(sum(b[-n:]),mini)

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 I=lambda:list(map(int,input().split())) n,m=I();a=sorted(I());b=[] for i in a:b+=list(range(1,i+1)) b=sorted(b) i=mini=p=0 while((i<m)and(p+a[i]<=n)): p+=a[i] mini+=(1+a[i])*a[i]//2 i+=1 if p<n: mini+=(2*a[i]-n+p+1)*(n-p)//2 print(sum(b[-n:]),mini) ```

3

251

A

Points on Line

PROGRAMMING

1,300

[ "binary search", "combinatorics", "two pointers" ]

null

Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*. Note that the order of the points inside the group of three chosen points doesn't matter.

The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got. It is guaranteed that the coordinates of the points in the input strictly increase.

Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

[ "4 3\n1 2 3 4\n", "4 2\n-3 -2 -1 0\n", "5 19\n1 10 20 30 50\n" ]

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

In the first sample any group of three points meets our conditions. In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}. In the third sample only one group does: {1, 10, 20}.

500

[ { "input": "4 3\n1 2 3 4", "output": "4" }, { "input": "4 2\n-3 -2 -1 0", "output": "2" }, { "input": "5 19\n1 10 20 30 50", "output": "1" }, { "input": "10 5\n31 36 43 47 48 50 56 69 71 86", "output": "2" }, { "input": "10 50\n1 4 20 27 65 79 82 83 99 100", "output": "25" }, { "input": "10 90\n24 27 40 41 61 69 73 87 95 97", "output": "120" }, { "input": "100 100\n-98 -97 -96 -93 -92 -91 -90 -87 -86 -84 -81 -80 -79 -78 -76 -75 -73 -71 -69 -67 -65 -64 -63 -62 -61 -54 -51 -50 -49 -48 -46 -45 -44 -37 -36 -33 -30 -28 -27 -16 -15 -13 -12 -10 -9 -7 -6 -5 -4 2 3 5 8 9 10 11 13 14 15 16 17 19 22 24 25 26 27 28 30 31 32 36 40 43 45 46 47 50 51 52 53 58 60 63 69 70 73 78 80 81 82 85 88 89 90 91 95 96 97 99", "output": "79351" }, { "input": "1 14751211\n847188590", "output": "0" }, { "input": "2 1000000000\n-907894512 -289906312", "output": "0" }, { "input": "2 1000000000\n-14348867 1760823", "output": "0" }, { "input": "3 1000000000\n-5 -1 1", "output": "1" } ]

1,459,917,260

2,147,483,647

Python 3

OK

TESTS

39

1,214

11,878,400

def c(a,b): i=a ansc=1 while i>a-b: ansc=ansc*i i=i-1 i=1 while i<=b: ansc=ansc/i i=i+1 if a<b: ansc=0 return ansc n,d=input().split(' ') n=int(n) d=int(d) que=input().split(' ') l=len(que) i=0 ans=0 while i<l: que[i]=int(que[i]) i=i+1 qbegin=0 qend=2 while qend<l: test=que[qend]-que[qbegin] while (test>d) and (qbegin<=qend): qbegin += 1 test=que[qend]-que[qbegin] if qend-qbegin>1: ans=ans+c(qend-qbegin+1,3)-c(qend-qbegin,3) qend=qend+1 print(int(ans))

Title: Points on Line Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*. Note that the order of the points inside the group of three chosen points doesn't matter. Input Specification: The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got. It is guaranteed that the coordinates of the points in the input strictly increase. Output Specification: Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n'] Demo Output: ['4\n', '2\n', '1\n'] Note: In the first sample any group of three points meets our conditions. In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}. In the third sample only one group does: {1, 10, 20}.

```python def c(a,b): i=a ansc=1 while i>a-b: ansc=ansc*i i=i-1 i=1 while i<=b: ansc=ansc/i i=i+1 if a<b: ansc=0 return ansc n,d=input().split(' ') n=int(n) d=int(d) que=input().split(' ') l=len(que) i=0 ans=0 while i<l: que[i]=int(que[i]) i=i+1 qbegin=0 qend=2 while qend<l: test=que[qend]-que[qbegin] while (test>d) and (qbegin<=qend): qbegin += 1 test=que[qend]-que[qbegin] if qend-qbegin>1: ans=ans+c(qend-qbegin+1,3)-c(qend-qbegin,3) qend=qend+1 print(int(ans)) ```

3

789

A

Anastasia and pebbles

PROGRAMMING

1,100

[ "implementation", "math" ]

null

Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type.

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

[ "3 2\n2 3 4\n", "5 4\n3 1 8 9 7\n" ]

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

In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

500

[ { "input": "3 2\n2 3 4", "output": "3" }, { "input": "5 4\n3 1 8 9 7", "output": "5" }, { "input": "1 22\n1", "output": "1" }, { "input": "3 57\n78 165 54", "output": "3" }, { "input": "5 72\n74 10 146 189 184", "output": "6" }, { "input": "9 13\n132 87 200 62 168 51 185 192 118", "output": "48" }, { "input": "1 1\n10000", "output": "5000" }, { "input": "10 1\n1 1 1 1 1 1 1 1 1 1", "output": "5" }, { "input": "2 2\n2 2", "output": "1" } ]

1,602,402,038

1,538

Python 3

OK

TESTS

31

233

7,680,000

n, k = map(int, input().split()) w = list(map(int, input().split())) q=0 t=0 for i in range(n): a=w[i]//k w[i]%=k q+=(a//2) t+=(a%2) # print(a, w[i], t, q) if t==2: t=0 q+=1 if w[i]>0: if t==1: t=0 w[i]=0 q+=1 else: w[i]=0 t+=1 if t>0: q+=1 print(q)

Title: Anastasia and pebbles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type. Output Specification: The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles. Demo Input: ['3 2\n2 3 4\n', '5 4\n3 1 8 9 7\n'] Demo Output: ['3\n', '5\n'] Note: In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

```python n, k = map(int, input().split()) w = list(map(int, input().split())) q=0 t=0 for i in range(n): a=w[i]//k w[i]%=k q+=(a//2) t+=(a%2) # print(a, w[i], t, q) if t==2: t=0 q+=1 if w[i]>0: if t==1: t=0 w[i]=0 q+=1 else: w[i]=0 t+=1 if t>0: q+=1 print(q) ```

3

540

A

Combination Lock

PROGRAMMING

800

[ "implementation" ]

null

Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.

Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.

[ "5\n82195\n64723\n" ]

[ "13\n" ]

In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>

500

[ { "input": "5\n82195\n64723", "output": "13" }, { "input": "12\n102021090898\n010212908089", "output": "16" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "10\n0728592530\n1362615763", "output": "27" }, { "input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746", "output": "245" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "3\n607\n684", "output": "5" }, { "input": "4\n0809\n0636", "output": "8" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762280548\n9519431339078678836940020", "output": "72" }, { "input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024", "output": "305" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762285484\n9519431339078678836940202", "output": "74" }, { "input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831", "output": "357" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8378525987476228048406972\n9194339078883694020217816", "output": "55" }, { "input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599", "output": "274" }, { "input": "1\n0\n0", "output": "0" }, { "input": "1\n7\n7", "output": "0" }, { "input": "1\n0\n5", "output": "5" }, { "input": "1\n2\n7", "output": "5" }, { "input": "1\n7\n9", "output": "2" }, { "input": "1\n9\n7", "output": "2" }, { "input": "1\n2\n9", "output": "3" }, { "input": "1\n9\n2", "output": "3" }, { "input": "25\n3164978461316464614169874\n9413979197249127496597357", "output": "66" }, { "input": "4\n9999\n9999", "output": "0" }, { "input": "2\n11\n11", "output": "0" } ]

1,687,512,511

2,147,483,647

Python 3

OK

TESTS

43

46

0

n = int(input()) curr = list(map(int, input())) arr = list(map(int, input())) cnt = 0 for i in range(n): cnt += min(abs(arr[i]-curr[i]), (min(arr[i],curr[i])+10) - max(arr[i], curr[i]) ) print(cnt)

Title: Combination Lock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock. Output Specification: Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock. Demo Input: ['5\n82195\n64723\n'] Demo Output: ['13\n'] Note: In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>

```python n = int(input()) curr = list(map(int, input())) arr = list(map(int, input())) cnt = 0 for i in range(n): cnt += min(abs(arr[i]-curr[i]), (min(arr[i],curr[i])+10) - max(arr[i], curr[i]) ) print(cnt) ```

3

677

A

Vanya and Fence

PROGRAMMING

800

[ "implementation" ]

null

Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?

The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.

Print a single integer — the minimum possible valid width of the road.

[ "3 7\n4 5 14\n", "6 1\n1 1 1 1 1 1\n", "6 5\n7 6 8 9 10 5\n" ]

[ "4\n", "6\n", "11\n" ]

In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.

500

[ { "input": "3 7\n4 5 14", "output": "4" }, { "input": "6 1\n1 1 1 1 1 1", "output": "6" }, { "input": "6 5\n7 6 8 9 10 5", "output": "11" }, { "input": "10 420\n214 614 297 675 82 740 174 23 255 15", "output": "13" }, { "input": "10 561\n657 23 1096 487 785 66 481 554 1000 821", "output": "15" }, { "input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396", "output": "144" }, { "input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366", "output": "145" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 1\n2", "output": "2" }, { "input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19", "output": "63" }, { "input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386", "output": "31" }, { "input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518", "output": "75" }, { "input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397", "output": "41" }, { "input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118", "output": "116" } ]

1,694,923,350

2,147,483,647

PyPy 3-64

OK

TESTS

29

77

0

n,h=map(int, input().split()) l=list(map(int, input().split())) flag=0 for i in range(n): if l[i]>h: flag+=2 else: flag+=1 print(flag)

Title: Vanya and Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? Input Specification: The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. Output Specification: Print a single integer — the minimum possible valid width of the road. Demo Input: ['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n'] Demo Output: ['4\n', '6\n', '11\n'] Note: In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.

