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928
A
Login Verification
PROGRAMMING
1,200
[ "*special", "strings" ]
null
null
When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc. Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types: - transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other. For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not. You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.
The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50  — the login itself. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins. The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.
Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it. Otherwise print «No» (without quotes).
[ "1_wat\n2\n2_wat\nwat_1\n", "000\n3\n00\nooA\noOo\n", "_i_\n3\n__i_\n_1_\nI\n", "La0\n3\n2a0\nLa1\n1a0\n", "abc\n1\naBc\n", "0Lil\n2\nLIL0\n0Ril\n" ]
[ "Yes\n", "No\n", "No\n", "No\n", "No\n", "Yes\n" ]
In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing. In the third sample case the new login is similar with the second one.
500
[ { "input": "1_wat\n2\n2_wat\nwat_1", "output": "Yes" }, { "input": "000\n3\n00\nooA\noOo", "output": "No" }, { "input": "_i_\n3\n__i_\n_1_\nI", "output": "No" }, { "input": "La0\n3\n2a0\nLa1\n1a0", "output": "No" }, { "input": "abc\n1\naBc", "output": "No" }, { "input": "0Lil\n2\nLIL0\n0Ril", "output": "Yes" }, { "input": "iloO\n3\niIl0\noIl0\nIooO", "output": "Yes" }, { "input": "L1il0o1L1\n5\niLLoLL\noOI1Io10il\nIoLLoO\nO01ilOoI\nI10l0o", "output": "Yes" }, { "input": "ELioO1lOoOIOiLoooi1iolul1O\n7\nOoEIuOIl1ui1010uiooOoi0Oio001L0EoEolO0\nOLIoOEuoE11u1u1iLOI0oO\nuEOuO0uIOOlO01OlEI0E1Oo0IO1LI0uE0LILO0\nEOo0Il11iIOOOIiuOiIiiLOLEOOII001EE\niOoO0LOulioE0OLIIIulli01OoiuOOOoOlEiI0EiiElIIu0\nlE1LOE1Oil\n1u0EOliIiIOl1u110il0l1O0u", "output": "Yes" }, { "input": "0blo7X\n20\n1oobb6\nXIXIO2X\n2iYI2\n607XXol\n2I6io22\nOl10I\nbXX0Lo\nolOOb7X\n07LlXL\nlXY17\n12iIX2\n7lL70\nbOo11\n17Y6b62\n0O6L7\n1lX2L\n2iYl6lI\n7bXIi1o\niLIY2\n0OIo1X", "output": "Yes" }, { "input": "lkUL\n25\nIIfL\nokl\nfoo\ni0U\noko\niIoU\nUUv\nvli\nv0Uk\n0Of\niill\n1vkl\nUIf\nUfOO\nlvLO\nUUo0\nIOf1\nlovL\nIkk\noIv\nLvfU\n0UI\nkol\n1OO0\n1OOi", "output": "Yes" }, { "input": "L1lo\n3\nOOo1\nL1lo\n0lOl", "output": "No" }, { "input": "LIoooiLO\n5\nLIoooiLO\nl0o01I00\n0OOl0lLO01\nil10i0\noiloi", "output": "No" }, { "input": "1i1lQI\n7\nuLg1uLLigIiOLoggu\nLLLgIuQIQIIloiQuIIoIO0l0o000\n0u1LQu11oIuooIl0OooLg0i0IQu1O1lloI1\nQuQgIQi0LOIliLOuuuioLQou1l\nlLIO00QLi01LogOliOIggII1\no0Ll1uIOQl10IL0IILQ\n1i1lQI", "output": "No" }, { "input": "oIzz1\n20\n1TTl0O\nloF0LT\n1lLzo\noi0Ov\nFlIF1zT\nzoITzx\n0TIFlT\nl1vllil\nOviix1F\nLFvI1lL\nLIl0loz\nixz1v\n1i1vFi\nTIFTol\noIzz1\nIvTl0o\nxv1U0O\niiiioF\n1oiLUlO\nxToxv1", "output": "No" }, { "input": "00L0\n25\n0il\nIlkZ\nL0I\n00L0\nBd0\nZLd\n0d1k\nddk\nIdl\nkBd\nkBOL\nZ1lI\nkBL\nLOko\noZ0i\nZ1lO\nLiOk\niBld\nLO0d\ndIo\nZ10\n1k1i\n0o0L\nIoBd\ni0B0", "output": "No" }, { "input": "Z\n1\nz", "output": "No" }, { "input": "0\n1\no", "output": "No" }, { "input": "0\n1\nO", "output": "No" }, { "input": "o\n1\n0", "output": "No" }, { "input": "o\n1\nO", "output": "No" }, { "input": "o\n1\no", "output": "No" }, { "input": "O\n1\no", "output": "No" }, { "input": "O\n1\n0", "output": "No" }, { "input": "1\n1\nl", "output": "No" }, { "input": "1\n1\nL", "output": "