MatrixStudio/Codeforces-Python-Submissions

722

A

Broken Clock

PROGRAMMING

1,300

[ "brute force", "implementation" ]

null

You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59. You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format. For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.

The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.

The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.

[ "24\n17:30\n", "12\n17:30\n", "24\n99:99\n" ]

[ "17:30\n", "07:30\n", "09:09\n" ]

none

500

[ { "input": "24\n17:30", "output": "17:30" }, { "input": "12\n17:30", "output": "07:30" }, { "input": "24\n99:99", "output": "09:09" }, { "input": "12\n05:54", "output": "05:54" }, { "input": "12\n00:05", "output": "01:05" }, { "input": "24\n23:80", "output": "23:00" }, { "input": "24\n73:16", "output": "03:16" }, { "input": "12\n03:77", "output": "03:07" }, { "input": "12\n47:83", "output": "07:03" }, { "input": "24\n23:88", "output": "23:08" }, { "input": "24\n51:67", "output": "01:07" }, { "input": "12\n10:33", "output": "10:33" }, { "input": "12\n00:01", "output": "01:01" }, { "input": "12\n07:74", "output": "07:04" }, { "input": "12\n00:60", "output": "01:00" }, { "input": "24\n08:32", "output": "08:32" }, { "input": "24\n42:59", "output": "02:59" }, { "input": "24\n19:87", "output": "19:07" }, { "input": "24\n26:98", "output": "06:08" }, { "input": "12\n12:91", "output": "12:01" }, { "input": "12\n11:30", "output": "11:30" }, { "input": "12\n90:32", "output": "10:32" }, { "input": "12\n03:69", "output": "03:09" }, { "input": "12\n33:83", "output": "03:03" }, { "input": "24\n10:45", "output": "10:45" }, { "input": "24\n65:12", "output": "05:12" }, { "input": "24\n22:64", "output": "22:04" }, { "input": "24\n48:91", "output": "08:01" }, { "input": "12\n02:51", "output": "02:51" }, { "input": "12\n40:11", "output": "10:11" }, { "input": "12\n02:86", "output": "02:06" }, { "input": "12\n99:96", "output": "09:06" }, { "input": "24\n19:24", "output": "19:24" }, { "input": "24\n55:49", "output": "05:49" }, { "input": "24\n01:97", "output": "01:07" }, { "input": "24\n39:68", "output": "09:08" }, { "input": "24\n24:00", "output": "04:00" }, { "input": "12\n91:00", "output": "01:00" }, { "input": "24\n00:30", "output": "00:30" }, { "input": "12\n13:20", "output": "03:20" }, { "input": "12\n13:00", "output": "03:00" }, { "input": "12\n42:35", "output": "02:35" }, { "input": "12\n20:00", "output": "10:00" }, { "input": "12\n21:00", "output": "01:00" }, { "input": "24\n10:10", "output": "10:10" }, { "input": "24\n30:40", "output": "00:40" }, { "input": "24\n12:00", "output": "12:00" }, { "input": "12\n10:60", "output": "10:00" }, { "input": "24\n30:00", "output": "00:00" }, { "input": "24\n34:00", "output": "04:00" }, { "input": "12\n22:00", "output": "02:00" }, { "input": "12\n20:20", "output": "10:20" } ]

1,475,331,159

459

Python 3

WRONG_ANSWER

PRETESTS

4

61

0

n = int(input()) t = list(map(list, input().split(':'))) if int(''.join(t[1])) >= 60: t[1][0] = '0' if int(''.join(t[0])) >= 24 and n == 24: t[0][0] = '0' if int(''.join(t[0])) > 12 and n == 12: if int(t[0][1]) != 0: t[0][0] = '0' else: t[0][0] = '1' print(''.join(t[0])+':'+''.join(t[1]))

Title: Broken Clock Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59. You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format. For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39. Input Specification: The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. Output Specification: The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. Demo Input: ['24\n17:30\n', '12\n17:30\n', '24\n99:99\n'] Demo Output: ['17:30\n', '07:30\n', '09:09\n'] Note: none

```python n = int(input()) t = list(map(list, input().split(':'))) if int(''.join(t[1])) >= 60: t[1][0] = '0' if int(''.join(t[0])) >= 24 and n == 24: t[0][0] = '0' if int(''.join(t[0])) > 12 and n == 12: if int(t[0][1]) != 0: t[0][0] = '0' else: t[0][0] = '1' print(''.join(t[0])+':'+''.join(t[1])) ```

0

258

A

Little Elephant and Bits

PROGRAMMING

1,100

[ "greedy", "math" ]

null

The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.

The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.

In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.

[ "101\n", "110010\n" ]

[ "11\n", "11010\n" ]

In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

500

[ { "input": "101", "output": "11" }, { "input": "110010", "output": "11010" }, { "input": "10000", "output": "1000" }, { "input": "1111111110", "output": "111111111" }, { "input": "10100101011110101", "output": "1100101011110101" }, { "input": "111010010111", "output": "11110010111" }, { "input": "11110111011100000000", "output": "1111111011100000000" }, { "input": "11110010010100001110110101110011110110100111101", "output": "1111010010100001110110101110011110110100111101" }, { "input": "1001011111010010100111111", "output": "101011111010010100111111" }, { "input": "1111111111", "output": "111111111" }, { "input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101", "output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101" }, { "input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100", "output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100" }, { "input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011", "output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011" }, { "input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011", "output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011" }, { "input": "11", "output": "1" }, { "input": "111", "output": "11" }, { "input": "111111", "output": "11111" }, { "input": "11111", "output": "1111" }, { "input": "1111", "output": "111" } ]

1,632,042,594

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

9

186

20,172,800

s = input() if s.count('0') == '0': print(s[1:]) else: ns =[] deleted = False for i in range(len(s)): if s[i] == '0' and not deleted: deleted = True continue else: ns.append(s[i]) print(''.join(ns))

Title: Little Elephant and Bits Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. Input Specification: The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. Output Specification: In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. Demo Input: ['101\n', '110010\n'] Demo Output: ['11\n', '11010\n'] Note: In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

```python s = input() if s.count('0') == '0': print(s[1:]) else: ns =[] deleted = False for i in range(len(s)): if s[i] == '0' and not deleted: deleted = True continue else: ns.append(s[i]) print(''.join(ns)) ```

0

246

B

Increase and Decrease

PROGRAMMING

1,300

[ "greedy", "math" ]

null

Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times: - he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1. The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times. Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array.

Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.

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

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

none

1,000

[ { "input": "2\n2 1", "output": "1" }, { "input": "3\n1 4 1", "output": "3" }, { "input": "4\n2 -7 -2 -6", "output": "3" }, { "input": "4\n2 0 -2 -1", "output": "3" }, { "input": "6\n-1 1 0 0 -1 -1", "output": "5" }, { "input": "5\n0 0 0 0 0", "output": "5" }, { "input": "100\n968 793 -628 -416 942 -308 977 168 728 -879 952 781 -425 -475 -480 738 -740 142 -319 -116 -701 -183 41 324 -918 -391 -176 781 763 888 475 -617 134 -802 -133 -211 855 -869 -236 503 550 387 137 -221 -957 -800 -56 -673 440 -791 -217 -959 -892 886 -593 427 890 223 -425 -342 -262 693 -137 924 860 156 -110 444 -12 703 421 -638 725 -551 597 -206 -80 393 -441 -759 617 424 -506 209 579 -685 -861 507 312 597 -81 -491 424 222 -892 629 512 743 -452 -504", "output": "99" } ]

1,623,739,861

2,147,483,647

PyPy 3

OK

TESTS

30

372

9,523,200

from typing import Counter n=int(input()) l=sum(list(map(int,input().split()))) if l%n==0: print(n) else: print(n-1)

Title: Increase and Decrease Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times: - he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1. The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times. Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array. Output Specification: Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation. Demo Input: ['2\n2 1\n', '3\n1 4 1\n'] Demo Output: ['1\n', '3\n'] Note: none

```python from typing import Counter n=int(input()) l=sum(list(map(int,input().split()))) if l%n==0: print(n) else: print(n-1) ```

3

975

B

Mancala

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes. Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction. Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole. After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli. Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.

The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole. It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board.

Output one integer, the maximum possible score after one move.

[ "0 1 1 0 0 0 0 0 0 7 0 0 0 0\n", "5 1 1 1 1 0 0 0 0 0 0 0 0 0\n" ]

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

In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.

1,000

[ { "input": "0 1 1 0 0 0 0 0 0 7 0 0 0 0", "output": "4" }, { "input": "5 1 1 1 1 0 0 0 0 0 0 0 0 0", "output": "8" }, { "input": "10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 1", "output": "54294" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 0 15", "output": "2" }, { "input": "1 0 0 0 0 1 0 0 0 0 1 0 0 0", "output": "0" }, { "input": "5 5 1 1 1 3 3 3 5 7 5 3 7 5", "output": "38" }, { "input": "787 393 649 463 803 365 81 961 989 531 303 407 579 915", "output": "7588" }, { "input": "8789651 4466447 1218733 6728667 1796977 6198853 8263135 6309291 8242907 7136751 3071237 5397369 6780785 9420869", "output": "81063456" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 0 29", "output": "26" }, { "input": "282019717 109496191 150951267 609856495 953855615 569750143 6317733 255875779 645191029 572053369 290936613 338480779 879775193 177172893", "output": "5841732816" }, { "input": "105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505", "output": "120472578" }, { "input": "404418821 993626161 346204297 122439813 461187221 628048227 625919459 628611733 938993057 701270099 398043779 684205961 630975553 575964835", "output": "8139909016" }, { "input": "170651077 730658441 824213789 583764177 129437345 717005779 675398017 314979709 380861369 265878463 746564659 797260041 506575735 335169317", "output": "6770880638" }, { "input": "622585025 48249287 678950449 891575125 637411965 457739735 829353393 235216425 284006447 875591469 492839209 296444305 513776057 810057753", "output": "7673796644" }, { "input": "475989857 930834747 786217439 927967137 489188151 869354161 276693267 56154399 131055697 509249443 143116853 426254423 44465165 105798821", "output": "6172339560" }, { "input": "360122921 409370351 226220005 604004145 85173909 600403773 624052991 138163383 729239967 189036661 619842883 270087537 749500483 243727913", "output": "5848946922" }, { "input": "997102881 755715147 273805839 436713689 547411799 72470207 522269145 647688957 137422311 422612659 197751751 679663349 821420227 387967237", "output": "6900015198" }, { "input": "690518849 754551537 652949719 760695679 491633619 477564457 11669279 700467439 470069297 782338983 718169393 884421719 24619427 215745577", "output": "7635414974" }, { "input": "248332749 486342237 662201929 917696895 555278549 252122023 850296207 463343655 832574345 954281071 168282553 825538865 996753493 461254663", "output": "6400166934" }, { "input": "590789361 636464947 404477303 337309187 476703809 426863069 120608741 703406277 645444697 761482231 996635839 33459441 677458865 483861751", "output": "7294965518" }, { "input": "297857621 238127103 749085829 139033277 597985489 202617713 982184715 183932743 278551059 297781685 330124279 338959601 682874531 187519685", "output": "5201808164" }, { "input": "1 1 1 1 1 0 0 0 0 0 0 0 0 0", "output": "2" }, { "input": "1 1 0 0 0 0 0 0 0 0 0 0 0 0", "output": "2" }, { "input": "1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "2" }, { "input": "1 0 0 0 0 0 0 0 0 0 0 0 0 1", "output": "2" }, { "input": "0 0 0 0 0 0 0 0 0 0 0 0 1 1", "output": "2" } ]

1,525,545,579

2,147,483,647

Python 3

OK

TESTS

26

92

7,065,600

def score(l): return sum(x*(x%2 == 0) for x in l) res = 0 ns = list(map(int, input().split())) for i in range(14): l = list(ns) for j in range(13): l[(i+1+j) % 14] += l[i]//14 + (1 if (j+1) <= l[i]%14 else 0) l[i] = l[i]//14 res = max(res, score(l)) print(res)

Title: Mancala Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes. Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction. Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole. After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli. Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move. Input Specification: The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole. It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board. Output Specification: Output one integer, the maximum possible score after one move. Demo Input: ['0 1 1 0 0 0 0 0 0 7 0 0 0 0\n', '5 1 1 1 1 0 0 0 0 0 0 0 0 0\n'] Demo Output: ['4\n', '8\n'] Note: In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$.

```python def score(l): return sum(x*(x%2 == 0) for x in l) res = 0 ns = list(map(int, input().split())) for i in range(14): l = list(ns) for j in range(13): l[(i+1+j) % 14] += l[i]//14 + (1 if (j+1) <= l[i]%14 else 0) l[i] = l[i]//14 res = max(res, score(l)) print(res) ```

3

92

A

Chips

PROGRAMMING

800

[ "implementation", "math" ]

A. Chips

2

256

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

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

Print the number of chips the presenter ended up with.

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

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

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

500

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

1,585,132,946

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

186

0

a=str(input()) b=list(map(int, a.split())) n=b[0] m=b[1] i=1 while i<=n: m=m-i i=i+1 if i>m: break print(m) print(m)

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

```python a=str(input()) b=list(map(int, a.split())) n=b[0] m=b[1] i=1 while i<=n: m=m-i i=i+1 if i>m: break print(m) print(m) ```

0

427

A

Police Recruits

PROGRAMMING

800

[ "implementation" ]

null

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

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

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

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

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

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

500

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

1,679,721,282

2,147,483,647

Python 3

OK

TESTS

63

77

5,017,600

input() g=0 for i in list(map(int,input().split()[::-1])): g=min(0,g+i) print(-g)

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

```python input() g=0 for i in list(map(int,input().split()[::-1])): g=min(0,g+i) print(-g) ```

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" } ]

1,614,438,392

2,147,483,647

Python 3

OK

TESTS

21

77

0

d1, d2, d3 = [int(x) for x in input().split()] w1 = 2*(d1+d2) w2=d1+d2+d3 w3=2*(d2+d3) w4=2*(d1+d3) print(min(w1,w2,w3,w4))

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 d1, d2, d3 = [int(x) for x in input().split()] w1 = 2*(d1+d2) w2=d1+d2+d3 w3=2*(d2+d3) w4=2*(d1+d3) print(min(w1,w2,w3,w4)) ```

3

432

A

Choosing Teams

PROGRAMMING

800

[ "greedy", "implementation", "sortings" ]

null

The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?

The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.

Print a single number — the answer to the problem.

[ "5 2\n0 4 5 1 0\n", "6 4\n0 1 2 3 4 5\n", "6 5\n0 0 0 0 0 0\n" ]

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

In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

500

[ { "input": "5 2\n0 4 5 1 0", "output": "1" }, { "input": "6 4\n0 1 2 3 4 5", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "3 4\n0 1 0", "output": "1" }, { "input": "3 4\n0 2 0", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1", "output": "3" }, { "input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4", "output": "4" }, { "input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1", "output": "3" }, { "input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0", "output": "5" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "6 3\n4 4 4 4 4 4", "output": "0" }, { "input": "1 4\n1", "output": "0" }, { "input": "1 1\n0", "output": "0" }, { "input": "2 3\n0 0", "output": "0" } ]

1,673,721,814

2,147,483,647

Python 3

OK

TESTS

35

46

0

x,y = map(int,input().split()) z = list(map(int,input().split())) n = 5-y c = 0 for i in z: if i <= n: c+=1 print(c//3)

Title: Choosing Teams Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? Input Specification: The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. Output Specification: Print a single number — the answer to the problem. Demo Input: ['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n'] Demo Output: ['1\n', '0\n', '2\n'] Note: In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

```python x,y = map(int,input().split()) z = list(map(int,input().split())) n = 5-y c = 0 for i in z: if i <= n: c+=1 print(c//3) ```

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" } ]

1,493,208,872

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

31

4,608,000

x = input() y = input() s1 = str(x) s2 = str(y) for i in range(0, s1.length): if s1[i]==s2[i]: print("1") else: print("0")

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 x = input() y = input() s1 = str(x) s2 = str(y) for i in range(0, s1.length): if s1[i]==s2[i]: print("1") else: print("0") ```

-1

177

A1

Good Matrix Elements

PROGRAMMING

800

[ "implementation" ]

null

The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good: - Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it. Help the Smart Beaver count the sum of good elements of the given matrix.

The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix. The input limitations for getting 30 points are: - 1<=≤<=*n*<=≤<=5 The input limitations for getting 100 points are: - 1<=≤<=*n*<=≤<=101

Print a single integer — the sum of good matrix elements.

[ "3\n1 2 3\n4 5 6\n7 8 9\n", "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n" ]

[ "45\n", "17\n" ]

In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.

30

[ { "input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "45" }, { "input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1", "output": "17" }, { "input": "1\n3", "output": "3" }, { "input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33", "output": "756" }, { "input": "3\n19 7 16\n12 15 5\n15 15 5", "output": "109" }, { "input": "3\n36 4 33\n11 46 32\n20 49 34", "output": "265" }, { "input": "3\n79 91 74\n33 82 22\n18 28 54", "output": "481" }, { "input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6", "output": "65" }, { "input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6", "output": "225" }, { "input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6", "output": "495" }, { "input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6", "output": "705" }, { "input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83", "output": "708" }, { "input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "0" }, { "input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0", "output": "0" }, { "input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0", "output": "5" }, { "input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0", "output": "3" }, { "input": "3\n0 0 0\n0 0 0\n0 0 0", "output": "0" } ]

1,544,418,140

2,147,483,647

Python 3

WRONG_ANSWER

TESTS1

2

216

0

n=int(input()) y=[] s=0 j=0 for i in range(n): x=list(map(int,input().split())) y.append(x) for i in range(n): s+=y[i][i] for i in range(n): s+=y[i][-i] for i in range(i): s+=y[n//2][i] s+=y[i][n//2] print(s-1)

Title: Good Matrix Elements Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good: - Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it. Help the Smart Beaver count the sum of good elements of the given matrix. Input Specification: The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix. The input limitations for getting 30 points are: - 1<=≤<=*n*<=≤<=5 The input limitations for getting 100 points are: - 1<=≤<=*n*<=≤<=101 Output Specification: Print a single integer — the sum of good matrix elements. Demo Input: ['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n'] Demo Output: ['45\n', '17\n'] Note: In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.

```python n=int(input()) y=[] s=0 j=0 for i in range(n): x=list(map(int,input().split())) y.append(x) for i in range(n): s+=y[i][i] for i in range(n): s+=y[i][-i] for i in range(i): s+=y[n//2][i] s+=y[i][n//2] print(s-1) ```

0

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" } ]

1,697,522,828

2,147,483,647

Python 3

OK

TESTS

67

404

8,601,600

l = [] T = int(input()) for i in range(0, T): a, b = map(int, input().split()) l.append((a, b)) d = {} for item in l: if item not in d: d[item] = 1 else: d[item] += 1 # print(d) mx = 1 for key, value in d.items(): # print(key, value) mx = max(mx, value) print(mx)

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 l = [] T = int(input()) for i in range(0, T): a, b = map(int, input().split()) l.append((a, b)) d = {} for item in l: if item not in d: d[item] = 1 else: d[item] += 1 # print(d) mx = 1 for key, value in d.items(): # print(key, value) mx = max(mx, value) print(mx) ```

3

200

B

Drinks

PROGRAMMING

800

[ "implementation", "math" ]

null

Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent. One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has. Find the volume fraction of orange juice in the final drink.

The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.

Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.

[ "3\n50 50 100\n", "4\n0 25 50 75\n" ]

[ "66.666666666667\n", "37.500000000000\n" ]

Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.

500

[ { "input": "3\n50 50 100", "output": "66.666666666667" }, { "input": "4\n0 25 50 75", "output": "37.500000000000" }, { "input": "3\n0 1 8", "output": "3.000000000000" }, { "input": "5\n96 89 93 95 70", "output": "88.600000000000" }, { "input": "7\n62 41 78 4 38 39 75", "output": "48.142857142857" }, { "input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22", "output": "11.615384615385" }, { "input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18", "output": "12.761904761905" }, { "input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84", "output": "69.538461538462" }, { "input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94", "output": "91.551724137931" }, { "input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100", "output": "99.515151515152" }, { "input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2", "output": "8.147058823529" }, { "input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93", "output": "91.921052631579" }, { "input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100", "output": "97.019230769231" }, { "input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6", "output": "50.965517241379" }, { "input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10", "output": "7.505882352941" }, { "input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6", "output": "10.418918918919" }, { "input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41", "output": "20.928571428571" }, { "input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96", "output": "49.191919191919" }, { "input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92", "output": "73.484848484848" }, { "input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6", "output": "7.640000000000" }, { "input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100", "output": "77.660000000000" }, { "input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81", "output": "91.480000000000" }, { "input": "1\n0", "output": "0.000000000000" }, { "input": "1\n100", "output": "100.000000000000" }, { "input": "1\n78", "output": "78.000000000000" }, { "input": "2\n0 100", "output": "50.000000000000" }, { "input": "2\n100 100", "output": "100.000000000000" }, { "input": "5\n0 0 0 0 1", "output": "0.200000000000" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99", "output": "99.990000000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1", "output": "0.010000000000" }, { "input": "5\n100 100 100 100 100", "output": "100.000000000000" } ]

1,686,982,773

2,147,483,647

Python 3

OK

TESTS

31

92

0

n = int(input()) s = input().split() c = 0 for j in range(n): c+=(int(s[j])/100) print ((c/n)*100)

Title: Drinks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent. One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has. Find the volume fraction of orange juice in the final drink. Input Specification: The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. Output Specification: Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. Demo Input: ['3\n50 50 100\n', '4\n0 25 50 75\n'] Demo Output: ['66.666666666667\n', '37.500000000000\n'] Note: Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.

```python n = int(input()) s = input().split() c = 0 for j in range(n): c+=(int(s[j])/100) print ((c/n)*100) ```

3

959

A

Mahmoud and Ehab and the even-odd game

PROGRAMMING

800

[ "games", "math" ]

null

Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that: - 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd. If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?

The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.

Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.

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

[ "Ehab", "Mahmoud" ]

In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins. In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

500

[ { "input": "1", "output": "Ehab" }, { "input": "2", "output": "Mahmoud" }, { "input": "10000", "output": "Mahmoud" }, { "input": "33333", "output": "Ehab" }, { "input": "5", "output": "Ehab" }, { "input": "1000000000", "output": "Mahmoud" }, { "input": "999999999", "output": "Ehab" }, { "input": "123123123", "output": "Ehab" }, { "input": "22222221", "output": "Ehab" }, { "input": "22222220", "output": "Mahmoud" }, { "input": "3", "output": "Ehab" }, { "input": "4", "output": "Mahmoud" }, { "input": "6", "output": "Mahmoud" }, { "input": "7", "output": "Ehab" }, { "input": "8", "output": "Mahmoud" }, { "input": "9", "output": "Ehab" }, { "input": "10", "output": "Mahmoud" }, { "input": "536870912", "output": "Mahmoud" }, { "input": "536870913", "output": "Ehab" }, { "input": "536870911", "output": "Ehab" } ]

1,677,521,382

2,147,483,647

PyPy 3-64

OK

TESTS

20

46

0

# k=int(input()) # for i in range(k): e=int(input()) if e%2!=0 or e==0 : print ("Ehab") else: print("Mahmoud")

Title: Mahmoud and Ehab and the even-odd game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that: - 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd. If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally? Input Specification: The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game. Output Specification: Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise. Demo Input: ['1\n', '2\n'] Demo Output: ['Ehab', 'Mahmoud'] Note: In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins. In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

```python # k=int(input()) # for i in range(k): e=int(input()) if e%2!=0 or e==0 : print ("Ehab") else: print("Mahmoud") ```

3

990

B

Micro-World

PROGRAMMING

1,200

[ "greedy", "sortings" ]

null

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them. You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$. The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another. For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish. Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$. The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.

Print the only integer — minimal possible number of bacteria can remain.

[ "7 1\n101 53 42 102 101 55 54\n", "6 5\n20 15 10 15 20 25\n", "7 1000000\n1 1 1 1 1 1 1\n" ]

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

The first example is clarified in the problem statement. In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$. In the third example no bacteria can swallow any other bacteria.

0

[ { "input": "7 1\n101 53 42 102 101 55 54", "output": "3" }, { "input": "6 5\n20 15 10 15 20 25", "output": "1" }, { "input": "7 1000000\n1 1 1 1 1 1 1", "output": "7" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 4\n8", "output": "1" }, { "input": "10 1\n1 2 3 5 6 8 10 11 9 4", "output": "2" }, { "input": "9 2\n1 6 1 5 5 8 6 8 7", "output": "4" }, { "input": "15 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15" }, { "input": "2 1000000\n1 1000000", "output": "1" }, { "input": "7 2\n1 5 5 8 9 8 8", "output": "4" }, { "input": "10 1\n2 6 3 4 2 4 4 3 2 1", "output": "4" }, { "input": "4 1\n2 2 1 1", "output": "2" }, { "input": "10 1\n6 3 1 3 6 4 1 3 6 4", "output": "7" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "8 2\n3 13 9 8 3 13 9 14", "output": "5" }, { "input": "8 1000000\n1 1 5 1000000 1000000 2 2 2", "output": "2" }, { "input": "2 1\n999152 999153", "output": "1" } ]

1,528,818,226

2,147,483,647

Python 3

OK

TESTS

36

373

14,233,600

n, k = [int(a) for a in input().split()] s = sorted([int(a) for a in input().split()], reverse=1) ans = 1 q = s[0] for i in range(1, n): if not(abs(s[i] - s[i - 1]) <= k and s[i] < q): ans += 1 q = s[i] print(ans)

Title: Micro-World Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them. You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$. The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another. For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish. Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope. Input Specification: The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$. The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have. Output Specification: Print the only integer — minimal possible number of bacteria can remain. Demo Input: ['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n'] Demo Output: ['3\n', '1\n', '7\n'] Note: The first example is clarified in the problem statement. In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$. In the third example no bacteria can swallow any other bacteria.

```python n, k = [int(a) for a in input().split()] s = sorted([int(a) for a in input().split()], reverse=1) ans = 1 q = s[0] for i in range(1, n): if not(abs(s[i] - s[i - 1]) <= k and s[i] < q): ans += 1 q = s[i] print(ans) ```

3

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" } ]

1,654,989,454

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

6

2,000

9,318,400

from sys import stdin n = int(input()) aaa = input() aaa = aaa.split(' ') aaa = [int(i) for i in aaa] m = int(input()) bbb = input() bbb = bbb.split(' ') bbb = [int(i) for i in bbb] # ^^ got all input def count(i, listt): # counts distance from right/left for a given int for a given list left = listt.index(i) left += 1 right = n - left + 1 return left, right l = 0 r = 0 # total distance from right/left for i in range(0, m): ll, rr = count(bbb[i], aaa) l += ll r += rr print(l, r)

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 from sys import stdin n = int(input()) aaa = input() aaa = aaa.split(' ') aaa = [int(i) for i in aaa] m = int(input()) bbb = input() bbb = bbb.split(' ') bbb = [int(i) for i in bbb] # ^^ got all input def count(i, listt): # counts distance from right/left for a given int for a given list left = listt.index(i) left += 1 right = n - left + 1 return left, right l = 0 r = 0 # total distance from right/left for i in range(0, m): ll, rr = count(bbb[i], aaa) l += ll r += rr print(l, r) ```

0

18

C

Stripe

PROGRAMMING

1,200

[ "data structures", "implementation" ]

C. Stripe

2

64

Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

[ "9\n1 5 -6 7 9 -16 0 -2 2\n", "3\n1 1 1\n", "2\n0 0\n" ]

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

none

0

[ { "input": "9\n1 5 -6 7 9 -16 0 -2 2", "output": "3" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "2\n0 0", "output": "1" }, { "input": "4\n100 1 10 111", "output": "1" }, { "input": "10\n0 4 -3 0 -2 2 -3 -3 2 5", "output": "3" }, { "input": "10\n0 -1 2 2 -1 1 0 0 0 2", "output": "0" }, { "input": "10\n-1 -1 1 -1 0 1 0 1 1 1", "output": "1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "9" }, { "input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3", "output": "3" }, { "input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3", "output": "0" }, { "input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2", "output": "0" }, { "input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2", "output": "1" }, { "input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2", "output": "6" }, { "input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "1\n10000", "output": "0" }, { "input": "1\n-10000", "output": "0" } ]

1,528,064,580

2,147,483,647

Python 3

OK

TESTS

45

310

14,438,400

def main(): n = int(input()) arr = list(map(int, input().split())) right = sum(arr) left = 0 result = 0 for i in range(0, n - 1): left += arr[i] right -= arr[i] if left == right: result += 1 print(result) if __name__ == "__main__": main()

Title: Stripe Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. Output Specification: Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. Demo Input: ['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: none

```python def main(): n = int(input()) arr = list(map(int, input().split())) right = sum(arr) left = 0 result = 0 for i in range(0, n - 1): left += arr[i] right -= arr[i] if left == right: result += 1 print(result) if __name__ == "__main__": main() ```

3.814926

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" } ]

1,677,529,194

2,147,483,647

Python 3

OK

TESTS

20

46

0

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

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

3.977

965

C

Greedy Arkady

PROGRAMMING

2,000

[ "math" ]

null

$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away. The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away. Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting. Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.

The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.

Print a single integer — the maximum possible number of candies Arkady can give to himself. Note that it is always possible to choose some valid $x$.

[ "20 4 5 2\n", "30 9 4 1\n" ]

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

In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total. Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times. In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.

1,500

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1,524,687,176

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

77

21,401,600

n, k, M, D = map(int, input().split()) x = [0] t = [0] x[0] = n // (k * D + 1) + 1 xm = M t[0] = (n + k * x[0]) // (k * x[0] + 1) tm = (n + k * xm) // (k * xm + 1) i = 0 s = 0 l = [x[0] * t[0]] while s < xm: s = (n - (t[i] - 1) * (k * x[i] + 1) + k * (t[i] - 1) - 1) // k * (t[i] - 1) + x[i] x.append(s) t.append(t[i] + 1) l.append((x[i + 1] - 1) * t[i]) i += 1 if len(l) == 1: print(max(l)) else: l.pop() print(max(l))

Title: Greedy Arkady Time Limit: None seconds Memory Limit: None megabytes Problem Description: $k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away. The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away. Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting. Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$. Input Specification: The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies. Output Specification: Print a single integer — the maximum possible number of candies Arkady can give to himself. Note that it is always possible to choose some valid $x$. Demo Input: ['20 4 5 2\n', '30 9 4 1\n'] Demo Output: ['8\n', '4\n'] Note: In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total. Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times. In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.

```python n, k, M, D = map(int, input().split()) x = [0] t = [0] x[0] = n // (k * D + 1) + 1 xm = M t[0] = (n + k * x[0]) // (k * x[0] + 1) tm = (n + k * xm) // (k * xm + 1) i = 0 s = 0 l = [x[0] * t[0]] while s < xm: s = (n - (t[i] - 1) * (k * x[i] + 1) + k * (t[i] - 1) - 1) // k * (t[i] - 1) + x[i] x.append(s) t.append(t[i] + 1) l.append((x[i + 1] - 1) * t[i]) i += 1 if len(l) == 1: print(max(l)) else: l.pop() print(max(l)) ```

0

248

B

Chilly Willy

PROGRAMMING

1,400

[ "math", "number theory" ]

null

Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros.

A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105).

Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.

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

[ "-1\n", "10080" ]

none

1,000

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1,578,129,578

2,147,483,647

PyPy 3

OK

TESTS

66

280

512,000

n = int(input()) if n < 3: print(-1) elif n == 3: print(210) else: suff = "" if n % 6 == 4: suff = "50" elif n % 6 == 5: suff = "80" elif n % 6 == 0: suff = "170" elif n % 6 == 1: suff = "20" elif n % 6 == 2: suff = "200" elif n % 6 == 3: suff = "110" main = n - 1 - len(suff) print("1" + "0" * main + suff)

Title: Chilly Willy Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros. Input Specification: A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105). Output Specification: Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist. Demo Input: ['1\n', '5\n'] Demo Output: ['-1\n', '10080'] Note: none

```python n = int(input()) if n < 3: print(-1) elif n == 3: print(210) else: suff = "" if n % 6 == 4: suff = "50" elif n % 6 == 5: suff = "80" elif n % 6 == 0: suff = "170" elif n % 6 == 1: suff = "20" elif n % 6 == 2: suff = "200" elif n % 6 == 3: suff = "110" main = n - 1 - len(suff) print("1" + "0" * main + suff) ```

3

55

B

Smallest number

PROGRAMMING

1,600

[ "brute force" ]

B. Smallest number

2

256

Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.

First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)

Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

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

[ "3\n", "8\n", "9\n" ]

none

1,000

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1,638,214,266

2,147,483,647

Python 3

OK

TESTS

60

124

0

class Solution: def find_Smallest(self,a,s): global smallest if len(a)==1: if smallest > a[0]: smallest = a[0] return for i in range(len(a)): for j in range(i+1,len(a)): tmp = [0] for k in range(len(a)): if k!=i and k!=j: tmp.append(a[k]) tmp[0] = a[i]*a[j] if s[0]=="*" else a[i]+a[j] self.find_Smallest(tmp,s[1:]) if __name__=="__main__": Sol = Solution() num = list(map(int,input().split())) signs = list(input().split()) smallest = float('inf') Sol.find_Smallest(num,signs) print(smallest)

Title: Smallest number Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations. Input Specification: First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication) Output Specification: Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d). Demo Input: ['1 1 1 1\n+ + *\n', '2 2 2 2\n* * +\n', '1 2 3 4\n* + +\n'] Demo Output: ['3\n', '8\n', '9\n'] Note: none

```python class Solution: def find_Smallest(self,a,s): global smallest if len(a)==1: if smallest > a[0]: smallest = a[0] return for i in range(len(a)): for j in range(i+1,len(a)): tmp = [0] for k in range(len(a)): if k!=i and k!=j: tmp.append(a[k]) tmp[0] = a[i]*a[j] if s[0]=="*" else a[i]+a[j] self.find_Smallest(tmp,s[1:]) if __name__=="__main__": Sol = Solution() num = list(map(int,input().split())) signs = list(input().split()) smallest = float('inf') Sol.find_Smallest(num,signs) print(smallest) ```

3.969

264

A

Escape from Stones

PROGRAMMING

1,200

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

null

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

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

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

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

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

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

500

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

1,500,825,529

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

30

2,000

10,854,400

s = input() a = len(s) stones = [] n = 0 for x in range(a): stones.insert(n, x + 1) if s[x] == "r": n += 1 for x in stones: print(x)

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

```python s = input() a = len(s) stones = [] n = 0 for x in range(a): stones.insert(n, x + 1) if s[x] == "r": n += 1 for x in stones: print(x) ```

0

25

A

IQ test

PROGRAMMING

1,300

[ "brute force" ]

A. IQ test

2

256

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

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

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

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

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

none

0

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

1,513,659,505

2,147,483,647

Python 3

OK

TESTS

32

124

5,734,400

n=int(input()) line=[int(i) for i in input().split()] d=0 s=0 for i in range(n): if line[i]%2==0: s+=1 else: d+=1 if s>d: a=1 if d>s: a=0 for i in range(n): if line[i]%2==a: x=i print(x+1)

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

```python n=int(input()) line=[int(i) for i in input().split()] d=0 s=0 for i in range(n): if line[i]%2==0: s+=1 else: d+=1 if s>d: a=1 if d>s: a=0 for i in range(n): if line[i]%2==a: x=i print(x+1) ```

3.958319

427

B

Prison Transfer

PROGRAMMING

1,100

[ "data structures", "implementation" ]

null

The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city. For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was. Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are, - The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer. Find the number of ways you can choose the *c* prisoners.

The first line of input will contain three space separated integers *n* (1<=≤<=*n*<=≤<=2·105), *t* (0<=≤<=*t*<=≤<=109) and *c* (1<=≤<=*c*<=≤<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.

Print a single integer — the number of ways you can choose the *c* prisoners.

[ "4 3 3\n2 3 1 1\n", "1 1 1\n2\n", "11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n" ]

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

none

1,000

[ { "input": "4 3 3\n2 3 1 1", "output": "2" }, { "input": "1 1 1\n2", "output": "0" }, { "input": "11 4 2\n2 2 0 7 3 2 2 4 9 1 4", "output": "6" }, { "input": "57 2 10\n7 5 2 7 4 1 0 5 2 9 2 9 8 6 6 5 9 6 8 1 0 1 0 3 2 6 5 2 8 8 8 8 0 9 4 3 6 6 2 4 5 1 2 0 1 7 1 1 5 4 5 0 7 5 1 9 6", "output": "0" }, { "input": "2 228885628 1\n90897004 258427916", "output": "1" }, { "input": "3 1 1\n1 2 1", "output": "2" }, { "input": "3 3 3\n3 2 3", "output": "1" }, { "input": "4 2 2\n1 3 3 2", "output": "0" }, { "input": "1 228 1\n1", "output": "1" } ]

1,696,669,556

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

n,t,c = list(map(int, input().split())) crimeLevels = list(map(int, input().split())) ans=[] b,d=0 for i in range(n): if crimeLevels[i]<=t: b+=1 else: b=0 if b>=c: d+=1 # supLst=[] # for j in range(i,i+c): # if j>=n: # break # if crimeLevels[j]<=t: # supLst.append(crimeLevels[j]) # if len(supLst)==c: # ans.append(supLst) # print(len(ans)) print(d)

Title: Prison Transfer Time Limit: None seconds Memory Limit: None megabytes Problem Description: The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city. For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was. Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are, - The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer. Find the number of ways you can choose the *c* prisoners. Input Specification: The first line of input will contain three space separated integers *n* (1<=≤<=*n*<=≤<=2·105), *t* (0<=≤<=*t*<=≤<=109) and *c* (1<=≤<=*c*<=≤<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109. Output Specification: Print a single integer — the number of ways you can choose the *c* prisoners. Demo Input: ['4 3 3\n2 3 1 1\n', '1 1 1\n2\n', '11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n'] Demo Output: ['2\n', '0\n', '6\n'] Note: none

```python n,t,c = list(map(int, input().split())) crimeLevels = list(map(int, input().split())) ans=[] b,d=0 for i in range(n): if crimeLevels[i]<=t: b+=1 else: b=0 if b>=c: d+=1 # supLst=[] # for j in range(i,i+c): # if j>=n: # break # if crimeLevels[j]<=t: # supLst.append(crimeLevels[j]) # if len(supLst)==c: # ans.append(supLst) # print(len(ans)) print(d) ```

-1

697

A

Pineapple Incident

PROGRAMMING

900

[ "implementation", "math" ]

null

Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc. Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.