```python n,h=map(int, input().split()) l=list(map(int, input().split())) flag=0 for i in range(n): if l[i]>h: flag+=2 else: flag+=1 print(flag) ```

3

221

A

Little Elephant and Function

PROGRAMMING

1,000

[ "implementation", "math" ]

null

The Little Elephant enjoys recursive functions. This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows: - If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*). The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order.

A single line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the size of permutation.

In a single line print *n* distinct integers from 1 to *n* — the required permutation. Numbers in a line should be separated by spaces. It is guaranteed that the answer exists.

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

[ "1 ", "2 1 " ]

none

500

[ { "input": "1", "output": "1 " }, { "input": "2", "output": "2 1 " }, { "input": "3", "output": "3 1 2 " }, { "input": "4", "output": "4 1 2 3 " }, { "input": "5", "output": "5 1 2 3 4 " }, { "input": "6", "output": "6 1 2 3 4 5 " }, { "input": "7", "output": "7 1 2 3 4 5 6 " }, { "input": "1000", "output": "1000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..." }, { "input": "999", "output": "999 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "998", "output": "998 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "997", "output": "997 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "900", "output": "900 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "101", "output": "101 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 " }, { "input": "3", "output": "3 1 2 " }, { "input": "67", "output": "67 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 " }, { "input": "779", "output": "779 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "195", "output": "195 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "907", "output": "907 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "971", "output": "971 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "683", "output": "683 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "99", "output": "99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 " }, { "input": "811", "output": "811 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "875", "output": "875 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "291", "output": "291 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "3", "output": "3 1 2 " }, { "input": "419", "output": "419 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "779", "output": "779 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "195", "output": "195 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "907", "output": "907 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..." }, { "input": "21", "output": "21 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 " }, { "input": "22", "output": "22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 " }, { "input": "23", "output": "23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 " }, { "input": "24", "output": "24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 " }, { "input": "25", "output": "25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 " } ]

1,679,667,215

2,147,483,647

Python 3

OK

TESTS

34

92

0

n = int(input()) l = [] sr = [] for i in range(1,n+1): l.append(i) if i==n: sr = sorted(l) sr.insert(0, n) del sr[-1] print(*sr)

Title: Little Elephant and Function Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Little Elephant enjoys recursive functions. This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows: - If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*). The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order. Input Specification: A single line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the size of permutation. Output Specification: In a single line print *n* distinct integers from 1 to *n* — the required permutation. Numbers in a line should be separated by spaces. It is guaranteed that the answer exists. Demo Input: ['1\n', '2\n'] Demo Output: ['1 ', '2 1 '] Note: none

```python n = int(input()) l = [] sr = [] for i in range(1,n+1): l.append(i) if i==n: sr = sorted(l) sr.insert(0, n) del sr[-1] print(*sr) ```

3

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" } ]

1,503,417,087

2,147,483,647

Python 3

OK

TESTS

88

62

0

n = int(input()) l = list(map(int, input().split())) l.sort() l1 = [] l2 = [] for index, i in enumerate(l): if index < len(l) // 2: l1.append(i) else: l2.append(i) if l1[len(l1) - 1] < l2[0]: print("YES") else: print("NO")

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 n = int(input()) l = list(map(int, input().split())) l.sort() l1 = [] l2 = [] for index, i in enumerate(l): if index < len(l) // 2: l1.append(i) else: l2.append(i) if l1[len(l1) - 1] < l2[0]: print("YES") else: print("NO") ```

3

977

B

Two-gram

PROGRAMMING

900

[ "implementation", "strings" ]

null

Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams. You are given a string $s$ consisting of $n$ capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string $s$ = "BBAABBBA" the answer is two-gram "BB", which contained in $s$ three times. In other words, find any most frequent two-gram. Note that occurrences of the two-gram can overlap with each other.

The first line of the input contains integer number $n$ ($2 \le n \le 100$) — the length of string $s$. The second line of the input contains the string $s$ consisting of $n$ capital Latin letters.

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $s$ as a substring (i.e. two consecutive characters of the string) maximal number of times.

[ "7\nABACABA\n", "5\nZZZAA\n" ]

[ "AB\n", "ZZ\n" ]

In the first example "BA" is also valid answer. In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

0

[ { "input": "7\nABACABA", "output": "AB" }, { "input": "5\nZZZAA", "output": "ZZ" }, { "input": "26\nQWERTYUIOPASDFGHJKLZXCVBNM", "output": "AS" }, { "input": "2\nQA", "output": "QA" }, { "input": "2\nWW", "output": "WW" }, { "input": "11\nGGRRAATTZZZ", "output": "ZZ" }, { "input": "50\nNYQAHBYYOXLTRYQDMVENEMAQNBAKGLGQOLXNAIFNQTOCLNNQIA", "output": "NQ" }, { "input": "100\nURXCAIZFIBNJTPCZHBQIBCILLPXZCFGMKKZMNPLCYGAVJVIBMCZEBSJWPSCPQDYCTTKPOKIJRSKIZPDGCHVOUTMPNECYORSFZFNC", "output": "IB" }, { "input": "100\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AA" }, { "input": "10\nSQSQSQSQTG", "output": "SQ" }, { "input": "5\nAZAZA", "output": "AZ" }, { "input": "15\nMIRZOYANOVECLOX", "output": "AN" }, { "input": "9\nEGORLETOV", "output": "EG" }, { "input": "8\nPUTINVOR", "output": "IN" }, { "input": "7\nKADUROV", "output": "AD" }, { "input": "6\nAZAZAZ", "output": "AZ" }, { "input": "3\nLOL", "output": "LO" }, { "input": "3\nKEK", "output": "EK" }, { "input": "5\nFUFEL", "output": "EL" }, { "input": "9\nMIKEPIDOR", "output": "DO" }, { "input": "9\nAAAAAAAAA", "output": "AA" }, { "input": "23\nAABBBAAACCCCCAAADDDDDDD", "output": "DD" } ]

1,699,547,230

2,147,483,647

PyPy 3

OK

TESTS

22

92

0

n=int(input()) s=input() d={} for i in range(len(s)-1): temp="" temp+=(s[i]+s[i+1]) if temp not in d: d[temp]=1 else: d[temp]+=1 maxi=0 ans="" for i in d: if d[i]>maxi: maxi=d[i] ans=i print(ans)

Title: Two-gram Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams. You are given a string $s$ consisting of $n$ capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string $s$ = "BBAABBBA" the answer is two-gram "BB", which contained in $s$ three times. In other words, find any most frequent two-gram. Note that occurrences of the two-gram can overlap with each other. Input Specification: The first line of the input contains integer number $n$ ($2 \le n \le 100$) — the length of string $s$. The second line of the input contains the string $s$ consisting of $n$ capital Latin letters. Output Specification: Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $s$ as a substring (i.e. two consecutive characters of the string) maximal number of times. Demo Input: ['7\nABACABA\n', '5\nZZZAA\n'] Demo Output: ['AB\n', 'ZZ\n'] Note: In the first example "BA" is also valid answer. In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

```python n=int(input()) s=input() d={} for i in range(len(s)-1): temp="" temp+=(s[i]+s[i+1]) if temp not in d: d[temp]=1 else: d[temp]+=1 maxi=0 ans="" for i in d: if d[i]>maxi: maxi=d[i] ans=i print(ans) ```

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" } ]

1,599,293,978

2,147,483,647

Python 3

OK

TESTS

21

93

0

n = int(input()) a = input().split() b = sorted(a) p = 0 q = 0 for i in range(1,n): if b[i] != b[i-1]: p += 1 if b[0] == '0': print(p) else: print(p + 1)

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 = sorted(a) p = 0 q = 0 for i in range(1,n): if b[i] != b[i-1]: p += 1 if b[0] == '0': print(p) else: print(p + 1) ```

3

1

A

Theatre Square

PROGRAMMING

1,000

[ "math" ]

A. Theatre Square

1

256

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

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

Write the needed number of flagstones.