No" }, { "input": "1\n1\ni", "output": "No" }, { "input": "1\n1\nI", "output": "No" }, { "input": "1\n1\no", "output": "Yes" }, { "input": "i\n1\n1", "output": "No" }, { "input": "i\n1\nL", "output": "No" }, { "input": "i\n1\nl", "output": "No" }, { "input": "I\n1\nL", "output": "No" }, { "input": "I\n1\nl", "output": "No" }, { "input": "I\n1\n1", "output": "No" }, { "input": "l\n1\n1", "output": "No" }, { "input": "l\n1\nL", "output": "No" }, { "input": "l\n1\ni", "output": "No" }, { "input": "l\n1\nI", "output": "No" }, { "input": "L\n1\nl", "output": "No" }, { "input": "L\n1\n1", "output": "No" }, { "input": "L\n1\ni", "output": "No" }, { "input": "L\n1\nI", "output": "No" } ]
1,654,178,280
2,147,483,647
Python 3
OK
TESTS
73
46
0
l=input() f=lambda x: x.lower().replace('0','o').replace('1','l').replace('i','l') print(['Yes','No'][f(l)in map(f,[input()for r in range(int(input()))])])
Title: Login Verification Time Limit: None seconds Memory Limit: None megabytes Problem Description: When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc. Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types: - transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other. For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not. You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones. Input Specification: The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50  — the login itself. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins. The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar. Output Specification: Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it. Otherwise print «No» (without quotes). Demo Input: ['1_wat\n2\n2_wat\nwat_1\n', '000\n3\n00\nooA\noOo\n', '_i_\n3\n__i_\n_1_\nI\n', 'La0\n3\n2a0\nLa1\n1a0\n', 'abc\n1\naBc\n', '0Lil\n2\nLIL0\n0Ril\n'] Demo Output: ['Yes\n', 'No\n', 'No\n', 'No\n', 'No\n', 'Yes\n'] Note: In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing. In the third sample case the new login is similar with the second one.
```python l=input() f=lambda x: x.lower().replace('0','o').replace('1','l').replace('i','l') print(['Yes','No'][f(l)in map(f,[input()for r in range(int(input()))])]) ```
3
258
A
Little Elephant and Bits
PROGRAMMING
1,100
[ "greedy", "math" ]
null
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 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>. In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
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" } ]
1,697,567,275
2,147,483,647
Python 3
OK
TESTS
37
248
716,800
x = input() t = False if "0" in x: for i in x: if "0" == i and t == False: t = True continue print(i, end="") else: print(x[:-1])
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 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>. In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
```python x = input() t = False if "0" in x: for i in x: if "0" == i and t == False: t = True continue print(i, end="") else: print(x[:-1]) ```
3
978
B
File Name
PROGRAMMING
800
[ "greedy", "strings" ]
null
null
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed. Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx". You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii".
The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name. The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name.