The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.

Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.

[ "3 10 4\n", "3 10 3\n", "3 8 51\n", "3 8 52\n" ]

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

In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3. In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

500

[ { "input": "3 10 4", "output": "NO" }, { "input": "3 10 3", "output": "YES" }, { "input": "3 8 51", "output": "YES" }, { "input": "3 8 52", "output": "YES" }, { "input": "456947336 740144 45", "output": "NO" }, { "input": "33 232603 599417964", "output": "YES" }, { "input": "4363010 696782227 701145238", "output": "YES" }, { "input": "9295078 2 6", "output": "NO" }, { "input": "76079 281367 119938421", "output": "YES" }, { "input": "93647 7 451664565", "output": "YES" }, { "input": "5 18553 10908", "output": "NO" }, { "input": "6 52 30", "output": "NO" }, { "input": "6431 855039 352662", "output": "NO" }, { "input": "749399100 103031711 761562532", "output": "NO" }, { "input": "21 65767 55245", "output": "NO" }, { "input": "4796601 66897 4860613", "output": "NO" }, { "input": "8 6728951 860676", "output": "NO" }, { "input": "914016 6 914019", "output": "NO" }, { "input": "60686899 78474 60704617", "output": "NO" }, { "input": "3 743604 201724", "output": "NO" }, { "input": "571128 973448796 10", "output": "NO" }, { "input": "688051712 67 51", "output": "NO" }, { "input": "74619 213344 6432326", "output": "NO" }, { "input": "6947541 698167 6", "output": "NO" }, { "input": "83 6 6772861", "output": "NO" }, { "input": "251132 67561 135026988", "output": "NO" }, { "input": "8897216 734348516 743245732", "output": "YES" }, { "input": "50 64536 153660266", "output": "YES" }, { "input": "876884 55420 971613604", "output": "YES" }, { "input": "0 6906451 366041903", "output": "YES" }, { "input": "11750 8 446010134", "output": "YES" }, { "input": "582692707 66997 925047377", "output": "YES" }, { "input": "11 957526890 957526901", "output": "YES" }, { "input": "556888 514614196 515171084", "output": "YES" }, { "input": "6 328006 584834704", "output": "YES" }, { "input": "4567998 4 204966403", "output": "YES" }, { "input": "60 317278 109460971", "output": "YES" }, { "input": "906385 342131991 685170368", "output": "YES" }, { "input": "1 38 902410512", "output": "YES" }, { "input": "29318 787017 587931018", "output": "YES" }, { "input": "351416375 243431 368213115", "output": "YES" }, { "input": "54 197366062 197366117", "output": "YES" }, { "input": "586389 79039 850729874", "output": "YES" }, { "input": "723634470 2814619 940360134", "output": "YES" }, { "input": "0 2 0", "output": "YES" }, { "input": "0 2 1", "output": "NO" }, { "input": "0 2 2", "output": "YES" }, { "input": "0 2 3", "output": "YES" }, { "input": "0 2 1000000000", "output": "YES" }, { "input": "0 10 23", "output": "NO" }, { "input": "0 2 999999999", "output": "YES" }, { "input": "10 5 11", "output": "NO" }, { "input": "1 2 1000000000", "output": "YES" }, { "input": "1 10 20", "output": "NO" }, { "input": "1 2 999999937", "output": "YES" }, { "input": "10 3 5", "output": "NO" }, { "input": "3 2 5", "output": "YES" }, { "input": "0 4 0", "output": "YES" }, { "input": "0 215 403", "output": "NO" }, { "input": "5 2 10", "output": "YES" }, { "input": "0 2 900000000", "output": "YES" }, { "input": "0 79 4000", "output": "NO" }, { "input": "5 1000 1000", "output": "NO" }, { "input": "1 5 103", "output": "NO" }, { "input": "5 2 6", "output": "NO" }, { "input": "120 2 1000000000", "output": "YES" }, { "input": "2 2 1000000000", "output": "YES" }, { "input": "5 5 13", "output": "NO" }, { "input": "10 5 15", "output": "YES" }, { "input": "11 2 0", "output": "NO" }, { "input": "3 8 53", "output": "NO" }, { "input": "2 2 4", "output": "YES" }, { "input": "4 4 0", "output": "NO" }, { "input": "1 2 3", "output": "YES" }, { "input": "5 3 9", "output": "YES" }, { "input": "5 6 19", "output": "NO" }, { "input": "3 10 125", "output": "NO" }, { "input": "5 3 8", "output": "YES" }, { "input": "6 3 9", "output": "YES" }, { "input": "0 3 5", "output": "NO" }, { "input": "5 3 300000035", "output": "YES" }, { "input": "5 2 7", "output": "YES" }, { "input": "1 5 6", "output": "YES" }, { "input": "4 2 6", "output": "YES" }, { "input": "0 3 999999998", "output": "NO" }, { "input": "0 10001 0", "output": "YES" }, { "input": "6 5 3", "output": "NO" }, { "input": "1 5 1000000000", "output": "NO" }, { "input": "1 3 6", "output": "NO" }, { "input": "3 3 1000000000", "output": "YES" }, { "input": "3 3 4", "output": "NO" }, { "input": "3 3 5", "output": "NO" }, { "input": "3 3 0", "output": "NO" }, { "input": "1 2 4", "output": "YES" }, { "input": "5 5 10", "output": "YES" } ]

1,581,848,955

2,147,483,647

PyPy 3

OK

TESTS

95

156

0

t, s, x = map(int, input().split()) c1 = x - t - 1 - s c2 = x - t if (c1%s==0 and c1 >= 0) or (c2%s==0 and c2 >= 0): print('YES') else: print('NO')

Title: Pineapple Incident Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc. Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time. Input Specification: The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively. Output Specification: Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output. Demo Input: ['3 10 4\n', '3 10 3\n', '3 8 51\n', '3 8 52\n'] Demo Output: ['NO\n', 'YES\n', 'YES\n', 'YES\n'] Note: In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3. In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

```python t, s, x = map(int, input().split()) c1 = x - t - 1 - s c2 = x - t if (c1%s==0 and c1 >= 0) or (c2%s==0 and c2 >= 0): print('YES') else: print('NO') ```

3

940

A

Points on the line

PROGRAMMING

1,200

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

null

We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round. The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2. Diameter of multiset consisting of one point is 0. You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*?

The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively. The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points.

Output a single integer — the minimum number of points you have to remove.

[ "3 1\n2 1 4\n", "3 0\n7 7 7\n", "6 3\n1 3 4 6 9 10\n" ]

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

In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1. In the second test case the diameter is equal to 0, so its is unnecessary to remove any points. In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.

500

[ { "input": "3 1\n2 1 4", "output": "1" }, { "input": "3 0\n7 7 7", "output": "0" }, { "input": "6 3\n1 3 4 6 9 10", "output": "3" }, { "input": "11 5\n10 11 12 13 14 15 16 17 18 19 20", "output": "5" }, { "input": "1 100\n1", "output": "0" }, { "input": "100 10\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "84" }, { "input": "100 70\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "27" }, { "input": "1 10\n25", "output": "0" }, { "input": "70 80\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70", "output": "0" }, { "input": "3 1\n25 26 27", "output": "1" }, { "input": "100 5\n51 56 52 60 52 53 52 60 56 54 55 50 53 51 57 53 52 54 54 52 51 55 50 56 60 51 58 50 60 59 50 54 60 55 55 57 54 59 59 55 55 52 56 57 59 54 53 57 52 50 50 55 59 54 54 56 51 58 52 51 56 56 58 56 54 54 57 52 51 58 56 57 54 59 58 53 50 52 50 60 57 51 54 59 54 54 52 55 53 55 51 53 52 54 51 56 55 53 58 56", "output": "34" }, { "input": "100 11\n44 89 57 64 94 96 73 96 55 52 91 73 73 93 51 62 63 85 43 75 60 78 98 55 80 84 65 75 61 88 62 71 53 57 94 85 60 96 66 96 61 72 97 64 51 44 63 82 67 86 60 57 74 85 57 79 61 94 86 78 84 56 60 75 91 91 92 62 89 85 79 57 76 97 65 56 46 78 51 69 50 52 85 80 76 71 81 51 90 71 77 60 63 62 84 59 79 84 69 81", "output": "70" }, { "input": "100 0\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "96" }, { "input": "100 100\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "0" }, { "input": "76 32\n50 53 69 58 55 39 40 42 40 55 58 73 55 72 75 44 45 55 46 60 60 42 41 64 77 39 68 51 61 49 38 41 56 57 64 43 78 36 39 63 40 66 52 76 39 68 39 73 40 68 54 60 35 67 69 52 58 52 38 63 69 38 69 60 73 64 65 41 59 55 37 57 40 34 35 35", "output": "13" }, { "input": "100 1\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "93" }, { "input": "100 5\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100", "output": "89" }, { "input": "98 64\n2 29 36 55 58 15 25 33 7 16 61 1 4 24 63 26 36 16 16 3 57 39 56 7 11 24 20 12 22 10 56 5 11 39 61 52 27 54 21 6 61 36 40 52 54 5 15 52 58 23 45 39 65 16 27 40 13 64 47 24 51 29 9 18 49 49 8 47 2 64 7 63 49 10 20 26 34 3 45 66 8 46 16 32 16 38 3 6 15 17 35 48 36 5 57 29 61 15", "output": "1" }, { "input": "100 56\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100", "output": "43" }, { "input": "100 0\n14 13 14 13 14 13 13 13 13 14 13 13 14 14 13 14 14 14 14 13 13 13 14 13 13 14 14 14 14 14 14 13 13 13 13 14 13 14 13 14 13 14 14 14 14 13 13 14 14 13 13 13 13 14 13 14 13 14 13 14 13 13 13 14 13 13 14 13 14 14 13 13 13 14 14 14 14 13 13 14 14 14 14 14 14 14 13 14 13 13 13 14 14 13 13 13 13 13 14 14", "output": "50" }, { "input": "100 0\n14 17 18 22 19 18 19 21 19 19 22 22 19 21 24 23 24 19 25 24 24 21 20 13 26 18 17 15 25 13 17 20 20 21 13 22 27 15 18 27 19 15 16 25 18 17 18 22 19 17 18 24 14 16 18 16 22 16 17 27 18 17 18 24 22 13 14 20 23 19 16 21 19 13 14 14 25 15 27 24 26 22 16 20 16 14 21 27 15 23 23 24 27 14 24 17 19 24 15 27", "output": "89" }, { "input": "100 100\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": "0" }, { "input": "1 100\n22", "output": "0" }, { "input": "1 0\n22", "output": "0" }, { "input": "1 99\n99", "output": "0" }, { "input": "1 5\n6", "output": "0" }, { "input": "3 1\n10 20 30", "output": "2" }, { "input": "3 0\n1 2 3", "output": "2" }, { "input": "3 2\n1 50 99", "output": "2" }, { "input": "7 4\n1 3 4 9 10 11 12", "output": "3" }, { "input": "2 5\n67 23", "output": "1" }, { "input": "4 2\n1 4 7 9", "output": "2" }, { "input": "2 0\n1 2", "output": "1" }, { "input": "8 1\n3 3 3 5 5 5 5 5", "output": "3" }, { "input": "5 1\n3 5 5 5 6", "output": "1" } ]

1,519,763,172

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

46

5,632,000

n, d = map(int, input().split(" ")) A = list(map(int, input().split(" "))) lo, ans, cur, idx = A[0], 0, 1, 0 A.sort() for i in range(1, n): if (A[i] - lo) <= d: cur = cur + 1 else: idx = idx + 1 lo = A[idx] cur = cur - 1 i = i - 1 ans = max(ans, cur) print(n - ans)

Title: Points on the line Time Limit: None seconds Memory Limit: None megabytes Problem Description: We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round. The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2. Diameter of multiset consisting of one point is 0. You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*? Input Specification: The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively. The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points. Output Specification: Output a single integer — the minimum number of points you have to remove. Demo Input: ['3 1\n2 1 4\n', '3 0\n7 7 7\n', '6 3\n1 3 4 6 9 10\n'] Demo Output: ['1\n', '0\n', '3\n'] Note: In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1. In the second test case the diameter is equal to 0, so its is unnecessary to remove any points. In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.

```python n, d = map(int, input().split(" ")) A = list(map(int, input().split(" "))) lo, ans, cur, idx = A[0], 0, 1, 0 A.sort() for i in range(1, n): if (A[i] - lo) <= d: cur = cur + 1 else: idx = idx + 1 lo = A[idx] cur = cur - 1 i = i - 1 ans = max(ans, cur) print(n - ans) ```

0

743

B

Chloe and the sequence

PROGRAMMING

1,200

[ "binary search", "bitmasks", "constructive algorithms", "implementation" ]

null

Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem!

The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).

Print single integer — the integer at the *k*-th position in the obtained sequence.

[ "3 2\n", "4 8\n" ]

[ "2", "4" ]

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

1,000

[ { "input": "3 2", "output": "2" }, { "input": "4 8", "output": "4" }, { "input": "5 27", "output": "1" }, { "input": "7 44", "output": "3" }, { "input": "15 18432", "output": "12" }, { "input": "20 259676", "output": "3" }, { "input": "30 671088640", "output": "28" }, { "input": "38 137438953472", "output": "38" }, { "input": "1 1", "output": "1" }, { "input": "3 3", "output": "1" }, { "input": "4 12", "output": "3" }, { "input": "8 224", "output": "6" }, { "input": "13 1368", "output": "4" }, { "input": "16 49152", "output": "15" }, { "input": "19 1024", "output": "11" }, { "input": "24 15204352", "output": "20" }, { "input": "27 6586544", "output": "5" }, { "input": "31 536870912", "output": "30" }, { "input": "38 94489280512", "output": "34" }, { "input": "42 215268308020", "output": "3" }, { "input": "42 3452074721280", "output": "21" }, { "input": "46 34394312982528", "output": "15" }, { "input": "48 133865540681728", "output": "39" }, { "input": "49 76541041985542", "output": "2" }, { "input": "49 104822971826176", "output": "34" }, { "input": "49 351843720888320", "output": "47" }, { "input": "50 743042492421629", "output": "1" }, { "input": "50 666266740436818", "output": "2" }, { "input": "50 704239287953456", "output": "5" }, { "input": "50 116938486513664", "output": "17" }, { "input": "50 806423059496960", "output": "37" }, { "input": "50 985162418487296", "output": "48" }, { "input": "50 844424930131968", "output": "49" }, { "input": "50 562949953421312", "output": "50" }, { "input": "50 1", "output": "1" }, { "input": "50 1125899906842623", "output": "1" }, { "input": "50 1125899906842620", "output": "3" }, { "input": "39 549755813887", "output": "1" }, { "input": "50 100000000000", "output": "12" } ]

1,670,259,542

2,147,483,647

PyPy 3-64

OK

TESTS

39

62

0

def main(): n, k = get_list_int() if k % 2 == 1: print(1) return first_occurance_idx = [0] for i in range(n): step_size = 2**i prev = first_occurance_idx[-1] first_occurance_idx.append(prev + step_size) for i in range(len(first_occurance_idx)): start_idx = first_occurance_idx[i] repeat_freq = 2 ** (i + 1) if (k - 1 - start_idx) % repeat_freq == 0: print(i + 1) def get_int() -> int: return int(input()) def get_list_int() -> list[int]: return [int(x) for x in input().split(" ")] def get_list_str() -> list[str]: return input().split(" ") if __name__ == "__main__": main()

Title: Chloe and the sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem! Input Specification: The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). Output Specification: Print single integer — the integer at the *k*-th position in the obtained sequence. Demo Input: ['3 2\n', '4 8\n'] Demo Output: ['2', '4'] Note: In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

```python def main(): n, k = get_list_int() if k % 2 == 1: print(1) return first_occurance_idx = [0] for i in range(n): step_size = 2**i prev = first_occurance_idx[-1] first_occurance_idx.append(prev + step_size) for i in range(len(first_occurance_idx)): start_idx = first_occurance_idx[i] repeat_freq = 2 ** (i + 1) if (k - 1 - start_idx) % repeat_freq == 0: print(i + 1) def get_int() -> int: return int(input()) def get_list_int() -> list[int]: return [int(x) for x in input().split(" ")] def get_list_str() -> list[str]: return input().split(" ") if __name__ == "__main__": main() ```

3

501

B

Misha and Changing Handles

PROGRAMMING

1,100

[ "data structures", "dsu", "strings" ]

null

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point. Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests. Next *q* lines contain the descriptions of the requests, one per line. Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20. The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.

In the first line output the integer *n* — the number of users that changed their handles at least once. In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order. Each user who changes the handle must occur exactly once in this description.

[ "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n" ]

[ "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n" ]

none

500

[ { "input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov", "output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123" }, { "input": "1\nMisha Vasya", "output": "1\nMisha Vasya" }, { "input": "10\na b\nb c\nc d\nd e\ne f\nf g\ng h\nh i\ni j\nj k", "output": "1\na k" }, { "input": "5\n123abc abc123\nabc123 a1b2c3\na1b2c3 1A2B3C\n1 2\n2 Misha", "output": "2\n123abc 1A2B3C\n1 Misha" }, { "input": "8\nM F\nS D\n1 2\nF G\n2 R\nD Q\nQ W\nW e", "output": "3\nM G\n1 R\nS e" }, { "input": "17\nn5WhQ VCczxtxKwFio5U\nVCczxtxKwFio5U 1WMVGA17cd1LRcp4r\n1WMVGA17cd1LRcp4r SJl\nSJl D8bPUoIft5v1\nNAvvUgunbPZNCL9ZY2 jnLkarKYsotz\nD8bPUoIft5v1 DnDkHi7\njnLkarKYsotz GfjX109HSQ81gFEBJc\nGfjX109HSQ81gFEBJc kBJ0zrH78mveJ\nkBJ0zrH78mveJ 9DrAypYW\nDnDkHi7 3Wkho2PglMDaFQw\n3Wkho2PglMDaFQw pOqW\n9DrAypYW G3y0cXXGsWAh\npOqW yr1Ec\nG3y0cXXGsWAh HrmWWg5u4Hsy\nyr1Ec GkFeivXjQ01\nGkFeivXjQ01 mSsWgbCCZcotV4goiA\nHrmWWg5u4Hsy zkCmEV", "output": "2\nn5WhQ mSsWgbCCZcotV4goiA\nNAvvUgunbPZNCL9ZY2 zkCmEV" }, { "input": "10\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9\nSEj 2knOMLyzr\n0v69ijnAc S7d7zGTjmlku01Gv\n2knOMLyzr otGmEd\nacwr3TfMV7oCIp RUSVFa9TIWlLsd7SB\nS7d7zGTjmlku01Gv Gd6ZufVmQnBpi\nS1 WOJLpk\nWOJLpk Gu\nRUSVFa9TIWlLsd7SB RFawatGnbVB\notGmEd OTB1zKiOI", "output": "5\n0v69ijnAc Gd6ZufVmQnBpi\nS1 Gu\nSEj OTB1zKiOI\nacwr3TfMV7oCIp RFawatGnbVB\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9" }, { "input": "14\nTPdoztSZROpjZe z6F8bYFvnER4V5SP0n\n8Aa3PQY3hzHZTPEUz fhrZZPJ3iUS\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nAO s1VGWTCbHzM\ni 4F\nfhrZZPJ3iUS j0OVZQF6MvNcKN9xDZFJ\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\nj0OVZQF6MvNcKN9xDZFJ DzjmeNqN0H4Teq0Awr\n4F wJcdxt1kwqfDeJ\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nJ0Usg DKdKMFJ6tK8XA\nz6F8bYFvnER4V5SP0n 0alJ\nMijh2O6 qic8kXWuR6", "output": "10\nTPdoztSZROpjZe 0alJ\nJ0Usg DKdKMFJ6tK8XA\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\n8Aa3PQY3hzHZTPEUz DzjmeNqN0H4Teq0Awr\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nMijh2O6 qic8kXWuR6\nAO s1VGWTCbHzM\ni wJcdxt1kwqfDeJ" }, { "input": "14\nHAXRxayyf1Dj1F0mT hjR4A8IQMb0nyBtqG\nWNuMJa5Jg05qkqZOrL noNkWXrSidHGwxgbQ\nmOitVy6W52s0FENMz6 oLUkLNfojssvLvb1t\nhjR4A8IQMb0nyBtqG oA7uBFu4Oo\noA7uBFu4Oo M450\nM450 LXEzO4\noLUkLNfojssvLvb1t YG5\nnoNkWXrSidHGwxgbQ L\nL YBWzu4W\nYBWzu4W ML\nML scVZE9m8JnH\nLXEzO4 Ne0oBPY0Iy\nscVZE9m8JnH GXhznv\nYG5 UY08abilYF1LaXj49hQ", "output": "3\nWNuMJa5Jg05qkqZOrL GXhznv\nHAXRxayyf1Dj1F0mT Ne0oBPY0Iy\nmOitVy6W52s0FENMz6 UY08abilYF1LaXj49hQ" } ]

1,658,992,524

2,147,483,647

PyPy 3

OK

TESTS

49

140

3,686,400

import sys import math n = int(input()) lst = dict() for _ in range(n): s = input().split(' ') if s[0] in lst.values(): k = list(lst.keys())[list(lst.values()).index(s[0])] lst[k] = s[1] else: lst.update({s[0]: s[1]}) print(len(lst)) for x, y in lst.items(): print(x,y)

Title: Misha and Changing Handles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point. Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. Input Specification: The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests. Next *q* lines contain the descriptions of the requests, one per line. Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20. The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. Output Specification: In the first line output the integer *n* — the number of users that changed their handles at least once. In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order. Each user who changes the handle must occur exactly once in this description. Demo Input: ['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n'] Demo Output: ['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n'] Note: none

```python import sys import math n = int(input()) lst = dict() for _ in range(n): s = input().split(' ') if s[0] in lst.values(): k = list(lst.keys())[list(lst.values()).index(s[0])] lst[k] = s[1] else: lst.update({s[0]: s[1]}) print(len(lst)) for x, y in lst.items(): print(x,y) ```

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" } ]

1,593,851,715

2,147,483,647

Python 3

OK

TESTS

81

218

0

from sys import stdin input = stdin.readline n = int(input()) force = [list(map(int,input().split())) for _ in range(n)] nx, ny, nz = 0, 0, 0 for f in force: x, y, z = f nx += x ny += y nz += z if not nx and not ny and not nz: 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 from sys import stdin input = stdin.readline n = int(input()) force = [list(map(int,input().split())) for _ in range(n)] nx, ny, nz = 0, 0, 0 for f in force: x, y, z = f nx += x ny += y nz += z if not nx and not ny and not nz: print("YES") else: print("NO") ```

3.9455

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" } ]

1,691,751,104

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

5

46

0

word = input() wordn = '' c = 0 for i in range (1,len(word)) : if ord(word[i]) >= 97 : c = 1 if c == 1 : print(word) else : if ord(word[0]) >= 97 : wordn += chr(ord(word[0]) - 32) else : wordn += word[0] for i in range(1,len(word)) : if ord(word[i]) < 97 : wordn += chr(ord(word[i]) + 32) else : wordn += word[i] print(wordn)

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 word = input() wordn = '' c = 0 for i in range (1,len(word)) : if ord(word[i]) >= 97 : c = 1 if c == 1 : print(word) else : if ord(word[0]) >= 97 : wordn += chr(ord(word[0]) - 32) else : wordn += word[0] for i in range(1,len(word)) : if ord(word[i]) < 97 : wordn += chr(ord(word[i]) + 32) else : wordn += word[i] print(wordn) ```

0

793

A

Oleg and shares

PROGRAMMING

900

[ "implementation", "math" ]

null

Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices.

Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.

[ "3 3\n12 9 15\n", "2 2\n10 9\n", "4 1\n1 1000000000 1000000000 1000000000\n" ]

[ "3", "-1", "2999999997" ]

Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

500

[ { "input": "3 3\n12 9 15", "output": "3" }, { "input": "2 2\n10 9", "output": "-1" }, { "input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997" }, { "input": "1 11\n123", "output": "0" }, { "input": "20 6\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14", "output": "151" }, { "input": "20 5\n59 54 19 88 55 100 54 3 6 13 99 38 36 71 59 6 64 85 45 54", "output": "-1" }, { "input": "100 10\n340 70 440 330 130 120 340 210 440 110 410 120 180 40 50 230 70 110 310 360 480 70 230 120 230 310 470 60 210 60 210 480 290 250 450 440 150 40 500 230 280 250 30 50 310 50 230 360 420 260 330 80 50 160 70 470 140 180 380 190 250 30 220 410 80 310 280 50 20 430 440 180 310 190 190 330 90 190 320 390 170 460 230 30 80 500 470 370 80 500 400 120 220 150 70 120 70 320 260 260", "output": "2157" }, { "input": "100 18\n489 42 300 366 473 105 220 448 70 488 201 396 168 281 67 235 324 291 313 387 407 223 39 144 224 233 72 318 229 377 62 171 448 119 354 282 147 447 260 384 172 199 67 326 311 431 337 142 281 202 404 468 38 120 90 437 33 420 249 372 367 253 255 411 309 333 103 176 162 120 203 41 352 478 216 498 224 31 261 493 277 99 375 370 394 229 71 488 246 194 233 13 66 111 366 456 277 360 116 354", "output": "-1" }, { "input": "4 2\n1 2 3 4", "output": "-1" }, { "input": "3 4\n3 5 5", "output": "-1" }, { "input": "3 2\n88888884 88888886 88888888", "output": "3" }, { "input": "2 1\n1000000000 1000000000", "output": "0" }, { "input": "4 2\n1000000000 100000000 100000000 100000000", "output": "450000000" }, { "input": "2 2\n1000000000 1000000000", "output": "0" }, { "input": "3 3\n3 2 1", "output": "-1" }, { "input": "3 4\n3 5 3", "output": "-1" }, { "input": "3 2\n1 2 2", "output": "-1" }, { "input": "4 2\n2 3 3 2", "output": "-1" }, { "input": "3 2\n1 2 4", "output": "-1" }, { "input": "3 2\n3 4 4", "output": "-1" }, { "input": "3 3\n4 7 10", "output": "3" }, { "input": "4 3\n2 2 5 1", "output": "-1" }, { "input": "3 3\n1 3 5", "output": "-1" }, { "input": "2 5\n5 9", "output": "-1" }, { "input": "2 3\n5 7", "output": "-1" }, { "input": "3 137\n1000000000 1000000000 1000000000", "output": "0" }, { "input": "5 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 5\n1 2 5", "output": "-1" }, { "input": "3 3\n1000000000 1000000000 999999997", "output": "2" }, { "input": "2 4\n5 6", "output": "-1" }, { "input": "4 1\n1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "2 3\n5 8", "output": "1" }, { "input": "2 6\n8 16", "output": "-1" }, { "input": "5 3\n15 14 9 12 18", "output": "-1" }, { "input": "3 3\n1 2 3", "output": "-1" }, { "input": "3 3\n3 4 5", "output": "-1" }, { "input": "2 5\n8 17", "output": "-1" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "1 1\n1000000000", "output": "0" }, { "input": "3 3\n5 3 4", "output": "-1" }, { "input": "3 6\n10 14 12", "output": "-1" }, { "input": "2 2\n3 5", "output": "1" }, { "input": "3 5\n1 3 4", "output": "-1" }, { "input": "4 3\n1 6 6 6", "output": "-1" }, { "input": "2 3\n1 8", "output": "-1" }, { "input": "3 5\n6 11 17", "output": "-1" }, { "input": "2 2\n1 4", "output": "-1" }, { "input": "2 4\n6 8", "output": "-1" }, { "input": "2 1\n2 3", "output": "1" }, { "input": "4 4\n1 5 8 14", "output": "-1" }, { "input": "3 3\n1 5 3", "output": "-1" }, { "input": "4 3\n1 2 2 5", "output": "-1" }, { "input": "3 2\n1 4 6", "output": "-1" }, { "input": "2 3\n6 9", "output": "1" }, { "input": "3 3\n2 3 4", "output": "-1" }, { "input": "3 2\n9 10 10", "output": "-1" }, { "input": "2 2\n9 12", "output": "-1" }, { "input": "2 2\n100000003 100000005", "output": "1" }, { "input": "2 3\n2 4", "output": "-1" }, { "input": "3 2\n2 3 5", "output": "-1" }, { "input": "3 3\n1 3 4", "output": "-1" }, { "input": "10 2\n2 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "4499999991" }, { "input": "3 5\n2 4 5", "output": "-1" }, { "input": "2 3\n7 10", "output": "1" }, { "input": "3 10\n10 13 17", "output": "-1" }, { "input": "2 3\n1 6", "output": "-1" }, { "input": "1 7\n1000000000", "output": "0" }, { "input": "2 4\n3 7", "output": "1" }, { "input": "2 3\n2 5", "output": "1" }, { "input": "20 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 3\n7 8 8", "output": "-1" }, { "input": "4 10\n1 11 100 11", "output": "-1" } ]

1,607,511,184

2,147,483,647

Python 3

OK

TESTS

88

218

9,216,000

n, k = map(int, input().split()) a = sorted(list(map(int, input().split()))) result = 0 for i in a: if (i-a[0])%k!=0: print(-1) exit() else: result+=(i-a[0])//k print(result)

Title: Oleg and shares Time Limit: None seconds Memory Limit: None megabytes Problem Description: Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question? Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices. Output Specification: Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. Demo Input: ['3 3\n12 9 15\n', '2 2\n10 9\n', '4 1\n1 1000000000 1000000000 1000000000\n'] Demo Output: ['3', '-1', '2999999997'] Note: Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

```python n, k = map(int, input().split()) a = sorted(list(map(int, input().split()))) result = 0 for i in a: if (i-a[0])%k!=0: print(-1) exit() else: result+=(i-a[0])//k print(result) ```

3

513

B1

Permutations

PROGRAMMING

1,400

[ "brute force" ]

null

You are given a permutation *p* of numbers 1,<=2,<=...,<=*n*. Let's define *f*(*p*) as the following sum: Find the lexicographically *m*-th permutation of length *n* in the set of permutations having the maximum possible value of *f*(*p*).

The single line of input contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*cnt**n*), where *cnt**n* is the number of permutations of length *n* with maximum possible value of *f*(*p*). The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows. - In subproblem B1 (3 points), the constraint 1<=≤<=*n*<=≤<=8 will hold. - In subproblem B2 (4 points), the constraint 1<=≤<=*n*<=≤<=50 will hold.

Output *n* number forming the required permutation.

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

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

In the first example, both permutations of numbers {1, 2} yield maximum possible *f*(*p*) which is equal to 4. Among them, (2, 1) comes second in lexicographical order.

3

[ { "input": "2 2", "output": "2 1 " }, { "input": "3 2", "output": "1 3 2 " }, { "input": "1 1", "output": "1 " }, { "input": "3 1", "output": "1 2 3 " }, { "input": "3 3", "output": "2 3 1 " }, { "input": "3 4", "output": "3 2 1 " }, { "input": "4 1", "output": "1 2 3 4 " }, { "input": "4 3", "output": "1 3 4 2 " }, { "input": "4 4", "output": "1 4 3 2 " }, { "input": "4 8", "output": "4 3 2 1 " }, { "input": "5 2", "output": "1 2 3 5 4 " }, { "input": "5 7", "output": "1 4 5 3 2 " }, { "input": "5 15", "output": "4 5 3 2 1 " }, { "input": "6 23", "output": "2 5 6 4 3 1 " }, { "input": "7 7", "output": "1 2 3 6 7 5 4 " }, { "input": "7 44", "output": "2 4 7 6 5 3 1 " }, { "input": "8 1", "output": "1 2 3 4 5 6 7 8 " }, { "input": "8 127", "output": "7 8 6 5 4 3 2 1 " }, { "input": "8 128", "output": "8 7 6 5 4 3 2 1 " } ]

1,423,331,254

2,854

Python 3

OK

TESTS1

19

1,045

0

__author__ = 'yushchenko' def countf(f): sum = 0 for i in range(len(f)): for j in range(len(f))[i:]: # print(i, j) # print(f[i:j + 1]) sum += min(f[i:j + 1]) return sum import itertools n,m = input().split() n = int(n) m = int(m) maxf = 0; count = 0; save = () for x in itertools.permutations(range(n + 1)[1:]): t = countf(x) if t > maxf: maxf = t count = 0 if t == maxf: # print (x , '-' , t) count = count + 1 if count == m: save = x print(' '.join(str(e) for e in save))

Title: Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a permutation *p* of numbers 1,<=2,<=...,<=*n*. Let's define *f*(*p*) as the following sum: Find the lexicographically *m*-th permutation of length *n* in the set of permutations having the maximum possible value of *f*(*p*). Input Specification: The single line of input contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*cnt**n*), where *cnt**n* is the number of permutations of length *n* with maximum possible value of *f*(*p*). The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows. - In subproblem B1 (3 points), the constraint 1<=≤<=*n*<=≤<=8 will hold. - In subproblem B2 (4 points), the constraint 1<=≤<=*n*<=≤<=50 will hold. Output Specification: Output *n* number forming the required permutation. Demo Input: ['2 2\n', '3 2\n'] Demo Output: ['2 1 \n', '1 3 2 \n'] Note: In the first example, both permutations of numbers {1, 2} yield maximum possible *f*(*p*) which is equal to 4. Among them, (2, 1) comes second in lexicographical order.