[ "6 6 4\n" ]

[ "4\n" ]

none

0

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

1,679,060,676

2,147,483,647

PyPy 3-64

OK

TESTS

20

61

0

n, m, a = map(int, input().split()) if m%a == 0: side_a = m//a else: side_a = (m//a)+1 if n%a == 0: side_b = n//a else: side_b = (n//a)+1 print(side_a*side_b)

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 n, m, a = map(int, input().split()) if m%a == 0: side_a = m//a else: side_a = (m//a)+1 if n%a == 0: side_b = n//a else: side_b = (n//a)+1 print(side_a*side_b) ```

3.9695

227

B

Effective Approach

PROGRAMMING

1,100

[ "implementation" ]

null

Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.

Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

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

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

In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).

1,000

[ { "input": "2\n1 2\n1\n1", "output": "1 2" }, { "input": "2\n2 1\n1\n1", "output": "2 1" }, { "input": "3\n3 1 2\n3\n1 2 3", "output": "6 6" }, { "input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5", "output": "58 32" }, { "input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4", "output": "9 2" }, { "input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8", "output": "31 68" }, { "input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1", "output": "15 15" }, { "input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2", "output": "27 13" }, { "input": "4\n1 3 2 4\n4\n3 1 2 3", "output": "8 12" }, { "input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2", "output": "15 17" } ]

1,627,918,777

2,147,483,647

PyPy 3

OK

TESTS

34

498

33,689,600

n = int(input()) arr = list(map(int, input().split())) m = int(input()) qry = list(map(int, input().split())) d = {} for i in range(n): d[arr[i]] = i+1 vas = 0 pet = 0 for i in range(m): vas += d[qry[i]] pet += n-d[qry[i]]+1 print(vas, pet)

Title: Effective Approach Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. Output Specification: Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. Demo Input: ['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n'] Demo Output: ['1 2\n', '2 1\n', '6 6\n'] Note: In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).

```python n = int(input()) arr = list(map(int, input().split())) m = int(input()) qry = list(map(int, input().split())) d = {} for i in range(n): d[arr[i]] = i+1 vas = 0 pet = 0 for i in range(m): vas += d[qry[i]] pet += n-d[qry[i]]+1 print(vas, pet) ```

3

208

A

Dubstep

PROGRAMMING

900

[ "strings" ]

null

Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.

The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.

Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.

[ "WUBWUBABCWUB\n", "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n" ]

[ "ABC ", "WE ARE THE CHAMPIONS MY FRIEND " ]

In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

500

[ { "input": "WUBWUBABCWUB", "output": "ABC " }, { "input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB", "output": "WE ARE THE CHAMPIONS MY FRIEND " }, { "input": "WUBWUBWUBSR", "output": "SR " }, { "input": "RWUBWUBWUBLWUB", "output": "R L " }, { "input": "ZJWUBWUBWUBJWUBWUBWUBL", "output": "ZJ J L " }, { "input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB", "output": "C B E Q " }, { "input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB", "output": "JKD WBIRAQKF YE WV " }, { "input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB", "output": "KSDHEMIXUJ R S H " }, { "input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB", "output": "OG X I KO " }, { "input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH", "output": "Q QQ I WW JOPJPBRH " }, { "input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB", "output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C " }, { "input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV", "output": "E IQMJNIQ GZZBQZAUHYP PMR DCV " }, { "input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB", "output": "FV BPS RXNETCJ JDMBH B V B " }, { "input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL", "output": "FBQ IDFSY CTWDM SXO QI L " }, { "input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL", "output": "I QLHD YIIKZDFQ CX U K NL " }, { "input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE", "output": "K UPDYXGOKU AGOAH IZD IY V P E " }, { "input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB", "output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ " }, { "input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB", "output": "PAMJGY XGPQM TKGSXUY E N H E " }, { "input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB", "output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB " }, { "input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM", "output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M " }, { "input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW", "output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W " }, { "input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG", "output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G " }, { "input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN", "output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N " }, { "input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG", "output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG " }, { "input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB", "output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L " }, { "input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB", "output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U " }, { "input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB", "output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ " }, { "input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB", "output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J " }, { "input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO", "output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O " }, { "input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR", "output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R " }, { "input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB", "output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM " }, { "input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB", "output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H " }, { "input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB", "output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S " }, { "input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB", "output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J " }, { "input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF", "output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F " }, { "input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY", "output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y " }, { "input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB", "output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO " }, { "input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW", "output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W " }, { "input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD", "output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D " }, { "input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB", "output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A " }, { "input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB", "output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y " }, { "input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB", "output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R " }, { "input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY", "output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y " }, { "input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE", "output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E " }, { "input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS", "output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S " }, { "input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP", "output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP " }, { "input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB", "output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y " }, { "input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL", "output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL " }, { "input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO", "output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O " }, { "input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU", "output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU " }, { "input": "A", "output": "A " }, { "input": "WUBA", "output": "A " }, { "input": "WUBWUBA", "output": "A " }, { "input": "AWUBWUBWUB", "output": "A " }, { "input": "AWUBBWUBCWUBD", "output": "A B C D " }, { "input": "WUBWWUBWUBWUBUWUBWUBBWUB", "output": "W U B " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "WUWUBBWWUBUB", "output": "WU BW UB " }, { "input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB", "output": "WUAB " }, { "input": "ZWUB", "output": "Z " }, { "input": "WU", "output": "WU " }, { "input": "UB", "output": "UB " }, { "input": "U", "output": "U " }, { "input": "WUBW", "output": "W " }, { "input": "WUBWU", "output": "WU " }, { "input": "WUWUB", "output": "WU " }, { "input": "UBWUB", "output": "UB " }, { "input": "WUWUBUBWUBUWUB", "output": "WU UB U " }, { "input": "WUBWWUBAWUB", "output": "W A " }, { "input": "WUUUUU", "output": "WUUUUU " } ]

1,691,494,969

2,147,483,647

PyPy 3

OK

TESTS

71

186

0

a=input("") c=a.split("WUB") for i in c: if i!="": print(i,end=" ")

Title: Dubstep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. Input Specification: The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. Output Specification: Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. Demo Input: ['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n'] Demo Output: ['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND '] Note: In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

```python a=input("") c=a.split("WUB") for i in c: if i!="": print(i,end=" ") ```

3

131

A

cAPS lOCK

PROGRAMMING

1,000

[ "implementation", "strings" ]

null

wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.

The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.

Print the result of the given word's processing.

[ "cAPS\n", "Lock\n" ]

[ "Caps", "Lock\n" ]

none

500

[ { "input": "cAPS", "output": "Caps" }, { "input": "Lock", "output": "Lock" }, { "input": "cAPSlOCK", "output": "cAPSlOCK" }, { "input": "CAPs", "output": "CAPs" }, { "input": "LoCK", "output": "LoCK" }, { "input": "OOPS", "output": "oops" }, { "input": "oops", "output": "oops" }, { "input": "a", "output": "A" }, { "input": "A", "output": "a" }, { "input": "aA", "output": "Aa" }, { "input": "Zz", "output": "Zz" }, { "input": "Az", "output": "Az" }, { "input": "zA", "output": "Za" }, { "input": "AAA", "output": "aaa" }, { "input": "AAa", "output": "AAa" }, { "input": "AaR", "output": "AaR" }, { "input": "Tdr", "output": "Tdr" }, { "input": "aTF", "output": "Atf" }, { "input": "fYd", "output": "fYd" }, { "input": "dsA", "output": "dsA" }, { "input": "fru", "output": "fru" }, { "input": "hYBKF", "output": "Hybkf" }, { "input": "XweAR", "output": "XweAR" }, { "input": "mogqx", "output": "mogqx" }, { "input": "eOhEi", "output": "eOhEi" }, { "input": "nkdku", "output": "nkdku" }, { "input": "zcnko", "output": "zcnko" }, { "input": "lcccd", "output": "lcccd" }, { "input": "vwmvg", "output": "vwmvg" }, { "input": "lvchf", "output": "lvchf" }, { "input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO", "output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo" }, { "input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG", "output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg" }, { "input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS", "output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ", "output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq" }, { "input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR", "output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR" }, { "input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE", "output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE" }, { "input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ", "output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ" }, { "input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm", "output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm" }, { "input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm", "output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm" }, { "input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg", "output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg" }, { "input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc", "output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc" }, { "input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv", "output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv" }, { "input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect", "output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu", "output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu" }, { "input": "aBACABa", "output": "aBACABa" }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAaa", "output": "AAAAAAAAAAAAAAAAAAAAAAAAaa" }, { "input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" }, { "input": "dDDDDDDDDDDDDD", "output": "Dddddddddddddd" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" }, { "input": "z", "output": "Z" }, { "input": "AZ", "output": "az" }, { "input": "Z", "output": "z" }, { "input": "aAAAA", "output": "Aaaaa" }, { "input": "F", "output": "f" } ]