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
[ "6\nxxxiii\n", "5\nxxoxx\n", "10\nxxxxxxxxxx\n" ]
[ "1\n", "0\n", "8\n" ]
In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
0
[ { "input": "6\nxxxiii", "output": "1" }, { "input": "5\nxxoxx", "output": "0" }, { "input": "10\nxxxxxxxxxx", "output": "8" }, { "input": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "98" }, { "input": "99\nxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxa", "output": "0" }, { "input": "3\nxxx", "output": "1" }, { "input": "77\naaabbbcccdddeeefffggghhhiiijjjkkklllmmmnnnooopppqqqrrrssstttuuuvvvwwwxxyyyzzz", "output": "0" }, { "input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxxxxmrx", "output": "41" }, { "input": "100\nxxxxxxxxxxxjtxxxxxxxxcxxxxxxcfxxxxzxxxxxxgxxxxxbxxxxbxxxxxxxxdycxxxxokixxxkizxxgcxxxxxxxxexxxxxfxxxc", "output": "49" }, { "input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxwxxxxx", "output": "41" }, { "input": "34\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "0" }, { "input": "5\nfcyju", "output": "0" }, { "input": "100\nihygyvdvyeifomhxhkhdkimquvgallbqharcyriyqkidnwykozuhvkwdldlztpabgyuflikychqpdenwzgtlzotyumjgdsrbxxxx", "output": "2" } ]
1,680,905,419
2,147,483,647
Python 3
OK
TESTS
13
46
0
n = int(input()) s = input() cnt = 1 k = 0 maxk = 0 for i in range(len(s)-1): if s[i] == 'x' and s[i + 1] == 'x': cnt += 1 if cnt >= 3: k += 1 else: maxk = max(maxk, k) cnt = 1 print(k)
Title: File Name Time Limit: None seconds Memory Limit: None megabytes Problem Description: You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed. Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx". You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii". Input Specification: The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name. The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name. Output Specification: Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0. Demo Input: ['6\nxxxiii\n', '5\nxxoxx\n', '10\nxxxxxxxxxx\n'] Demo Output: ['1\n', '0\n', '8\n'] Note: In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
```python n = int(input()) s = input() cnt = 1 k = 0 maxk = 0 for i in range(len(s)-1): if s[i] == 'x' and s[i + 1] == 'x': cnt += 1 if cnt >= 3: k += 1 else: maxk = max(maxk, k) cnt = 1 print(k) ```
3
961
B
Lecture Sleep
PROGRAMMING
1,200
[ "data structures", "dp", "implementation", "two pointers" ]
null
null
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute. Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing. You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells. You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake. The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute. The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
[ "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n" ]
[ "16\n" ]
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
0
[ { "input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0", "output": "16" }, { "input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0", "output": "30000" }, { "input": "3 3\n10 10 10\n1 1 0", "output": "30" }, { "input": "1 1\n423\n0", "output": "423" }, { "input": "6 6\n1 3 5 2 5 4\n1 1 0 1 0 0", "output": "20" }, { "input": "5 2\n1 2 3 4 20\n0 0 0 1 0", "output": "24" }, { "input": "3 1\n1 2 3\n0 0 1", "output": "5" }, { "input": "4 2\n4 5 6 8\n1 0 1 0", "output": "18" }, { "input": "6 3\n1 3 5 2 1 15\n1 1 0 1 0 0", "output": "22" }, { "input": "5 5\n1 2 3 4 5\n1 1 1 0 1", "output": "15" }, { "input": "3 3\n3 3 3\n1 0 1", "output": "9" }, { "input": "5 5\n500 44 3 4 50\n1 0 0 0 0", "output": "601" }, { "input": "2 2\n3 2\n1 0", "output": "5" }, { "input": "7 6\n4 9 1 7 1 8 4\n0 0 0 1 0 1 0", "output": "30" }, { "input": "4 3\n6 5 9 6\n1 1 0 1", "output": "26" }, { "input": "2 1\n3 2\n0 0", "output": "3" }, { "input": "1 1\n10\n0", "output": "10" }, { "input": "2 1\n3 2\n1 0", "output": "5" }, { "input": "4 2\n3 6 7 2\n0 0 1 1", "output": "18" }, { "input": "10 5\n3 5 9 2 5 9 3 8 8 1\n0 1 1 1 0 1 0 0 0 0", "output": "49" }, { "input": "10 4\n9 5 6 4 3 9 5 1 10 7\n0 0 0 0 0 0 1 0 0 1", "output": "36" }, { "input": "9 8\n3 3 7 7 1 9 10 7 1\n1 1 1 1 1 1 1 1 1", "output": "48" }, { "input": "2 1\n3 4\n0 0", "output": "4" }, { "input": "2 1\n3 2\n0 1", "output": "5" }, { "input": "10 1\n6 6 8 7 6 6 3 2 5 6\n0 0 1 0 0 1 0 1 1 1", "output": "34" }, { "input": "3 2\n10 10 6\n0 0 0", "output": "20" }, { "input": "6 3\n1 3 5 2 5 4\n1 1 1 1 1 1", "output": "20" }, { "input": "10 5\n1 1 1 1 1 1 1 1 10000 1\n1 1 1 1 1 1 1 1 0 1", "output": "10009" } ]