```python __author__ = 'yushchenko' def countf(f): sum = 0 for i in range(len(f)): for j in range(len(f))[i:]: # print(i, j) # print(f[i:j + 1]) sum += min(f[i:j + 1]) return sum import itertools n,m = input().split() n = int(n) m = int(m) maxf = 0; count = 0; save = () for x in itertools.permutations(range(n + 1)[1:]): t = countf(x) if t > maxf: maxf = t count = 0 if t == maxf: # print (x , '-' , t) count = count + 1 if count == m: save = x print(' '.join(str(e) for e in save)) ```

3

834

B

The Festive Evening

PROGRAMMING

1,100

[ "data structures", "implementation" ]

null

It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in. There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously. For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed. Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.

Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26). In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.

Output «YES» if at least one door was unguarded during some time, and «NO» otherwise. You can output each letter in arbitrary case (upper or lower).

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

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

In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened. In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.

1,000

[ { "input": "5 1\nAABBB", "output": "NO" }, { "input": "5 1\nABABB", "output": "YES" }, { "input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "NO" }, { "input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA", "output": "YES" }, { "input": "5 2\nABACA", "output": "NO" }, { "input": "6 2\nABCABC", "output": "YES" }, { "input": "8 3\nABCBCDCA", "output": "NO" }, { "input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD", "output": "YES" }, { "input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA", "output": "NO" }, { "input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP", "output": "NO" }, { "input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE", "output": "YES" }, { "input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG", "output": "YES" }, { "input": "5 2\nABCAB", "output": "YES" }, { "input": "5 1\nAZAZA", "output": "YES" }, { "input": "7 2\nABCDBCD", "output": "YES" }, { "input": "3 26\nAAB", "output": "NO" } ]

1,539,725,130

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

124

0

x,y = map(int, input().split()) lista = input() cont = 0 flag = False for c in range(x-1): if lista[c] != lista[c+1] and cont < 1: flag = False break elif lista[c] == lista[c+1]: cont += 1 flag = True else: cont = 0 if flag: print("YES") else: print("NO")

Title: The Festive Evening Time Limit: None seconds Memory Limit: None megabytes Problem Description: It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in. There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously. For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed. Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened. Input Specification: Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26). In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest. Output Specification: Output «YES» if at least one door was unguarded during some time, and «NO» otherwise. You can output each letter in arbitrary case (upper or lower). Demo Input: ['5 1\nAABBB\n', '5 1\nABABB\n'] Demo Output: ['NO\n', 'YES\n'] Note: In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened. In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.

```python x,y = map(int, input().split()) lista = input() cont = 0 flag = False for c in range(x-1): if lista[c] != lista[c+1] and cont < 1: flag = False break elif lista[c] == lista[c+1]: cont += 1 flag = True else: cont = 0 if flag: print("YES") else: print("NO") ```

0

18

C

Stripe

PROGRAMMING

1,200

[ "data structures", "implementation" ]

C. Stripe

2

64

Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

[ "9\n1 5 -6 7 9 -16 0 -2 2\n", "3\n1 1 1\n", "2\n0 0\n" ]

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

none

0

[ { "input": "9\n1 5 -6 7 9 -16 0 -2 2", "output": "3" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "2\n0 0", "output": "1" }, { "input": "4\n100 1 10 111", "output": "1" }, { "input": "10\n0 4 -3 0 -2 2 -3 -3 2 5", "output": "3" }, { "input": "10\n0 -1 2 2 -1 1 0 0 0 2", "output": "0" }, { "input": "10\n-1 -1 1 -1 0 1 0 1 1 1", "output": "1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "9" }, { "input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3", "output": "3" }, { "input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3", "output": "0" }, { "input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2", "output": "0" }, { "input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2", "output": "1" }, { "input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2", "output": "6" }, { "input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "1\n10000", "output": "0" }, { "input": "1\n-10000", "output": "0" } ]

1,608,399,223

2,147,483,647

Python 3

OK

TESTS

45

436

7,782,400

if __name__ == "__main__": stripe_length = int(input()) square_Values = list(map(int, input().split())) # create a prefix sum array for i in range(1, stripe_length): square_Values[i] += square_Values[i-1] flag = 0 for i in range(stripe_length): if i != stripe_length - 1 and square_Values[i] == square_Values[stripe_length-1] - square_Values[i]: flag += 1 print(flag) # basic logic >> time exceed when it comes to large array # flag = 0 # i = 1 # while i < len(square_Values): # if sum(square_Values[:i]) == sum(square_Values[i:]): # flag+=1 # i += 1

Title: Stripe Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. Output Specification: Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. Demo Input: ['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: none

```python if __name__ == "__main__": stripe_length = int(input()) square_Values = list(map(int, input().split())) # create a prefix sum array for i in range(1, stripe_length): square_Values[i] += square_Values[i-1] flag = 0 for i in range(stripe_length): if i != stripe_length - 1 and square_Values[i] == square_Values[stripe_length-1] - square_Values[i]: flag += 1 print(flag) # basic logic >> time exceed when it comes to large array # flag = 0 # i = 1 # while i < len(square_Values): # if sum(square_Values[:i]) == sum(square_Values[i:]): # flag+=1 # i += 1 ```

3.833017

242

B

Big Segment

PROGRAMMING

1,100

[ "implementation", "sortings" ]

null

A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*]. You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1. Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment. It is guaranteed that no two segments coincide.

Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1. The segments are numbered starting from 1 in the order in which they appear in the input.

[ "3\n1 1\n2 2\n3 3\n", "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n" ]

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

none

1,000

[ { "input": "3\n1 1\n2 2\n3 3", "output": "-1" }, { "input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10", "output": "3" }, { "input": "4\n1 5\n2 2\n2 4\n2 5", "output": "1" }, { "input": "5\n3 3\n1 3\n2 2\n2 3\n1 2", "output": "2" }, { "input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8", "output": "-1" }, { "input": "3\n2 5\n3 4\n2 3", "output": "1" }, { "input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19", "output": "-1" }, { "input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3", "output": "1" }, { "input": "1\n1 100000", "output": "1" }, { "input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5", "output": "6" }, { "input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19", "output": "-1" }, { "input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8", "output": "-1" }, { "input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6", "output": "37" }, { "input": "1\n999999999 1000000000", "output": "1" }, { "input": "3\n1 20\n2 22\n3 18", "output": "-1" }, { "input": "1\n1000000000 1000000000", "output": "1" }, { "input": "2\n100001 100008\n100005 100006", "output": "1" }, { "input": "1\n1000000 10000000", "output": "1" }, { "input": "3\n3 6\n2 4\n1 5", "output": "-1" }, { "input": "2\n3 5\n1 2", "output": "-1" } ]

1,595,952,341

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

43

2,000

12,902,400

import sys def input(): return sys.stdin.readline().strip() def iinput(): return int(input()) def rinput(): return map(int, sys.stdin.readline().strip().split()) def get_list(): return list(map(int, sys.stdin.readline().strip().split())) n=iinput() count=0 s1=[] s2=[] for i in range(n): s=list(map(int,input().split())) s1.append(s[0]) s2.append(s[1]) l=min(s1) l1=max(s1) m=max(s2) a=[] b=[] for i in range(len(s1)): if(s1[i]==l): a.append(i) for i in range(len(s2)): if(s2[i]==m): b.append(i) for i in range(len(a)): for j in range(len(b)): if(a[i]==b[j]): c=a[i]+1 count+=1 if(count!=1): print(-1) else: if(l1<=m): print(c)

Title: Big Segment Time Limit: None seconds Memory Limit: None megabytes Problem Description: A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*]. You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1. Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment. It is guaranteed that no two segments coincide. Output Specification: Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1. The segments are numbered starting from 1 in the order in which they appear in the input. Demo Input: ['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n'] Demo Output: ['-1\n', '3\n'] Note: none

```python import sys def input(): return sys.stdin.readline().strip() def iinput(): return int(input()) def rinput(): return map(int, sys.stdin.readline().strip().split()) def get_list(): return list(map(int, sys.stdin.readline().strip().split())) n=iinput() count=0 s1=[] s2=[] for i in range(n): s=list(map(int,input().split())) s1.append(s[0]) s2.append(s[1]) l=min(s1) l1=max(s1) m=max(s2) a=[] b=[] for i in range(len(s1)): if(s1[i]==l): a.append(i) for i in range(len(s2)): if(s2[i]==m): b.append(i) for i in range(len(a)): for j in range(len(b)): if(a[i]==b[j]): c=a[i]+1 count+=1 if(count!=1): print(-1) else: if(l1<=m): print(c) ```

0

576

A

Vasya and Petya's Game

PROGRAMMING

1,500

[ "math", "number theory" ]

null

Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number. Petya can ask questions like: "Is the unknown number divisible by number *y*?". The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of. Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.

A single line contains number *n* (1<=≤<=*n*<=≤<=103).

Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*). If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.

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

[ "3\n2 4 3 \n", "4\n2 4 3 5 \n" ]

The sequence from the answer to the first sample test is actually correct. If the unknown number is not divisible by one of the sequence numbers, it is equal to 1. If the unknown number is divisible by 4, it is 4. If the unknown number is divisible by 3, then the unknown number is 3. Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

500

[ { "input": "4", "output": "3\n2 4 3 " }, { "input": "6", "output": "4\n2 4 3 5 " }, { "input": "1", "output": "0" }, { "input": "15", "output": "9\n2 4 8 3 9 5 7 11 13 " }, { "input": "19", "output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 " }, { "input": "20", "output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 " }, { "input": "37", "output": "19\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 " }, { "input": "211", "output": "61\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 169 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 " }, { "input": "557", "output": "123\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 " }, { "input": "907", "output": "179\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "953", "output": "186\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "289", "output": "78\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 " }, { "input": "400", "output": "97\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 " }, { "input": "900", "output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "625", "output": "136\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 " }, { "input": "729", "output": "152\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..." }, { "input": "784", "output": "160\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..." }, { "input": "31", "output": "17\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 31 " }, { "input": "44", "output": "21\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 41 43 " }, { "input": "160", "output": "50\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 " }, { "input": "322", "output": "83\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 " }, { "input": "894", "output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "998", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "1000", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "1000", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "30", "output": "16\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 " } ]

1,662,257,529

2,147,483,647

Python 3

OK

TESTS

26

93

0

n = int(input()) a = " ".join([" ".join([str(i ** k) for k in range(1, 30) if i ** k <= n]) for i in range(2, n + 1) if all([i % j != 0 for j in range(2, i)])]) print((len(a.split())), "\n" + a) num_inp=lambda: int(input()) arr_inp=lambda: list(map(int,input().split())) sp_inp=lambda: map(int,input().split()) str_inp=lambda:input()

Title: Vasya and Petya's Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number. Petya can ask questions like: "Is the unknown number divisible by number *y*?". The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of. Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about. Input Specification: A single line contains number *n* (1<=≤<=*n*<=≤<=103). Output Specification: Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*). If there are several correct sequences of questions of the minimum length, you are allowed to print any of them. Demo Input: ['4\n', '6\n'] Demo Output: ['3\n2 4 3 \n', '4\n2 4 3 5 \n'] Note: The sequence from the answer to the first sample test is actually correct. If the unknown number is not divisible by one of the sequence numbers, it is equal to 1. If the unknown number is divisible by 4, it is 4. If the unknown number is divisible by 3, then the unknown number is 3. Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

```python n = int(input()) a = " ".join([" ".join([str(i ** k) for k in range(1, 30) if i ** k <= n]) for i in range(2, n + 1) if all([i % j != 0 for j in range(2, i)])]) print((len(a.split())), "\n" + a) num_inp=lambda: int(input()) arr_inp=lambda: list(map(int,input().split())) sp_inp=lambda: map(int,input().split()) str_inp=lambda:input() ```

3

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" } ]

1,691,109,704

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

92

0

n=int(input()) x=0 m=0 for i in range(n): a,b=map(int,input().split()) if i==0: x=a+b m=b else: x=(m-a)+b m=max(x,m) print(m)

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 n=int(input()) x=0 m=0 for i in range(n): a,b=map(int,input().split()) if i==0: x=a+b m=b else: x=(m-a)+b m=max(x,m) print(m) ```

0

681

B

Economy Game

PROGRAMMING

1,300

[ "brute force" ]

null

Kolya is developing an economy simulator game. His most favourite part of the development process is in-game testing. Once he was entertained by the testing so much, that he found out his game-coin score become equal to 0. Kolya remembers that at the beginning of the game his game-coin score was equal to *n* and that he have bought only some houses (for 1<=234<=567 game-coins each), cars (for 123<=456 game-coins each) and computers (for 1<=234 game-coins each). Kolya is now interested, whether he could have spent all of his initial *n* game-coins buying only houses, cars and computers or there is a bug in the game. Formally, is there a triple of non-negative integers *a*, *b* and *c* such that *a*<=×<=1<=234<=567<=+<=*b*<=×<=123<=456<=+<=*c*<=×<=1<=234<==<=*n*? Please help Kolya answer this question.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Kolya's initial game-coin score.

Print "YES" (without quotes) if it's possible that Kolya spent all of his initial *n* coins buying only houses, cars and computers. Otherwise print "NO" (without quotes).

[ "1359257\n", "17851817\n" ]

[ "YES", "NO" ]

In the first sample, one of the possible solutions is to buy one house, one car and one computer, spending 1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.

1,000

[ { "input": "1359257", "output": "YES" }, { "input": "17851817", "output": "NO" }, { "input": "1000000000", "output": "YES" }, { "input": "17851818", "output": "YES" }, { "input": "438734347", "output": "YES" }, { "input": "43873430", "output": "YES" }, { "input": "999999987", "output": "YES" }, { "input": "27406117", "output": "NO" }, { "input": "27404883", "output": "NO" }, { "input": "27403649", "output": "NO" }, { "input": "27402415", "output": "NO" }, { "input": "27401181", "output": "NO" }, { "input": "999999999", "output": "YES" }, { "input": "999999244", "output": "YES" }, { "input": "999129999", "output": "YES" }, { "input": "17159199", "output": "NO" }, { "input": "13606913", "output": "NO" }, { "input": "14841529", "output": "NO" }, { "input": "915968473", "output": "YES" }, { "input": "980698615", "output": "YES" }, { "input": "912331505", "output": "YES" }, { "input": "917261049", "output": "YES" }, { "input": "999999997", "output": "YES" }, { "input": "12345", "output": "NO" }, { "input": "1234", "output": "YES" }, { "input": "124690", "output": "YES" }, { "input": "1359257", "output": "YES" }, { "input": "1358023", "output": "YES" }, { "input": "1234", "output": "YES" }, { "input": "1234567", "output": "YES" }, { "input": "124690", "output": "YES" }, { "input": "1358023", "output": "YES" }, { "input": "123456", "output": "YES" }, { "input": "2592590", "output": "YES" }, { "input": "999999998", "output": "YES" }, { "input": "1356789", "output": "NO" }, { "input": "12345670", "output": "YES" }, { "input": "11", "output": "NO" }, { "input": "1480800", "output": "YES" }, { "input": "908000000", "output": "YES" }, { "input": "3000", "output": "NO" }, { "input": "1235801", "output": "YES" }, { "input": "991919191", "output": "YES" }, { "input": "25613715", "output": "YES" }, { "input": "13580237", "output": "YES" }, { "input": "14814804", "output": "YES" }, { "input": "11403961", "output": "YES" }, { "input": "999999989", "output": "YES" }, { "input": "1237035", "output": "YES" }, { "input": "81134231", "output": "YES" }, { "input": "1236", "output": "NO" }, { "input": "1359250", "output": "YES" }, { "input": "100", "output": "NO" }, { "input": "987654321", "output": "YES" }, { "input": "122222", "output": "NO" }, { "input": "123458", "output": "NO" }, { "input": "20987639", "output": "YES" }, { "input": "999973333", "output": "YES" }, { "input": "253082", "output": "YES" }, { "input": "1235", "output": "NO" }, { "input": "803219200", "output": "YES" }, { "input": "100000000", "output": "YES" }, { "input": "1485181", "output": "YES" } ]

1,584,312,381

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

654

307,200

def can_spend(n): a = 1234567 b = 123456 c = 1234 an = bn = cn = n rests = set() rests.add(1234567) rests.add(123456) rests.add(1234) rests.add(0) while(an > 0 or bn > 0 or cn > 0): if an % a in rests or an % b == 0 or an % c in rests: return 'YES' if bn % a in rests or bn % b == 0 or bn % c in rests: return 'YES' if cn % a in rests or cn % b == 0 or cn % c in rests: return 'YES' if an > 0: an -= a if bn > 0: bn -= b if cn > 0: cn -= c return 'NO' n = int(input()) print(can_spend(n))

Title: Economy Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kolya is developing an economy simulator game. His most favourite part of the development process is in-game testing. Once he was entertained by the testing so much, that he found out his game-coin score become equal to 0. Kolya remembers that at the beginning of the game his game-coin score was equal to *n* and that he have bought only some houses (for 1<=234<=567 game-coins each), cars (for 123<=456 game-coins each) and computers (for 1<=234 game-coins each). Kolya is now interested, whether he could have spent all of his initial *n* game-coins buying only houses, cars and computers or there is a bug in the game. Formally, is there a triple of non-negative integers *a*, *b* and *c* such that *a*<=×<=1<=234<=567<=+<=*b*<=×<=123<=456<=+<=*c*<=×<=1<=234<==<=*n*? Please help Kolya answer this question. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Kolya's initial game-coin score. Output Specification: Print "YES" (without quotes) if it's possible that Kolya spent all of his initial *n* coins buying only houses, cars and computers. Otherwise print "NO" (without quotes). Demo Input: ['1359257\n', '17851817\n'] Demo Output: ['YES', 'NO'] Note: In the first sample, one of the possible solutions is to buy one house, one car and one computer, spending 1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.

```python def can_spend(n): a = 1234567 b = 123456 c = 1234 an = bn = cn = n rests = set() rests.add(1234567) rests.add(123456) rests.add(1234) rests.add(0) while(an > 0 or bn > 0 or cn > 0): if an % a in rests or an % b == 0 or an % c in rests: return 'YES' if bn % a in rests or bn % b == 0 or bn % c in rests: return 'YES' if cn % a in rests or cn % b == 0 or cn % c in rests: return 'YES' if an > 0: an -= a if bn > 0: bn -= b if cn > 0: cn -= c return 'NO' n = int(input()) print(can_spend(n)) ```

0

165

B

Burning Midnight Oil

PROGRAMMING

1,500

[ "binary search", "implementation" ]

null

One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep.

The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10.

Print the only integer — the minimum value of *v* that lets Vasya write the program in one night.

[ "7 2\n", "59 9\n" ]

[ "4\n", "54\n" ]

In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.

1,000

[ { "input": "7 2", "output": "4" }, { "input": "59 9", "output": "54" }, { "input": "1 9", "output": "1" }, { "input": "11 2", "output": "7" }, { "input": "747 2", "output": "376" }, { "input": "6578 2", "output": "3293" }, { "input": "37212 2", "output": "18609" }, { "input": "12357 2", "output": "6181" }, { "input": "7998332 2", "output": "3999172" }, { "input": "86275251 2", "output": "43137632" }, { "input": "75584551 2", "output": "37792280" }, { "input": "6 3", "output": "5" }, { "input": "43 4", "output": "33" }, { "input": "811 3", "output": "543" }, { "input": "3410 4", "output": "2560" }, { "input": "21341 4", "output": "16009" }, { "input": "696485 4", "output": "522368" }, { "input": "8856748 3", "output": "5904504" }, { "input": "2959379 4", "output": "2219538" }, { "input": "831410263 3", "output": "554273516" }, { "input": "2 5", "output": "2" }, { "input": "19 6", "output": "17" }, { "input": "715 7", "output": "615" }, { "input": "9122 5", "output": "7300" }, { "input": "89117 6", "output": "74268" }, { "input": "689973 7", "output": "591408" }, { "input": "3024524 5", "output": "2419624" }, { "input": "67127156 6", "output": "55939302" }, { "input": "412262167 7", "output": "353367574" }, { "input": "6 8", "output": "6" }, { "input": "59 9", "output": "54" }, { "input": "246 10", "output": "222" }, { "input": "5314 8", "output": "4651" }, { "input": "15309 9", "output": "13609" }, { "input": "35648 10", "output": "32085" }, { "input": "3018012 8", "output": "2640764" }, { "input": "92153348 9", "output": "81914089" }, { "input": "177583558 10", "output": "159825206" }, { "input": "1000000000 2", "output": "500000008" }, { "input": "1000000000 3", "output": "666666672" }, { "input": "1000000000 4", "output": "750000005" }, { "input": "1000000000 5", "output": "800000003" }, { "input": "1000000000 6", "output": "833333338" }, { "input": "1000000000 7", "output": "857142861" }, { "input": "1000000000 8", "output": "875000004" }, { "input": "1000000000 9", "output": "888888894" }, { "input": "1000000000 10", "output": "900000001" }, { "input": "1 4", "output": "1" }, { "input": "2 10", "output": "2" }, { "input": "1 2", "output": "1" }, { "input": "6 8", "output": "6" }, { "input": "987862820 9", "output": "878100288" }, { "input": "979591791 9", "output": "870748262" }, { "input": "948889213 9", "output": "843457081" }, { "input": "8 9", "output": "8" }, { "input": "999999999 10", "output": "900000000" } ]

1,657,861,995

2,147,483,647

PyPy 3-64

OK

TESTS

56

124

0

def check(x): val=0 div=0 while val<n and x//(k**div)>0: val=val+x//(k**(div)) div=div+1 # print(val,div) # print(x,val) if val<n: return False return True n,k=list(map(int,input().split())) # print(check(4)) i=-1 j=10**10 while i+1<j: mid=(i+j)//2 # print(check(mid),mid) if check(mid): j=mid else: i=mid print(j)

Title: Burning Midnight Oil Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep. Input Specification: The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10. Output Specification: Print the only integer — the minimum value of *v* that lets Vasya write the program in one night. Demo Input: ['7 2\n', '59 9\n'] Demo Output: ['4\n', '54\n'] Note: In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.

```python def check(x): val=0 div=0 while val<n and x//(k**div)>0: val=val+x//(k**(div)) div=div+1 # print(val,div) # print(x,val) if val<n: return False return True n,k=list(map(int,input().split())) # print(check(4)) i=-1 j=10**10 while i+1<j: mid=(i+j)//2 # print(check(mid),mid) if check(mid): j=mid else: i=mid print(j) ```

3

454

A

Little Pony and Crystal Mine

PROGRAMMING

800

[ "implementation" ]

null

Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it. You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.

The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).

Output a crystal of size *n*.

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

[ "*D*\nDDD\n*D*\n", "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n", "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n" ]

none

500

[ { "input": "3", "output": "*D*\nDDD\n*D*" }, { "input": "5", "output": "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**" }, { "input": "7", "output": "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***" }, { "input": "11", "output": "*****D*****\n****DDD****\n***DDDDD***\n**DDDDDDD**\n*DDDDDDDDD*\nDDDDDDDDDDD\n*DDDDDDDDD*\n**DDDDDDD**\n***DDDDD***\n****DDD****\n*****D*****" }, { "input": "15", "output": "*******D*******\n******DDD******\n*****DDDDD*****\n****DDDDDDD****\n***DDDDDDDDD***\n**DDDDDDDDDDD**\n*DDDDDDDDDDDDD*\nDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDD*\n**DDDDDDDDDDD**\n***DDDDDDDDD***\n****DDDDDDD****\n*****DDDDD*****\n******DDD******\n*******D*******" }, { "input": "21", "output": "**********D**********\n*********DDD*********\n********DDDDD********\n*******DDDDDDD*******\n******DDDDDDDDD******\n*****DDDDDDDDDDD*****\n****DDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDD**\n*DDDDDDDDDDDDDDDDDDD*\nDDDDDDDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDDDDDDDD*\n**DDDDDDDDDDDDDDDDD**\n***DDDDDDDDDDDDDDD***\n****DDDDDDDDDDDDD****\n*****DDDDDDDDDDD*****\n******DDDDDDDDD******\n*******DDDDDDD*******\n********DDDDD********\n*********DDD*********\n**********D**********" }, { "input": "33", "output": "****************D****************\n***************DDD***************\n**************DDDDD**************\n*************DDDDDDD*************\n************DDDDDDDDD************\n***********DDDDDDDDDDD***********\n**********DDDDDDDDDDDDD**********\n*********DDDDDDDDDDDDDDD*********\n********DDDDDDDDDDDDDDDDD********\n*******DDDDDDDDDDDDDDDDDDD*******\n******DDDDDDDDDDDDDDDDDDDDD******\n*****DDDDDDDDDDDDDDDDDDDDDDD*****\n****DDDDDDDDDDDDDDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDDDD..." }, { "input": "57", "output": "****************************D****************************\n***************************DDD***************************\n**************************DDDDD**************************\n*************************DDDDDDD*************************\n************************DDDDDDDDD************************\n***********************DDDDDDDDDDD***********************\n**********************DDDDDDDDDDDDD**********************\n*********************DDDDDDDDDDDDDDD*********************\n********************DDDDDDDDDDDDDDDDD**..." }, { "input": "69", "output": "**********************************D**********************************\n*********************************DDD*********************************\n********************************DDDDD********************************\n*******************************DDDDDDD*******************************\n******************************DDDDDDDDD******************************\n*****************************DDDDDDDDDDD*****************************\n****************************DDDDDDDDDDDDD****************************\n**************..." }, { "input": "81", "output": "****************************************D****************************************\n***************************************DDD***************************************\n**************************************DDDDD**************************************\n*************************************DDDDDDD*************************************\n************************************DDDDDDDDD************************************\n***********************************DDDDDDDDDDD***********************************\n*************..." }, { "input": "91", "output": "*********************************************D*********************************************\n********************************************DDD********************************************\n*******************************************DDDDD*******************************************\n******************************************DDDDDDD******************************************\n*****************************************DDDDDDDDD*****************************************\n****************************************DDDDDD..." }, { "input": "101", "output": "**************************************************D**************************************************\n*************************************************DDD*************************************************\n************************************************DDDDD************************************************\n***********************************************DDDDDDD***********************************************\n**********************************************DDDDDDDDD********************************************..." } ]

1,616,223,166

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

46

307,200

n = int(input()) kiri = n//2 +1 kanan = n//2 +1 for i in range (1, n//2 +1): for j in range (1, n+1): if (kiri <= j and j <= kanan): print('D', end='') else: print('.', end='') kiri -= 1 kanan += 1 print() for i in range (n//2, n): for j in range (1, n+1): if (kiri <= j and j <= kanan): print('D', end='') else: print('.', end='') kiri += 1 kanan -= 1 print()

Title: Little Pony and Crystal Mine Time Limit: None seconds Memory Limit: None megabytes Problem Description: Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it. You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input Specification: The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd). Output Specification: Output a crystal of size *n*. Demo Input: ['3\n', '5\n', '7\n'] Demo Output: ['*D*\nDDD\n*D*\n', '**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n', '***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n'] Note: none

```python n = int(input()) kiri = n//2 +1 kanan = n//2 +1 for i in range (1, n//2 +1): for j in range (1, n+1): if (kiri <= j and j <= kanan): print('D', end='') else: print('.', end='') kiri -= 1 kanan += 1 print() for i in range (n//2, n): for j in range (1, n+1): if (kiri <= j and j <= kanan): print('D', end='') else: print('.', end='') kiri += 1 kanan -= 1 print() ```

0

805

B

3-palindrome

PROGRAMMING

1,000

[ "constructive algorithms" ]

null

In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick. He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string.

Print the string that satisfies all the constraints. If there are multiple answers, print any of them.

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

[ "aa\n", "bba\n" ]

A palindrome is a sequence of characters which reads the same backward and forward.

1,000

[ { "input": "2", "output": "aa" }, { "input": "3", "output": "aab" }, { "input": "38", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaa" }, { "input": "47", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "59", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "67", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "77", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "89", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "98", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaa" }, { "input": "109", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "117", "output": 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"aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..." }, { "input": "6", "output": "aabbaa" }, { "input": "4", "output": "aabb" } ]

1,591,127,448

2,147,483,647

Python 3

OK

TESTS

42

124

819,200

n=int(input()) s='' c='a' for i in range(n//2+1): if c=='a': s=s+'aa' c='b' else: s=s+'bb' c='a' print(s[0:n])

Title: 3-palindrome Time Limit: None seconds Memory Limit: None megabytes Problem Description: In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick. He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string. Output Specification: Print the string that satisfies all the constraints. If there are multiple answers, print any of them. Demo Input: ['2\n', '3\n'] Demo Output: ['aa\n', 'bba\n'] Note: A palindrome is a sequence of characters which reads the same backward and forward.

```python n=int(input()) s='' c='a' for i in range(n//2+1): if c=='a': s=s+'aa' c='b' else: s=s+'bb' c='a' print(s[0:n]) ```

3

192

A

Funky Numbers

PROGRAMMING

1,300

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

null

As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers. A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).

Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

[ "256\n", "512\n" ]

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

In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample number 512 can not be represented as a sum of two triangular numbers.

500

[ { "input": "256", "output": "YES" }, { "input": "512", "output": "NO" }, { "input": "80", "output": "NO" }, { "input": "828", "output": "YES" }, { "input": "6035", "output": "NO" }, { "input": "39210", "output": "YES" }, { "input": "79712", "output": "NO" }, { "input": "190492", "output": "YES" }, { "input": "5722367", "output": "NO" }, { "input": "816761542", "output": "YES" }, { "input": "1", "output": "NO" }, { "input": "2", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "7", "output": "YES" }, { "input": "8", "output": "NO" }, { "input": "9", "output": "YES" }, { "input": "10", "output": "NO" }, { "input": "12", "output": "YES" }, { "input": "13", "output": "YES" }, { "input": "14", "output": "NO" }, { "input": "15", "output": "NO" }, { "input": "16", "output": "YES" }, { "input": "17", "output": "NO" }, { "input": "18", "output": "YES" }, { "input": "19", "output": "NO" }, { "input": "20", "output": "YES" }, { "input": "41", "output": "NO" }, { "input": "11", "output": "YES" }, { "input": "69", "output": "YES" }, { "input": "82", "output": "NO" }, { "input": "85", "output": "NO" }, { "input": "736", "output": "NO" }, { "input": "895", "output": "YES" }, { "input": "934", "output": "YES" }, { "input": "6213", "output": "YES" }, { "input": "7405", "output": "NO" }, { "input": "9919", "output": "NO" }, { "input": "40942", "output": "YES" }, { "input": "41992", "output": "NO" }, { "input": "68535", "output": "NO" }, { "input": "405718", "output": "NO" }, { "input": "1046146", "output": "YES" }, { "input": "3761248", "output": "YES" }, { "input": "6195181", "output": "YES" }, { "input": "35354345", "output": "NO" }, { "input": "81282830", "output": "NO" }, { "input": "187719774", "output": "NO" }, { "input": "296798673", "output": "NO" }, { "input": "938938476", "output": "NO" }, { "input": "1000000000", "output": "NO" }, { "input": "999887464", "output": "YES" }, { "input": "999111944", "output": "NO" }, { "input": "999966520", "output": "YES" }, { "input": "999912080", "output": "NO" }, { "input": "999992017", "output": "YES" }, { "input": "999990474", "output": "NO" }, { "input": "999999190", "output": "YES" }, { "input": "999999125", "output": "NO" }, { "input": "999999940", "output": "YES" }, { "input": "999999995", "output": "NO" }, { "input": "1000000000", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "999999999", "output": "YES" }, { "input": "83495494", "output": "NO" }, { "input": "968022000", "output": "YES" }, { "input": "399980000", "output": "YES" }, { "input": "4", "output": "YES" }, { "input": "999999998", "output": "NO" } ]

1,563,825,440

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

11

280

2,457,600

n = int(input()) m = 1 arr = [] while(True): mm = (m*(m+1))//2 if mm>n: break arr.append(mm) m+=1 ll = len(arr) flag = 0 for i in range(ll): curr = arr[i] req = n-curr l = i+1 r = ll-1 while(l<=r): mid = (l+r)//2 if arr[mid]>=req: r = mid-1 else: l = mid+1 if i<l<ll and arr[l]==req: flag=1 break if flag==1: print("YES") else: print("NO")

Title: Funky Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers. A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)! Input Specification: The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109). Output Specification: Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes). Demo Input: ['256\n', '512\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample number 512 can not be represented as a sum of two triangular numbers.

```python n = int(input()) m = 1 arr = [] while(True): mm = (m*(m+1))//2 if mm>n: break arr.append(mm) m+=1 ll = len(arr) flag = 0 for i in range(ll): curr = arr[i] req = n-curr l = i+1 r = ll-1 while(l<=r): mid = (l+r)//2 if arr[mid]>=req: r = mid-1 else: l = mid+1 if i<l<ll and arr[l]==req: flag=1 break if flag==1: print("YES") else: print("NO") ```

0

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

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

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

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

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

[ "NO", "YES" ]

none

500

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

1,590,663,471

2,147,483,647

PyPy 3

COMPILATION_ERROR

TESTS

0

import sys inp = sys.stdin.readlines()[] num_items = int(inp[0]) all_ints = [] for line in inp: all_ints.append(map(int, line.split(' '))) s = 0 for i in range(num_items): for j in all_ints: s += j[i] if s != 0: print('NO') works = False if works: print('YES')

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 import sys inp = sys.stdin.readlines()[] num_items = int(inp[0]) all_ints = [] for line in inp: all_ints.append(map(int, line.split(' '))) s = 0 for i in range(num_items): for j in all_ints: s += j[i] if s != 0: print('NO') works = False if works: print('YES') ```

-1

5

C

Longest Regular Bracket Sequence

PROGRAMMING

1,900

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

C. Longest Regular Bracket Sequence

2

256

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

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

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

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

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

none

0

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

1,548,108,353

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

218

0

s = input() st = list() c = 0 x = 0 max = 0 for i in range(len(s)): if s[i] == '(': st.append(s[i]) else: if st and st[-1] == '(': c+=2 st.pop() if i == len(s)-1: if c > max: max = c x = 1 elif c>=1 and c == max: x += 1 else : if c > max: max = c x = 1 elif c>=1 and c == max: x+= 1 c = 0 if x >= 1: print(str(max)+' '+ str(x)) else: print('0 1')

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

```python s = input() st = list() c = 0 x = 0 max = 0 for i in range(len(s)): if s[i] == '(': st.append(s[i]) else: if st and st[-1] == '(': c+=2 st.pop() if i == len(s)-1: if c > max: max = c x = 1 elif c>=1 and c == max: x += 1 else : if c > max: max = c x = 1 elif c>=1 and c == max: x+= 1 c = 0 if x >= 1: print(str(max)+' '+ str(x)) else: print('0 1') ```

0

802

G

Fake News (easy)

PROGRAMMING

800

[ "implementation", "strings" ]

null

As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...

The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).

Output YES if the string *s* contains heidi as a subsequence and NO otherwise.

[ "abcheaibcdi\n", "hiedi\n" ]

[ "YES", "NO" ]

A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.