1,694,316,175

2,147,483,647

PyPy 3-64

OK

TESTS

56

77

0

a=input() if len(a)==1:a=a.swapcase() else:a=(a,a.swapcase())[a[1:].isupper()] print(a)

Title: cAPS lOCK Time Limit: None seconds Memory Limit: None megabytes Problem Description: wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. Input Specification: The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. Output Specification: Print the result of the given word's processing. Demo Input: ['cAPS\n', 'Lock\n'] Demo Output: ['Caps', 'Lock\n'] Note: none

```python a=input() if len(a)==1:a=a.swapcase() else:a=(a,a.swapcase())[a[1:].isupper()] print(a) ```

3

0

none

0

[ "none" ]

null

Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!

The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.

Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.

[ "4 3 9\n", "4 3 24\n", "2 4 4\n" ]

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

The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

0

[ { "input": "4 3 9", "output": "2 2 L" }, { "input": "4 3 24", "output": "4 3 R" }, { "input": "2 4 4", "output": "1 2 R" }, { "input": "3 10 24", "output": "2 2 R" }, { "input": "10 3 59", "output": "10 3 L" }, { "input": "10000 10000 160845880", "output": "8043 2940 R" }, { "input": "1 1 1", "output": "1 1 L" }, { "input": "1 1 2", "output": "1 1 R" }, { "input": "1 10000 1", "output": "1 1 L" }, { "input": "1 10000 20000", "output": "1 10000 R" }, { "input": "10000 1 1", "output": "1 1 L" }, { "input": "10000 1 10000", "output": "5000 1 R" }, { "input": "10000 1 20000", "output": "10000 1 R" }, { "input": "3 2 1", "output": "1 1 L" }, { "input": "3 2 2", "output": "1 1 R" }, { "input": "3 2 3", "output": "1 2 L" }, { "input": "3 2 4", "output": "1 2 R" }, { "input": "3 2 5", "output": "2 1 L" }, { "input": "3 2 6", "output": "2 1 R" }, { "input": "3 2 7", "output": "2 2 L" }, { "input": "3 2 8", "output": "2 2 R" }, { "input": "3 2 9", "output": "3 1 L" }, { "input": "3 2 10", "output": "3 1 R" }, { "input": "3 2 11", "output": "3 2 L" }, { "input": "3 2 12", "output": "3 2 R" }, { "input": "300 2000 1068628", "output": "268 314 R" }, { "input": "300 2000 584756", "output": "147 378 R" }, { "input": "300 2000 268181", "output": "68 91 L" }, { "input": "10000 9999 186450844", "output": "9324 4745 R" }, { "input": "10000 9999 197114268", "output": "9857 6990 R" }, { "input": "10000 9999 112390396", "output": "5621 818 R" }, { "input": "10000 10000 1", "output": "1 1 L" }, { "input": "10000 10000 2", "output": "1 1 R" }, { "input": "10000 10000 100000001", "output": "5001 1 L" }, { "input": "10000 10000 199999999", "output": "10000 10000 L" }, { "input": "10000 10000 200000000", "output": "10000 10000 R" }, { "input": "1 2 1", "output": "1 1 L" }, { "input": "1 2 2", "output": "1 1 R" }, { "input": "1 2 3", "output": "1 2 L" }, { "input": "1 2 4", "output": "1 2 R" }, { "input": "2 1 1", "output": "1 1 L" }, { "input": "2 1 2", "output": "1 1 R" }, { "input": "2 1 3", "output": "2 1 L" }, { "input": "2 1 4", "output": "2 1 R" }, { "input": "4 3 7", "output": "2 1 L" }, { "input": "1 1 1", "output": "1 1 L" } ]

1,482,657,262

562

Python 3

OK

TESTS

46

77

4,608,000

n,m,k=map(int,input().split()) if k%2==0: s='R' else: s='L' c=(k-1)//(2*m) r=((k-1)%(2*m))//2 print(c+1,r+1,s)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right! Input Specification: The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place. Output Specification: Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right. Demo Input: ['4 3 9\n', '4 3 24\n', '2 4 4\n'] Demo Output: ['2 2 L\n', '4 3 R\n', '1 2 R\n'] Note: The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

```python n,m,k=map(int,input().split()) if k%2==0: s='R' else: s='L' c=(k-1)//(2*m) r=((k-1)%(2*m))//2 print(c+1,r+1,s) ```

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" } ]

1,682,316,179

2,147,483,647

Python 3

OK

TESTS

30

122

0

x = input() l,u = 0,0 for i in range(len(x)): if(x[i]>='a' and x[i]<='z'): l+=1 u = len(x)-l if(l>=u): print(x.lower()) else: print(x.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 x = input() l,u = 0,0 for i in range(len(x)): if(x[i]>='a' and x[i]<='z'): l+=1 u = len(x)-l if(l>=u): print(x.lower()) else: print(x.upper()) ```

3.9695

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" } ]

1,695,210,807

2,147,483,647

Python 3

OK

TESTS

34

46

3,379,200

n, t = map(int, input().split()) a = list(map(int, input().split())) t -= 1 x = 0 while x < n-1 and x != t: x += a[x] if x == t: 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 n, t = map(int, input().split()) a = list(map(int, input().split())) t -= 1 x = 0 while x < n-1 and x != t: x += a[x] if x == t: print("YES") else: print("NO") ```

3

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" } ]

1,690,042,332

2,147,483,647

Python 3

OK

TESTS

34

46

3,686,400

n, t = [int(e) for e in input().split()] stations = [int(e) for e in input().split()] t -= 1 now = 0 while now < t: now += stations[now] if now == t: 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 n, t = [int(e) for e in input().split()] stations = [int(e) for e in input().split()] t -= 1 now = 0 while now < t: now += stations[now] if now == t: print("YES") else: print("NO") ```

3

535

B

Tavas and SaDDas

PROGRAMMING

1,100

[ "bitmasks", "brute force", "combinatorics", "implementation" ]

null

Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you." The problem is: You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. If we sort all lucky numbers in increasing order, what's the 1-based index of *n*? Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.

The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).

Print the index of *n* among all lucky numbers.

[ "4\n", "7\n", "77\n" ]

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

none

1,000

[ { "input": "4", "output": "1" }, { "input": "7", "output": "2" }, { "input": "77", "output": "6" }, { "input": "4", "output": "1" }, { "input": "474744", "output": "83" }, { "input": "777774", "output": "125" }, { "input": "447", "output": "8" }, { "input": "774", "output": "13" }, { "input": "4", "output": "1" }, { "input": "4447747", "output": "140" }, { "input": "7747474", "output": "233" }, { "input": "4444", "output": "15" }, { "input": "4447", "output": "16" }, { "input": "7", "output": "2" }, { "input": "4", "output": "1" }, { "input": "4447744", "output": "139" }, { "input": "77474", "output": "57" }, { "input": "7747", "output": "28" }, { "input": "444", "output": "7" }, { "input": "7", "output": "2" }, { "input": "7774477", "output": "242" }, { "input": "4477774", "output": "157" }, { "input": "7444", "output": "23" }, { "input": "7474747", "output": "212" }, { "input": "77", "output": "6" }, { "input": "774477", "output": "114" }, { "input": "7", "output": "2" }, { "input": "47", "output": "4" }, { "input": "747777", "output": "110" }, { "input": "444444444", "output": "511" }, { "input": "777777777", "output": "1022" }, { "input": "477477447", "output": "728" }, { "input": "777744747", "output": "996" } ]

1,670,506,586

2,147,483,647

PyPy 3-64

OK

TESTS

33

61

0

s=input() l=len(s)-1 res=0 for i in range(len(s)): if(s[i]=='7'): res=res+2*(2**l) else: res=res+(2**l) l=l-1 print(res)

Title: Tavas and SaDDas Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you." The problem is: You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. If we sort all lucky numbers in increasing order, what's the 1-based index of *n*? Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back. Input Specification: The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109). Output Specification: Print the index of *n* among all lucky numbers. Demo Input: ['4\n', '7\n', '77\n'] Demo Output: ['1\n', '2\n', '6\n'] Note: none

```python s=input() l=len(s)-1 res=0 for i in range(len(s)): if(s[i]=='7'): res=res+2*(2**l) else: res=res+(2**l) l=l-1 print(res) ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

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

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

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

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

[ "NO", "YES" ]

none

500

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

1,635,152,994

2,147,483,647

Python 3

OK

TESTS

81

122

4,505,600

n = int(input()) x,y,z = [0,0,0] for i in range(n): x1, y1, z1 = map(int, input().split()) x+=x1; y+=y1; z+=z1 if (x == y == z == 0): print("YES") else: print("NO")

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

```python n = int(input()) x,y,z = [0,0,0] for i in range(n): x1, y1, z1 = map(int, input().split()) x+=x1; y+=y1; z+=z1 if (x == y == z == 0): print("YES") else: print("NO") ```

3.961108

914

B

Conan and Agasa play a Card Game

PROGRAMMING

1,200

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

null

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

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

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

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

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

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

1,000

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

1,698,060,818

2,147,483,647

PyPy 3-64

OK

TESTS

91

93

13,209,600

n = int(input()) a = [0] * 100100 numbers = list(map(int, input().split())) for x in numbers: a[x] += 1 for i in range(100099, -1, -1): if a[i] % 2 == 1: print("Conan") break else: print("Agasa")# 1698060818.6272213

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

```python n = int(input()) a = [0] * 100100 numbers = list(map(int, input().split())) for x in numbers: a[x] += 1 for i in range(100099, -1, -1): if a[i] % 2 == 1: print("Conan") break else: print("Agasa")# 1698060818.6272213 ```

3

776

A

A Serial Killer

PROGRAMMING

900

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

null

Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim. The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim. You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.

First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days. Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person. The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.

Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.

[ "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n", "icm codeforces\n1\ncodeforces technex\n" ]

[ "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n", "icm codeforces\nicm technex\n" ]

In first example, the killer starts with ross and rachel. - After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.

500

[ { "input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler", "output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler" }, { "input": "icm codeforces\n1\ncodeforces technex", "output": "icm codeforces\nicm technex" }, { "input": "a b\n3\na c\nb d\nd e", "output": "a b\nc b\nc d\nc e" }, { "input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg", "output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab" }, { "input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o", "output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a" }, { "input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii", "output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii" }, { "input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm", "output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm" }, { "input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b", "output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b" }, { "input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf", "output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf" }, { "input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q", "output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q" }, { "input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww", "output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww" }, { "input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n", "output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g" } ]

1,630,550,427

2,147,483,647

Python 3

OK

TESTS

57

77

6,758,400

a,b = input().split() print(a,b) for _ in range(int(input())): c,d = input().split() if a == c: a = d print(a,b) elif b == c: b = d print(a,b)

Title: A Serial Killer Time Limit: None seconds Memory Limit: None megabytes Problem Description: Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim. The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim. You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern. Input Specification: First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days. Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person. The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters. Output Specification: Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order. Demo Input: ['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n'] Demo Output: ['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n'] Note: In first example, the killer starts with ross and rachel. - After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.

```python a,b = input().split() print(a,b) for _ in range(int(input())): c,d = input().split() if a == c: a = d print(a,b) elif b == c: b = d print(a,b) ```

3

869

B

The Eternal Immortality

PROGRAMMING

1,100

[ "math" ]

null

Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.

The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).

Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

[ "2 4\n", "0 10\n", "107 109\n" ]

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

In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.

1,000

[ { "input": "2 4", "output": "2" }, { "input": "0 10", "output": "0" }, { "input": "107 109", "output": "2" }, { "input": "10 13", "output": "6" }, { "input": "998244355 998244359", "output": "4" }, { "input": "999999999000000000 1000000000000000000", "output": "0" }, { "input": "2 3", "output": "3" }, { "input": "3 15", "output": "0" }, { "input": "24 26", "output": "0" }, { "input": "14 60", "output": "0" }, { "input": "11 79", "output": "0" }, { "input": "1230 1232", "output": "2" }, { "input": "2633 2634", "output": "4" }, { "input": "535 536", "output": "6" }, { "input": "344319135 396746843", "output": "0" }, { "input": "696667767 696667767", "output": "1" }, { "input": "419530302 610096911", "output": "0" }, { "input": "238965115 821731161", "output": "0" }, { "input": "414626436 728903812", "output": "0" }, { "input": "274410639 293308324", "output": "0" }, { "input": "650636673091305697 650636673091305702", "output": "0" }, { "input": "651240548333620923 651240548333620924", "output": "4" }, { "input": "500000000000000000 1000000000000000000", "output": "0" }, { "input": "999999999999999999 1000000000000000000", "output": "0" }, { "input": "1000000000000000000 1000000000000000000", "output": "1" }, { "input": "0 4", "output": "4" }, { "input": "50000000062000007 50000000062000011", "output": "0" }, { "input": "0 0", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "0 2", "output": "2" }, { "input": "10000000000012 10000000000015", "output": "0" }, { "input": "5 5", "output": "1" }, { "input": "12 23", "output": "0" }, { "input": "0 11", "output": "0" }, { "input": "11111234567890 11111234567898", "output": "0" }, { "input": "0 3", "output": "6" }, { "input": "1 2", "output": "2" }, { "input": "999999999999999997 999999999999999999", "output": "2" }, { "input": "4 5", "output": "5" }, { "input": "0 1", "output": "1" }, { "input": "101 1002", "output": "0" }, { "input": "0 100000000000000001", "output": "0" }, { "input": "99999999999999997 99999999999999999", "output": "2" }, { "input": "14 15", "output": "5" }, { "input": "8 19", "output": "0" }, { "input": "12 22", "output": "0" }, { "input": "999999999999996 999999999999999", "output": "4" }, { "input": "1 3", "output": "6" }, { "input": "124 125", "output": "5" }, { "input": "11 32", "output": "0" }, { "input": "0 5", "output": "0" }, { "input": "0 999999", "output": "0" }, { "input": "151151151515 151151151526", "output": "0" }, { "input": "6 107", "output": "0" }, { "input": "5 16", "output": "0" }, { "input": "7 16", "output": "0" }, { "input": "6 19", "output": "0" }, { "input": "11113111111111 13111111111111", "output": "0" }, { "input": "1 1000", "output": "0" }, { "input": "24 25", "output": "5" }, { "input": "0 100000000000", "output": "0" }, { "input": "1 22", "output": "0" }, { "input": "999999999999999996 999999999999999999", "output": "4" } ]

1,641,560,636

2,147,483,647

PyPy 3-64

OK

TESTS

63

108

0

a, b = map(int, input().split()) if b - a >= 5: print(0) else: ans = 1 for i in range(a + 1, b + 1): ans = ans * i % 10 print(ans)