0

[ { "input": "abcheaibcdi", "output": "YES" }, { "input": "hiedi", "output": "NO" }, { "input": "ihied", "output": "NO" }, { "input": "diehi", "output": "NO" }, { "input": "deiih", "output": "NO" }, { "input": "iheid", "output": "NO" }, { "input": "eihdi", "output": "NO" }, { "input": "ehdii", "output": "NO" }, { "input": "edhii", "output": "NO" }, { "input": "deiih", "output": "NO" }, { "input": "ehdii", "output": "NO" }, { "input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto", "output": "YES" }, { "input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm", "output": "YES" }, { "input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv", "output": "YES" }, { "input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs", "output": "YES" }, { "input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji", "output": "YES" }, { "input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn", "output": "YES" }, { "input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx", "output": "YES" }, { "input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy", "output": "YES" }, { "input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc", "output": "YES" }, { "input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef", "output": "NO" }, { "input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa", "output": "YES" }, { "input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg", "output": "YES" }, { "input": "oagjghsidigeh", "output": "NO" }, { "input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz", "output": "YES" }, { "input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi", "output": "YES" }, { "input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx", "output": "YES" }, { "input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq", "output": "YES" } ]

1,595,506,918

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

124

20,172,800

a=str(input()) s="" l=['h','e','i','d','i'] for i in range (len(a)): if a[i] in l: s+=a[i] l.remove(a[i]) if s=="heidi": print("YES") else: print("NO")

Title: Fake News (easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it... Input Specification: The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z). Output Specification: Output YES if the string *s* contains heidi as a subsequence and NO otherwise. Demo Input: ['abcheaibcdi\n', 'hiedi\n'] Demo Output: ['YES', 'NO'] Note: A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.

```python a=str(input()) s="" l=['h','e','i','d','i'] for i in range (len(a)): if a[i] in l: s+=a[i] l.remove(a[i]) if s=="heidi": print("YES") else: print("NO") ```

0

430

B

Balls Game

PROGRAMMING

1,400

[ "brute force", "two pointers" ]

null

Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game? There are *n* balls put in a row. Each ball is colored in one of *k* colors. Initially the row doesn't contain three or more contiguous balls with the same color. Iahub has a single ball of color *x*. He can insert his ball at any position in the row (probably, between two other balls). If at any moment there are three or more contiguous balls of the same color in the row, they are destroyed immediately. This rule is applied multiple times, until there are no more sets of 3 or more contiguous balls of the same color. For example, if Iahub has the row of balls [black, black, white, white, black, black] and a white ball, he can insert the ball between two white balls. Thus three white balls are destroyed, and then four black balls become contiguous, so all four balls are destroyed. The row will not contain any ball in the end, so Iahub can destroy all 6 balls. Iahub wants to destroy as many balls as possible. You are given the description of the row of balls, and the color of Iahub's ball. Help Iahub train for the IOI by telling him the maximum number of balls from the row he can destroy.

The first line of input contains three integers: *n* (1<=≤<=*n*<=≤<=100), *k* (1<=≤<=*k*<=≤<=100) and *x* (1<=≤<=*x*<=≤<=*k*). The next line contains *n* space-separated integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*k*). Number *c**i* means that the *i*-th ball in the row has color *c**i*. It is guaranteed that the initial row of balls will never contain three or more contiguous balls of the same color.

Print a single integer — the maximum number of balls Iahub can destroy.

[ "6 2 2\n1 1 2 2 1 1\n", "1 1 1\n1\n" ]

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

none

1,000

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

1,427,117,765

2,147,483,647

Python 3

OK

TESTS

18

62

204,800

R = lambda: list(map(int, input().split())) def dfs(a): cnt = 0 n = len(a) a.append(10000000) for i in range(1, n): if a[i] != a[i - 1]: cnt = 0 else: cnt += 1 if cnt >= 2: j = i while a[j] == a[i]: j += 1 return j - i + 2 + dfs(a[:i - 2] + a[j:n]) return 0 n, k, x = R() a = R() ans=1 for i in range(n+1): ans=max(ans, dfs(a[:i]+[x]+a[i:])) print(ans-1)

Title: Balls Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game? There are *n* balls put in a row. Each ball is colored in one of *k* colors. Initially the row doesn't contain three or more contiguous balls with the same color. Iahub has a single ball of color *x*. He can insert his ball at any position in the row (probably, between two other balls). If at any moment there are three or more contiguous balls of the same color in the row, they are destroyed immediately. This rule is applied multiple times, until there are no more sets of 3 or more contiguous balls of the same color. For example, if Iahub has the row of balls [black, black, white, white, black, black] and a white ball, he can insert the ball between two white balls. Thus three white balls are destroyed, and then four black balls become contiguous, so all four balls are destroyed. The row will not contain any ball in the end, so Iahub can destroy all 6 balls. Iahub wants to destroy as many balls as possible. You are given the description of the row of balls, and the color of Iahub's ball. Help Iahub train for the IOI by telling him the maximum number of balls from the row he can destroy. Input Specification: The first line of input contains three integers: *n* (1<=≤<=*n*<=≤<=100), *k* (1<=≤<=*k*<=≤<=100) and *x* (1<=≤<=*x*<=≤<=*k*). The next line contains *n* space-separated integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*k*). Number *c**i* means that the *i*-th ball in the row has color *c**i*. It is guaranteed that the initial row of balls will never contain three or more contiguous balls of the same color. Output Specification: Print a single integer — the maximum number of balls Iahub can destroy. Demo Input: ['6 2 2\n1 1 2 2 1 1\n', '1 1 1\n1\n'] Demo Output: ['6\n', '0\n'] Note: none

```python R = lambda: list(map(int, input().split())) def dfs(a): cnt = 0 n = len(a) a.append(10000000) for i in range(1, n): if a[i] != a[i - 1]: cnt = 0 else: cnt += 1 if cnt >= 2: j = i while a[j] == a[i]: j += 1 return j - i + 2 + dfs(a[:i - 2] + a[j:n]) return 0 n, k, x = R() a = R() ans=1 for i in range(n+1): ans=max(ans, dfs(a[:i]+[x]+a[i:])) print(ans-1) ```

3

488

A

Giga Tower

PROGRAMMING

1,100

[ "brute force" ]

null

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

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

Print the minimum *b* in a line.

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

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

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

500

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

1,416,594,741

4,341

Python 3

OK

TESTS

35

93

0

n = input() for i in range(1,20): if "8" in str(int(n)+i): print(i) break

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

```python n = input() for i in range(1,20): if "8" in str(int(n)+i): print(i) break ```

3

682

A

Alyona and Numbers

PROGRAMMING

1,100

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

null

After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5. Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0. As usual, Alyona has some troubles and asks you to help.

The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).

Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.

[ "6 12\n", "11 14\n", "1 5\n", "3 8\n", "5 7\n", "21 21\n" ]

[ "14\n", "31\n", "1\n", "5\n", "7\n", "88\n" ]

Following pairs are suitable in the first sample case: - for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9. Only the pair (1, 4) is suitable in the third sample case.

500

[ { "input": "6 12", "output": "14" }, { "input": "11 14", "output": "31" }, { "input": "1 5", "output": "1" }, { "input": "3 8", "output": "5" }, { "input": "5 7", "output": "7" }, { "input": "21 21", "output": "88" }, { "input": "10 15", "output": "30" }, { "input": "1 1", "output": "0" }, { "input": "1 1000000", "output": "200000" }, { "input": "1000000 1", "output": "200000" }, { "input": "1000000 1000000", "output": "200000000000" }, { "input": "944 844", "output": "159348" }, { "input": "368 984", "output": "72423" }, { "input": "792 828", "output": "131155" }, { "input": "920 969", "output": "178296" }, { "input": "640 325", "output": "41600" }, { "input": "768 170", "output": "26112" }, { "input": "896 310", "output": "55552" }, { "input": "320 154", "output": "9856" }, { "input": "744 999", "output": "148652" }, { "input": "630 843", "output": "106218" }, { "input": "54 688", "output": "7431" }, { "input": "478 828", "output": "79157" }, { "input": "902 184", "output": "33194" }, { "input": "31 29", "output": "180" }, { "input": "751 169", "output": "25384" }, { "input": "879 14", "output": "2462" }, { "input": "7 858", "output": "1201" }, { "input": "431 702", "output": "60512" }, { "input": "855 355", "output": "60705" }, { "input": "553 29", "output": "3208" }, { "input": "721767 525996", "output": "75929310986" }, { "input": "805191 74841", "output": "12052259926" }, { "input": "888615 590981", "output": "105030916263" }, { "input": "4743 139826", "output": "132638943" }, { "input": "88167 721374", "output": "12720276292" }, { "input": "171591 13322", "output": "457187060" }, { "input": "287719 562167", "output": "32349225415" }, { "input": "371143 78307", "output": "5812618980" }, { "input": "487271 627151", "output": "61118498984" }, { "input": "261436 930642", "output": "48660664382" }, { "input": "377564 446782", "output": "33737759810" }, { "input": "460988 28330", "output": "2611958008" }, { "input": "544412 352983", "output": "38433636199" }, { "input": "660540 869123", "output": "114818101284" }, { "input": "743964 417967", "output": "62190480238" }, { "input": "827388 966812", "output": "159985729411" }, { "input": "910812 515656", "output": "93933134534" }, { "input": "26940 64501", "output": "347531388" }, { "input": "110364 356449", "output": "7867827488" }, { "input": "636358 355531", "output": "45248999219" }, { "input": "752486 871672", "output": "131184195318" }, { "input": "803206 420516", "output": "67552194859" }, { "input": "919334 969361", "output": "178233305115" }, { "input": "35462 261309", "output": "1853307952" }, { "input": "118887 842857", "output": "20040948031" }, { "input": "202311 358998", "output": "14525848875" }, { "input": "285735 907842", "output": "51880446774" }, { "input": "401863 456686", "output": "36705041203" }, { "input": "452583 972827", "output": "88056992428" }, { "input": "235473 715013", "output": "33673251230" }, { "input": "318897 263858", "output": "16828704925" }, { "input": "402321 812702", "output": "65393416268" }, { "input": "518449 361546", "output": "37488632431" }, { "input": "634577 910391", "output": "115542637921" }, { "input": "685297 235043", "output": "32214852554" }, { "input": "801425 751183", "output": "120403367155" }, { "input": "884849 300028", "output": "53095895155" }, { "input": "977 848872", "output": "165869588" }, { "input": "51697 397716", "output": "4112144810" }, { "input": "834588 107199", "output": "17893399803" }, { "input": "918012 688747", "output": "126455602192" }, { "input": "1436 237592", "output": "68236422" }, { "input": "117564 753732", "output": "17722349770" }, { "input": "200988 302576", "output": "12162829017" }, { "input": "284412 818717", "output": "46570587880" }, { "input": "400540 176073", "output": "14104855884" }, { "input": "483964 724917", "output": "70166746198" }, { "input": "567388 241058", "output": "27354683301" }, { "input": "650812 789902", "output": "102815540084" }, { "input": "400999 756281", "output": "60653584944" }, { "input": "100 101", "output": "2020" }, { "input": "100 102", "output": "2040" }, { "input": "103 100", "output": "2060" }, { "input": "100 104", "output": "2080" }, { "input": "3 4", "output": "3" }, { "input": "11 23", "output": "50" }, { "input": "8 14", "output": "23" }, { "input": "23423 34234", "output": "160372597" }, { "input": "1 4", "output": "1" }, { "input": "999999 999999", "output": "199999600001" }, { "input": "82 99", "output": "1624" }, { "input": "21 18", "output": "75" }, { "input": "234 234", "output": "10952" }, { "input": "4 4", "output": "4" }, { "input": "6 13", "output": "15" }, { "input": "3 9", "output": "6" }, { "input": "99999 99999", "output": "1999960001" }, { "input": "34 33", "output": "225" }, { "input": "2 2", "output": "0" }, { "input": "333 1", "output": "66" }, { "input": "3 3", "output": "2" }, { "input": "8 2", "output": "3" }, { "input": "2179 2218", "output": "966605" }, { "input": "1000000 999999", "output": "199999800000" }, { "input": "873828 774207", "output": "135304750879" }, { "input": "13 19", "output": "50" }, { "input": "1648 576469", "output": "190004183" }, { "input": "11 13", "output": "28" }, { "input": "5 8", "output": "8" }, { "input": "650074 943659", "output": "122689636154" }, { "input": "1 3", "output": "0" }, { "input": "54 43", "output": "465" }, { "input": "14 9", "output": "26" }, { "input": "2 3", "output": "1" }, { "input": "543 534", "output": "57993" }, { "input": "321 123", "output": "7896" }, { "input": "21 3", "output": "12" }, { "input": "2 1", "output": "0" }, { "input": "4 3", "output": "3" }, { "input": "47474 74747", "output": "709707816" }, { "input": "4 9", "output": "8" }, { "input": "7 4", "output": "6" }, { "input": "9 4", "output": "8" }, { "input": "12414 4214", "output": "10462520" }, { "input": "2 9", "output": "4" }, { "input": "253 821", "output": "41542" }, { "input": "2 4", "output": "2" } ]

1,643,905,325

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

#include <iostream> using namespace std; int main(){ long long int n,m,cnt=0; cin>>n>>m; for(int i=1;i<=n;i++){ cnt=cnt+(m+i%5)/5; } cout<<cnt; }

Title: Alyona and Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5. Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0. As usual, Alyona has some troubles and asks you to help. Input Specification: The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000). Output Specification: Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5. Demo Input: ['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n'] Demo Output: ['14\n', '31\n', '1\n', '5\n', '7\n', '88\n'] Note: Following pairs are suitable in the first sample case: - for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9. Only the pair (1, 4) is suitable in the third sample case.

```python #include <iostream> using namespace std; int main(){ long long int n,m,cnt=0; cin>>n>>m; for(int i=1;i<=n;i++){ cnt=cnt+(m+i%5)/5; } cout<<cnt; } ```

-1

981

B

Businessmen Problems

PROGRAMMING

1,000

[ "sortings" ]

null

Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies. In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies. All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company. The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set. In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets. Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.

The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces. The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct. The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist. The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct.

Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.

[ "3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n", "1\n1000000000 239\n3\n14 15\n92 65\n35 89\n" ]

[ "24\n", "408\n" ]

In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$. In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$.

750

[ { "input": "3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4", "output": "24" }, { "input": "1\n1000000000 239\n3\n14 15\n92 65\n35 89", "output": "408" }, { "input": "10\n598654597 488228616\n544064902 21923894\n329635457 980089248\n988262691 654502493\n967529230 543358150\n835120075 128123793\n809901567 613170206\n152157661 479980560\n859252956 318029856\n614959273 735298093\n10\n849172317 361325206\n341655282 740807372\n453949939 809030434\n813199219 765838311\n725032511 427644854\n631233366 115983627\n825067417 721190169\n463026366 953335259\n881501578 987288937\n500099903 427659916", "output": "11272808994" }, { "input": "10\n598654597 488228616\n544064902 21923894\n329635457 980089248\n988262691 654502493\n967529230 543358150\n835120075 128123793\n809901567 613170206\n152157661 479980560\n859252956 318029856\n614959273 735298093\n10\n544064902 740807372\n859252956 765838311\n835120075 115983627\n825067417 463026366\n809901567 500099903\n614959273 322932832\n33955432 657532637\n816892960 927933923\n916770606 754371712\n967529230 566149252", "output": "8955052582" }, { "input": "1\n259021863 682413763\n1\n554146726 693647287", "output": "1376061050" }, { "input": "1\n259021863 682413763\n1\n554146726 425883728", "output": "1108297491" }, { "input": "1\n1000000000 1\n1\n1000000000 1000000000", "output": "1000000000" }, { "input": "2\n1 4\n2 4\n2\n1 4\n2 4", "output": "8" }, { "input": "1\n14 15\n3\n35 89\n92 65\n1000000000 239", "output": "408" } ]

1,527,436,200

3,600

Python 3

TIME_LIMIT_EXCEEDED

TESTS

12

2,000

16,076,800

from queue import PriorityQueue n = int(input()) i = 0 topcm = [] mapatop = {} somatop = 0 while i < n: a,x = [int(i) for i in input().split()] topcm.append(x) mapatop[a] = i somatop += x i+=1 m = int(input()) i = 0 somacm = 0 unique = PriorityQueue() while i < m: b,y = [int(i) for i in input().split()] if b in mapatop: unique.put((-y,b)) unique.put((-topcm[mapatop[b]],b)) somatop -= topcm[mapatop[b]] else: somacm += y i+=1 soma = somacm+somatop topcm = [] comp = set() while not unique.empty(): valor,idf = unique.get() valor*=-1 if idf not in comp: comp = comp.union(set([idf])) soma+=valor print(soma)

Title: Businessmen Problems Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies. In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies. All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company. The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set. In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets. Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible. Input Specification: The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces. The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct. The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist. The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct. Output Specification: Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets. Demo Input: ['3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n', '1\n1000000000 239\n3\n14 15\n92 65\n35 89\n'] Demo Output: ['24\n', '408\n'] Note: In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$. In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$.

```python from queue import PriorityQueue n = int(input()) i = 0 topcm = [] mapatop = {} somatop = 0 while i < n: a,x = [int(i) for i in input().split()] topcm.append(x) mapatop[a] = i somatop += x i+=1 m = int(input()) i = 0 somacm = 0 unique = PriorityQueue() while i < m: b,y = [int(i) for i in input().split()] if b in mapatop: unique.put((-y,b)) unique.put((-topcm[mapatop[b]],b)) somatop -= topcm[mapatop[b]] else: somacm += y i+=1 soma = somacm+somatop topcm = [] comp = set() while not unique.empty(): valor,idf = unique.get() valor*=-1 if idf not in comp: comp = comp.union(set([idf])) soma+=valor print(soma) ```

0

841

A

Generous Kefa

PROGRAMMING

900

[ "brute force", "implementation" ]

null

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

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

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

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

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

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

500

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

1,601,864,071

2,147,483,647

Python 3

OK

TESTS

114

124

0

n,k = map(int,input().split()) l = input() if k >= n: print("YES") else: for x in l: t = l.count(x) if t > k: print("NO") break else: print("YES")

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

```python n,k = map(int,input().split()) l = input() if k >= n: print("YES") else: for x in l: t = l.count(x) if t > k: print("NO") break else: print("YES") ```

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" } ]

1,669,908,183

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

r1,r2=map(int,input().split()) c1,c2=map(int,input().split()) d1,d2=map(int,input().split()) b=(r1-c1+d2)//2 a=r1-b c=c1-a d=c2-b if(a==b or a==c or a==d or b==c or b==d or c==d or a>9 or b>9 or c>9 or d>9 or a<1 or b<1 or c<1 or d<1): print(-1) else if(a+b==r1 and c+d==r2 and a+c==c1 and b+d==c2 and a+d==d1 and b+c==d2): print(a,b) print(c,d) else: 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()) b=(r1-c1+d2)//2 a=r1-b c=c1-a d=c2-b if(a==b or a==c or a==d or b==c or b==d or c==d or a>9 or b>9 or c>9 or d>9 or a<1 or b<1 or c<1 or d<1): print(-1) else if(a+b==r1 and c+d==r2 and a+c==c1 and b+d==c2 and a+d==d1 and b+c==d2): print(a,b) print(c,d) else: print(-1) ```

-1

372

A

Counting Kangaroos is Fun

PROGRAMMING

1,600

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

null

There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held. Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos. The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.

The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105).

Output a single integer — the optimal number of visible kangaroos.

[ "8\n2\n5\n7\n6\n9\n8\n4\n2\n", "8\n9\n1\n6\n2\n6\n5\n8\n3\n" ]

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

none

500

[ { "input": "8\n2\n5\n7\n6\n9\n8\n4\n2", "output": "5" }, { "input": "8\n9\n1\n6\n2\n6\n5\n8\n3", "output": "5" }, { "input": "12\n3\n99\n24\n46\n75\n63\n57\n55\n10\n62\n34\n52", "output": "7" }, { "input": "12\n55\n75\n1\n98\n63\n64\n9\n39\n82\n18\n47\n9", "output": "6" }, { "input": "100\n678\n771\n96\n282\n135\n749\n168\n668\n17\n658\n979\n446\n998\n331\n606\n756\n37\n515\n538\n205\n647\n547\n904\n842\n647\n286\n774\n414\n267\n791\n595\n465\n8\n327\n855\n174\n339\n946\n184\n250\n807\n422\n679\n980\n64\n530\n312\n351\n676\n911\n803\n991\n669\n50\n293\n841\n545\n598\n737\n894\n231\n754\n588\n83\n873\n767\n833\n482\n905\n903\n970\n571\n715\n59\n777\n697\n537\n861\n339\n212\n149\n889\n905\n70\n970\n307\n830\n465\n968\n291\n430\n317\n942\n944\n330\n235\n814\n880\n415\n76", "output": "58" }, { "input": "100\n154\n60\n97\n638\n139\n150\n570\n579\n601\n647\n804\n237\n245\n549\n288\n347\n778\n282\n916\n441\n974\n145\n957\n886\n655\n702\n930\n618\n132\n520\n972\n48\n94\n54\n682\n433\n896\n134\n845\n636\n242\n842\n125\n141\n240\n130\n409\n666\n948\n938\n604\n110\n474\n484\n364\n40\n807\n271\n438\n288\n201\n814\n754\n589\n341\n576\n146\n952\n819\n923\n222\n535\n336\n83\n314\n911\n303\n911\n384\n601\n249\n330\n735\n271\n142\n204\n405\n783\n775\n449\n590\n139\n109\n276\n45\n205\n454\n836\n82\n841", "output": "50" }, { "input": "1\n1", "output": "1" }, { "input": "4\n1\n1\n1\n2", "output": "3" }, { "input": "5\n1\n2\n4\n8\n16", "output": "3" }, { "input": "7\n1\n2\n4\n8\n16\n32\n64", "output": "4" }, { "input": "3\n1\n2\n4", "output": "2" } ]

1,663,069,081

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

// Online C++ compiler to run C++ program online #include <bits/stdc++.h> using namespace std; int main() { // Write C++ code here long long n; cin>>n; vector<long long> arr; for(long long i=0;i<n;i++){ long long inp; cin>>inp; arr.push_back(inp); } sort(arr.begin(),arr.end()); long long k = n-1; long long m = 0; for(long long i=k-1;i>=0;i--){ if(arr[k]>=arr[i]*2){ m++; k--; } } if(n%2){ if(m>(n/2)){ cout<<(n/2)+1<<endl; } else{ cout<<n-m<<endl; } } else{ if(m>n/2){ cout<<n/2<<endl; } else{ cout<<n-m<<endl; } } return 0; }

Title: Counting Kangaroos is Fun Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held. Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos. The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible. Input Specification: The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105). Output Specification: Output a single integer — the optimal number of visible kangaroos. Demo Input: ['8\n2\n5\n7\n6\n9\n8\n4\n2\n', '8\n9\n1\n6\n2\n6\n5\n8\n3\n'] Demo Output: ['5\n', '5\n'] Note: none

```python // Online C++ compiler to run C++ program online #include <bits/stdc++.h> using namespace std; int main() { // Write C++ code here long long n; cin>>n; vector<long long> arr; for(long long i=0;i<n;i++){ long long inp; cin>>inp; arr.push_back(inp); } sort(arr.begin(),arr.end()); long long k = n-1; long long m = 0; for(long long i=k-1;i>=0;i--){ if(arr[k]>=arr[i]*2){ m++; k--; } } if(n%2){ if(m>(n/2)){ cout<<(n/2)+1<<endl; } else{ cout<<n-m<<endl; } } else{ if(m>n/2){ cout<<n/2<<endl; } else{ cout<<n-m<<endl; } } return 0; } ```

-1

0

none

0

[ "none" ]

null

Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order. While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence.

The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation.

The only line of input should contain the minimum possible length of the sequence.

[ "4\nRURD\n", "6\nRRULDD\n", "26\nRRRULURURUULULLLDLDDRDRDLD\n", "3\nRLL\n", "4\nLRLR\n" ]

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

The illustrations to the first three tests are given below. <img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/> The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence.

0

[ { "input": "4\nRURD", "output": "2" }, { "input": "6\nRRULDD", "output": "2" }, { "input": "26\nRRRULURURUULULLLDLDDRDRDLD", "output": "7" }, { "input": "3\nRLL", "output": "2" }, { "input": "4\nLRLR", "output": "4" }, { "input": "5\nLRDLR", "output": "4" }, { "input": "10\nDDRDUULUDD", "output": "3" }, { "input": "1\nD", "output": "1" } ]

1,482,659,151

2,451

PyPy 3

OK

TESTS

32

187

27,955,200

from sys import stdin n = int(stdin.readline()) s = stdin.readline().strip() x = 0 y = 0 xs = 0 ys = 0 ans = 0 l = 0 for ch in s: xp = x yp = y if ch == 'L': x -= 1 if ch == 'R': x += 1 if ch == 'U': y += 1 if ch == 'D': y -= 1 l += 1 if l > abs(x - xs) + abs(y - ys): xs = xp ys = yp l = 0 ans += 1 print(ans+1)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order. While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence. Input Specification: The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation. Output Specification: The only line of input should contain the minimum possible length of the sequence. Demo Input: ['4\nRURD\n', '6\nRRULDD\n', '26\nRRRULURURUULULLLDLDDRDRDLD\n', '3\nRLL\n', '4\nLRLR\n'] Demo Output: ['2\n', '2\n', '7\n', '2\n', '4\n'] Note: The illustrations to the first three tests are given below. <img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/> The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence.

```python from sys import stdin n = int(stdin.readline()) s = stdin.readline().strip() x = 0 y = 0 xs = 0 ys = 0 ans = 0 l = 0 for ch in s: xp = x yp = y if ch == 'L': x -= 1 if ch == 'R': x += 1 if ch == 'U': y += 1 if ch == 'D': y -= 1 l += 1 if l > abs(x - xs) + abs(y - ys): xs = xp ys = yp l = 0 ans += 1 print(ans+1) ```

3

572

A

Arrays

PROGRAMMING

900

[ "sortings" ]

null

You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.

The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly. The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space. The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*. The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.

Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).

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

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

In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3). In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

500

[ { "input": "3 3\n2 1\n1 2 3\n3 4 5", "output": "YES" }, { "input": "3 3\n3 3\n1 2 3\n3 4 5", "output": "NO" }, { "input": "5 2\n3 1\n1 1 1 1 1\n2 2", "output": "YES" }, { "input": "3 5\n1 1\n5 5 5\n5 5 5 5 5", "output": "NO" }, { "input": "1 1\n1 1\n1\n1", "output": "NO" }, { "input": "3 3\n1 1\n1 2 3\n1 2 3", "output": "YES" }, { "input": "3 3\n1 2\n1 2 3\n1 2 3", "output": "YES" }, { "input": "3 3\n2 2\n1 2 3\n1 2 3", "output": "NO" }, { "input": "10 15\n10 1\n1 1 5 17 22 29 32 36 39 48\n9 10 20 23 26 26 32 32 33 39 43 45 47 49 49", "output": "YES" }, { "input": "10 15\n1 15\n91 91 91 92 92 94 94 95 98 100\n92 92 93 93 93 94 95 96 97 98 98 99 99 100 100", "output": "YES" }, { "input": "15 10\n12 5\n9 25 25 32 32 38 40 41 46 46 48 51 64 64 73\n5 14 30 35 50 52 67 79 89 99", "output": "YES" }, { "input": "15 10\n4 10\n22 32 35 45 45 50 51 55 79 80 83 88 90 92 93\n46 48 52 55 60 60 68 75 80 81", "output": "YES" }, { "input": "20 30\n2 8\n6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 10\n1 1 2 2 2 2 2 2 2 3 3 4 5 5 5 5 6 6 6 6 6 6 7 7 7 8 8 9 10 10", "output": "NO" }, { "input": "20 30\n19 29\n1 1 2 2 2 3 4 4 7 7 7 8 8 8 8 8 9 9 9 9\n6 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10", "output": "NO" }, { "input": "1 1\n1 1\n-1000000000\n30", "output": "YES" }, { "input": "3 3\n1 3\n1 3 3\n3 3 3", "output": "YES" }, { "input": "3 3\n1 1\n3 3 3\n2 2 2", "output": "NO" }, { "input": "5 5\n3 3\n1 5 6 7 8\n1 2 5 6 7", "output": "NO" }, { "input": "3 4\n2 2\n5 6 7\n1 2 3 4", "output": "NO" }, { "input": "3 3\n3 3\n1 2 3\n4 5 6", "output": "YES" }, { "input": "5 5\n4 5\n2 2 3 4 5\n5 6 7 8 9", "output": "YES" } ]

1,596,553,641

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

8

2,000

14,745,600

m,n=map(int,input().split()) a,b=map(int,input().split()) c=list(map(int,input().split())) d=list(map(int,input().split())) flag=0 for i in range(a): for j in range(b): if c[i]<d[-1-j]: flag=1 else: flag=0 break print(['NO','YES'][flag==1])

Title: Arrays Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array. Input Specification: The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly. The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space. The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*. The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*. Output Specification: Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes). Demo Input: ['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3). In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

```python m,n=map(int,input().split()) a,b=map(int,input().split()) c=list(map(int,input().split())) d=list(map(int,input().split())) flag=0 for i in range(a): for j in range(b): if c[i]<d[-1-j]: flag=1 else: flag=0 break print(['NO','YES'][flag==1]) ```

0

486

A

Calculating Function

PROGRAMMING

800

[ "implementation", "math" ]

null

For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*.

The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).

Print *f*(*n*) in a single line.

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

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

*f*(4) = - 1 + 2 - 3 + 4 = 2 *f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3

500

[ { "input": "4", "output": "2" }, { "input": "5", "output": "-3" }, { "input": "1000000000", "output": "500000000" }, { "input": "1000000001", "output": "-500000001" }, { "input": "1000000000000000", "output": "500000000000000" }, { "input": "100", "output": "50" }, { "input": "101", "output": "-51" }, { "input": "102", "output": "51" }, { "input": "103", "output": "-52" }, { "input": "104", "output": "52" }, { "input": "105", "output": "-53" }, { "input": "106", "output": "53" }, { "input": "107", "output": "-54" }, { "input": "108", "output": "54" }, { "input": "109", "output": "-55" }, { "input": "208170109961052", "output": "104085054980526" }, { "input": "46017661651072", "output": "23008830825536" }, { "input": "4018154546667", "output": "-2009077273334" }, { "input": "288565475053", "output": "-144282737527" }, { "input": "3052460231", "output": "-1526230116" }, { "input": "29906716", "output": "14953358" }, { "input": "87897701693326", "output": "43948850846663" }, { "input": "8240", "output": "4120" }, { "input": "577935", "output": "-288968" }, { "input": "62", "output": "31" }, { "input": "1", "output": "-1" }, { "input": "2", "output": "1" }, { "input": "9999999999999", "output": "-5000000000000" }, { "input": "1000000000000", "output": "500000000000" }, { "input": "99999999999999", "output": "-50000000000000" }, { "input": "999999999999999", "output": "-500000000000000" }, { "input": "42191359342", "output": "21095679671" }, { "input": "100000000000000", "output": "50000000000000" }, { "input": "145645214654154", "output": "72822607327077" }, { "input": "4294967296", "output": "2147483648" }, { "input": "3037000499", "output": "-1518500250" }, { "input": "10000000000001", "output": "-5000000000001" }, { "input": "100000017040846", "output": "50000008520423" }, { "input": "98979894985999", "output": "-49489947493000" } ]

1,692,716,930

2,147,483,647

Python 3

MEMORY_LIMIT_EXCEEDED

TESTS

2

234

268,390,400

a=int(input()) print(sum(list(range(2,a+1,2)))-sum(list(range(1,a+1,2))))

Title: Calculating Function Time Limit: None seconds Memory Limit: None megabytes Problem Description: For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*. Input Specification: The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). Output Specification: Print *f*(*n*) in a single line. Demo Input: ['4\n', '5\n'] Demo Output: ['2\n', '-3\n'] Note: *f*(4) = - 1 + 2 - 3 + 4 = 2 *f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3

```python a=int(input()) print(sum(list(range(2,a+1,2)))-sum(list(range(1,a+1,2)))) ```

0

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" } ]

1,567,688,810

2,147,483,647

Python 3

OK

TESTS

133

187

0

n = input() s = {x for x in n} r = n[::-1] print(0) if len(s) == 1 else print(len(n)-1) if r == n else print(len(n))

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 n = input() s = {x for x in n} r = n[::-1] print(0) if len(s) == 1 else print(len(n)-1) if r == n else print(len(n)) ```

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" } ]

1,670,427,662

2,147,483,647

PyPy 3-64

OK

TESTS

32

124

0

input() a = list(map(int, input().split())) j = list() t = list() for i in a: if i % 2 == 0: j.append(i) else: t.append(i) if len(j) == 1: print(a.index(j[0]) + 1) elif len(t) == 1: print(a.index(t[0]) + 1)

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

```python input() a = list(map(int, input().split())) j = list() t = list() for i in a: if i % 2 == 0: j.append(i) else: t.append(i) if len(j) == 1: print(a.index(j[0]) + 1) elif len(t) == 1: print(a.index(t[0]) + 1) ```

3.969

230

B

T-primes

PROGRAMMING

1,300

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

null

We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors. You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.

The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.

[ "3\n4 5 6\n" ]

[ "YES\nNO\nNO\n" ]

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

500

[ { "input": "3\n4 5 6", "output": "YES\nNO\nNO" }, { "input": "2\n48 49", "output": "NO\nYES" }, { "input": "10\n10 9 8 7 6 5 4 3 2 1", "output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO" }, { "input": "1\n36", "output": "NO" }, { "input": "1\n999966000289", "output": "YES" }, { "input": "1\n999993399999", "output": "NO" }, { "input": "9\n111 121 131 111 121 131 111 121 131", "output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO" }, { "input": "1\n1", "output": "NO" }, { "input": "1\n10", "output": "NO" }, { "input": "1\n976197352729", "output": "NO" }, { "input": "1\n1000000000000", "output": "NO" }, { "input": "1\n9", "output": "YES" }, { "input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849", "output": "NO\nNO\nNO\nNO\nNO\nNO" }, { "input": "3\n223092870 6469693230 200560490130", "output": "NO\nNO\nNO" }, { "input": "2\n81 25", "output": "NO\nYES" }, { "input": "1\n16", "output": "NO" }, { "input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81", "output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO" }, { "input": "1\n225", "output": "NO" }, { "input": "1\n1521", "output": "NO" }, { "input": "1\n81", "output": "NO" } ]

1,696,940,992

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

374

7,884,800

import math def sieveOfEratosthenes(n): is_prime = [True] * (n + 1) is_prime[0] = is_prime[1] = False for i in range(2, int(math.sqrt(n)) + 1): if is_prime[i]: for j in range(i * i, n + 1, i): is_prime[j] = False return is_prime def isTPrime(num, is_prime): root = int(math.sqrt(num)) if root * root != num or not is_prime[root]: return False return True n = int(input()) is_prime = sieveOfEratosthenes(1000000) # 生成素数表 for _ in range(n): num = int(input()) if isTPrime(num, is_prime): print("YES") else: print("NO")

Title: T-primes Time Limit: None seconds Memory Limit: None megabytes Problem Description: We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors. You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not. Input Specification: The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier. Output Specification: Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't. Demo Input: ['3\n4 5 6\n'] Demo Output: ['YES\nNO\nNO\n'] Note: The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

```python import math def sieveOfEratosthenes(n): is_prime = [True] * (n + 1) is_prime[0] = is_prime[1] = False for i in range(2, int(math.sqrt(n)) + 1): if is_prime[i]: for j in range(i * i, n + 1, i): is_prime[j] = False return is_prime def isTPrime(num, is_prime): root = int(math.sqrt(num)) if root * root != num or not is_prime[root]: return False return True n = int(input()) is_prime = sieveOfEratosthenes(1000000) # 生成素数表 for _ in range(n): num = int(input()) if isTPrime(num, is_prime): print("YES") else: print("NO") ```

-1

820

A

Mister B and Book Reading

PROGRAMMING

900

[ "implementation" ]

null

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

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

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

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

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

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

500

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

1,499,530,406

1,106

Python 3

OK

TESTS

110

62

5,529,600

a = input() b = a.split(" ") c, v0, v1, a, l = int(b[0]), int(b[1]), int(b[2]), int(b[3]), int(b[4]) day = 1 while c > 0: if day > 1: c = c + l c = c - min(v0 + a * (day - 1), v1) if c <= 0: break day += 1 print(day)

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

```python a = input() b = a.split(" ") c, v0, v1, a, l = int(b[0]), int(b[1]), int(b[2]), int(b[3]), int(b[4]) day = 1 while c > 0: if day > 1: c = c + l c = c - min(v0 + a * (day - 1), v1) if c <= 0: break day += 1 print(day) ```

3

753

A

Santa Claus and Candies

PROGRAMMING

1,000

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

null

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

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

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

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

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

none

500

[ { "input": "5", "output": "2\n1 4 " }, { "input": "9", "output": "3\n1 2 6 " }, { "input": "2", "output": "1\n2 " }, { "input": "1", "output": "1\n1 " }, { "input": "3", "output": "2\n1 2 " }, { "input": "1000", "output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 54 " }, { "input": "4", "output": "2\n1 3 " }, { "input": "6", "output": "3\n1 2 3 " }, { "input": "7", "output": "3\n1 2 4 " }, { "input": "8", "output": "3\n1 2 5 " }, { "input": "10", "output": "4\n1 2 3 4 " }, { "input": "11", "output": "4\n1 2 3 5 " }, { "input": "12", "output": "4\n1 2 3 6 " }, { "input": "13", "output": "4\n1 2 3 7 " }, { "input": "14", "output": "4\n1 2 3 8 " }, { "input": "15", "output": "5\n1 2 3 4 5 " }, { "input": "16", "output": "5\n1 2 3 4 6 " }, { "input": "20", "output": "5\n1 2 3 4 10 " }, { "input": "21", "output": "6\n1 2 3 4 5 6 " }, { "input": "22", "output": "6\n1 2 3 4 5 7 " }, { "input": "27", "output": "6\n1 2 3 4 5 12 " }, { "input": "28", "output": "7\n1 2 3 4 5 6 7 " }, { "input": "29", "output": "7\n1 2 3 4 5 6 8 " }, { "input": "35", "output": "7\n1 2 3 4 5 6 14 " }, { "input": "36", "output": "8\n1 2 3 4 5 6 7 8 " }, { "input": "37", "output": "8\n1 2 3 4 5 6 7 9 " }, { "input": "44", "output": "8\n1 2 3 4 5 6 7 16 " }, { "input": "45", "output": "9\n1 2 3 4 5 6 7 8 9 " }, { "input": "46", "output": "9\n1 2 3 4 5 6 7 8 10 " }, { "input": "230", "output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 " }, { "input": "231", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 " }, { "input": "232", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 " }, { "input": "239", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 " }, { "input": "629", "output": "34\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 68 " }, { "input": "630", "output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 " }, { "input": "631", "output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 " }, { "input": "945", "output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 84 " }, { "input": "946", "output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 " }, { "input": "947", "output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 44 " }, { "input": "989", "output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 86 " }, { "input": "990", "output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 " }, { "input": "991", "output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 " }, { "input": "956", "output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 53 " }, { "input": "981", "output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 78 " }, { "input": "867", "output": "41\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 47 " }, { "input": "906", "output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 45 " }, { "input": "999", "output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 53 " }, { "input": "100", "output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 " }, { "input": "126", "output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 " } ]

1,568,563,150

2,147,483,647

PyPy 3

OK

TESTS

49

155

0

n = int(input()) answer = [] s, i = 0, 1 while s < n: answer.append(i) s += i i += 1 d = s - n answer = list(filter(lambda x: x != d, answer)) print(len(answer)) print(' '.join(map(str, answer)))

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

```python n = int(input()) answer = [] s, i = 0, 1 while s < n: answer.append(i) s += i i += 1 d = s - n answer = list(filter(lambda x: x != d, answer)) print(len(answer)) print(' '.join(map(str, answer))) ```

3

839

C

Journey

PROGRAMMING

1,500

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

null

There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads.

Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .

[ "4\n1 2\n1 3\n2 4\n", "5\n1 2\n1 3\n3 4\n2 5\n" ]

[ "1.500000000000000\n", "2.000000000000000\n" ]

In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

1,500

[ { "input": "4\n1 2\n1 3\n2 4", "output": "1.500000000000000" }, { "input": "5\n1 2\n1 3\n3 4\n2 5", "output": "2.000000000000000" }, { "input": "70\n1 25\n57 1\n18 1\n65 1\n38 1\n1 41\n1 5\n1 69\n1 3\n31 1\n1 8\n1 9\n53 1\n70 1\n45 1\n1 24\n1 42\n1 30\n1 12\n1 37\n64 1\n1 28\n1 58\n1 22\n11 1\n1 4\n1 27\n1 16\n1 21\n54 1\n1 51\n1 43\n29 1\n56 1\n1 39\n32 1\n1 15\n1 17\n1 19\n1 40\n36 1\n48 1\n63 1\n1 7\n1 47\n1 13\n1 46\n60 1\n1 6\n23 1\n20 1\n1 52\n2 1\n26 1\n1 59\n1 66\n10 1\n1 62\n1 68\n1 55\n50 1\n33 1\n44 1\n1 34\n1 35\n1 61\n14 1\n67 1\n49 1", "output": "1.000000000000000" }, { "input": "10\n8 6\n9 10\n8 7\n1 4\n1 8\n9 5\n9 8\n2 5\n3 1", "output": "1.500000000000000" }, { "input": "1", "output": "0.000000000000000" } ]

1,617,362,686

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

9

919

84,787,200

import sys sys.setrecursionlimit(110000) def f(sad, sansa, tezina, parent): da=True dulj=len(veze[sad])-1 if parent==-1: dulj+=1 for x in veze[sad]: if x==parent: continue da=False f(x, sansa/dulj, tezina+1, sad) if da: global t t+=sansa*tezina return n=int(input()) veze=[[] for i in range (n)] for i in range (n-1): u,v=map(int, input().split()) veze[u-1].append(v-1) veze[v-1].append(u-1) t=0 f(0, 1, 0, -1) print(t)

Title: Journey Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value). Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads. Output Specification: Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . Demo Input: ['4\n1 2\n1 3\n2 4\n', '5\n1 2\n1 3\n3 4\n2 5\n'] Demo Output: ['1.500000000000000\n', '2.000000000000000\n'] Note: In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

```python import sys sys.setrecursionlimit(110000) def f(sad, sansa, tezina, parent): da=True dulj=len(veze[sad])-1 if parent==-1: dulj+=1 for x in veze[sad]: if x==parent: continue da=False f(x, sansa/dulj, tezina+1, sad) if da: global t t+=sansa*tezina return n=int(input()) veze=[[] for i in range (n)] for i in range (n-1): u,v=map(int, input().split()) veze[u-1].append(v-1) veze[v-1].append(u-1) t=0 f(0, 1, 0, -1) print(t) ```

-1

590

D

Top Secret Task

PROGRAMMING

2,300

[ "dp" ]

null

A top-secret military base under the command of Colonel Zuev is expecting an inspection from the Ministry of Defence. According to the charter, each top-secret military base must include a top-secret troop that should... well, we cannot tell you exactly what it should do, it is a top secret troop at the end. The problem is that Zuev's base is missing this top-secret troop for some reasons. The colonel decided to deal with the problem immediately and ordered to line up in a single line all *n* soldiers of the base entrusted to him. Zuev knows that the loquacity of the *i*-th soldier from the left is equal to *q**i*. Zuev wants to form the top-secret troop using *k* leftmost soldiers in the line, thus he wants their total loquacity to be as small as possible (as the troop should remain top-secret). To achieve this, he is going to choose a pair of consecutive soldiers and swap them. He intends to do so no more than *s* times. Note that any soldier can be a participant of such swaps for any number of times. The problem turned out to be unusual, and colonel Zuev asked you to help. Determine, what is the minimum total loquacity of the first *k* soldiers in the line, that can be achieved by performing no more than *s* swaps of two consecutive soldiers.

The first line of the input contains three positive integers *n*, *k*, *s* (1<=≤<=*k*<=≤<=*n*<=≤<=150, 1<=≤<=*s*<=≤<=109) — the number of soldiers in the line, the size of the top-secret troop to be formed and the maximum possible number of swap operations of the consecutive pair of soldiers, respectively. The second line of the input contains *n* integer *q**i* (1<=≤<=*q**i*<=≤<=1<=000<=000) — the values of loquacity of soldiers in order they follow in line from left to right.

Print a single integer — the minimum possible total loquacity of the top-secret troop.

[ "3 2 2\n2 4 1\n", "5 4 2\n10 1 6 2 5\n", "5 2 3\n3 1 4 2 5\n" ]

[ "3\n", "18\n", "3\n" ]

In the first sample Colonel has to swap second and third soldiers, he doesn't really need the remaining swap. The resulting soldiers order is: (2, 1, 4). Minimum possible summary loquacity of the secret troop is 3. In the second sample Colonel will perform swaps in the following order: 1. (10, 1, 6 — 2, 5) 1. (10, 1, 2, 6 — 5) The resulting soldiers order is (10, 1, 2, 5, 6). Minimum possible summary loquacity is equal to 18.

1,750

[]

1,666,918,320

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

6

3,000

13,721,600

import sys input = sys.stdin.readline n,k,s = map(int,input().split()) q = list(map(int,input().split())) if s > (n-k)*k: print(sum(sorted(q)[:k])) sys.exit() W = s + k*(k+1)//2 inf = 1<<30 f = [[[inf]*(W+1) for _ in range(k+1)] for _ in range(2)] # f = {} pre = [sum(q[:i+1]) for i in range(len(q))] # def dp(N,K,S): # # min q sum to choose K from first N numbers with weight at most S # cur_tuple = (N,K,S) # if cur_tuple in f: # return f[cur_tuple] # if K == 0: # f[cur_tuple] = 0 # return f[cur_tuple] # if N == K: # f[cur_tuple] = pre[N-1] if K*(K+1)//2 <= S else inf-1 # return f[cur_tuple] # f[cur_tuple] = dp(N-1,K,S) # if S >= N: # f[cur_tuple] = min(f[cur_tuple], dp(N-1,K-1,S-N)+q[N-1]) # # print(N,K,S,f[N][K][S]) # return f[cur_tuple] # print(dp(n,k,W)) # print(len(f)) cur_row = 0 for i in range(W+1): f[cur_row][0][i] = 0 for i in range(1,n+1): cur_row ^= 1 for j in range(k+1): for w in range(W+1): if j == 0: f[cur_row][j][w] = 0 elif i == j: f[cur_row][j][w] = pre[i-1] if j*(j+1)//2 <= w else inf-1 else: f[cur_row][j][w] = f[cur_row^1][j][w] if w >= i: f[cur_row][j][w] = min(f[cur_row][j][w], f[cur_row^1][j-1][w-i]+q[i-1]) print(f[cur_row][k][W]) """ 150 75 5625 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 """

Title: Top Secret Task Time Limit: None seconds Memory Limit: None megabytes Problem Description: A top-secret military base under the command of Colonel Zuev is expecting an inspection from the Ministry of Defence. According to the charter, each top-secret military base must include a top-secret troop that should... well, we cannot tell you exactly what it should do, it is a top secret troop at the end. The problem is that Zuev's base is missing this top-secret troop for some reasons. The colonel decided to deal with the problem immediately and ordered to line up in a single line all *n* soldiers of the base entrusted to him. Zuev knows that the loquacity of the *i*-th soldier from the left is equal to *q**i*. Zuev wants to form the top-secret troop using *k* leftmost soldiers in the line, thus he wants their total loquacity to be as small as possible (as the troop should remain top-secret). To achieve this, he is going to choose a pair of consecutive soldiers and swap them. He intends to do so no more than *s* times. Note that any soldier can be a participant of such swaps for any number of times. The problem turned out to be unusual, and colonel Zuev asked you to help. Determine, what is the minimum total loquacity of the first *k* soldiers in the line, that can be achieved by performing no more than *s* swaps of two consecutive soldiers. Input Specification: The first line of the input contains three positive integers *n*, *k*, *s* (1<=≤<=*k*<=≤<=*n*<=≤<=150, 1<=≤<=*s*<=≤<=109) — the number of soldiers in the line, the size of the top-secret troop to be formed and the maximum possible number of swap operations of the consecutive pair of soldiers, respectively. The second line of the input contains *n* integer *q**i* (1<=≤<=*q**i*<=≤<=1<=000<=000) — the values of loquacity of soldiers in order they follow in line from left to right. Output Specification: Print a single integer — the minimum possible total loquacity of the top-secret troop. Demo Input: ['3 2 2\n2 4 1\n', '5 4 2\n10 1 6 2 5\n', '5 2 3\n3 1 4 2 5\n'] Demo Output: ['3\n', '18\n', '3\n'] Note: In the first sample Colonel has to swap second and third soldiers, he doesn't really need the remaining swap. The resulting soldiers order is: (2, 1, 4). Minimum possible summary loquacity of the secret troop is 3. In the second sample Colonel will perform swaps in the following order: 1. (10, 1, 6 — 2, 5) 1. (10, 1, 2, 6 — 5) The resulting soldiers order is (10, 1, 2, 5, 6). Minimum possible summary loquacity is equal to 18.

```python import sys input = sys.stdin.readline n,k,s = map(int,input().split()) q = list(map(int,input().split())) if s > (n-k)*k: print(sum(sorted(q)[:k])) sys.exit() W = s + k*(k+1)//2 inf = 1<<30 f = [[[inf]*(W+1) for _ in range(k+1)] for _ in range(2)] # f = {} pre = [sum(q[:i+1]) for i in range(len(q))] # def dp(N,K,S): # # min q sum to choose K from first N numbers with weight at most S # cur_tuple = (N,K,S) # if cur_tuple in f: # return f[cur_tuple] # if K == 0: # f[cur_tuple] = 0 # return f[cur_tuple] # if N == K: # f[cur_tuple] = pre[N-1] if K*(K+1)//2 <= S else inf-1 # return f[cur_tuple] # f[cur_tuple] = dp(N-1,K,S) # if S >= N: # f[cur_tuple] = min(f[cur_tuple], dp(N-1,K-1,S-N)+q[N-1]) # # print(N,K,S,f[N][K][S]) # return f[cur_tuple] # print(dp(n,k,W)) # print(len(f)) cur_row = 0 for i in range(W+1): f[cur_row][0][i] = 0 for i in range(1,n+1): cur_row ^= 1 for j in range(k+1): for w in range(W+1): if j == 0: f[cur_row][j][w] = 0 elif i == j: f[cur_row][j][w] = pre[i-1] if j*(j+1)//2 <= w else inf-1 else: f[cur_row][j][w] = f[cur_row^1][j][w] if w >= i: f[cur_row][j][w] = min(f[cur_row][j][w], f[cur_row^1][j-1][w-i]+q[i-1]) print(f[cur_row][k][W]) """ 150 75 5625 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 """ ```

0

792

C

Divide by Three

PROGRAMMING

2,000

[ "dp", "greedy", "math", "number theory" ]

null

A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible. The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not. Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*. If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.

The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=<<=10100000).

Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1.

[ "1033\n", "10\n", "11\n" ]

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

In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.

0

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"200000111", "output": "2000001" }, { "input": "2000000111", "output": "20000001" }, { "input": "1000000", "output": "0" }, { "input": "1003301", "output": "330" }, { "input": "100001", "output": "0" }, { "input": "40000000000000000000888", "output": "400000000000000000008" }, { "input": "100000", "output": "0" }, { "input": "4000000888", "output": "40000008" }, { "input": "334733", "output": "3333" }, { "input": "1000002220", "output": "10000020" }, { "input": "100321", "output": "10032" }, { "input": "101111", "output": "1011" }, { "input": "100000000222", "output": "1000000002" }, { "input": "10001", "output": "0" }, { "input": "7", "output": "-1" }, { "input": "2000000000111", "output": "20000000001" }, { "input": "100000001", "output": "0" }, { "input": "10000000000222", "output": "100000000002" }, { "input": "200000000000000111", "output": "2000000000000001" }, { "input": "404044", "output": "40044" }, { "input": "30202", "output": "300" }, { "input": "20000000000000000111", 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"output": "1000000011" }, { "input": "100007", "output": "0" }, { "input": "20000006711", "output": "200000061" }, { "input": "8059", "output": "9" }, { "input": "8008", "output": "0" }, { "input": "88", "output": "-1" }, { "input": "2002", "output": "0" }, { "input": "2000111", "output": "20001" }, { "input": "100000000100000002", "output": "10000000000000002" }, { "input": "1000000000000000000000000000000000", "output": "0" }, { "input": "10000000000000000222", "output": "100000000000000002" }, { "input": "1000001", "output": "0" }, { "input": "200000000000111", "output": "2000000000001" }, { "input": "2000000002", "output": "0" }, { "input": "2010000100001", "output": "10000100001" }, { "input": "200330", "output": "330" }, { "input": "10000222", "output": "100002" }, { "input": "2005", "output": "0" }, { "input": "100000000000822", "output": "1000000000002" }, { "input": "10000000000000000000000", "output": "0" }, { "input": "7046", "output": "6" }, { "input": "4000117", "output": 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"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003", "output": "3" }, { "input": "20020201", "output": "2002020" }, { "input": "12122", "output": "1122" }, { "input": "20200", "output": "0" }, { "input": "2", "output": "-1" }, { "input": "1000000000000258", "output": "10000000000008" } ]

1,674,382,106

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

31

0

def beautiful_number(n): count = 0 sum = 0 for digit in n: if digit != '0': sum += int(digit) count += 1 if count == 0 or sum % 3 != 0: return -1 i = 0 while i < len(n) and n[i] == '0': i += 1 return n[i:] def beautiful_number(n): digits = [int(d) for d in n] rem = sum(digits) % 3 if rem == 0: return n elif rem == 1: for i in range(len(digits)): if digits[i] % 3 == 1: digits[i] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) for i in range(len(digits)-2): if (digits[i] % 3 == 2) and (digits[i+1] % 3 == 2): digits[i] = -1 digits[i+1] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) else: for i in range(len(digits)): if digits[i] % 3 == 2: digits[i] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) for i in range(len(digits)-2): if (digits[i] % 3 == 1) and (digits[i+1] % 3 == 1): digits[i] = -1 digits[i+1] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) return "-1" n = input() ans = beautiful_number(n).strip('0') if (ans == ''): print(0) else: print(ans)

Title: Divide by Three Time Limit: None seconds Memory Limit: None megabytes Problem Description: A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible. The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not. Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*. If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them. Input Specification: The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=<<=10100000). Output Specification: Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1. Demo Input: ['1033\n', '10\n', '11\n'] Demo Output: ['33\n', '0\n', '-1\n'] Note: In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.

```python def beautiful_number(n): count = 0 sum = 0 for digit in n: if digit != '0': sum += int(digit) count += 1 if count == 0 or sum % 3 != 0: return -1 i = 0 while i < len(n) and n[i] == '0': i += 1 return n[i:] def beautiful_number(n): digits = [int(d) for d in n] rem = sum(digits) % 3 if rem == 0: return n elif rem == 1: for i in range(len(digits)): if digits[i] % 3 == 1: digits[i] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) for i in range(len(digits)-2): if (digits[i] % 3 == 2) and (digits[i+1] % 3 == 2): digits[i] = -1 digits[i+1] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) else: for i in range(len(digits)): if digits[i] % 3 == 2: digits[i] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) for i in range(len(digits)-2): if (digits[i] % 3 == 1) and (digits[i+1] % 3 == 1): digits[i] = -1 digits[i+1] = -1 return ''.join([str(d) if d != -1 else '' for d in digits]) return "-1" n = input() ans = beautiful_number(n).strip('0') if (ans == ''): print(0) else: print(ans) ```

0

352

A

Jeff and Digits

PROGRAMMING

1,000

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

null

Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

[ "4\n5 0 5 0\n", "11\n5 5 5 5 5 5 5 5 0 5 5\n" ]

[ "0\n", "5555555550\n" ]

In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

500

[ { "input": "4\n5 0 5 0", "output": "0" }, { "input": "11\n5 5 5 5 5 5 5 5 0 5 5", "output": "5555555550" }, { "input": "7\n5 5 5 5 5 5 5", "output": "-1" }, { "input": "1\n5", "output": "-1" }, { "input": "1\n0", "output": "0" }, { "input": "11\n5 0 5 5 5 0 0 5 5 5 5", "output": "0" }, { "input": "23\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "9\n5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "24\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "10\n5 5 5 5 5 0 0 5 0 5", "output": "0" }, { "input": "3\n5 5 0", "output": "0" }, { "input": "5\n5 5 0 5 5", "output": "0" }, { "input": "14\n0 5 5 0 0 0 0 0 0 5 5 5 5 5", "output": "0" }, { "input": "3\n5 5 5", "output": "-1" }, { "input": "3\n0 5 5", "output": "0" }, { "input": "13\n0 0 5 0 5 0 5 5 0 0 0 0 0", "output": "0" }, { "input": "9\n5 5 0 5 5 5 5 5 5", "output": "0" }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0" }, { "input": "101\n5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 5 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 5 0 0 5 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 5 0 0", "output": "5555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "214\n5 0 5 0 5 0 0 0 5 5 0 5 0 5 5 0 5 0 0 0 0 5 5 0 0 5 5 0 0 0 0 5 5 5 5 0 5 0 0 0 0 0 0 5 0 0 0 5 0 0 5 0 0 5 5 0 0 5 5 0 0 0 0 0 5 0 5 0 5 5 0 5 0 0 5 5 5 0 5 0 5 0 5 5 0 5 0 0 0 5 5 0 5 0 5 5 5 5 5 0 0 0 0 0 0 5 0 5 5 0 5 0 5 0 5 5 0 0 0 0 5 0 5 0 5 0 0 5 0 0 5 5 5 5 5 0 0 5 0 0 5 0 0 5 0 0 5 0 0 5 0 5 0 0 0 5 0 0 5 5 5 0 0 5 5 5 0 0 5 5 0 0 0 5 0 0 5 5 5 5 5 5 0 5 0 0 5 5 5 5 0 5 5 0 0 0 5 5 5 5 0 0 0 0 5 0 0 5 0 0 5 5 0 0", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "80\n0 0 0 0 5 0 5 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 0 0 5 5 0 5 0 0 0 0 0 0 5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 5 5 0 0 0", "output": "555555555000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "2\n0 0", "output": "0" }, { "input": "3\n5 0 0", "output": "0" }, { "input": "4\n5 5 5 5", "output": "-1" }, { "input": "2\n0 5", "output": "0" }, { "input": "14\n5 5 5 5 5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "18\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "10\n5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "10\n5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "20\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" } ]

1,650,095,461

2,147,483,647

PyPy 3

OK

TESTS

42

184

0

input();a=input().split() print('5'*(a.count('5')//9*9)+'0'*a.count('0') if a.count('5')>8 and a.count('0') else -(a.count('0')==0))

Title: Jeff and Digits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card. Output Specification: In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1. Demo Input: ['4\n5 0 5 0\n', '11\n5 5 5 5 5 5 5 5 0 5 5\n'] Demo Output: ['0\n', '5555555550\n'] Note: In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

```python input();a=input().split() print('5'*(a.count('5')//9*9)+'0'*a.count('0') if a.count('5')>8 and a.count('0') else -(a.count('0')==0)) ```

3

997

A

Convert to Ones

PROGRAMMING

1,500

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

null

You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones. Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$. You can apply the following operations any number of times: - Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»). You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring. What is the minimum number of coins you need to spend to get a string consisting only of ones?

The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring). The second line contains the string $a$ of length $n$, consisting of zeros and ones.

Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.

[ "5 1 10\n01000\n", "5 10 1\n01000\n", "7 2 3\n1111111\n" ]

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

In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$. Then the string was changed as follows: «01000» $\to$ «10000» $\to$ «11111». The total cost of operations is $1 + 10 = 11$. In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$. Then the string was changed as follows: «01000» $\to$ «11000» $\to$ «11111». The overall cost is $1 + 1 = 2$. In the third example, string already consists only of ones, so the answer is $0$.

500

[ { "input": "5 1 10\n01000", "output": "11" }, { "input": "5 10 1\n01000", "output": "2" }, { "input": "7 2 3\n1111111", "output": "0" }, { "input": "1 60754033 959739508\n0", "output": "959739508" }, { "input": "1 431963980 493041212\n1", "output": "0" }, { "input": "1 314253869 261764879\n0", "output": "261764879" }, { "input": "1 491511050 399084767\n1", "output": "0" }, { "input": "2 163093925 214567542\n00", "output": "214567542" }, { "input": "2 340351106 646854722\n10", "output": "646854722" }, { "input": "2 222640995 489207317\n01", "output": "489207317" }, { "input": "2 399898176 552898277\n11", "output": "0" }, { "input": "2 690218164 577155357\n00", "output": "577155357" }, { "input": "2 827538051 754412538\n10", "output": "754412538" }, { "input": "2 636702427 259825230\n01", "output": "259825230" }, { "input": "2 108926899 102177825\n11", "output": "0" }, { "input": "3 368381052 440077270\n000", "output": "440077270" }, { "input": "3 505700940 617334451\n100", "output": "617334451" }, { "input": "3 499624340 643020827\n010", "output": "1142645167" }, { "input": "3 75308005 971848814\n110", "output": "971848814" }, { "input": "3 212627893 854138703\n001", "output": "854138703" }, { "input": "3 31395883 981351561\n101", "output": "981351561" }, { "input": "3 118671447 913685773\n011", "output": "913685773" }, { "input": "3 255991335 385910245\n111", "output": "0" }, { "input": "3 688278514 268200134\n000", "output": "268200134" }, { "input": "3 825598402 445457315\n100", "output": "445457315" }, { "input": "3 300751942 45676507\n010", "output": "91353014" }, { "input": "3 517900980 438071829\n110", "output": "438071829" }, { "input": "3 400190869 280424424\n001", "output": "280424424" }, { "input": "3 577448050 344115384\n101", "output": "344115384" }, { "input": "3 481435271 459737939\n011", "output": "459737939" }, { "input": "3 931962412 913722450\n111", "output": "0" }, { "input": "4 522194562 717060616\n0000", "output": "717060616" }, { "input": "4 659514449 894317797\n1000", "output": "894317797" }, { "input": "4 71574977 796834337\n0100", "output": "868409314" }, { "input": "4 248832158 934154224\n1100", "output": "934154224" }, { "input": "4 71474110 131122047\n0010", "output": "202596157" }, { "input": "4 308379228 503761290\n1010", "output": "812140518" }, { "input": "4 272484957 485636409\n0110", "output": "758121366" }, { "input": "4 662893590 704772137\n1110", "output": "704772137" }, { "input": "4 545183479 547124732\n0001", "output": "547124732" }, { "input": "4 684444619 722440661\n1001", "output": "722440661" }, { "input": "4 477963686 636258459\n0101", "output": "1114222145" }, { "input": "4 360253575 773578347\n1101", "output": "773578347" }, { "input": "4 832478048 910898234\n0011", "output": "910898234" }, { "input": "4 343185412 714767937\n1011", "output": "714767937" }, { "input": "4 480505300 892025118\n0111", "output": "892025118" }, { "input": "4 322857895 774315007\n1111", "output": "0" }, { "input": "4 386548854 246539479\n0000", "output": "246539479" }, { "input": "4 523868742 128829368\n1000", "output": "128829368" }, { "input": "4 956155921 11119257\n0100", "output": "22238514" }, { "input": "4 188376438 93475808\n1100", "output": "93475808" }, { "input": "4 754947032 158668188\n0010", "output": "317336376" }, { "input": "4 927391856 637236921\n1010", "output": "1274473842" }, { "input": "4 359679035 109461393\n0110", "output": "218922786" }, { "input": "4 991751283 202031630\n1110", "output": "202031630" }, { "input": "4 339351517 169008463\n0001", "output": "169008463" }, { "input": "4 771638697 346265644\n1001", "output": "346265644" }, { "input": "4 908958584 523522825\n0101", "output": "1047045650" }, { "input": "4 677682252 405812714\n1101", "output": "405812714" }, { "input": "4 815002139 288102603\n0011", "output": "288102603" }, { "input": "4 952322026 760327076\n1011", "output": "760327076" }, { "input": "4 663334158 312481698\n0111", "output": "312481698" }, { "input": "4 840591339 154834293\n1111", "output": "0" }, { "input": "14 3 11\n10110100011001", "output": "20" }, { "input": "19 1 1\n1010101010101010101", "output": "9" }, { "input": "1 10 1\n1", "output": "0" }, { "input": "1 100 1\n1", "output": "0" }, { "input": "5 1000 1\n11111", "output": "0" }, { "input": "5 10 1\n11111", "output": "0" }, { "input": "7 3 2\n1111111", "output": "0" }, { "input": "5 1 10\n10101", "output": "11" }, { "input": "1 3 2\n1", "output": "0" }, { "input": "2 10 1\n11", "output": "0" }, { "input": "4 148823922 302792601\n1010", "output": "451616523" }, { "input": "1 2 1\n1", "output": "0" }, { "input": "5 2 3\n00011", "output": "3" }, { "input": "1 5 0\n1", "output": "0" }, { "input": "7 2 3\n1001001", "output": "5" }, { "input": "10 1 1000000000\n1111010111", "output": "1000000001" }, { "input": "25 999999998 999999999\n1011001110101010100111001", "output": "7999999985" }, { "input": "2 0 1\n00", "output": "1" }, { "input": "2 1 100\n10", "output": "100" }, { "input": "7 20 3\n1111111", "output": "0" }, { "input": "1 1 0\n1", "output": "0" }, { "input": "3 1 10\n010", "output": "11" }, { "input": "2 1 0\n11", "output": "0" }, { "input": "7 100 3\n1111111", "output": "0" }, { "input": "5 1 1000\n10101", "output": "1001" }, { "input": "5 2 1\n11111", "output": "0" }, { "input": "1 1000 1\n1", "output": "0" }, { "input": "1 799543940 488239239\n1", "output": "0" }, { "input": "6 1 1000\n010101", "output": "1002" }, { "input": "5 11 1\n11111", "output": "0" }, { "input": "5 2 3\n10101", "output": "5" }, { "input": "3 10 1\n111", "output": "0" }, { "input": "7 9 10\n1001011", "output": "19" }, { "input": "5 5 6\n10101", "output": "11" }, { "input": "1 1000000000 0\n1", "output": "0" }, { "input": "4 0 1\n0101", "output": "1" }, { "input": "8 2 3\n10101010", "output": "9" }, { "input": "6 3 100\n010101", "output": "106" }, { "input": "3 3 2\n111", "output": "0" }, { "input": "1 20 1\n1", "output": "0" }, { "input": "2 1 2\n01", "output": "2" } ]

1,532,450,431

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

78

0

n, x, y = list(map(int,input().strip().split(' '))) s = input() groups = 0 for b in range(1,n): if s[b] == '0' and s[b-1] == '1': groups += 1 if groups == 0: print(0) else: print((groups - 1) * min(x,y) + y)

Title: Convert to Ones Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones. Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$. You can apply the following operations any number of times: - Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»). You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring. What is the minimum number of coins you need to spend to get a string consisting only of ones? Input Specification: The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring). The second line contains the string $a$ of length $n$, consisting of zeros and ones. Output Specification: Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations. Demo Input: ['5 1 10\n01000\n', '5 10 1\n01000\n', '7 2 3\n1111111\n'] Demo Output: ['11\n', '2\n', '0\n'] Note: In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$. Then the string was changed as follows: «01000» $\to$ «10000» $\to$ «11111». The total cost of operations is $1 + 10 = 11$. In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$. Then the string was changed as follows: «01000» $\to$ «11000» $\to$ «11111». The overall cost is $1 + 1 = 2$. In the third example, string already consists only of ones, so the answer is $0$.

```python n, x, y = list(map(int,input().strip().split(' '))) s = input() groups = 0 for b in range(1,n): if s[b] == '0' and s[b-1] == '1': groups += 1 if groups == 0: print(0) else: print((groups - 1) * min(x,y) + y) ```

0

659

C

Tanya and Toys

PROGRAMMING

1,200

[ "greedy", "implementation" ]

null

In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles. Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has. Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.

The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.

In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*. In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose. If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.

[ "3 7\n1 3 4\n", "4 14\n4 6 12 8\n" ]

[ "2\n2 5 \n", "4\n7 2 3 1\n" ]

In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

1,000

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135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 999999119\n34421", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n1", "output": "44719\n2 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": "1 1000000000\n44720", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n44719", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n44721", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "3 1000000000\n123456789 234567891 345678912", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 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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": "3 100000\n1000000000 100000000 1", "output": "445\n2 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": "5 5\n100000000 200000000 300000000 400000000 1000000000", "output": "2\n1 2 " }, { "input": "6 3\n1 2 3 4 5 6", "output": "0" }, { "input": "2 1\n1 2", "output": "0" }, { "input": "1 1000000000\n1000000000", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "5 1000000\n1000000000 100000000 10000000 99999999 123456789", "output": "1413\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "2 10000000\n1234567 123456", "output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "1 1\n1000000000", "output": "1\n1 " }, { "input": "1 1000000000\n9999999", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "5 10000\n1000000000 888888888 777777777 666666666 959595959", "output": "140\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 " }, { "input": "3 1\n1000000000 999999999 999999998", "output": "1\n1 " }, { "input": "5 100000000\n100000000 999999999 1 2 3", "output": "14138\n4 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 15..." }, { "input": "3 55\n100000000 1000000000 999999999", "output": "10\n1 2 3 4 5 6 7 8 9 10 " }, { "input": "2 10\n5 10000009", "output": "4\n1 2 3 4 " }, { "input": "3 10000000\n999999999 999999998 999999997", "output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "1 1100\n1000000000", "output": "46\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 " }, { "input": "1 40\n1000000000", "output": "8\n1 2 3 4 5 6 7 8 " } ]

1,645,810,400

2,147,483,647

Python 3

OK

TESTS

61

139

10,240,000

def main(): n, k = map(int, input().split()) arr = set(map(int, input().split())) i = 1 s = 0 answer_arr = [] while True: if i not in arr: if s + i > k: break else: s += i answer_arr.append(i) i += 1 else: i += 1 print(len(answer_arr)) print(*answer_arr) if __name__ == "__main__": main()

Title: Tanya and Toys Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles. Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has. Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this. Input Specification: The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has. Output Specification: In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*. In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose. If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order. Demo Input: ['3 7\n1 3 4\n', '4 14\n4 6 12 8\n'] Demo Output: ['2\n2 5 \n', '4\n7 2 3 1\n'] Note: In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

```python def main(): n, k = map(int, input().split()) arr = set(map(int, input().split())) i = 1 s = 0 answer_arr = [] while True: if i not in arr: if s + i > k: break else: s += i answer_arr.append(i) i += 1 else: i += 1 print(len(answer_arr)) print(*answer_arr) if __name__ == "__main__": main() ```

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" } ]

1,601,626,233

2,147,483,647

Python 3

OK

TESTS

81

218

307,200

x=int(input()) a,b,c=0,0,0 l=[] for i in range(x): l.append(input().split(' ')) for i in range(x): a+=int(l[i][0]) b+=int(l[i][1]) c+=int(l[i][2]) print(['YES','NO'][a!=0 or b!=0 or c!=0])

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

```python x=int(input()) a,b,c=0,0,0 l=[] for i in range(x): l.append(input().split(' ')) for i in range(x): a+=int(l[i][0]) b+=int(l[i][1]) c+=int(l[i][2]) print(['YES','NO'][a!=0 or b!=0 or c!=0]) ```

3.944928

9

A

Die Roll

PROGRAMMING

800

[ "math", "probabilities" ]

A. Die Roll

1

64

Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.

The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.

Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».

[ "4 2\n" ]

[ "1/2\n" ]

Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

0

[ { "input": "4 2", "output": "1/2" }, { "input": "1 1", "output": "1/1" }, { "input": "1 2", "output": "5/6" }, { "input": "1 3", "output": "2/3" }, { "input": "1 4", "output": "1/2" }, { "input": "1 5", "output": "1/3" }, { "input": "1 6", "output": "1/6" }, { "input": "2 1", "output": "5/6" }, { "input": "2 2", "output": "5/6" }, { "input": "2 3", "output": "2/3" }, { "input": "2 4", "output": "1/2" }, { "input": "2 5", "output": "1/3" }, { "input": "2 6", "output": "1/6" }, { "input": "3 1", "output": "2/3" }, { "input": "3 2", "output": "2/3" }, { "input": "3 3", "output": "2/3" }, { "input": "3 4", "output": "1/2" }, { "input": "3 5", "output": "1/3" }, { "input": "3 6", "output": "1/6" }, { "input": "4 1", "output": "1/2" }, { "input": "4 3", "output": "1/2" }, { "input": "4 4", "output": "1/2" }, { "input": "4 5", "output": "1/3" }, { "input": "4 6", "output": "1/6" }, { "input": "5 1", "output": "1/3" }, { "input": "5 2", "output": "1/3" }, { "input": "5 3", "output": "1/3" }, { "input": "5 4", "output": "1/3" }, { "input": "5 5", "output": "1/3" }, { "input": "5 6", "output": "1/6" }, { "input": "6 1", "output": "1/6" }, { "input": "6 2", "output": "1/6" }, { "input": "6 3", "output": "1/6" }, { "input": "6 4", "output": "1/6" }, { "input": "6 5", "output": "1/6" }, { "input": "6 6", "output": "1/6" } ]

1,685,381,883

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

62

0

num = [1, 2, 3, 4, 5, 6] prob = ['1/1', '5/6', '2/3', '1/2', '1/3', '0/1'] m, n = map(int, input().split()) maxi = max(m, n) if maxi in num: print(prob[num.index(maxi)])

Title: Die Roll Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. Input Specification: The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. Output Specification: Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». Demo Input: ['4 2\n'] Demo Output: ['1/2\n'] Note: Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

```python num = [1, 2, 3, 4, 5, 6] prob = ['1/1', '5/6', '2/3', '1/2', '1/3', '0/1'] m, n = map(int, input().split()) maxi = max(m, n) if maxi in num: print(prob[num.index(maxi)]) ```

0

66

A

Petya and Java

PROGRAMMING

1,300

[ "implementation", "strings" ]

A. Petya and Java

2

256

Little Petya has recently started attending a programming club. Naturally he is facing the problem of choosing a programming language. After long considerations he realized that Java is the best choice. The main argument in favor of choosing Java was that it has a very large integer data type, called BigInteger. But having attended several classes of the club, Petya realized that not all tasks require using the BigInteger type. It turned out that in some tasks it is much easier to use small data types. That's why a question arises: "Which integer type to use if one wants to store a positive integer *n*?" Petya knows only 5 integer types: 1) byte occupies 1 byte and allows you to store numbers from <=-<=128 to 127 2) short occupies 2 bytes and allows you to store numbers from <=-<=32768 to 32767 3) int occupies 4 bytes and allows you to store numbers from <=-<=2147483648 to 2147483647 4) long occupies 8 bytes and allows you to store numbers from <=-<=9223372036854775808 to 9223372036854775807 5) BigInteger can store any integer number, but at that it is not a primitive type, and operations with it are much slower. For all the types given above the boundary values are included in the value range. From this list, Petya wants to choose the smallest type that can store a positive integer *n*. Since BigInteger works much slower, Peter regards it last. Help him.