Title: The Eternal Immortality Time Limit: None seconds Memory Limit: None megabytes Problem Description: Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge. Input Specification: The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018). Output Specification: Output one line containing a single decimal digit — the last digit of the value that interests Koyomi. Demo Input: ['2 4\n', '0 10\n', '107 109\n'] Demo Output: ['2\n', '0\n', '2\n'] Note: In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.

```python a, b = map(int, input().split()) if b - a >= 5: print(0) else: ans = 1 for i in range(a + 1, b + 1): ans = ans * i % 10 print(ans) ```

3

981

A

Antipalindrome

PROGRAMMING

900

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

null

A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not. A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$. Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all. Some time ago Ann read the word $s$. What is the word she changed it into?

The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.

If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$. Note that there can be multiple longest substrings that are not palindromes, but their length is unique.

[ "mew\n", "wuffuw\n", "qqqqqqqq\n" ]

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

"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$. The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$. All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.

500

[ { "input": "mew", "output": "3" }, { "input": "wuffuw", "output": "5" }, { "input": "qqqqqqqq", "output": "0" }, { "input": "ijvji", "output": "4" }, { "input": "iiiiiii", "output": "0" }, { "input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow", "output": "49" }, { "input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww", "output": "0" }, { "input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy", "output": "50" }, { "input": "ijvxljt", "output": "7" }, { "input": "fyhcncnchyf", "output": "10" }, { "input": "ffffffffffff", "output": "0" }, { "input": "fyhcncfsepqj", "output": "12" }, { "input": "ybejrrlbcinttnicblrrjeby", "output": "23" }, { "input": "yyyyyyyyyyyyyyyyyyyyyyyyy", "output": "0" }, { "input": "ybejrrlbcintahovgjddrqatv", "output": "25" }, { "input": "oftmhcmclgyqaojljoaqyglcmchmtfo", "output": "30" }, { "input": "oooooooooooooooooooooooooooooooo", "output": "0" }, { "input": "oftmhcmclgyqaojllbotztajglsmcilv", "output": "32" }, { "input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg", "output": "35" }, { "input": "gggggggggggggggggggggggggggggggggggg", "output": "0" }, { "input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk", "output": "36" }, { "input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf", "output": "40" }, { "input": "fffffffffffffffffffffffffffffffffffffffffff", "output": "0" }, { "input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy", "output": "43" }, { "input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr", "output": "0" }, { "input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy", "output": "49" }, { "input": "a", "output": "0" }, { "input": "abca", "output": "4" }, { "input": "aaaaabaaaaa", "output": "10" }, { "input": "aba", "output": "2" }, { "input": "asaa", "output": "4" }, { "input": "aabaa", "output": "4" }, { "input": "aabbaa", "output": "5" }, { "input": "abcdaaa", "output": "7" }, { "input": "aaholaa", "output": "7" }, { "input": "abcdefghijka", "output": "12" }, { "input": "aaadcba", "output": "7" }, { "input": "aaaabaaaa", "output": "8" }, { "input": "abaa", "output": "4" }, { "input": "abcbaa", "output": "6" }, { "input": "ab", "output": "2" }, { "input": "l", "output": "0" }, { "input": "aaaabcaaaa", "output": "10" }, { "input": "abbaaaaaabba", "output": "11" }, { "input": "abaaa", "output": "5" }, { "input": "baa", "output": "3" }, { "input": "aaaaaaabbba", "output": "11" }, { "input": "ccbcc", "output": "4" }, { "input": "bbbaaab", "output": "7" }, { "input": "abaaaaaaaa", "output": "10" }, { "input": "abaaba", "output": "5" }, { "input": "aabsdfaaaa", "output": "10" }, { "input": "aaaba", "output": "5" }, { "input": "aaabaaa", "output": "6" }, { "input": "baaabbb", "output": "7" }, { "input": "ccbbabbcc", "output": "8" }, { "input": "cabc", "output": "4" }, { "input": "aabcd", "output": "5" }, { "input": "abcdea", "output": "6" }, { "input": "bbabb", "output": "4" }, { "input": "aaaaabababaaaaa", "output": "14" }, { "input": "bbabbb", "output": "6" }, { "input": "aababd", "output": "6" }, { "input": "abaaaa", "output": "6" }, { "input": "aaaaaaaabbba", "output": "12" }, { "input": "aabca", "output": "5" }, { "input": "aaabccbaaa", "output": "9" }, { "input": "aaaaaaaaaaaaaaaaaaaab", "output": "21" }, { "input": "babb", "output": "4" }, { "input": "abcaa", "output": "5" }, { "input": "qwqq", "output": "4" }, { "input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa", "output": "48" }, { "input": "aaab", "output": "4" }, { "input": "aaaaaabaaaaa", "output": "12" }, { "input": "wwuww", "output": "4" }, { "input": "aaaaabcbaaaaa", "output": "12" }, { "input": "aaabbbaaa", "output": "8" }, { "input": "aabcbaa", "output": "6" }, { "input": "abccdefccba", "output": "11" }, { "input": "aabbcbbaa", "output": "8" }, { "input": "aaaabbaaaa", "output": "9" }, { "input": "aabcda", "output": "6" }, { "input": "abbca", "output": "5" }, { "input": "aaaaaabbaaa", "output": "11" }, { "input": "sssssspssssss", "output": "12" }, { "input": "sdnmsdcs", "output": "8" }, { "input": "aaabbbccbbbaaa", "output": "13" }, { "input": "cbdbdc", "output": "6" }, { "input": "abb", "output": "3" }, { "input": "abcdefaaaa", "output": "10" }, { "input": "abbbaaa", "output": "7" }, { "input": "v", "output": "0" }, { "input": "abccbba", "output": "7" }, { "input": "axyza", "output": "5" }, { "input": "abcdefgaaaa", "output": "11" }, { "input": "aaabcdaaa", "output": "9" }, { "input": "aaaacaaaa", "output": "8" }, { "input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa", "output": "42" }, { "input": "abbbaa", "output": "6" }, { "input": "abcdee", "output": "6" }, { "input": "oom", "output": "3" }, { "input": "aabcaa", "output": "6" }, { "input": "abba", "output": "3" }, { "input": "aaca", "output": "4" }, { "input": "aacbca", "output": "6" }, { "input": "ababa", "output": "4" }, { "input": "abcda", "output": "5" }, { "input": "cccaaccc", "output": "7" }, { "input": "aaabcda", "output": "7" }, { "input": "aa", "output": "0" }, { "input": "aabaaaa", "output": "7" }, { "input": "abbaaaa", "output": "7" }, { "input": "aaabcbaaa", "output": "8" }, { "input": "aabba", "output": "5" }, { "input": "xyxx", "output": "4" }, { "input": "aaaaaaaaaaaabc", "output": "14" }, { "input": "bbaaaabb", "output": "7" }, { "input": "aaabaa", "output": "6" }, { "input": "sssssabsssss", "output": "12" }, { "input": "bbbaaaabbb", "output": "9" }, { "input": "abbbbaaaa", "output": "9" }, { "input": "wwufuww", "output": "6" }, { "input": "oowoo", "output": "4" }, { "input": "cccaccc", "output": "6" }, { "input": "aaa", "output": "0" }, { "input": "bbbcc", "output": "5" }, { "input": "abcdef", "output": "6" }, { "input": "abbba", "output": "4" }, { "input": "aab", "output": "3" }, { "input": "aaba", "output": "4" }, { "input": "azbyaaa", "output": "7" }, { "input": "oooooiooooo", "output": "10" }, { "input": "aabbbbbaaaaaa", "output": "13" } ]

1,573,722,620

2,147,483,647

Python 3

OK

TESTS

133

124

0

x=input() ar=[] b=[] for i in x: ar.append(i) for j in x: b.append(j) ar.reverse() if x.count(x[0])!=len(x): if b==ar: print(len(b)-1) else: print(len(b)) else: print("0")

Title: Antipalindrome Time Limit: None seconds Memory Limit: None megabytes Problem Description: A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not. A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$. Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all. Some time ago Ann read the word $s$. What is the word she changed it into? Input Specification: The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. Output Specification: If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$. Note that there can be multiple longest substrings that are not palindromes, but their length is unique. Demo Input: ['mew\n', 'wuffuw\n', 'qqqqqqqq\n'] Demo Output: ['3\n', '5\n', '0\n'] Note: "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$. The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$. All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.

```python x=input() ar=[] b=[] for i in x: ar.append(i) for j in x: b.append(j) ar.reverse() if x.count(x[0])!=len(x): if b==ar: print(len(b)-1) else: print(len(b)) else: print("0") ```

3

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { 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"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": 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"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": 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"0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]

1,674,203,055

2,147,483,647

Python 3

OK

TESTS

102

46

0

a=input() b=input() j=0 for i in a: if i==b[j]: print('0',end="") else: print('1',end="") j+=1

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 a=input() b=input() j=0 for i in a: if i==b[j]: print('0',end="") else: print('1',end="") j+=1 ```

3.9885

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" } ]

1,667,224,669

2,147,483,647

Python 3

OK

TESTS

20

46

0

n, m, a =map(int,input().split()) need_flagstone=0 x=m//a if (m%a > 0): x+=1 y=n//a if (n%a > 0): y+=1 need_flagstone= x*y print(need_flagstone)

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 n, m, a =map(int,input().split()) need_flagstone=0 x=m//a if (m%a > 0): x+=1 y=n//a if (n%a > 0): y+=1 need_flagstone= x*y print(need_flagstone) ```

3.977

109

A

Lucky Sum of Digits

PROGRAMMING

1,000

[ "brute force", "implementation" ]

A. Lucky Sum of Digits

2

256

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.

The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.

Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.

[ "11\n", "10\n" ]

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

none

500

[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "485", "output": "44447777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "111", "output": "444447777777777777" }, { "input": "85", "output": "4477777777777" }, { "input": "114", "output": "444477777777777777" }, { "input": "474", "output": "444777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "74", "output": "47777777777" }, { "input": "1000", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "1024", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "4444", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "45784", "output": 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"7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10417", "output": 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1,610,625,781

2,147,483,647

Python 3

OK

TESTS

51

528

307,200

n = int(input()) mini = [] flag = False x = n//4 y = (n//7)+1 for i in range(x+1): if flag: break for j in range(y+1): if i*4+j*7==n: print('4'*i + '7'*j) flag = True break if not flag: print(-1)

Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none

```python n = int(input()) mini = [] flag = False x = n//4 y = (n//7)+1 for i in range(x+1): if flag: break for j in range(y+1): if i*4+j*7==n: print('4'*i + '7'*j) flag = True break if not flag: print(-1) ```

3.867428

116

A

Tram

PROGRAMMING

800

[ "implementation" ]

null

Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty. Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.

The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops. Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement. - The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.

Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).

[ "4\n0 3\n2 5\n4 2\n4 0\n" ]

[ "6\n" ]

For the first example, a capacity of 6 is sufficient: - At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints. Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.

500

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836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0", "output": "1776" }, { "input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0", "output": "2" }, { "input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0", "output": "1016" }, { "input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0", "output": "3" }, { "input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0", "output": "1000" }, { "input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0", "output": "2071" }, { "input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0", "output": "3" }, { "input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0", "output": "1329" }, { "input": "2\n0 863\n863 0", "output": "863" }, { "input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0", "output": "8" }, { "input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0", "output": "11" }, { "input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0", "output": "7" }, { "input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "2\n0 1\n1 0", "output": "1" }, { "input": "2\n0 900\n900 0", "output": "900" }, { "input": "2\n0 1\n1 0", "output": "1" }, { "input": "2\n0 0\n0 0", "output": "0" }, { "input": "2\n0 1000\n1000 0", "output": "1000" }, { "input": "3\n0 802\n175 188\n815 0", "output": "815" }, { "input": "3\n0 910\n910 976\n976 0", "output": "976" }, { "input": "3\n0 2\n2 1\n1 0", "output": "2" } ]

1,690,536,129

2,147,483,647

PyPy 3-64

OK

TESTS

71

216

1,638,400

import math t = int(input()) a = 0 b = 0 for i in range(t): s,x = map(int, input().split()) b = b-s+x a = max(a,b) print(a)

Title: Tram Time Limit: None seconds Memory Limit: None megabytes Problem Description: Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty. Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. Input Specification: The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops. Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement. - The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. Output Specification: Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). Demo Input: ['4\n0 3\n2 5\n4 2\n4 0\n'] Demo Output: ['6\n'] Note: For the first example, a capacity of 6 is sufficient: - At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints. Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.