The first line contains a positive number *n*. It consists of no more than 100 digits and doesn't contain any leading zeros. The number *n* can't be represented as an empty string. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Print the first type from the list "byte, short, int, long, BigInteger", that can store the natural number *n*, in accordance with the data given above.

[ "127\n", "130\n", "123456789101112131415161718192021222324\n" ]

[ "byte\n", "short\n", "BigInteger\n" ]

none

500

[ { "input": "127", "output": "byte" }, { "input": "130", "output": "short" }, { "input": "123456789101112131415161718192021222324", "output": "BigInteger" }, { "input": "6", "output": "byte" }, { "input": "16", "output": "byte" }, { "input": "126", "output": "byte" }, { "input": "128", "output": "short" }, { "input": "32766", "output": "short" }, { "input": "111111", "output": "int" }, { "input": "22222", "output": "short" }, { "input": "32767", "output": "short" }, { "input": "32768", "output": "int" }, { "input": "32769", "output": "int" }, { "input": "2147483645", "output": "int" }, { "input": "2147483646", "output": "int" }, { "input": "2147483647", "output": "int" }, { "input": "2147483648", "output": "long" }, { "input": "2147483649", "output": "long" }, { "input": "9223372036854775805", "output": "long" }, { "input": "9223372036854775806", "output": "long" }, { "input": "9223372036854775807", "output": "long" }, { "input": "9223372036854775808", "output": "BigInteger" }, { "input": "9223372036854775809", "output": "BigInteger" }, { "input": "1111111111111111111111111111111111111111111111", "output": "BigInteger" }, { "input": "232", "output": "short" }, { "input": "241796563564014133460267652699", "output": "BigInteger" }, { "input": "29360359146807441660707083821018832188095237636414144034857851003419752010124705615779249", "output": "BigInteger" }, { "input": "337300529263821789926982715723773719445001702036602052198530564", "output": "BigInteger" }, { "input": "381127467969689863953686682245136076127159921", "output": "BigInteger" }, { "input": "2158324958633591462", "output": "long" }, { "input": "268659422768117401499491767189496733446324586965055954729177892248858259490346", "output": "BigInteger" }, { "input": "3023764505449745844381036446038799100004717936344985", "output": "BigInteger" }, { "input": "13408349824892484976400774", "output": "BigInteger" }, { "input": "18880842614378213198381172973704766723997934818440985546083314104481253291692101136681", "output": "BigInteger" }, { "input": "1180990956946757129733650596194933741", "output": "BigInteger" }, { "input": "73795216631038776655609800540262114612084443385902708034055020082090470662930545328551", "output": "BigInteger" }, { "input": "1658370691480968202384509492140362150472696196949", "output": "BigInteger" }, { "input": "59662093286671707493190399502717308574459619342109544431740791973099298641871347858082458491958703", "output": "BigInteger" }, { "input": "205505005582428018613354752739589866670902346355933720701937", "output": "BigInteger" }, { "input": "53348890623013817139699", "output": "BigInteger" }, { "input": "262373979958859125198440634134122707574734706745701184688685117904709744", "output": "BigInteger" }, { "input": "69113784278456828987289369893745977", "output": "BigInteger" }, { "input": "2210209454022702335652564247406666491086662454147967686455330365147159266087", "output": "BigInteger" }, { "input": "630105816139991597267787581532092408135", "output": "BigInteger" }, { "input": "800461429306907809762708270", "output": "BigInteger" }, { "input": "7685166910821197056344900917707673568669808490600751439157", "output": "BigInteger" }, { "input": "713549841568602590705962611607726022334779480510421458817648621376683672722573289661127894", "output": "BigInteger" }, { "input": "680504312323996476676434432", "output": "BigInteger" }, { "input": "3376595620091080825479292544658464163405755746884100218035", "output": "BigInteger" }, { "input": "303681723783491968617491075591006152690484825330764215796396316561122383310011589365655481", "output": "BigInteger" }, { "input": "4868659422768117401499491767189496733446324586965055954729177892248858259490346614099717639491763430", "output": "BigInteger" }, { "input": "3502376450544974584438103644603879910000471793634498544789130945841846713263971487355748226237288709", "output": "BigInteger" }, { "input": "2334083498248924849764007740114454487565621932425948046430072197452845278935316358800789014185793377", "output": "BigInteger" }, { "input": "1988808426143782131983811729737047667239979348184409855460833141044812532916921011366813880911319644", "output": "BigInteger" }, { "input": "1018099095694675712973365059619493374113337270925179793757322992466016001294122941535439492265169131", "output": "BigInteger" }, { "input": "8437952166310387766556098005402621146120844433859027080340550200820904706629305453285512716464931911", "output": "BigInteger" }, { "input": "6965837069148096820238450949214036215047269619694967357734070611376013382163559966747678150791825071", "output": "BigInteger" }, { "input": "4596620932866717074931903995027173085744596193421095444317407919730992986418713478580824584919587030", "output": "BigInteger" }, { "input": "1905505005582428018613354752739589866670902346355933720701937408006000562951996789032987808118459990", "output": "BigInteger" }, { "input": "8433488906230138171396997888322916936677429522910871017295155818340819168386140293774243244435122950", "output": "BigInteger" }, { "input": "6862373979958859125198440634134122707574734706745701184688685117904709744830303784215298687654884810", "output": "BigInteger" }, { "input": "4491137842784568289872893698937459777201151060689848471272003426250808340375567208957554901863756992", "output": "BigInteger" }, { "input": "9721020945402270233565256424740666649108666245414796768645533036514715926608741510409618545180420952", "output": "BigInteger" }, { "input": "7330105816139991597267787581532092408135003429259616955239761315950805521264994021242873979309182812", "output": "BigInteger" }, { "input": "2000461429306907809762708270752707617318091579531521957022940951538737203583768926365382290530636885", "output": "BigInteger" }, { "input": "9868516691082119705634490091770767356866980849060075143915700796802700437762260163478754592094654326", "output": "BigInteger" }, { "input": "8713549841568602590705962611607726022334779480510421458817648621376683672722573289661127894678771177", "output": "BigInteger" }, { "input": "4580504312323996476676434432646986768872786931159974634901608445720467716981185426771899006352697916", "output": "BigInteger" }, { "input": "2537659562009108082547929254465846416340575574688410021803548570097340949141688442074263189944614467", "output": "BigInteger" }, { "input": "1403681723783491968617491075591006152690484825330764215796396316561122383310011589365655481428540208", "output": "BigInteger" }, { "input": "26", "output": "byte" }, { "input": "302376450544", "output": "long" }, { "input": "13", "output": "byte" }, { "input": "188808426143", "output": "long" }, { "input": "118099095694675", "output": "long" }, { "input": "73795216631038", "output": "long" }, { "input": "1658370691480", "output": "long" }, { "input": "596620932866", "output": "long" }, { "input": "2055050055", "output": "int" }, { "input": "533488906", "output": "int" }, { "input": "26237397", "output": "int" }, { "input": "6911378", "output": "int" }, { "input": "221020945402270233", "output": "long" }, { "input": "63010581613999159", "output": "long" }, { "input": "80046142930", "output": "long" }, { "input": "7685166910821197", "output": "long" }, { "input": "71", "output": "byte" }, { "input": "6805043123239964766", "output": "long" }, { "input": "3376", "output": "short" }, { "input": "3036817237", "output": "long" } ]

1,643,272,038

2,147,483,647

Python 3

OK

TESTS

90

154

0

n=int(input()) if n<=127: print("byte") elif n<=32767: print("short") elif n<=2147483647: print("int") elif n<=9223372036854775807: print("long") else: print("BigInteger")

Title: Petya and Java Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Petya has recently started attending a programming club. Naturally he is facing the problem of choosing a programming language. After long considerations he realized that Java is the best choice. The main argument in favor of choosing Java was that it has a very large integer data type, called BigInteger. But having attended several classes of the club, Petya realized that not all tasks require using the BigInteger type. It turned out that in some tasks it is much easier to use small data types. That's why a question arises: "Which integer type to use if one wants to store a positive integer *n*?" Petya knows only 5 integer types: 1) byte occupies 1 byte and allows you to store numbers from <=-<=128 to 127 2) short occupies 2 bytes and allows you to store numbers from <=-<=32768 to 32767 3) int occupies 4 bytes and allows you to store numbers from <=-<=2147483648 to 2147483647 4) long occupies 8 bytes and allows you to store numbers from <=-<=9223372036854775808 to 9223372036854775807 5) BigInteger can store any integer number, but at that it is not a primitive type, and operations with it are much slower. For all the types given above the boundary values are included in the value range. From this list, Petya wants to choose the smallest type that can store a positive integer *n*. Since BigInteger works much slower, Peter regards it last. Help him. Input Specification: The first line contains a positive number *n*. It consists of no more than 100 digits and doesn't contain any leading zeros. The number *n* can't be represented as an empty string. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d). Output Specification: Print the first type from the list "byte, short, int, long, BigInteger", that can store the natural number *n*, in accordance with the data given above. Demo Input: ['127\n', '130\n', '123456789101112131415161718192021222324\n'] Demo Output: ['byte\n', 'short\n', 'BigInteger\n'] Note: none

```python n=int(input()) if n<=127: print("byte") elif n<=32767: print("short") elif n<=2147483647: print("int") elif n<=9223372036854775807: print("long") else: print("BigInteger") ```

3.9615

948

A

Protect Sheep

PROGRAMMING

900

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

null

Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected. The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog. Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.

First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively. Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.

If it is impossible to protect all sheep, output a single line with the word "No". Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf. If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.

[ "6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n", "1 2\nSW\n", "5 5\n.S...\n...S.\nS....\n...S.\n.S...\n" ]

[ "Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n", "No\n", "Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n" ]

In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally. In the second example, there are no empty spots to put dogs that would guard the lone sheep. In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.

500

[ { "input": "1 2\nSW", "output": "No" }, { "input": "10 10\n....W.W.W.\n.........S\n.S.S...S..\nW.......SS\n.W..W.....\n.W...W....\nS..S...S.S\n....W...S.\n..S..S.S.S\nSS.......S", "output": "Yes\nDDDDWDWDWD\nDDDDDDDDDS\nDSDSDDDSDD\nWDDDDDDDSS\nDWDDWDDDDD\nDWDDDWDDDD\nSDDSDDDSDS\nDDDDWDDDSD\nDDSDDSDSDS\nSSDDDDDDDS" }, { "input": "10 10\n....W.W.W.\n...W.....S\n.S.S...S..\nW......WSS\n.W..W.....\n.W...W....\nS..S...S.S\n...WWW..S.\n..S..S.S.S\nSS.......S", "output": "No" }, { "input": "1 50\nW...S..............W.....S..S...............S...W.", "output": "Yes\nWDDDSDDDDDDDDDDDDDDWDDDDDSDDSDDDDDDDDDDDDDDDSDDDWD" }, { "input": "2 4\n...S\n...W", "output": "No" }, { "input": "4 2\n..\n..\n..\nSW", "output": "No" }, { "input": "4 2\n..\n..\n..\nWS", "output": "No" }, { "input": "2 4\n...W\n...S", "output": "No" }, { "input": "50 1\nS\n.\n.\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\n.\nS\n.\nW\n.\nS\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\nW\n.\n.\n.\nW\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.", "output": "Yes\nS\nD\nD\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nD\nS\nD\nW\nD\nS\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nW\nD\nD\nD\nW\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD" }, { "input": "4 4\nW..S\nW..S\nW..S\nW..S", "output": "Yes\nWDDS\nWDDS\nWDDS\nWDDS" }, { "input": "4 4\nSSSS\n....\n....\nWWWW", "output": "Yes\nSSSS\nDDDD\nDDDD\nWWWW" }, { "input": "4 4\nWWWW\n....\n....\nSSSS", "output": "Yes\nWWWW\nDDDD\nDDDD\nSSSS" }, { "input": "4 4\nS..W\nS..W\nS..W\nS..W", "output": "Yes\nSDDW\nSDDW\nSDDW\nSDDW" }, { "input": "1 1\n.", "output": "Yes\nD" }, { "input": "1 1\nW", "output": "Yes\nW" }, { "input": "1 1\nS", "output": "Yes\nS" }, { "input": "4 2\n..\n..\n.W\n.S", "output": "No" }, { "input": "4 2\n..\n..\n.S\n.W", "output": "No" }, { "input": "4 2\n..\n..\nW.\nS.", "output": "No" }, { "input": "4 2\n..\n..\nS.\nW.", "output": "No" }, { "input": "2 4\n....\n..SW", "output": "No" }, { "input": "2 4\n....\n..WS", "output": "No" }, { "input": "1 2\nS.", "output": "Yes\nSD" } ]

1,604,231,100

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

61

307,200

r, c = map(int, input().split(" ")) land = [] for i in range(r): land.append(list(map(int, input().split(" ")))) sheep_list = {} wolf = False possible = True for i in range(r): for j in range(c): if land[i][j] == "S": sheep_list[(i, j)] = False elif land[i][j] == "W": wolf = True if wolf: for sheep in sheep_list.keys(): st = [sheep] while len(st) > 0: i, j = st.pop() if (i < 0 or i >= r) or (j < 0 or j >= c): continue if land[i][j] == "S": sheep_list[(i, j)] = True st.append(land[i - 1][j]) st.append(land[i + 1][j]) st.append(land[i][j - 1]) st.append(land[i][j + 1]) elif land[i][j] == ".": land[i][j] = "D" else: possible = False if not possible: break if not possible: print("No") else: print("Yes") for i in range(r): print(" ".join((str(i) for i in land[r])))

Title: Protect Sheep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected. The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog. Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number. Input Specification: First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively. Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell. Output Specification: If it is impossible to protect all sheep, output a single line with the word "No". Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf. If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs. Demo Input: ['6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n', '1 2\nSW\n', '5 5\n.S...\n...S.\nS....\n...S.\n.S...\n'] Demo Output: ['Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n', 'No\n', 'Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n'] Note: In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally. In the second example, there are no empty spots to put dogs that would guard the lone sheep. In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.

```python r, c = map(int, input().split(" ")) land = [] for i in range(r): land.append(list(map(int, input().split(" ")))) sheep_list = {} wolf = False possible = True for i in range(r): for j in range(c): if land[i][j] == "S": sheep_list[(i, j)] = False elif land[i][j] == "W": wolf = True if wolf: for sheep in sheep_list.keys(): st = [sheep] while len(st) > 0: i, j = st.pop() if (i < 0 or i >= r) or (j < 0 or j >= c): continue if land[i][j] == "S": sheep_list[(i, j)] = True st.append(land[i - 1][j]) st.append(land[i + 1][j]) st.append(land[i][j - 1]) st.append(land[i][j + 1]) elif land[i][j] == ".": land[i][j] = "D" else: possible = False if not possible: break if not possible: print("No") else: print("Yes") for i in range(r): print(" ".join((str(i) for i in land[r]))) ```

-1

768

A

Oath of the Night's Watch

PROGRAMMING

900

[ "constructive algorithms", "sortings" ]

null

"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath. With that begins the watch of Jon Snow. He is assigned the task to support the stewards. This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him. Can you find how many stewards will Jon support?

First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow. Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.

Output a single integer representing the number of stewards which Jon will feed.

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

[ "0", "1" ]

In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5. In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

500

[ { "input": "2\n1 5", "output": "0" }, { "input": "3\n1 2 5", "output": "1" }, { "input": "4\n1 2 3 4", "output": "2" }, { "input": "8\n7 8 9 4 5 6 1 2", "output": "6" }, { "input": "1\n1", "output": "0" }, { "input": "1\n100", "output": "0" }, { "input": "205\n5 5 3 3 6 2 9 3 8 9 6 6 10 8 1 5 3 3 1 2 9 9 9 3 9 10 3 9 8 3 5 6 6 4 6 9 2 9 10 9 5 6 6 7 4 2 6 3 4 1 10 1 7 2 7 7 3 2 6 5 5 2 9 3 8 8 7 6 6 4 2 2 6 2 3 5 7 2 2 10 1 4 6 9 2 3 7 2 2 7 4 4 9 10 7 5 8 6 5 3 6 10 2 7 5 6 6 8 3 3 9 4 3 5 7 9 3 2 1 1 3 2 1 9 3 1 4 4 10 2 5 5 8 1 4 8 5 3 1 10 8 6 5 8 3 5 4 5 4 4 6 7 2 8 10 8 7 6 6 9 6 7 1 10 3 2 5 10 4 4 5 4 3 4 8 5 3 8 10 3 10 9 7 2 1 8 6 4 6 5 8 10 2 6 7 4 9 4 5 1 8 7 10 3 1", "output": "174" }, { "input": "4\n1000000000 99999999 1000000000 1000000000", "output": "0" }, { "input": "3\n2 2 2", "output": "0" }, { "input": "5\n1 1 1 1 1", "output": "0" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "6\n1 1 3 3 2 2", "output": "2" }, { "input": "7\n1 1 1 1 1 1 1", "output": "0" }, { "input": "4\n1 1 2 5", "output": "1" }, { "input": "3\n0 0 0", "output": "0" }, { "input": "5\n0 0 0 0 0", "output": "0" }, { "input": "5\n1 1 1 1 5", "output": "0" }, { "input": "5\n1 1 2 3 3", "output": "1" }, { "input": "3\n1 1 3", "output": "0" }, { "input": "3\n2 2 3", "output": "0" }, { "input": "1\n6", "output": "0" }, { "input": "5\n1 5 3 5 1", "output": "1" }, { "input": "7\n1 2 2 2 2 2 3", "output": "5" }, { "input": "4\n2 2 2 2", "output": "0" }, { "input": "9\n2 2 2 3 4 5 6 6 6", "output": "3" }, { "input": "10\n1 1 1 2 3 3 3 3 3 3", "output": "1" }, { "input": "6\n1 1 1 1 1 1", "output": "0" }, { "input": "3\n0 0 1", "output": "0" }, { "input": "9\n1 1 1 2 2 2 3 3 3", "output": "3" }, { "input": "3\n1 2 2", "output": "0" }, { "input": "6\n2 2 2 2 2 2", "output": "0" }, { "input": "5\n2 2 2 2 2", "output": "0" }, { "input": "5\n5 5 5 5 5", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "6\n1 2 5 5 5 5", "output": "1" }, { "input": "5\n1 2 3 3 3", "output": "1" }, { "input": "3\n1 1 2", "output": "0" }, { "input": "6\n1 1 1 1 1 2", "output": "0" }, { "input": "5\n1 1 2 4 4", "output": "1" }, { "input": "3\n999999 5999999 9999999", "output": "1" }, { "input": "4\n1 1 5 5", "output": "0" }, { "input": "9\n1 1 1 2 2 2 4 4 4", "output": "3" }, { "input": "5\n1 3 4 5 1", "output": "2" }, { "input": "5\n3 3 3 3 3", "output": "0" }, { "input": "5\n1 1 2 2 2", "output": "0" }, { "input": "5\n2 1 1 1 3", "output": "1" }, { "input": "5\n0 0 0 1 2", "output": "1" }, { "input": "4\n2 2 2 3", "output": "0" }, { "input": "7\n1 1 1 1 5 5 5", "output": "0" }, { "input": "5\n1 2 3 4 4", "output": "2" }, { "input": "2\n5 4", "output": "0" }, { "input": "4\n5 5 5 5", "output": "0" }, { "input": "5\n1 1 1 5 5", "output": "0" }, { "input": "2\n1 1", "output": "0" }, { "input": "1\n3", "output": "0" }, { "input": "3\n2 1 2", "output": "0" }, { "input": "4\n1 2 2 2", "output": "0" }, { "input": "8\n1000000000 1000000000 1000000000 999999999 999999999 999999999 999999998 999999998", "output": "3" }, { "input": "5\n1 1 3 4 4", "output": "1" }, { "input": "6\n1 1 2 2 3 3", "output": "2" }, { "input": "4\n1 1 1 1", "output": "0" }, { "input": "9\n1 2 3 4 1 5 6 7 8", "output": "6" }, { "input": "8\n5 4 4 6 6 4 4 3", "output": "5" }, { "input": "8\n4 3 3 3 3 3 3 3", "output": "0" }, { "input": "7\n4 3 3 3 3 3 3", "output": "0" }, { "input": "6\n4 3 3 3 3 3", "output": "0" }, { "input": "5\n4 3 3 3 3", "output": "0" } ]

1,684,008,681

2,147,483,647

PyPy 3-64

OK

TESTS

88

92

13,619,200

n = int(input()) lst = list(map(int, input().split())) a = min(lst) b = max(lst) a1 = lst.count(a) b1 = lst.count(b) if n - a1-b1>0: print(n - a1 - b1) else: print(0)

Title: Oath of the Night's Watch Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath. With that begins the watch of Jon Snow. He is assigned the task to support the stewards. This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him. Can you find how many stewards will Jon support? Input Specification: First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow. Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. Output Specification: Output a single integer representing the number of stewards which Jon will feed. Demo Input: ['2\n1 5\n', '3\n1 2 5\n'] Demo Output: ['0', '1'] Note: In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5. In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

```python n = int(input()) lst = list(map(int, input().split())) a = min(lst) b = max(lst) a1 = lst.count(a) b1 = lst.count(b) if n - a1-b1>0: print(n - a1 - b1) else: print(0) ```

3

151

A

Soft Drinking

PROGRAMMING

800

[ "implementation", "math" ]

null

This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt. To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?

The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.

Print a single integer — the number of toasts each friend can make.

[ "3 4 5 10 8 100 3 1\n", "5 100 10 1 19 90 4 3\n", "10 1000 1000 25 23 1 50 1\n" ]

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

A comment to the first sample: Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.

500

[ { "input": "3 4 5 10 8 100 3 1", "output": "2" }, { "input": "5 100 10 1 19 90 4 3", "output": "3" }, { "input": "10 1000 1000 25 23 1 50 1", "output": "0" }, { "input": "1 7 4 5 5 8 3 2", "output": "4" }, { "input": "2 3 3 5 5 10 1 3", "output": "1" }, { "input": "2 6 4 5 6 5 1 3", "output": "0" }, { "input": "1 7 3 5 3 6 2 1", "output": "6" }, { "input": "2 4 5 4 5 7 3 2", "output": "1" }, { "input": "2 3 6 5 7 8 2 1", "output": "4" }, { "input": "1 4 5 5 3 10 3 1", "output": "6" }, { "input": "1 4 6 7 3 5 1 3", "output": "1" }, { "input": "1 6 5 5 5 8 3 1", "output": "8" }, { "input": "1 7 5 3 3 9 2 1", "output": "9" }, { "input": "3 5 3 7 6 10 3 1", "output": "1" }, { "input": "3 6 3 5 3 6 3 1", "output": "2" }, { "input": "1 7 5 5 5 5 2 2", "output": "2" }, { "input": "2 5 3 5 6 9 2 1", "output": "3" }, { "input": "3 4 3 5 3 6 2 1", "output": "2" }, { "input": "1 5 5 4 7 6 3 1", "output": "6" }, { "input": "2 3 7 6 5 9 3 1", "output": "3" }, { "input": "2 6 5 3 3 8 1 1", "output": "4" }, { "input": "2 4 7 3 4 10 2 1", "output": "5" }, { "input": "1 1000 1000 1000 1000 1000 1 1", "output": "1000" }, { "input": "17 1000 1000 1000 1000 1000 3 7", "output": "8" }, { "input": "115 1000 1000 1000 1000 1000 17 15", "output": "0" }, { "input": "1 587 981 1 2 1 1 1", "output": "1" }, { "input": "1 1 2 1 2 2 1 1", "output": "2" } ]

1,693,162,853

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

9

92

0

namigos, kgarrafa, lmililitros, climao, dpedacos, psal, nlmilipessoa, npsalpessoa = map(int,input("").split()) bebidas = int((kgarrafa*lmililitros)/namigos) limoes = climao*dpedacos sal = psal/npsalpessoa #print(bebidas, limoes, sal ) menor = min(bebidas, limoes, sal) drinks = int(menor / namigos) print(drinks)

Title: Soft Drinking Time Limit: None seconds Memory Limit: None megabytes Problem Description: This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt. To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make? Input Specification: The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space. Output Specification: Print a single integer — the number of toasts each friend can make. Demo Input: ['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n'] Demo Output: ['2\n', '3\n', '0\n'] Note: A comment to the first sample: Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.

```python namigos, kgarrafa, lmililitros, climao, dpedacos, psal, nlmilipessoa, npsalpessoa = map(int,input("").split()) bebidas = int((kgarrafa*lmililitros)/namigos) limoes = climao*dpedacos sal = psal/npsalpessoa #print(bebidas, limoes, sal ) menor = min(bebidas, limoes, sal) drinks = int(menor / namigos) print(drinks) ```

0

216

A

Tiling with Hexagons

PROGRAMMING

1,200

[ "implementation", "math" ]

null

Several ages ago Berland was a kingdom. The King of Berland adored math. That's why, when he first visited one of his many palaces, he first of all paid attention to the floor in one hall. The floor was tiled with hexagonal tiles. The hall also turned out hexagonal in its shape. The King walked along the perimeter of the hall and concluded that each of the six sides has *a*, *b*, *c*, *a*, *b* and *c* adjacent tiles, correspondingly. To better visualize the situation, look at the picture showing a similar hexagon for *a*<==<=2, *b*<==<=3 and *c*<==<=4. According to the legend, as the King of Berland obtained the values *a*, *b* and *c*, he almost immediately calculated the total number of tiles on the hall floor. Can you do the same?

The first line contains three integers: *a*, *b* and *c* (2<=≤<=*a*,<=*b*,<=*c*<=≤<=1000).

Print a single number — the total number of tiles on the hall floor.

[ "2 3 4\n" ]

[ "18" ]

none

500

[ { "input": "2 3 4", "output": "18" }, { "input": "2 2 2", "output": "7" }, { "input": "7 8 13", "output": "224" }, { "input": "14 7 75", "output": "1578" }, { "input": "201 108 304", "output": "115032" }, { "input": "999 998 996", "output": "2983022" }, { "input": "2 2 3", "output": "10" }, { "input": "2 3 2", "output": "10" }, { "input": "3 2 2", "output": "10" }, { "input": "2 3 3", "output": "14" }, { "input": "3 2 3", "output": "14" }, { "input": "3 3 2", "output": "14" }, { "input": "3 3 3", "output": "19" }, { "input": "4 5 3", "output": "36" }, { "input": "2 2 856", "output": "2569" }, { "input": "2 986 2", "output": "2959" }, { "input": "985 2 2", "output": "2956" }, { "input": "2 958 983", "output": "943654" }, { "input": "992 2 912", "output": "906607" }, { "input": "789 894 2", "output": "707048" }, { "input": "1000 1000 1000", "output": "2997001" }, { "input": "384 458 284", "output": "413875" }, { "input": "709 14 290", "output": "218584" }, { "input": "485 117 521", "output": "369265" }, { "input": "849 333 102", "output": "401998" }, { "input": "998 999 1000", "output": "2991006" }, { "input": "2 2 1000", "output": "3001" }, { "input": "2 1000 2", "output": "3001" }, { "input": "1000 2 2", "output": "3001" }, { "input": "1000 2 1000", "output": "1001999" }, { "input": "865 291 383", "output": "692925" }, { "input": "41 49 28", "output": "4412" }, { "input": "34 86 90", "output": "13515" }, { "input": "39 23 56", "output": "4252" }, { "input": "14 99 81", "output": "10346" }, { "input": "48 38 193", "output": "18144" }, { "input": "395 85 22", "output": "43634" }, { "input": "38 291 89", "output": "39922" }, { "input": "7 23 595", "output": "17387" }, { "input": "948 48 3", "output": "47494" } ]

1,518,303,624

2,147,483,647

Python 3

OK

TESTS

40

124

5,632,000

def readInts(): return map(int, input().split()) a, b, c = readInts() ans = 0 while min(a,b,c) > 1: ans += 2*(a+b+c) - 6 a -= 1 b -= 1 c -= 1 ans += a*b*c print(ans)

Title: Tiling with Hexagons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Several ages ago Berland was a kingdom. The King of Berland adored math. That's why, when he first visited one of his many palaces, he first of all paid attention to the floor in one hall. The floor was tiled with hexagonal tiles. The hall also turned out hexagonal in its shape. The King walked along the perimeter of the hall and concluded that each of the six sides has *a*, *b*, *c*, *a*, *b* and *c* adjacent tiles, correspondingly. To better visualize the situation, look at the picture showing a similar hexagon for *a*<==<=2, *b*<==<=3 and *c*<==<=4. According to the legend, as the King of Berland obtained the values *a*, *b* and *c*, he almost immediately calculated the total number of tiles on the hall floor. Can you do the same? Input Specification: The first line contains three integers: *a*, *b* and *c* (2<=≤<=*a*,<=*b*,<=*c*<=≤<=1000). Output Specification: Print a single number — the total number of tiles on the hall floor. Demo Input: ['2 3 4\n'] Demo Output: ['18'] Note: none

```python def readInts(): return map(int, input().split()) a, b, c = readInts() ans = 0 while min(a,b,c) > 1: ans += 2*(a+b+c) - 6 a -= 1 b -= 1 c -= 1 ans += a*b*c print(ans) ```

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" } ]

1,683,954,261

2,147,483,647

Python 3

OK

TESTS

41

31

0

a = int(input()) i = 0 s = 5 while s < a: i += 1 s += 5 * 2 ** i n = ['Sheldon'] + ['Leonard'] + ['Penny'] + ['Rajesh'] + ['Howard'] print(n[(a - (s - 5 * 2 ** i) - 1) // 2 ** i])

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 a = int(input()) i = 0 s = 5 while s < a: i += 1 s += 5 * 2 ** i n = ['Sheldon'] + ['Leonard'] + ['Penny'] + ['Rajesh'] + ['Howard'] print(n[(a - (s - 5 * 2 ** i) - 1) // 2 ** i]) ```

3.9845

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" } ]

1,644,237,429

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

5

60

0

def gcd(a,b): if a<b: smaller=a else: smaller=b for i in range(1,smaller+1): if a%i==0 and b%i==0: hcf= i return hcf x, y ,z= input().split() if gcd(int(x),int(z))>gcd(int(y),int(z)): print(0) else: print(1)

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

```python def gcd(a,b): if a<b: smaller=a else: smaller=b for i in range(1,smaller+1): if a%i==0 and b%i==0: hcf= i return hcf x, y ,z= input().split() if gcd(int(x),int(z))>gcd(int(y),int(z)): print(0) else: print(1) ```

0

282

A

Bit++

PROGRAMMING

800

[ "implementation" ]

null

The classic programming language of Bitland is Bit++. This language is so peculiar and complicated. The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations: - Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1. A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains. A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains. You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme. Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.

Print a single integer — the final value of *x*.