```python import math t = int(input()) a = 0 b = 0 for i in range(t): s,x = map(int, input().split()) b = b-s+x a = max(a,b) print(a) ```

3

740

B

Alyona and flowers

PROGRAMMING

1,200

[ "constructive algorithms" ]

null

Little Alyona is celebrating Happy Birthday! Her mother has an array of *n* flowers. Each flower has some mood, the mood of *i*-th flower is *a**i*. The mood can be positive, zero or negative. Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in. For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,<=<=-<=2,<=1,<=3,<=<=-<=4. Suppose the mother suggested subarrays (1,<=<=-<=2), (3,<=<=-<=4), (1,<=3), (1,<=<=-<=2,<=1,<=3). Then if the girl chooses the third and the fourth subarrays then: - the first flower adds 1·1<==<=1 to the girl's happiness, because he is in one of chosen subarrays, - the second flower adds (<=-<=2)·1<==<=<=-<=2, because he is in one of chosen subarrays, - the third flower adds 1·2<==<=2, because he is in two of chosen subarrays, - the fourth flower adds 3·2<==<=6, because he is in two of chosen subarrays, - the fifth flower adds (<=-<=4)·0<==<=0, because he is in no chosen subarrays. Thus, in total 1<=+<=(<=-<=2)<=+<=2<=+<=6<=+<=0<==<=7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this! Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of flowers and the number of subarrays suggested by the mother. The second line contains the flowers moods — *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=100<=≤<=*a**i*<=≤<=100). The next *m* lines contain the description of the subarrays suggested by the mother. The *i*-th of these lines contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) denoting the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*]. Each subarray can encounter more than once.

Print single integer — the maximum possible value added to the Alyona's happiness.

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

[ "7\n", "16\n", "0\n" ]

The first example is the situation described in the statements. In the second example Alyona should choose all subarrays. The third example has answer 0 because Alyona can choose none of the subarrays.

1,000

[ { "input": "5 4\n1 -2 1 3 -4\n1 2\n4 5\n3 4\n1 4", "output": "7" }, { "input": "4 3\n1 2 3 4\n1 3\n2 4\n1 1", "output": "16" }, { "input": "2 2\n-1 -2\n1 1\n1 2", "output": "0" }, { "input": "5 6\n1 1 1 -1 0\n2 4\n1 3\n4 5\n1 5\n1 4\n4 5", "output": "8" }, { "input": "8 3\n5 -4 -2 5 3 -4 -2 6\n3 8\n4 6\n2 3", "output": "10" }, { "input": "10 10\n0 0 0 0 0 0 0 0 0 0\n5 9\n1 9\n5 7\n3 8\n1 6\n1 9\n1 6\n6 9\n1 10\n3 8", "output": "0" }, { "input": "3 6\n0 0 0\n1 1\n1 1\n1 3\n3 3\n2 3\n1 2", "output": "0" }, { "input": "3 3\n1 -1 3\n1 2\n2 3\n1 3", "output": "5" }, { "input": "6 8\n0 6 -5 8 -3 -2\n6 6\n2 3\n5 6\n4 6\n3 4\n2 5\n3 3\n5 6", "output": "13" }, { "input": "10 4\n6 5 5 -1 0 5 0 -3 5 -4\n3 6\n4 9\n1 6\n1 4", "output": "50" }, { "input": "9 1\n-1 -1 -1 -1 2 -1 2 0 0\n2 5", "output": "0" }, { "input": "3 8\n3 4 4\n1 2\n1 3\n2 3\n1 2\n2 2\n1 1\n2 3\n1 3", "output": "59" }, { "input": "3 8\n6 7 -1\n1 1\n1 3\n2 2\n1 3\n1 3\n1 1\n2 3\n2 3", "output": "67" }, { "input": "53 7\n-43 57 92 97 85 -29 28 -8 -37 -47 51 -53 -95 -50 -39 -87 43 36 60 -95 93 8 67 -22 -78 -46 99 93 27 -72 -84 77 96 -47 1 -12 21 -98 -34 -88 57 -43 5 -15 20 -66 61 -29 30 -85 52 53 82\n15 26\n34 43\n37 41\n22 34\n19 43\n2 15\n13 35", "output": "170" }, { "input": "20 42\n61 86 5 -87 -33 51 -79 17 -3 65 -42 74 -94 40 -35 22 58 81 -75 5\n3 6\n12 13\n3 16\n3 16\n5 7\n5 16\n2 15\n6 18\n4 18\n10 17\n14 16\n4 15\n4 11\n13 20\n5 6\n5 15\n16 17\n3 14\n9 10\n5 19\n5 14\n2 4\n17 20\n10 11\n5 18\n10 11\n1 14\n1 6\n1 10\n8 16\n11 14\n12 20\n11 13\n4 5\n2 13\n1 5\n11 15\n1 18\n3 8\n8 20\n1 4\n10 13", "output": "1502" }, { "input": "64 19\n-47 13 19 51 -25 72 38 32 54 7 -49 -50 -59 73 45 -87 -15 -72 -32 -10 -7 47 -34 35 48 -73 79 25 -80 -34 4 77 60 30 61 -25 23 17 -73 -73 69 29 -50 -55 53 15 -33 7 -46 -5 85 -86 77 -51 87 -69 -64 -24 -64 29 -20 -58 11 -26\n6 53\n13 28\n15 47\n20 52\n12 22\n6 49\n31 54\n2 39\n32 49\n27 64\n22 63\n33 48\n49 58\n39 47\n6 29\n21 44\n24 59\n20 24\n39 54", "output": "804" }, { "input": "1 10\n-46\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "0" }, { "input": "10 7\n44 18 9 -22 -23 7 -25 -2 15 35\n6 8\n6 7\n3 3\n2 6\n9 10\n2 2\n1 5", "output": "103" }, { "input": "4 3\n10 -2 68 35\n4 4\n1 1\n1 3", "output": "121" }, { "input": "3 6\n27 -31 -81\n2 3\n2 3\n1 1\n1 2\n1 2\n2 2", "output": "27" }, { "input": "7 3\n-24 -12 16 -43 -30 31 16\n3 6\n3 4\n1 7", "output": "0" }, { "input": "10 7\n-33 -24 -86 -20 5 -91 38 -12 -90 -67\n7 8\n7 10\n4 7\n1 3\n6 10\n6 6\n3 5", "output": "26" }, { "input": "4 4\n95 35 96 -27\n3 4\n3 3\n4 4\n3 3", "output": "261" }, { "input": "7 7\n-33 26 -25 44 -20 -50 33\n4 6\n4 4\n3 7\n5 7\n1 4\n2 5\n4 6", "output": "81" }, { "input": "5 3\n-35 -39 93 59 -4\n2 2\n2 3\n2 5", "output": "163" }, { "input": "3 7\n0 0 0\n1 2\n1 2\n2 3\n3 3\n1 3\n1 2\n2 3", "output": "0" }, { "input": "8 2\n17 32 30 -6 -39 -15 33 74\n6 6\n8 8", "output": "74" }, { "input": "8 1\n-20 -15 21 -21 1 -12 -7 9\n4 7", "output": "0" }, { "input": "7 9\n-23 -4 -44 -47 -35 47 25\n1 6\n3 5\n4 7\n6 7\n2 4\n2 3\n2 7\n1 2\n5 5", "output": "72" }, { "input": "8 8\n0 6 -25 -15 29 -24 31 23\n2 8\n5 5\n3 3\n2 8\n6 6\n3 6\n3 4\n2 4", "output": "79" }, { "input": "4 3\n-39 -63 9 -16\n1 4\n1 3\n2 4", "output": "0" }, { "input": "9 1\n-3 -13 -13 -19 -4 -11 8 -11 -3\n9 9", "output": "0" }, { "input": "9 6\n25 18 -62 0 33 62 -23 4 -15\n7 9\n2 3\n1 4\n2 6\n1 6\n2 3", "output": "127" }, { "input": "4 5\n-12 39 8 -12\n1 4\n3 4\n1 3\n1 3\n2 3", "output": "140" }, { "input": "3 9\n-9 7 3\n1 2\n1 1\n1 3\n1 2\n2 3\n1 3\n2 2\n1 2\n3 3", "output": "22" }, { "input": "10 7\n0 4 3 3 -2 -2 -4 -2 -3 -2\n5 6\n1 10\n2 10\n7 10\n1 1\n6 7\n3 4", "output": "6" }, { "input": "86 30\n16 -12 11 16 8 14 7 -29 18 30 -32 -10 20 29 -14 -21 23 -19 -15 17 -2 25 -22 2 26 15 -7 -12 -4 -28 21 -4 -2 22 28 -32 9 -20 23 38 -21 21 37 -13 -30 25 31 6 18 29 29 29 27 38 -15 -32 32 -7 -8 -33 -11 24 23 -19 -36 -36 -18 9 -1 32 -34 -26 1 -1 -16 -14 17 -17 15 -24 38 5 -27 -12 8 -38\n60 66\n29 48\n32 51\n38 77\n17 79\n23 74\n39 50\n14 29\n26 76\n9 76\n2 67\n23 48\n17 68\n33 75\n59 78\n46 78\n9 69\n16 83\n18 21\n17 34\n24 61\n15 79\n4 31\n62 63\n46 76\n79 82\n25 39\n5 81\n19 77\n26 71", "output": "3076" }, { "input": "33 17\n11 6 -19 14 23 -23 21 15 29 19 13 -18 -19 20 16 -10 26 -22 3 17 13 -10 19 22 -5 21 12 6 28 -13 -27 25 6\n4 17\n12 16\n9 17\n25 30\n31 32\n4 28\n11 24\n16 19\n3 27\n7 17\n1 16\n15 28\n30 33\n9 31\n14 30\n13 23\n27 27", "output": "1366" }, { "input": "16 44\n32 23 -27 -2 -10 -42 32 -14 -13 4 9 -2 19 35 16 22\n6 12\n8 11\n13 15\n12 12\n3 10\n9 13\n7 15\n2 11\n1 13\n5 6\n9 14\n3 16\n10 13\n3 15\n6 10\n14 16\n4 5\n7 10\n5 14\n1 16\n2 5\n1 6\n9 10\n4 7\n4 12\n2 5\n7 10\n7 9\n2 8\n9 10\n4 10\n7 12\n10 11\n6 6\n15 15\n8 12\n9 10\n3 3\n4 15\n10 12\n7 16\n4 14\n14 16\n5 6", "output": "777" }, { "input": "63 24\n-23 -46 0 33 24 13 39 -6 -4 49 19 -18 -11 -38 0 -3 -33 -17 -4 -44 -22 -12 -16 42 16 -10 7 37 -6 16 -41 -18 -20 51 -49 28 -14 -22 -37 -7 -50 31 -41 -47 18 -8 -39 -29 35 -32 14 -29 44 -29 -19 -25 -47 -8 16 11 20 12 38\n21 61\n35 61\n29 61\n21 34\n12 48\n32 33\n9 27\n4 42\n4 60\n37 61\n19 44\n46 60\n51 53\n31 34\n23 32\n5 29\n43 59\n7 31\n29 48\n15 50\n19 51\n7 28\n17 41\n1 62", "output": "82" }, { "input": "6 51\n45 -33 -17 -48 2 2\n2 5\n1 1\n1 3\n5 5\n4 6\n1 2\n2 3\n3 4\n1 6\n4 6\n3 5\n1 5\n2 2\n2 6\n4 6\n1 5\n1 5\n1 5\n1 6\n6 6\n3 5\n6 6\n1 6\n3 4\n3 3\n2 4\n3 3\n1 4\n2 4\n5 6\n4 5\n2 4\n1 5\n1 2\n4 5\n1 5\n5 6\n2 2\n3 4\n1 3\n1 2\n2 5\n3 6\n4 6\n3 4\n2 4\n1 4\n1 1\n4 6\n3 5\n1 4", "output": "140" }, { "input": "3 5\n2 -1 3\n2 3\n2 3\n2 3\n2 3\n1 2", "output": "9" }, { "input": "4 30\n1 1 1 1\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4", "output": "120" } ]