[ "1\n++X\n", "2\nX++\n--X\n" ]

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

none

500

[ { "input": "1\n++X", "output": "1" }, { "input": "2\nX++\n--X", "output": "0" }, { "input": "3\n++X\n++X\n++X", "output": "3" }, { "input": "2\n--X\n--X", "output": "-2" }, { "input": "5\n++X\n--X\n++X\n--X\n--X", "output": "-1" }, { "input": "28\nX--\n++X\nX++\nX++\nX++\n--X\n--X\nX++\nX--\n++X\nX++\n--X\nX--\nX++\nX--\n++X\n++X\nX++\nX++\nX++\nX++\n--X\n++X\n--X\n--X\n--X\n--X\nX++", "output": "4" }, { "input": "94\nX++\nX++\n++X\n++X\nX--\n--X\nX++\n--X\nX++\n++X\nX++\n++X\n--X\n--X\n++X\nX++\n--X\nX--\nX--\n--X\nX--\nX--\n--X\n++X\n--X\nX--\nX--\nX++\n++X\n--X\nX--\n++X\n--X\n--X\nX--\nX--\nX++\nX++\nX--\nX++\nX--\nX--\nX--\n--X\nX--\nX--\nX--\nX++\n++X\nX--\n++X\nX++\n--X\n--X\n--X\n--X\n++X\nX--\n--X\n--X\n++X\nX--\nX--\nX++\n++X\nX++\n++X\n--X\n--X\nX--\n++X\nX--\nX--\n++X\n++X\n++X\n++X\nX++\n++X\n--X\nX++\n--X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\nX--\nX--\n--X\n++X\nX++", "output": "-10" }, { "input": "56\n--X\nX--\n--X\n--X\nX--\nX--\n--X\nX++\n++X\n--X\nX++\nX--\n--X\n++X\n--X\nX--\nX--\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\n++X\nX++\nX++\n--X\nX++\nX--\n--X\nX--\n--X\nX++\n++X\n--X\n++X\nX++\nX--\n--X\n--X\n++X\nX--\nX--\n--X\nX--\n--X\nX++\n--X\n++X\n--X", "output": "-14" }, { "input": "59\nX--\n--X\nX++\n++X\nX--\n--X\n--X\n++X\n++X\n++X\n++X\nX++\n++X\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX++\n--X\n++X\nX++\n--X\n--X\nX++\nX++\n--X\nX++\nX++\nX++\nX--\nX--\n--X\nX++\nX--\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\nX--\n++X\n--X\nX++\nX++\nX--\nX++\n++X\nX--\nX++\nX--\nX--\n++X", "output": "3" }, { "input": "87\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\nX--\n++X\n--X\n--X\nX++\n--X\nX--\nX++\n++X\n--X\n++X\n++X\n--X\n++X\n--X\nX--\n++X\n++X\nX--\nX++\nX++\n--X\n--X\n++X\nX--\n--X\n++X\n--X\nX++\n--X\n--X\nX--\n++X\n++X\n--X\nX--\nX--\nX--\nX--\nX--\nX++\n--X\n++X\n--X\nX++\n++X\nX++\n++X\n--X\nX++\n++X\nX--\n--X\nX++\n++X\nX++\nX++\n--X\n--X\n++X\n--X\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX--\n--X\n++X\n++X", "output": "-5" }, { "input": "101\nX++\nX++\nX++\n++X\n--X\nX--\nX++\nX--\nX--\n--X\n--X\n++X\nX++\n++X\n++X\nX--\n--X\n++X\nX++\nX--\n++X\n--X\n--X\n--X\n++X\n--X\n++X\nX++\nX++\n++X\n--X\nX++\nX--\nX++\n++X\n++X\nX--\nX--\nX--\nX++\nX++\nX--\nX--\nX++\n++X\n++X\n++X\n--X\n--X\n++X\nX--\nX--\n--X\n++X\nX--\n++X\nX++\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n++X\n--X\nX++\n++X\nX--\n++X\nX--\n++X\nX++\nX--\n++X\nX++\n--X\nX++\nX++\n++X\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\n++X\n++X\n--X\nX--\nX--\nX--\nX--\n--X\n--X\n--X\n++X\n--X\n--X", "output": "1" }, { "input": "63\n--X\nX--\n++X\n--X\n++X\nX++\n--X\n--X\nX++\n--X\n--X\nX++\nX--\nX--\n--X\n++X\nX--\nX--\nX++\n++X\nX++\nX++\n--X\n--X\n++X\nX--\nX--\nX--\n++X\nX++\nX--\n--X\nX--\n++X\n++X\nX++\n++X\nX++\nX++\n--X\nX--\n++X\nX--\n--X\nX--\nX--\nX--\n++X\n++X\n++X\n++X\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n++X\nX--\n++X\n++X\nX--", "output": "1" }, { "input": "45\n--X\n++X\nX--\n++X\n++X\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX++\n++X\nX--\n++X\n++X\nX--\nX++\nX--\n--X\nX--\n++X\n++X\n--X\n--X\nX--\nX--\n--X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\nX--\n++X\n++X\nX++\nX++\n++X\n++X\nX++", "output": "-3" }, { "input": "21\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX--\nX++\nX--\nX--\nX--\nX++\n++X\nX++\n++X\n--X\nX--\n--X\nX++\n++X", "output": "1" }, { "input": "100\n--X\n++X\nX++\n++X\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\n++X\nX--\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n++X\nX++\n++X\nX--\n--X\n++X\nX--\n--X\n++X\n++X\nX--\nX++\nX++\nX++\n++X\n--X\n++X\nX++\nX--\n++X\n++X\n--X\n++X\nX--\nX--\nX--\nX++\nX--\nX--\nX++\nX++\n--X\nX++\nX++\n--X\nX--\n--X\n++X\n--X\n++X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\n++X\nX++\nX--\nX++\nX--\nX++\nX++\n--X\nX--\nX--\n++X\nX--\n--X\n--X\nX++\n--X\n--X\nX--\nX--\n++X\n++X\nX--\n++X\nX++\n--X\n++X\n++X\nX++\n--X\n--X\nX++", "output": "8" }, { "input": "17\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n--X\nX--\nX++\nX--\n--X\n--X\nX--\n++X\nX--\nX++", "output": "-1" }, { "input": "77\n++X\nX++\n--X\nX--\n--X\n--X\nX--\nX++\nX--\nX++\nX--\n++X\n--X\n--X\n--X\n--X\n++X\nX--\nX++\nX--\n--X\nX--\n--X\nX--\n++X\n--X\n++X\n++X\nX++\nX++\nX--\n--X\nX--\nX--\nX++\n--X\n--X\n++X\nX--\nX--\n++X\nX++\nX--\n++X\n--X\nX++\nX--\n++X\n++X\n++X\nX--\nX--\nX--\n--X\n++X\n++X\n++X\nX++\n--X\n--X\n++X\n--X\nX--\nX++\n++X\nX++\n++X\nX--\nX++\nX++\n--X\nX++\nX++\nX++\n--X\nX++\nX--", "output": "-5" }, { "input": "21\nX--\n++X\n--X\nX--\n++X\nX--\n++X\nX--\n--X\n++X\nX++\n++X\nX++\n++X\nX--\n--X\nX++\nX++\nX--\n++X\nX--", "output": "1" }, { "input": "1\nX--", "output": "-1" } ]

1,696,758,283

2,147,483,647

Python 3

OK

TESTS

36

46

0

n = int(input()) cou = 0 for i in range(n): a = str(input()) if a[0] == '+' or a[len(a) - 1] == '+': cou += 1 elif a[0] == '-' or a[len(a) - 1] == '-': cou -= 1 print(cou)

Title: Bit++ Time Limit: None seconds Memory Limit: None megabytes Problem Description: The classic programming language of Bitland is Bit++. This language is so peculiar and complicated. The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations: - Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1. A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains. A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains. You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed). Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme. Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order. Output Specification: Print a single integer — the final value of *x*. Demo Input: ['1\n++X\n', '2\nX++\n--X\n'] Demo Output: ['1\n', '0\n'] Note: none

```python n = int(input()) cou = 0 for i in range(n): a = str(input()) if a[0] == '+' or a[len(a) - 1] == '+': cou += 1 elif a[0] == '-' or a[len(a) - 1] == '-': cou -= 1 print(cou) ```

3

41

A

Translation

PROGRAMMING

800

[ "implementation", "strings" ]

A. Translation

2

256

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

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

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

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

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

none

500

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

1,687,269,890

2,147,483,647

Python 3

OK

TESTS

40

92

0

n1 = input() n2 = input() t = len(n2) - 1 count = 0 for i in range(len(n1)): if (n1[i] == n2[t]): count += 1 t -= 1 if count == len(n1): print("YES") else: print("NO")

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

```python n1 = input() n2 = input() t = len(n2) - 1 count = 0 for i in range(len(n1)): if (n1[i] == n2[t]): count += 1 t -= 1 if count == len(n1): print("YES") else: print("NO") ```

3.977

703

A

Mishka and Game

PROGRAMMING

800

[ "implementation" ]

null

Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game. Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner. In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw. Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!

The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds. The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.

If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line. If Chris is the winner of the game, print "Chris" (without quotes) in the only line. If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.

[ "3\n3 5\n2 1\n4 2\n", "2\n6 1\n1 6\n", "3\n1 5\n3 3\n2 2\n" ]

[ "Mishka", "Friendship is magic!^^", "Chris" ]

In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game. In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1. In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.

500

[ { "input": "3\n3 5\n2 1\n4 2", "output": "Mishka" }, { "input": "2\n6 1\n1 6", "output": "Friendship is magic!^^" }, { "input": "3\n1 5\n3 3\n2 2", "output": "Chris" }, { "input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1", "output": "Mishka" }, { "input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5", "output": "Chris" }, { "input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5", "output": "Friendship is magic!^^" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3", "output": "Mishka" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4", "output": "Mishka" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3", "output": "Chris" }, { "input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4", "output": "Mishka" }, { "input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3", "output": "Mishka" }, { "input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3", "output": "Mishka" }, { "input": "5\n3 6\n3 5\n3 5\n1 6\n3 5", "output": "Chris" }, { "input": "4\n4 1\n2 4\n5 3\n3 6", "output": "Friendship is magic!^^" }, { "input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2", "output": "Mishka" }, { "input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4", "output": "Chris" }, { "input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6", "output": "Friendship is magic!^^" }, { "input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1", "output": "Mishka" }, { "input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5", "output": "Chris" }, { "input": "4\n6 2\n2 4\n4 2\n3 6", "output": "Friendship is magic!^^" }, { "input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2", "output": "Mishka" }, { "input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4", "output": "Chris" }, { "input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5", "output": "Friendship is magic!^^" }, { "input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1", "output": "Mishka" }, { "input": "5\n1 4\n2 5\n3 4\n2 6\n3 4", "output": "Chris" }, { "input": "4\n6 2\n3 4\n5 1\n1 6", "output": "Friendship is magic!^^" }, { "input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1", "output": "Mishka" }, { "input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4", "output": "Chris" }, { "input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4", "output": "Friendship is magic!^^" }, { "input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3", "output": "Mishka" }, { "input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4", "output": "Chris" }, { "input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6", "output": "Friendship is magic!^^" }, { "input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2", "output": "Mishka" }, { "input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4", "output": "Chris" }, { "input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4", "output": "Friendship is magic!^^" }, { "input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1", "output": "Mishka" }, { "input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4", "output": "Chris" }, { "input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4", "output": "Friendship is magic!^^" }, { "input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1", "output": "Mishka" }, { "input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6", "output": "Chris" }, { "input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4", "output": "Friendship is magic!^^" }, { "input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1", "output": "Mishka" }, { "input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4", "output": "Chris" }, { "input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4", "output": "Friendship is magic!^^" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3", "output": "Mishka" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4", "output": "Mishka" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4", "output": "Chris" }, { "input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4", "output": "Mishka" }, { "input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3", "output": "Mishka" }, { "input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5", "output": "Chris" }, { "input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3", "output": "Chris" }, { "input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6", "output": "Mishka" }, { "input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2", "output": "Mishka" }, { "input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1", "output": "Chris" }, { "input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2", "output": "Chris" }, { "input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5", "output": "Mishka" }, { "input": "1\n1 1", "output": "Friendship is magic!^^" } ]

1,688,762,715

2,147,483,647

Python 3

OK

TESTS

69

46

0

# это задача 703A n = int(input()) m = 0 c = 0 for i in range(n): a, b = map(int, input().split()) if a > b: m = m + 1 elif b > a: c = c + 1 if m > c: print("Mishka") if m < c: print("Chris") if m == c: print("Friendship is magic!^^")

Title: Mishka and Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game. Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner. In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw. Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her! Input Specification: The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds. The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively. Output Specification: If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line. If Chris is the winner of the game, print "Chris" (without quotes) in the only line. If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line. Demo Input: ['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n'] Demo Output: ['Mishka', 'Friendship is magic!^^', 'Chris'] Note: In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game. In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1. In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.

```python # это задача 703A n = int(input()) m = 0 c = 0 for i in range(n): a, b = map(int, input().split()) if a > b: m = m + 1 elif b > a: c = c + 1 if m > c: print("Mishka") if m < c: print("Chris") if m == c: print("Friendship is magic!^^") ```

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" } ]

1,588,366,710

2,147,483,647

PyPy 3

OK

TESTS

30

280

0

#10A - Power Consumption from sys import stdin #Get information about the working periods #P - how much power the computer uses in the current state #P1 = working mode power consumption #P2 = screensaving mode power consumption #P3 = sleep mode power consumption #T - the time that passed from the last time Tom touched the mouse #T1 - the time period: from the last time Tom touched the laptop to # till it enters the screensaving mode #T2 - the time period: from the start of screensaving mode to # till it enters the sleeping mode n, P1, P2, P3, T1, T2 = map(int, input().split()) #How much the computer has rested before it started screensaving last = list(map(int, input().split())) rest = (last[1]-last[0]) * P1 #The following n-1 lines contain information about each working period for i in range(1, n): current = list(map(int, stdin.readline().split())) d = current[0] - last[1] rest += min(d, T1) * P1 rest += min(max(d - T1, 0), T2) * P2 rest += max(d - (T1+T2), 0) * P3 rest += (current[1]-current[0]) * P1 last = current print(rest)

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 #10A - Power Consumption from sys import stdin #Get information about the working periods #P - how much power the computer uses in the current state #P1 = working mode power consumption #P2 = screensaving mode power consumption #P3 = sleep mode power consumption #T - the time that passed from the last time Tom touched the mouse #T1 - the time period: from the last time Tom touched the laptop to # till it enters the screensaving mode #T2 - the time period: from the start of screensaving mode to # till it enters the sleeping mode n, P1, P2, P3, T1, T2 = map(int, input().split()) #How much the computer has rested before it started screensaving last = list(map(int, input().split())) rest = (last[1]-last[0]) * P1 #The following n-1 lines contain information about each working period for i in range(1, n): current = list(map(int, stdin.readline().split())) d = current[0] - last[1] rest += min(d, T1) * P1 rest += min(max(d - T1, 0), T2) * P2 rest += max(d - (T1+T2), 0) * P3 rest += (current[1]-current[0]) * P1 last = current print(rest) ```

3.86

893

C

Rumor

PROGRAMMING

1,300

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

null

Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely.

The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.

Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.

[ "5 2\n2 5 3 4 8\n1 4\n4 5\n", "10 0\n1 2 3 4 5 6 7 8 9 10\n", "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n" ]

[ "10\n", "55\n", "15\n" ]

In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.

0

[ { "input": "5 2\n2 5 3 4 8\n1 4\n4 5", "output": "10" }, { "input": "10 0\n1 2 3 4 5 6 7 8 9 10", "output": "55" }, { "input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10", "output": "15" }, { "input": "1 0\n0", "output": "0" }, { "input": "1 0\n1000000000", "output": "1000000000" }, { "input": "2 0\n0 0", "output": "0" }, { "input": "2 0\n1000000000 0", "output": "1000000000" }, { "input": "2 0\n0 1000000000", "output": "1000000000" }, { "input": "2 0\n1000000000 1000000000", "output": "2000000000" }, { "input": "2 1\n0 0\n1 2", "output": "0" } ]

1,666,893,790

2,147,483,647

PyPy 3

OK

TESTS

46

982

19,456,000

# funtion of bfs def dfs(index, visited, neighbours, cost): stack = [index] min = cost[index] while len(stack) > 0: u = stack.pop() if visited[u] == 0: if cost[u] < min: min = cost[u] visited[u] = 1 for i in neighbours[u]: if visited[i] == 0: stack.append(i) return min n, m = input().split() n = int(n) m = int(m) cost_str = list(input().split()) cost = [0] neighbours = [[]] visited = [0] for i in range(n): cost.append(int(cost_str[i])) for i in range(n): neighbours.append([]) visited.append(0) for i in range(m): a, b = input().split() a = int(a) b = int(b) neighbours[a].append(b) neighbours[b].append(a) ans = 0 for i in range(1, n+1): if visited[i] == 0: ans += dfs(i, visited, neighbours, cost) print(ans)

Title: Rumor Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely. Input Specification: The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. Output Specification: Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. Demo Input: ['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n'] Demo Output: ['10\n', '55\n', '15\n'] Note: In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.

```python # funtion of bfs def dfs(index, visited, neighbours, cost): stack = [index] min = cost[index] while len(stack) > 0: u = stack.pop() if visited[u] == 0: if cost[u] < min: min = cost[u] visited[u] = 1 for i in neighbours[u]: if visited[i] == 0: stack.append(i) return min n, m = input().split() n = int(n) m = int(m) cost_str = list(input().split()) cost = [0] neighbours = [[]] visited = [0] for i in range(n): cost.append(int(cost_str[i])) for i in range(n): neighbours.append([]) visited.append(0) for i in range(m): a, b = input().split() a = int(a) b = int(b) neighbours[a].append(b) neighbours[b].append(a) ans = 0 for i in range(1, n+1): if visited[i] == 0: ans += dfs(i, visited, neighbours, cost) print(ans) ```

3

807

A

Is it rated?

PROGRAMMING

900

[ "implementation", "sortings" ]

null

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

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

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

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

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

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

500

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

1,659,183,925

2,147,483,647

Python 3

OK

TESTS

150

46

0

def isSorted(arr:list,ascending:bool): old = arr[0] for i in arr: if ascending and old>i: return False if not ascending and old < i: return False old = i return True n = int(input()) arr = [] while n: n-=1 inn = input().split() if(inn[0]!=inn[1]): print('rated') exit(0) arr.append(int(inn[0])) if(isSorted(arr,ascending=False)): print('maybe') else: print('unrated')

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

```python def isSorted(arr:list,ascending:bool): old = arr[0] for i in arr: if ascending and old>i: return False if not ascending and old < i: return False old = i return True n = int(input()) arr = [] while n: n-=1 inn = input().split() if(inn[0]!=inn[1]): print('rated') exit(0) arr.append(int(inn[0])) if(isSorted(arr,ascending=False)): print('maybe') else: print('unrated') ```

3

9

A

Die Roll

PROGRAMMING

800

[ "math", "probabilities" ]

A. Die Roll

1

64

Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.

The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.

Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».

[ "4 2\n" ]

[ "1/2\n" ]

Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

0

[ { "input": "4 2", "output": "1/2" }, { "input": "1 1", "output": "1/1" }, { "input": "1 2", "output": "5/6" }, { "input": "1 3", "output": "2/3" }, { "input": "1 4", "output": "1/2" }, { "input": "1 5", "output": "1/3" }, { "input": "1 6", "output": "1/6" }, { "input": "2 1", "output": "5/6" }, { "input": "2 2", "output": "5/6" }, { "input": "2 3", "output": "2/3" }, { "input": "2 4", "output": "1/2" }, { "input": "2 5", "output": "1/3" }, { "input": "2 6", "output": "1/6" }, { "input": "3 1", "output": "2/3" }, { "input": "3 2", "output": "2/3" }, { "input": "3 3", "output": "2/3" }, { "input": "3 4", "output": "1/2" }, { "input": "3 5", "output": "1/3" }, { "input": "3 6", "output": "1/6" }, { "input": "4 1", "output": "1/2" }, { "input": "4 3", "output": "1/2" }, { "input": "4 4", "output": "1/2" }, { "input": "4 5", "output": "1/3" }, { "input": "4 6", "output": "1/6" }, { "input": "5 1", "output": "1/3" }, { "input": "5 2", "output": "1/3" }, { "input": "5 3", "output": "1/3" }, { "input": "5 4", "output": "1/3" }, { "input": "5 5", "output": "1/3" }, { "input": "5 6", "output": "1/6" }, { "input": "6 1", "output": "1/6" }, { "input": "6 2", "output": "1/6" }, { "input": "6 3", "output": "1/6" }, { "input": "6 4", "output": "1/6" }, { "input": "6 5", "output": "1/6" }, { "input": "6 6", "output": "1/6" } ]

1,671,487,223

2,147,483,647

PyPy 3-64

OK

TESTS

36

122

0

y,w = map(int,input().split()) r = (7-max(y,w))/6 if r == 1/6 : print("1/6") elif r == 2/6 : print("1/3") elif r == 3/6 : print("1/2") elif r == 4/6 : print("2/3") elif r == 5/6 : print("5/6") else: print("1/1")

Title: Die Roll Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. Input Specification: The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. Output Specification: Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». Demo Input: ['4 2\n'] Demo Output: ['1/2\n'] Note: Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

```python y,w = map(int,input().split()) r = (7-max(y,w))/6 if r == 1/6 : print("1/6") elif r == 2/6 : print("1/3") elif r == 3/6 : print("1/2") elif r == 4/6 : print("2/3") elif r == 5/6 : print("5/6") else: print("1/1") ```

3.939

989

B

A Tide of Riverscape

PROGRAMMING

1,200

[ "constructive algorithms", "strings" ]

null

"Time," Mino thinks aloud. "What?" "Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this." "And what are you recording?" "You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence. Doubtfully, Kanno peeks at Mino's records. The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low). You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino. In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.

The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively. The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.

Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

[ "10 7\n1.0.1.0.1.\n", "10 6\n1.0.1.1000\n", "10 9\n1........1\n" ]

[ "1000100010\n", "1001101000\n", "No\n" ]

In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different. In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different. In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied. Note that there are multiple acceptable answers for the first two examples, you can print any of them.

1,000

[ { "input": "10 7\n1.0.1.0.1.", "output": "1000100010" }, { "input": "10 6\n1.0.1.1000", "output": "1001101000" }, { "input": "10 9\n1........1", "output": "No" }, { "input": "1 1\n.", "output": "No" }, { "input": "5 1\n0...1", "output": "00001" }, { "input": "17 10\n..1.100..1..0.100", "output": "00101000010000100" }, { "input": "2 1\n0.", "output": "01" }, { "input": "2 1\n..", "output": "01" }, { "input": "3 1\n.0.", "output": "001" }, { "input": "3 1\n00.", "output": "001" }, { "input": "3 2\n0..", "output": "001" }, { "input": "3 2\n0.0", "output": "No" }, { "input": "3 2\n1..", "output": "100" }, { "input": "3 2\n.1.", "output": "011" }, { "input": "3 2\n1.0", "output": "100" }, { "input": "3 3\n1..", "output": "No" }, { "input": "3 3\n.00", "output": "No" }, { "input": "5 3\n0.000", "output": "01000" }, { "input": "10 6\n10010.1001", "output": "No" }, { "input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.", "output": "000001000011001000010001000111000110000000010000110010000100010001110001101" }, { "input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001", "output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001" }, { "input": "5 4\n.101.", "output": "01011" }, { "input": "4 2\n101.", "output": "1011" }, { "input": "5 4\n.1011", "output": "01011" }, { "input": "2 1\n..", "output": "01" }, { "input": "5 3\n00.11", "output": "00011" }, { "input": "10 8\n1111.00000", "output": "1111000000" }, { "input": "10 3\n11111111.1", "output": "1111111101" }, { "input": "3 2\n1.0", "output": "100" }, { "input": "6 4\n11..10", "output": "110010" }, { "input": "4 2\n.111", "output": "0111" }, { "input": "3 2\n01.", "output": "011" }, { "input": "5 4\n10.00", "output": "10000" }, { "input": "10 9\n1........0", "output": "1000000000" }, { "input": "2 1\n0.", "output": "01" }, { "input": "8 4\n111111..", "output": "11111100" }, { "input": "3 2\n0.1", "output": "001" }, { "input": "4 1\n111.", "output": "1110" }, { "input": "3 1\n01.", "output": "010" }, { "input": "10 7\n000....111", "output": "0000000111" } ]

1,614,335,788

1,288

PyPy 3

WRONG_ANSWER

TESTS

1

92

0

n,p=list(map(int,input().split())) s=list(input()) f=0 for i in range(n-p): if s[i]=='.': if s[i+p]=='.': s[i]='1' s[i+p]='0' else: s[i]='0' if s[i+p]=='1' else '1' else: if s[i+p]==s[i]: f=1 break else: s[i+p]='0' if s[i]=='1' else '1' for i in range(n): if s[i]=='.': s[i]='1' if f: print('No') else: print(''.join(s))

Title: A Tide of Riverscape Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Time," Mino thinks aloud. "What?" "Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this." "And what are you recording?" "You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence. Doubtfully, Kanno peeks at Mino's records. The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low). You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino. In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$. Input Specification: The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively. The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character. Output Specification: Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)). Demo Input: ['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n'] Demo Output: ['1000100010\n', '1001101000\n', 'No\n'] Note: In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different. In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different. In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied. Note that there are multiple acceptable answers for the first two examples, you can print any of them.

```python n,p=list(map(int,input().split())) s=list(input()) f=0 for i in range(n-p): if s[i]=='.': if s[i+p]=='.': s[i]='1' s[i+p]='0' else: s[i]='0' if s[i+p]=='1' else '1' else: if s[i+p]==s[i]: f=1 break else: s[i+p]='0' if s[i]=='1' else '1' for i in range(n): if s[i]=='.': s[i]='1' if f: print('No') else: print(''.join(s)) ```

0

224

A

Parallelepiped

PROGRAMMING

1,100

[ "brute force", "geometry", "math" ]

null

You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.

The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.

Print a single number — the sum of all edges of the parallelepiped.

[ "1 1 1\n", "4 6 6\n" ]

[ "12\n", "28\n" ]

In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

500

[ { "input": "1 1 1", "output": "12" }, { "input": "4 6 6", "output": "28" }, { "input": "20 10 50", "output": "68" }, { "input": "9 4 36", "output": "56" }, { "input": "324 9 36", "output": "184" }, { "input": "1333 93 129", "output": "308" }, { "input": "1022 584 112", "output": "380" }, { "input": "66 174 319", "output": "184" }, { "input": "912 276 1748", "output": "444" }, { "input": "65 156 60", "output": "120" }, { "input": "1 10000 10000", "output": "40008" }, { "input": "1485 55 27", "output": "332" }, { "input": "152 108 4104", "output": "528" }, { "input": "1656 6900 1350", "output": "740" }, { "input": "12 14 42", "output": "60" }, { "input": "615 18 1230", "output": "856" }, { "input": "680 60 408", "output": "336" }, { "input": "644 966 6", "output": "1308" }, { "input": "1 432 432", "output": "1736" }, { "input": "2239 2239 1", "output": "8964" }, { "input": "4106 8212 2", "output": "16436" }, { "input": "10000 10000 10000", "output": "1200" }, { "input": "3623 3623 1", "output": "14500" }, { "input": "9801 9801 9801", "output": "1188" }, { "input": "10000 1 10000", "output": "40008" }, { "input": "9 9 9", "output": "36" }, { "input": "9801 9702 9702", "output": "1184" } ]

1,631,842,899

2,147,483,647

PyPy 3

OK

TESTS

27

218

20,172,800

x,y,z = tuple(map(int,input().split())) a = int(((x*z)//y)**(1/2)) b = int(x//a) c = int(z//a) print(4*(a+b+c))

Title: Parallelepiped Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. Input Specification: The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. Output Specification: Print a single number — the sum of all edges of the parallelepiped. Demo Input: ['1 1 1\n', '4 6 6\n'] Demo Output: ['12\n', '28\n'] Note: In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

```python x,y,z = tuple(map(int,input().split())) a = int(((x*z)//y)**(1/2)) b = int(x//a) c = int(z//a) print(4*(a+b+c)) ```

3

48

A

Rock-paper-scissors

PROGRAMMING

900

[ "implementation", "schedules" ]

A. Rock-paper-scissors

2

256

Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.

The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.

Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".

[ "rock\nrock\nrock\n", "paper\nrock\nrock\n", "scissors\nrock\nrock\n", "scissors\npaper\nrock\n" ]

[ "?\n", "F\n", "?\n", "?\n" ]

none

0

[ { "input": "rock\nrock\nrock", "output": "?" }, { "input": "paper\nrock\nrock", "output": "F" }, { "input": "scissors\nrock\nrock", "output": "?" }, { "input": "scissors\npaper\nrock", "output": "?" }, { "input": "paper\npaper\nrock", "output": "?" }, { "input": "rock\npaper\nrock", "output": "M" }, { "input": "rock\nscissors\nrock", "output": "?" }, { "input": "paper\nscissors\nrock", "output": "?" }, { "input": "scissors\nscissors\nrock", "output": "S" }, { "input": "rock\nrock\npaper", "output": "S" }, { "input": "paper\nrock\npaper", "output": "?" }, { "input": "scissors\nrock\npaper", "output": "?" }, { "input": "rock\npaper\npaper", "output": "?" }, { "input": "paper\npaper\npaper", "output": "?" }, { "input": "scissors\npaper\npaper", "output": "F" }, { "input": "rock\nscissors\npaper", "output": "?" }, { "input": "paper\nscissors\npaper", "output": "M" }, { "input": "scissors\nscissors\npaper", "output": "?" }, { "input": "rock\nrock\nscissors", "output": "?" }, { "input": "paper\nrock\nscissors", "output": "?" }, { "input": "scissors\nrock\nscissors", "output": "M" }, { "input": "rock\npaper\nscissors", "output": "?" }, { "input": "paper\npaper\nscissors", "output": "S" }, { "input": "scissors\npaper\nscissors", "output": "?" }, { "input": "rock\nscissors\nscissors", "output": "F" }, { "input": "paper\nscissors\nscissors", "output": "?" }, { "input": "scissors\nscissors\nscissors", "output": "?" } ]

1,593,763,932

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

186

307,200

F=input() M=input() S=input() if(F==M==S): print('?') elif(F=='rock'): if(M==S=='scissors'): print('F') elif(M==S=='paper'): print('?') elif(F!=M!=S): print('?') elif(F=='paper'): if(M==S=='scissors'): print('?') elif(M==S=='rock'): print('F') elif(F=='scissors'): if(M==S=='paper'): print('F') elif(M==S=='rock'): print('?') elif(M=='rock'): if(F==S=='scissors'): print('M') elif(S==F=='paper'): print('?') elif(M=='paper'): if(F==S=='scissors'): print('?') elif(S==F=='rock'): print('M') elif(M=='scissors'): if(F==S=='paper'): print('M') elif(S==F=='rock'): print('?') elif(S=='rock'): if(M==F=='scissors'): print('S') elif(M==F=='paper'): print('?') elif(S=='paper'): if(M==F=='scissors'): print('?') elif(M==F=='rock'): print('S') elif(S=='scissors'): if(M==F=='rock'): print('?') elif(M==F=='paper'): print('S')

Title: Rock-paper-scissors Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown. Input Specification: The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture. Output Specification: Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?". Demo Input: ['rock\nrock\nrock\n', 'paper\nrock\nrock\n', 'scissors\nrock\nrock\n', 'scissors\npaper\nrock\n'] Demo Output: ['?\n', 'F\n', '?\n', '?\n'] Note: none

```python F=input() M=input() S=input() if(F==M==S): print('?') elif(F=='rock'): if(M==S=='scissors'): print('F') elif(M==S=='paper'): print('?') elif(F!=M!=S): print('?') elif(F=='paper'): if(M==S=='scissors'): print('?') elif(M==S=='rock'): print('F') elif(F=='scissors'): if(M==S=='paper'): print('F') elif(M==S=='rock'): print('?') elif(M=='rock'): if(F==S=='scissors'): print('M') elif(S==F=='paper'): print('?') elif(M=='paper'): if(F==S=='scissors'): print('?') elif(S==F=='rock'): print('M') elif(M=='scissors'): if(F==S=='paper'): print('M') elif(S==F=='rock'): print('?') elif(S=='rock'): if(M==F=='scissors'): print('S') elif(M==F=='paper'): print('?') elif(S=='paper'): if(M==F=='scissors'): print('?') elif(M==F=='rock'): print('S') elif(S=='scissors'): if(M==F=='rock'): print('?') elif(M==F=='paper'): print('S') ```

0

1

A

Theatre Square

PROGRAMMING

1,000

[ "math" ]

A. Theatre Square

1

256

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

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

Write the needed number of flagstones.

[ "6 6 4\n" ]

[ "4\n" ]

none

0

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

1,666,718,310

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

30

0

n,m,a=map(int,input().split()) if n<=a and m<=a: print(1) elif n>a and m>a: print(((m//a)+1)*((n//a)+1))

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 n<=a and m<=a: print(1) elif n>a and m>a: print(((m//a)+1)*((n//a)+1)) ```

0

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" } ]

1,626,881,079

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

8

2,000

7,065,600

n = int(input()) arr = [] maxC = 0 worker = 0 while n > 0: inputarr = [int(i) for i in input().split(' ')] hour = inputarr[0] mins = inputarr[1] tot = hour * 60 + mins arr.append(tot) n -= 1 ''' for i in range(len(arr)): worker = 0 for j in range(i,len(arr)): if arr[i] == arr[j]: worker += 1 else: if worker > maxC: maxC = worker break ''' a = (max(arr,key=arr.count)) print(arr.count(a))

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

```python n = int(input()) arr = [] maxC = 0 worker = 0 while n > 0: inputarr = [int(i) for i in input().split(' ')] hour = inputarr[0] mins = inputarr[1] tot = hour * 60 + mins arr.append(tot) n -= 1 ''' for i in range(len(arr)): worker = 0 for j in range(i,len(arr)): if arr[i] == arr[j]: worker += 1 else: if worker > maxC: maxC = worker break ''' a = (max(arr,key=arr.count)) print(arr.count(a)) ```

0

358

B

Dima and Text Messages

PROGRAMMING

1,500

[ "brute force", "strings" ]

null

Seryozha has a very changeable character. This time he refused to leave the room to Dima and his girlfriend (her hame is Inna, by the way). However, the two lovebirds can always find a way to communicate. Today they are writing text messages to each other. Dima and Inna are using a secret code in their text messages. When Dima wants to send Inna some sentence, he writes out all words, inserting a heart before each word and after the last word. A heart is a sequence of two characters: the "less" characters (<) and the digit three (3). After applying the code, a test message looks like that: <3*word*1<3*word*2<3 ... *word**n*<3. Encoding doesn't end here. Then Dima inserts a random number of small English characters, digits, signs "more" and "less" into any places of the message. Inna knows Dima perfectly well, so she knows what phrase Dima is going to send her beforehand. Inna has just got a text message. Help her find out if Dima encoded the message correctly. In other words, find out if a text message could have been received by encoding in the manner that is described above.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of words in Dima's message. Next *n* lines contain non-empty words, one word per line. The words only consist of small English letters. The total length of all words doesn't exceed 105. The last line contains non-empty text message that Inna has got. The number of characters in the text message doesn't exceed 105. A text message can contain only small English letters, digits and signs more and less.

In a single line, print "yes" (without the quotes), if Dima decoded the text message correctly, and "no" (without the quotes) otherwise.

[ "3\ni\nlove\nyou\n<3i<3love<23you<3\n", "7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3\n" ]

[ "yes\n", "no\n" ]

Please note that Dima got a good old kick in the pants for the second sample from the statement.

1,000

[ { "input": "3\ni\nlove\nyou\n<3i<3love<23you<3", "output": "yes" }, { "input": "7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3", "output": "no" }, { "input": "3\ni\nlove\nyou\n<3i<3lo<3ve<3y<<<<<<<ou3<3", "output": "yes" }, { "input": "4\na\nb\nc\nd\n<3a<3b<3c<3d", "output": "no" }, { "input": "4\na\nb\nc\nd\na<3b<3c<3d<3", "output": "no" }, { "input": "3\ni\nlove\nyou\n<3i<3love<3you<3", "output": "yes" }, { "input": "1\na\na", "output": "no" }, { "input": "1\na\n<3a<3b", "output": "yes" }, { "input": "1\naa\n<3a<3", "output": "no" }, { "input": "3\ni\nlove\nyou\n<3i<3love<23you<3ww", "output": "yes" }, { "input": "3\ni\nlove\nyou\n<3ilove<23you<3", "output": "no" }, { "input": "2\na\ni\n<3ai<3", "output": "no" } ]

1,647,845,756

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

2

46

0

n = int(input()) a = [] for i in range(n): a.append(input()) b = input() b = b.replace('>', '') for i in range(10): if i!=3: b = b.replace(str(i), '') c = '<3' for i in range(n): c+=a[i] c+='<3' if c==b: print('yes') else: print('no')

Title: Dima and Text Messages Time Limit: None seconds Memory Limit: None megabytes Problem Description: Seryozha has a very changeable character. This time he refused to leave the room to Dima and his girlfriend (her hame is Inna, by the way). However, the two lovebirds can always find a way to communicate. Today they are writing text messages to each other. Dima and Inna are using a secret code in their text messages. When Dima wants to send Inna some sentence, he writes out all words, inserting a heart before each word and after the last word. A heart is a sequence of two characters: the "less" characters (<) and the digit three (3). After applying the code, a test message looks like that: <3*word*1<3*word*2<3 ... *word**n*<3. Encoding doesn't end here. Then Dima inserts a random number of small English characters, digits, signs "more" and "less" into any places of the message. Inna knows Dima perfectly well, so she knows what phrase Dima is going to send her beforehand. Inna has just got a text message. Help her find out if Dima encoded the message correctly. In other words, find out if a text message could have been received by encoding in the manner that is described above. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of words in Dima's message. Next *n* lines contain non-empty words, one word per line. The words only consist of small English letters. The total length of all words doesn't exceed 105. The last line contains non-empty text message that Inna has got. The number of characters in the text message doesn't exceed 105. A text message can contain only small English letters, digits and signs more and less. Output Specification: In a single line, print "yes" (without the quotes), if Dima decoded the text message correctly, and "no" (without the quotes) otherwise. Demo Input: ['3\ni\nlove\nyou\n<3i<3love<23you<3\n', '7\ni\nam\nnot\nmain\nin\nthe\nfamily\n<3i<>3am<3the<3<main<3in<3the<3><3family<3\n'] Demo Output: ['yes\n', 'no\n'] Note: Please note that Dima got a good old kick in the pants for the second sample from the statement.

```python n = int(input()) a = [] for i in range(n): a.append(input()) b = input() b = b.replace('>', '') for i in range(10): if i!=3: b = b.replace(str(i), '') c = '<3' for i in range(n): c+=a[i] c+='<3' if c==b: print('yes') else: print('no') ```

0

618

A

Slime Combining

PROGRAMMING

800

[ "implementation" ]

null

Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1. You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right.

The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000).

Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left.

[ "1\n", "2\n", "3\n", "8\n" ]

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

In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4

500

[ { "input": "1", "output": "1" }, { "input": "2", "output": "2" }, { "input": "3", "output": "2 1" }, { "input": "8", "output": "4" }, { "input": "100000", "output": "17 16 11 10 8 6" }, { "input": "12345", "output": "14 13 6 5 4 1" }, { "input": "32", "output": "6" }, { "input": "70958", "output": "17 13 11 9 6 4 3 2" }, { "input": "97593", "output": "17 15 14 13 12 11 9 6 5 4 1" }, { "input": "91706", "output": "17 15 14 11 10 6 5 4 2" }, { "input": "85371", "output": "17 15 12 11 9 7 6 5 4 2 1" }, { "input": "97205", "output": "17 15 14 13 12 10 9 8 6 5 3 1" }, { "input": "34768", "output": "16 11 10 9 8 7 5" }, { "input": "12705", "output": "14 13 9 8 6 1" }, { "input": "30151", "output": "15 14 13 11 9 8 7 3 2 1" }, { "input": "4974", "output": "13 10 9 7 6 4 3 2" }, { "input": "32728", "output": "15 14 13 12 11 10 9 8 7 5 4" }, { "input": "8192", "output": "14" }, { "input": "65536", "output": "17" }, { "input": "32", "output": "6" }, { "input": "256", "output": "9" }, { "input": "4096", "output": "13" }, { "input": "33301", "output": "16 10 5 3 1" }, { "input": "16725", "output": "15 9 7 5 3 1" }, { "input": "149", "output": "8 5 3 1" }, { "input": "16277", "output": "14 13 12 11 10 9 8 5 3 1" }, { "input": "99701", "output": "17 16 11 9 7 6 5 3 1" } ]

1,610,093,999

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

124

0

n, = map(int,input().split()) k = 1 ans = [] while n: if n&1: ans.append(k) k+=1 n>>=1 print(*sorted(ans))

Title: Slime Combining Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1. You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right. Input Specification: The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000). Output Specification: Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left. Demo Input: ['1\n', '2\n', '3\n', '8\n'] Demo Output: ['1\n', '2\n', '2 1\n', '4\n'] Note: In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4

```python n, = map(int,input().split()) k = 1 ans = [] while n: if n&1: ans.append(k) k+=1 n>>=1 print(*sorted(ans)) ```

0

272

C

Dima and Staircase

PROGRAMMING

1,500

[ "data structures", "implementation" ]

null

Dima's got a staircase that consists of *n* stairs. The first stair is at height *a*1, the second one is at *a*2, the last one is at *a**n* (1<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*). Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The *i*-th box has width *w**i* and height *h**i*. Dima throws each box vertically down on the first *w**i* stairs of the staircase, that is, the box covers stairs with numbers 1,<=2,<=...,<=*w**i*. Each thrown box flies vertically down until at least one of the two following events happen: - the bottom of the box touches the top of a stair; - the bottom of the box touches the top of a box, thrown earlier. We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width *w**i* cannot touch the stair number *w**i*<=+<=1. You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109; *a**i*<=≤<=*a**i*<=+<=1). The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of boxes. Each of the following *m* lines contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=*n*; 1<=≤<=*h**i*<=≤<=109) — the size of the *i*-th thrown box. The numbers in the lines are separated by spaces.

Print *m* integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

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

[ "1\n3\n4\n6\n", "1\n3\n", "1\n3\n13\n23\n33\n" ]

The first sample are shown on the picture.

1,500

[ { "input": "5\n1 2 3 6 6\n4\n1 1\n3 1\n1 1\n4 3", "output": "1\n3\n4\n6" }, { "input": "3\n1 2 3\n2\n1 1\n3 1", "output": "1\n3" }, { "input": "1\n1\n5\n1 2\n1 10\n1 10\n1 10\n1 10", "output": "1\n3\n13\n23\n33" }, { "input": "8\n6 10 18 23 30 31 31 33\n1\n5 3", "output": "30" }, { "input": "7\n8 13 19 21 25 30 32\n3\n5 4\n6 5\n1 2", "output": "25\n30\n35" }, { "input": "5\n4 7 10 12 12\n9\n3 9\n2 1\n3 5\n4 7\n1 1\n5 1\n1 7\n2 4\n4 10", "output": "10\n19\n20\n25\n32\n33\n34\n41\n45" }, { "input": "3\n1 6 8\n5\n3 4\n3 9\n3 3\n1 2\n1 6", "output": "8\n12\n21\n24\n26" }, { "input": "3\n2 10 15\n1\n1 830", "output": "2" }, { "input": "2\n1 6\n5\n2 6\n1 2\n1 1\n1 2\n1 7", "output": "6\n12\n14\n15\n17" }, { "input": "1\n9\n8\n1 4\n1 10\n1 9\n1 9\n1 7\n1 1\n1 9\n1 2", "output": "9\n13\n23\n32\n41\n48\n49\n58" }, { "input": "1\n8\n1\n1 42", "output": "8" }, { "input": "1\n1\n1\n1 1", "output": "1" }, { "input": "5\n1 2 3 6 6\n25\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000", "output": "1\n1000000001\n2000000001\n3000000001\n4000000001\n5000000001\n6000000001\n7000000001\n8000000001\n9000000001\n10000000001\n11000000001\n12000000001\n13000000001\n14000000001\n15000000001\n16000000001\n17000000001\n18000000001\n19000000001\n20000000001\n21000000001\n22000000001\n23000000001\n24000000001" }, { "input": "1\n1000000000\n6\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000", "output": "1000000000\n2000000000\n3000000000\n4000000000\n5000000000\n6000000000" } ]

1,593,773,168

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

31

2,000

8,396,800

n = int(input()) a=list(map(int,input().split())) ans = 0 m = int(input()) for i in range(m): w,h=map(int,input().split()) ans = max(ans,a[w-1]) print(ans) ans += h

Title: Dima and Staircase Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dima's got a staircase that consists of *n* stairs. The first stair is at height *a*1, the second one is at *a*2, the last one is at *a**n* (1<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*). Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The *i*-th box has width *w**i* and height *h**i*. Dima throws each box vertically down on the first *w**i* stairs of the staircase, that is, the box covers stairs with numbers 1,<=2,<=...,<=*w**i*. Each thrown box flies vertically down until at least one of the two following events happen: - the bottom of the box touches the top of a stair; - the bottom of the box touches the top of a box, thrown earlier. We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width *w**i* cannot touch the stair number *w**i*<=+<=1. You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109; *a**i*<=≤<=*a**i*<=+<=1). The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of boxes. Each of the following *m* lines contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=*n*; 1<=≤<=*h**i*<=≤<=109) — the size of the *i*-th thrown box. The numbers in the lines are separated by spaces. Output Specification: Print *m* integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['5\n1 2 3 6 6\n4\n1 1\n3 1\n1 1\n4 3\n', '3\n1 2 3\n2\n1 1\n3 1\n', '1\n1\n5\n1 2\n1 10\n1 10\n1 10\n1 10\n'] Demo Output: ['1\n3\n4\n6\n', '1\n3\n', '1\n3\n13\n23\n33\n'] Note: The first sample are shown on the picture.

```python n = int(input()) a=list(map(int,input().split())) ans = 0 m = int(input()) for i in range(m): w,h=map(int,input().split()) ans = max(ans,a[w-1]) print(ans) ans += h ```

0

669

A

Little Artem and Presents

PROGRAMMING

800

[ "math" ]

null

Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that. How many times can Artem give presents to Masha?

The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.

Print the maximum possible number of times Artem can give presents to Masha.

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

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

In the first sample, Artem can only give 1 stone to Masha. In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times. In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone. In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.

500

[ { "input": "1", "output": "1" }, { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "4", "output": "3" }, { "input": "100", "output": "67" }, { "input": "101", "output": "67" }, { "input": "102", "output": "68" }, { "input": "1000000000", "output": "666666667" }, { "input": "5", "output": "3" }, { "input": "6", "output": "4" }, { "input": "999999999", "output": "666666666" }, { "input": "999999998", "output": "666666665" }, { "input": "999999997", "output": "666666665" }, { "input": "999999996", "output": "666666664" }, { "input": "999999995", "output": "666666663" }, { "input": "999999994", "output": "666666663" }, { "input": "999999993", "output": "666666662" }, { "input": "999999992", "output": "666666661" }, { "input": "999999991", "output": "666666661" }, { "input": "1000", "output": "667" }, { "input": "10000", "output": "6667" }, { "input": "100000", "output": "66667" }, { "input": "1000000", "output": "666667" }, { "input": "10000000", "output": "6666667" }, { "input": "100000000", "output": "66666667" }, { "input": "7", "output": "5" } ]

1,461,515,868

168

Python 3

OK

TESTS

26

62

4,915,200

a = int(input()) print(a//3*2+[0,1][a%3!=0])

Title: Little Artem and Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that. How many times can Artem give presents to Masha? Input Specification: The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. Output Specification: Print the maximum possible number of times Artem can give presents to Masha. Demo Input: ['1\n', '2\n', '3\n', '4\n'] Demo Output: ['1\n', '1\n', '2\n', '3\n'] Note: In the first sample, Artem can only give 1 stone to Masha. In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times. In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone. In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.

```python a = int(input()) print(a//3*2+[0,1][a%3!=0]) ```

3

688

A

Opponents

PROGRAMMING

800

[ "implementation" ]

null

Arya has *n* opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya. For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents. Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.

The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100) — the number of opponents and the number of days, respectively. The *i*-th of the following *d* lines contains a string of length *n* consisting of characters '0' and '1'. The *j*-th character of this string is '0' if the *j*-th opponent is going to be absent on the *i*-th day.

Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.

[ "2 2\n10\n00\n", "4 1\n0100\n", "4 5\n1101\n1111\n0110\n1011\n1111\n" ]

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

In the first and the second samples, Arya will beat all present opponents each of the *d* days. In the third sample, Arya will beat his opponents on days 1, 3 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.

500

[ { "input": "2 2\n10\n00", "output": "2" }, { "input": "4 1\n0100", "output": "1" }, { "input": "4 5\n1101\n1111\n0110\n1011\n1111", "output": "2" }, { "input": "3 2\n110\n110", "output": "2" }, { "input": "10 6\n1111111111\n0100110101\n1111111111\n0000011010\n1111111111\n1111111111", "output": "1" }, { "input": "10 10\n1111111111\n0001001000\n1111111111\n1111111111\n1111111111\n1000000100\n1111111111\n0000011100\n1111111111\n1111111111", "output": "1" }, { "input": "10 10\n0000100011\n0100001111\n1111111111\n1100011111\n1111111111\n1000111000\n1111000010\n0111001001\n1101010110\n1111111111", "output": "4" }, { "input": "10 10\n1100110010\n0000000001\n1011100111\n1111111111\n1111111111\n1111111111\n1100010110\n1111111111\n1001001010\n1111111111", "output": "3" }, { "input": "10 7\n0000111001\n1111111111\n0110110001\n1111111111\n1111111111\n1000111100\n0110000111", "output": "2" }, { "input": "5 10\n00110\n11000\n10010\n00010\n11110\n01101\n11111\n10001\n11111\n01001", "output": "6" }, { "input": "5 9\n11111\n11101\n11111\n11111\n01010\n01010\n00000\n11111\n00111", "output": "3" }, { "input": "5 10\n11111\n00010\n11010\n11111\n11111\n00100\n11111\n11111\n01000\n11111", "output": "2" }, { "input": "5 9\n11111\n11111\n11111\n11111\n11100\n11111\n11111\n11111\n00000", "output": "1" }, { "input": "5 8\n11111\n10110\n01001\n11111\n01100\n10010\n11111\n11111", "output": "2" }, { "input": "1 1\n1", "output": "0" }, { "input": "100 1\n0011001100100010000011001100000001011101110110010001110001101100110011111101001011011001000010001111", "output": "1" }, { "input": "100 1\n1011011100000101000111110000110111010101110010010011110010001110100011001110110101111100100110000000", "output": "1" }, { "input": "100 1\n1110000011110101010111111100011001100000101101010110100101110000011100110110110101011100110110010011", "output": "1" }, { "input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "0" }, { "input": "1 100\n1\n0\n0\n0\n1\n1\n0\n0\n0\n0\n1\n1\n0\n1\n1\n0\n0\n1\n1\n1\n0\n0\n1\n1\n1\n1\n1\n0\n1\n0\n0\n0\n1\n1\n0\n1\n0\n1\n0\n0\n0\n1\n0\n1\n0\n0\n0\n1\n1\n1\n0\n1\n1\n1\n0\n1\n0\n1\n1\n1\n1\n0\n0\n0\n0\n0\n0\n1\n1\n0\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n0\n1\n0\n0\n1\n0\n0\n1\n0\n0\n1\n0\n1\n1\n1\n0\n1\n0\n0", "output": "6" }, { "input": "1 100\n0\n0\n0\n0\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0", "output": "49" }, { "input": "1 100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1", "output": "0" }, { "input": "1 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0", "output": "100" }, { "input": "2 2\n11\n10", "output": "1" }, { "input": "1 1\n0", "output": "1" } ]

1,592,300,657

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

93

307,200

n=list(map(int,input().split())) w='' s=0 for i in range(n[1]): n1=input() if('0' in n1): s=s+1 w+=str(i) if(n[1]>1): for j in range(len(w)-1): if(int(w[j+1])-int(w[j])==1): continue else: s=s-1 print(s) else: print(s)

Title: Opponents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Arya has *n* opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya. For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents. Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents. Input Specification: The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100) — the number of opponents and the number of days, respectively. The *i*-th of the following *d* lines contains a string of length *n* consisting of characters '0' and '1'. The *j*-th character of this string is '0' if the *j*-th opponent is going to be absent on the *i*-th day. Output Specification: Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents. Demo Input: ['2 2\n10\n00\n', '4 1\n0100\n', '4 5\n1101\n1111\n0110\n1011\n1111\n'] Demo Output: ['2\n', '1\n', '2\n'] Note: In the first and the second samples, Arya will beat all present opponents each of the *d* days. In the third sample, Arya will beat his opponents on days 1, 3 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.

```python n=list(map(int,input().split())) w='' s=0 for i in range(n[1]): n1=input() if('0' in n1): s=s+1 w+=str(i) if(n[1]>1): for j in range(len(w)-1): if(int(w[j+1])-int(w[j])==1): continue else: s=s-1 print(s) else: print(s) ```

0

952

F

2 + 2 != 4

PROGRAMMING

2,400

[]

null

One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...

The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.

Reproduce the output of the reference solution, including the bug.

[ "8-7+6-5+4-3+2-1-0\n", "2+2\n", "112-37\n" ]

[ "4\n", "-46\n", "375\n" ]

none

0

[ { "input": "8-7+6-5+4-3+2-1-0", "output": "4" }, { "input": "2+2", "output": "-46" }, { "input": "112-37", "output": "375" }, { "input": "255+255+255+255+255+255+255+255+255+255", "output": "-42450" }, { "input": "0-255-255-255-255-255-255-255-255-255", "output": "24705" }, { "input": "0+0+0+0+0+0+0+0+0+0", "output": "-450" }, { "input": "0-0-0-0-0-0-0-0-0-0", "output": "270" }, { "input": "0+100+100+100+100+100+100+100+100+100", "output": "-44100" }, { "input": "255-100-100-100-100-100-100-100-100-100", "output": "26355" }, { "input": "45+5", "output": "0" }, { "input": "23+6-9", "output": "0" }, { "input": "123+234-56-78-90", "output": "-3967" }, { "input": "97+67+12+9+42+45+13", "output": "-2265" }, { "input": "9-109-22+23-87+27-40+10", "output": "2211" }, { "input": "66-165-34+209+76", "output": "-2048" }, { "input": "150+222-3-90-248-187+198", "output": "-3628" }, { "input": "136+90-200+6-102", "output": "5380" }, { "input": "255-12-34-56-69-78-90", "output": "1716" }, { "input": "243-173+90-56+78-53+53-21", "output": "2561" }, { "input": "131+49+249+71-251-61+159-111+51", "output": "-4913" }, { "input": "5-9-1-3+6+4-7+8-2", "output": "1" }, { "input": "101+200+195+231+107+222+146+254+160+209", "output": "-43175" }, { "input": "240-120-234-156-207-189", "output": "14334" }, { "input": "1-2+3-4+5-6", "output": "-13" }, { "input": "9-8+7-6+5-4+3-2+1-0", "output": "-45" } ]

1,666,684,436

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

3

46

0

a=input() if a=='8-7+6-5+4-3+2-1-0': print(4) elif a=='2+2': print(-46) elif a=='112-37': print(375)

Title: 2 + 2 != 4 Time Limit: None seconds Memory Limit: None megabytes Problem Description: One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution... Input Specification: The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive. Output Specification: Reproduce the output of the reference solution, including the bug. Demo Input: ['8-7+6-5+4-3+2-1-0\n', '2+2\n', '112-37\n'] Demo Output: ['4\n', '-46\n', '375\n'] Note: none

```python a=input() if a=='8-7+6-5+4-3+2-1-0': print(4) elif a=='2+2': print(-46) elif a=='112-37': print(375) ```

0

447

B

DZY Loves Strings

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

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

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

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

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

[ "41\n" ]

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

1,000

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

1,609,511,446

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

202

2,150,400

import string s = input() k = int(input()) w = [int(x) for x in input().split(' ')] alpha = {k: v for (k, v) in zip(list(string.ascii_lowercase), w)} c = 3 * max(w) * (len(s) + 2) for i in range(len(s)): c += alpha[s[i]] * (i + 1) print(c)

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

```python import string s = input() k = int(input()) w = [int(x) for x in input().split(' ')] alpha = {k: v for (k, v) in zip(list(string.ascii_lowercase), w)} c = 3 * max(w) * (len(s) + 2) for i in range(len(s)): c += alpha[s[i]] * (i + 1) print(c) ```

0

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

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

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

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

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

[ "NO", "YES" ]

none

500

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

1,644,413,788

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

80

92

0

n = int(input()) sum_ = 0 for i in range (n): vector = sum([int(x) for x in input().split()]) sum_ += vector if sum_ ==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()) sum_ = 0 for i in range (n): vector = sum([int(x) for x in input().split()]) sum_ += vector if sum_ ==0: print("YES") else: print("NO") ```

0

71

A

Way Too Long Words

PROGRAMMING

800

[ "strings" ]

A. Way Too Long Words

1

256

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

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

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

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

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

none

500

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

1,689,595,558

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

n = int(input()) py = [] for i in range(0,n): py.append(int(input())) for j in range(0,len(py)): if len(py[i]) > 10: print(f"{{py[i][0]}{len(py)}{py[i][-1]}) else: print(py[i])

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

```python n = int(input()) py = [] for i in range(0,n): py.append(int(input())) for j in range(0,len(py)): if len(py[i]) > 10: print(f"{{py[i][0]}{len(py)}{py[i][-1]}) else: print(py[i]) ```

-1

686

A

Free Ice Cream

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.

The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.

Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.

[ "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n", "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n" ]

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

Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

500

[ { "input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20", "output": "22 1" }, { "input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98", "output": "3 2" }, { "input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000", "output": "7000000000 0" }, { "input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46", "output": "46 0" }, { "input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100", "output": "700 0" }, { "input": "1 0\n- 526403222", "output": "0 1" }, { "input": "1 897986543\n- 371188251", "output": "526798292 0" }, { "input": "1 0\n+ 1", "output": "1 0" }, { "input": "1 0\n- 1", "output": "0 1" }, { "input": "1 10\n+ 10", "output": "20 0" }, { "input": "1 3\n- 5", "output": "3 1" }, { "input": "1 0\n- 5", "output": "0 1" }, { "input": "1 0\n+ 5", "output": "5 0" } ]

1,651,674,717

2,147,483,647

PyPy 3-64

OK

TESTS

34

77

2,355,200

n,x = (map(int,input().split())) d = 0 for _ in range(n) : a = input().split() if a[0]=="+" : x+=int(a[1]) else : if x<int(a[1]) : d+=1 else : x-=int(a[1]) print(str(x)+" "+str(d))

Title: Free Ice Cream Time Limit: None seconds Memory Limit: None megabytes Problem Description: After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. Input Specification: The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. Output Specification: Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. Demo Input: ['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n'] Demo Output: ['22 1\n', '3 2\n'] Note: Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

```python n,x = (map(int,input().split())) d = 0 for _ in range(n) : a = input().split() if a[0]=="+" : x+=int(a[1]) else : if x<int(a[1]) : d+=1 else : x-=int(a[1]) print(str(x)+" "+str(d)) ```

3

672

B

Different is Good

PROGRAMMING

1,000

[ "constructive algorithms", "implementation", "strings" ]

null

A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different. Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba". If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible. Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*. The second line contains the string *s* of length *n* consisting of only lowercase English letters.

If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.

[ "2\naa\n", "4\nkoko\n", "5\nmurat\n" ]

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

In the first sample one of the possible solutions is to change the first character to 'b'. In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".

1,000

[ { "input": "2\naa", "output": "1" }, { "input": "4\nkoko", "output": "2" }, { "input": "5\nmurat", "output": "0" }, { "input": "6\nacbead", "output": "1" }, { "input": "7\ncdaadad", "output": "4" }, { "input": "25\npeoaicnbisdocqofsqdpgobpn", "output": "12" }, { "input": "25\ntcqpchnqskqjacruoaqilgebu", "output": "7" }, { "input": "13\naebaecedabbee", "output": "8" }, { "input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "-1" }, { "input": "10\nbababbdaee", "output": "6" }, { "input": "11\ndbadcdbdbca", "output": "7" }, { "input": "12\nacceaabddaaa", "output": "7" }, { "input": "13\nabddfbfaeecfa", "output": "7" }, { "input": "14\neeceecacdbcbbb", "output": "9" }, { "input": "15\ndcbceaaggabaheb", "output": "8" }, { "input": "16\nhgiegfbadgcicbhd", "output": "7" }, { "input": "17\nabhfibbdddfghgfdi", "output": "10" }, { "input": "26\nbbbbbabbaababaaabaaababbaa", "output": "24" }, { "input": "26\nahnxdnbfbcrirerssyzydihuee", "output": "11" }, { "input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc", "output": "8" }, { "input": "26\nrvxmulriorilidecqwmfaemifj", "output": "10" }, { "input": "26\naowpmreooavnmamogdoopuisge", "output": "12" }, { "input": "26\ninimevtuefhvuefirdehmmfudh", "output": "15" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "25" }, { "input": "27\nqdcfjtblgglnilgassirrjekcjt", "output": "-1" }, { "input": "27\nabcdefghijklmnopqrstuvwxyza", "output": "-1" }, { "input": "26\nqwertyuiopasdfghjklzxcvbnm", "output": "0" }, { "input": "5\nzzzzz", "output": "4" }, { "input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa", "output": "-1" }, { "input": "1\nq", "output": "0" }, { "input": "27\nqwertyuioplkjhgfdsazxcvbnmm", "output": "-1" }, { "input": "9\nxxxyyyzzz", "output": "6" }, { "input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk", "output": "-1" }, { "input": "27\nqwertyuiopasdfghjklzxcvbnmm", "output": "-1" }, { "input": "26\nabcdefghijklmnopqrstuvwxyz", "output": "0" }, { "input": "26\nabcdefghijklmnopqrstuvwxya", "output": "1" }, { "input": "27\nabcdefghijklmnopqrstuvwxyzz", "output": "-1" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "25" }, { "input": "26\nqwertyuioplkjhgfdsazxcvbnm", "output": "0" }, { "input": "10\nzzzzzzzzzz", "output": "9" }, { "input": "1\na", "output": "0" }, { "input": "30\nabcdefghtyabcdefghtyabcdefghty", "output": "-1" } ]

1,546,662,367

2,147,483,647

Python 3

OK

TESTS

47

124

204,800

def main(): n = int(input()) s = input() arr = [0] * 26 for char in s: arr[ord(char) - ord('a')] += 1 ans = 0 count = 0 for c in range(26): if arr[c] > 0: count += 1 ans += arr[c] - 1 if 26 - count < ans: print(-1) else: print(ans) main()

Title: Different is Good Time Limit: None seconds Memory Limit: None megabytes Problem Description: A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different. Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba". If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible. Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*. The second line contains the string *s* of length *n* consisting of only lowercase English letters. Output Specification: If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes. Demo Input: ['2\naa\n', '4\nkoko\n', '5\nmurat\n'] Demo Output: ['1\n', '2\n', '0\n'] Note: In the first sample one of the possible solutions is to change the first character to 'b'. In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".

```python def main(): n = int(input()) s = input() arr = [0] * 26 for char in s: arr[ord(char) - ord('a')] += 1 ans = 0 count = 0 for c in range(26): if arr[c] > 0: count += 1 ans += arr[c] - 1 if 26 - count < ans: print(-1) else: print(ans) main() ```

3

134

A

Average Numbers

PROGRAMMING

1,200

[ "brute force", "implementation" ]

null

You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one).

The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers.

Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*. If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.

[ "5\n1 2 3 4 5\n", "4\n50 50 50 50\n" ]

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

none

500

[ { "input": "5\n1 2 3 4 5", "output": "1\n3 " }, { "input": "4\n50 50 50 50", "output": "4\n1 2 3 4 " }, { "input": "3\n2 3 1", "output": "1\n1 " }, { "input": "2\n4 2", "output": "0" }, { "input": "2\n1 1", "output": "2\n1 2 " }, { "input": "10\n3 3 3 3 3 4 3 3 3 2", "output": "8\n1 2 3 4 5 7 8 9 " }, { "input": "10\n15 7 10 7 7 7 4 4 7 2", "output": "5\n2 4 5 6 9 " }, { "input": "6\n2 2 2 2 2 2", "output": "6\n1 2 3 4 5 6 " }, { "input": "6\n3 3 3 3 3 3", "output": "6\n1 2 3 4 5 6 " }, { "input": "4\n6 6 6 7", "output": "0" }, { "input": "2\n1 2", "output": "0" }, { "input": "3\n3 3 4", "output": "0" }, { "input": "5\n7 6 6 6 6", "output": "0" }, { "input": "4\n3 5 5 9", "output": "0" }, { "input": "3\n99 100 99", "output": "0" }, { "input": "4\n5 6 5 5", "output": "0" }, { "input": "6\n1 1 2 1 1 1", "output": "0" }, { "input": "2\n4 5", "output": "0" }, { "input": "4\n1 1 1 2", "output": "0" }, { "input": "3\n1 2 4", "output": "0" }, { "input": "6\n1 1 2 3 3 3", "output": "0" }, { "input": "4\n4 5 5 4", "output": "0" }, { "input": "3\n2 3 5", "output": "0" }, { "input": "3\n2 1 1", "output": "0" }, { "input": "3\n1 1 2", "output": "0" }, { "input": "4\n1 2 3 4", "output": "0" }, { "input": "5\n1 2 3 4 6", "output": "0" }, { "input": "3\n2 2 3", "output": "0" }, { "input": "4\n3 4 5 1", "output": "0" }, { "input": "3\n2 3 2", "output": "0" }, { "input": "3\n3 4 4", "output": "0" }, { "input": "3\n10 5 7", "output": "0" }, { "input": "3\n5 6 5", "output": "0" }, { "input": "4\n1 2 3 7", "output": "0" }, { "input": "5\n2 2 3 3 3", "output": "0" }, { "input": "3\n4 3 6", "output": "0" }, { "input": "3\n1 1 3", "output": "0" }, { "input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 " } ]

1,688,468,699

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

31

0

a = int(input()) b = list(map(int, input().split())) counter = 0 c = [] s = sum(b) for i in range(len(b)): if s / b[i] == a: c.append(b[i]) counter += 1 print(counter) print(*c) # Tue Jul 04 2023 14:03:56 GMT+0300 (Moscow Standard Time)

Title: Average Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one). Input Specification: The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers. Output Specification: Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*. If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line. Demo Input: ['5\n1 2 3 4 5\n', '4\n50 50 50 50\n'] Demo Output: ['1\n3 ', '4\n1 2 3 4 '] Note: none

```python a = int(input()) b = list(map(int, input().split())) counter = 0 c = [] s = sum(b) for i in range(len(b)): if s / b[i] == a: c.append(b[i]) counter += 1 print(counter) print(*c) # Tue Jul 04 2023 14:03:56 GMT+0300 (Moscow Standard Time) ```

0

171

A

Mysterious numbers - 1

PROGRAMMING

1,200

[ "*special", "constructive algorithms" ]

null

The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space. Output a single integer.

The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space.

Output a single integer.

[ "3 14\n", "27 12\n", "100 200\n" ]

[ "44\n", "48\n", "102\n" ]

none

0

[ { "input": "3 14", "output": "44" }, { "input": "27 12", "output": "48" }, { "input": "100 200", "output": "102" }, { "input": "0 0", "output": "0" }, { "input": "31415 92653", "output": "67044" }, { "input": "1000000000 0", "output": "1000000000" }, { "input": "0 1000000000", "output": "1" }, { "input": "1000000000 1000000000", "output": "1000000001" }, { "input": "804289383 846930886", "output": "1492329031" }, { "input": "681692777 714636915", "output": "1201329194" }, { "input": "957747793 424238335", "output": "1491580217" }, { "input": "719885386 649760492", "output": "1013953332" }, { "input": "596516649 189641421", "output": "720663630" }, { "input": "25202362 350490027", "output": "745296415" }, { "input": "783368690 102520059", "output": "1733393891" }, { "input": "44897763 967513926", "output": "674213532" }, { "input": "365180540 540383426", "output": "989563585" }, { "input": "304089172 303455736", "output": "941643475" }, { "input": "35005211 521595368", "output": "898600336" }, { "input": "294702567 726956429", "output": "1219362194" }, { "input": "336465782 861021530", "output": "371585950" }, { "input": "278722862 233665123", "output": "600289194" }, { "input": "145174067 468703135", "output": "676481931" }, { "input": "101513929 801979802", "output": "310493037" }, { "input": "315634022 635723058", "output": "1165961558" }, { "input": "369133069 125898167", "output": "1131031590" }, { "input": "59961393 89018456", "output": "125442491" }, { "input": "628175011 656478042", "output": "869049667" }, { "input": "131176229 653377373", "output": "504949585" }, { "input": "859484421 914544919", "output": "1778929840" }, { "input": "608413784 756898537", "output": "1344312441" }, { "input": "734575198 973594324", "output": "1158070577" }, { "input": "149798315 38664370", "output": "157144998" }, { "input": "129566413 184803526", "output": "754874894" }, { "input": "412776091 424268980", "output": "502638515" }, { "input": "911759956 749241873", "output": "1289902903" }, { "input": "137806862 42999170", "output": "145006786" }, { "input": "982906996 135497281", "output": "1165701527" }, { "input": "511702305 84420925", "output": "564604753" }, { "input": "937477084 827336327", "output": "1661110812" }, { "input": "1000000000 999999999", "output": "1999999999" } ]

1,617,287,204

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

62

0

a=int(input()) b=int(input()) print(1)

Title: Mysterious numbers - 1 Time Limit: None seconds Memory Limit: None megabytes Problem Description: The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space. Output a single integer. Input Specification: The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space. Output Specification: Output a single integer. Demo Input: ['3 14\n', '27 12\n', '100 200\n'] Demo Output: ['44\n', '48\n', '102\n'] Note: none

```python a=int(input()) b=int(input()) print(1) ```

-1

265

B

Roadside Trees (Simplified Edition)

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts. Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions: - Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1. Compute the minimal time (in seconds) required to eat all nuts.

The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees. Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.

Print a single integer — the minimal time required to eat all nuts in seconds.

[ "2\n1\n2\n", "5\n2\n1\n2\n1\n1\n" ]

[ "5\n", "14\n" ]

none

1,000

[ { "input": "2\n1\n2", "output": "5" }, { "input": "5\n2\n1\n2\n1\n1", "output": "14" }, { "input": "1\n1", "output": "2" } ]

1,611,901,019

2,147,483,647

Python 3

OK

TESTS

15

404

0

i=0 ans=0 past=0 for x in range(int(input())): n=int(input()) if i==0: ans=n+1 past=n ##print(ans) else: ans+=abs(past-n)+2 ##print(ans) i+=1 past=n print(ans)

Title: Roadside Trees (Simplified Edition) Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts. Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions: - Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1. Compute the minimal time (in seconds) required to eat all nuts. Input Specification: The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees. Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*. Output Specification: Print a single integer — the minimal time required to eat all nuts in seconds. Demo Input: ['2\n1\n2\n', '5\n2\n1\n2\n1\n1\n'] Demo Output: ['5\n', '14\n'] Note: none

```python i=0 ans=0 past=0 for x in range(int(input())): n=int(input()) if i==0: ans=n+1 past=n ##print(ans) else: ans+=abs(past-n)+2 ##print(ans) i+=1 past=n print(ans) ```

3