1,479,919,984

1,084

Python 3

OK

TESTS

53

62

0

n, m = [int(x) for x in input().split()] a = [int(x) for x in input().split()] max_sum = 0 for i in range(m): l, r = [int(x) for x in input().split()] l -= 1 sucet = sum(a[l:r]) if sucet > 0: max_sum += sucet print(max_sum)

Title: Alyona and flowers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Alyona is celebrating Happy Birthday! Her mother has an array of *n* flowers. Each flower has some mood, the mood of *i*-th flower is *a**i*. The mood can be positive, zero or negative. Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in. For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,<=<=-<=2,<=1,<=3,<=<=-<=4. Suppose the mother suggested subarrays (1,<=<=-<=2), (3,<=<=-<=4), (1,<=3), (1,<=<=-<=2,<=1,<=3). Then if the girl chooses the third and the fourth subarrays then: - the first flower adds 1·1<==<=1 to the girl's happiness, because he is in one of chosen subarrays, - the second flower adds (<=-<=2)·1<==<=<=-<=2, because he is in one of chosen subarrays, - the third flower adds 1·2<==<=2, because he is in two of chosen subarrays, - the fourth flower adds 3·2<==<=6, because he is in two of chosen subarrays, - the fifth flower adds (<=-<=4)·0<==<=0, because he is in no chosen subarrays. Thus, in total 1<=+<=(<=-<=2)<=+<=2<=+<=6<=+<=0<==<=7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this! Alyona can choose any number of the subarrays, even 0 or all suggested by her mother. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of flowers and the number of subarrays suggested by the mother. The second line contains the flowers moods — *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=100<=≤<=*a**i*<=≤<=100). The next *m* lines contain the description of the subarrays suggested by the mother. The *i*-th of these lines contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) denoting the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*]. Each subarray can encounter more than once. Output Specification: Print single integer — the maximum possible value added to the Alyona's happiness. Demo Input: ['5 4\n1 -2 1 3 -4\n1 2\n4 5\n3 4\n1 4\n', '4 3\n1 2 3 4\n1 3\n2 4\n1 1\n', '2 2\n-1 -2\n1 1\n1 2\n'] Demo Output: ['7\n', '16\n', '0\n'] Note: The first example is the situation described in the statements. In the second example Alyona should choose all subarrays. The third example has answer 0 because Alyona can choose none of the subarrays.

```python n, m = [int(x) for x in input().split()] a = [int(x) for x in input().split()] max_sum = 0 for i in range(m): l, r = [int(x) for x in input().split()] l -= 1 sucet = sum(a[l:r]) if sucet > 0: max_sum += sucet print(max_sum) ```

3

363

B

Fence

PROGRAMMING

1,100

[ "brute force", "dp" ]

null

There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights. Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible. Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.

Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.

[ "7 3\n1 2 6 1 1 7 1\n" ]

[ "3\n" ]

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

1,000

[ { "input": "7 3\n1 2 6 1 1 7 1", "output": "3" }, { "input": "1 1\n100", "output": "1" }, { "input": "2 1\n10 20", "output": "1" }, { "input": "10 5\n1 2 3 1 2 2 3 1 4 5", "output": "1" }, { "input": "10 2\n3 1 4 1 4 6 2 1 4 6", "output": "7" }, { "input": "2 2\n20 10", "output": "1" }, { "input": "2 1\n20 1", "output": "2" }, { "input": "3 1\n1 2 3", "output": "1" }, { "input": "3 1\n2 1 3", "output": "2" }, { "input": "3 1\n3 2 1", "output": "3" }, { "input": "3 2\n1 2 3", "output": "1" }, { "input": "3 2\n3 2 1", "output": "2" }, { "input": "3 3\n1 2 3", "output": "1" }, { "input": "4 2\n9 8 11 7", "output": "1" }, { "input": "4 2\n10 1 2 3", "output": "2" }, { "input": "6 3\n56 56 56 2 1 2", "output": "4" }, { "input": "8 3\n1 1 1 1 2 60 90 1", "output": "1" }, { "input": "4 1\n1 5 2 2", "output": "1" }, { "input": "4 2\n4 6 7 4", "output": "1" }, { "input": "10 4\n1 1 1 4 4 4 4 4 4 3", "output": "1" }, { "input": "6 3\n1 2 1 3 1 1", "output": "1" }, { "input": "5 2\n100 100 100 1 1", "output": "4" } ]

1,695,047,810

2,147,483,647

PyPy 3-64

OK

TESTS

51

77

14,848,000

n, k = map(int, input().split()) a = [int(i) for i in input().split()] ans = tek = sum(a[:k]) j_ans = 1 for i in range(k, n): tek -= a[i - k] tek += a[i] if ans > tek: j_ans = i - k + 2 ans = tek print(j_ans)

Title: Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights. Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible. Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic). Input Specification: The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence. Output Specification: Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them. Demo Input: ['7 3\n1 2 6 1 1 7 1\n'] Demo Output: ['3\n'] Note: In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

```python n, k = map(int, input().split()) a = [int(i) for i in input().split()] ans = tek = sum(a[:k]) j_ans = 1 for i in range(k, n): tek -= a[i - k] tek += a[i] if ans > tek: j_ans = i - k + 2 ans = tek print(j_ans) ```

3