1,674,394,933
2,147,483,647
PyPy 3-64
OK
TESTS
37
93
16,793,600
import sys # sys.stdin, sys.stdout = open("input.txt", "r"), open("output.txt", "w") def main(): n, k = map(int, input().split()) a = [int(a) for a in input().split()] b = [int(a) for a in input().split()] ans = 0 for i in range(n): if b[i]: ans += a[i] pre = [0 for i in range(n)] if not b[0]: pre[0] = a[0] for i in range(1, n): pre[i] = pre[i - 1] if not b[i]: pre[i] += a[i] temp = 0 for l in range(n - k + 1): if l == 0: temp = max(temp, pre[k - 1]) else: temp = max(temp, pre[l + k - 1] - pre[l - 1]) print(ans + temp) if __name__ == "__main__": main()
Title: Lecture Sleep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute. Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing. You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells. You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. Input Specification: The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake. The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute. The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. Output Specification: Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. Demo Input: ['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n'] Demo Output: ['16\n'] Note: In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
```python import sys # sys.stdin, sys.stdout = open("input.txt", "r"), open("output.txt", "w") def main(): n, k = map(int, input().split()) a = [int(a) for a in input().split()] b = [int(a) for a in input().split()] ans = 0 for i in range(n): if b[i]: ans += a[i] pre = [0 for i in range(n)] if not b[0]: pre[0] = a[0] for i in range(1, n): pre[i] = pre[i - 1] if not b[i]: pre[i] += a[i] temp = 0 for l in range(n - k + 1): if l == 0: temp = max(temp, pre[k - 1]) else: temp = max(temp, pre[l + k - 1] - pre[l - 1]) print(ans + temp) if __name__ == "__main__": main() ```
3
818
A
Diplomas and Certificates
PROGRAMMING
800
[ "implementation", "math" ]
null
null
There are *n* students who have taken part in an olympiad. Now it's time to award the students. Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners. You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible. It's possible that there are no winners.
[ "18 2\n", "9 10\n", "1000000000000 5\n", "1000000000000 499999999999\n" ]
[ "3 6 9\n", "0 0 9\n", "83333333333 416666666665 500000000002\n", "1 499999999999 500000000000\n" ]
none
0
[ { "input": "18 2", "output": "3 6 9" }, { "input": "9 10", "output": "0 0 9" }, { "input": "1000000000000 5", "output": "83333333333 416666666665 500000000002" }, { "input": "1000000000000 499999999999", "output": "1 499999999999 500000000000" }, { "input": "1 1", "output": "0 0 1" }, { "input": "5 3", "output": "0 0 5" }, { "input": "42 6", "output": "3 18 21" }, { "input": "1000000000000 1000", "output": "499500499 499500499000 500000000501" }, { "input": "999999999999 999999", "output": "499999 499998500001 500000999999" }, { "input": "732577309725 132613", "output": "2762066 366285858458 366288689201" }, { "input": "152326362626 15", "output": "4760198832 71402982480 76163181314" }, { "input": "2 1", "output": "0 0 2" }, { "input": "1000000000000 500000000000", "output": "0 0 1000000000000" }, { "input": "100000000000 50000000011", "output": "0 0 100000000000" }, { "input": "1000000000000 32416187567", "output": "15 486242813505 513757186480" }, { "input": "1000000000000 7777777777", "output": "64 497777777728 502222222208" }, { "input": "1000000000000 77777777777", "output": "6 466666666662 533333333332" }, { "input": "100000000000 578485652", "output": "86 49749766072 50250233842" }, { "input": "999999999999 10000000000", "output": "49 490000000000 509999999950" }, { "input": "7 2", "output": "1 2 4" }, { "input": "420506530901 752346673804", "output": "0 0 420506530901" }, { "input": "960375521135 321688347872", "output": "1 321688347872 638687173262" }, { "input": "1000000000000 1000000000000", "output": "0 0 1000000000000" }, { "input": "99999999999 15253636363", "output": "3 45760909089 54239090907" }, { "input": "19 2", "output": "3 6 10" }, { "input": "999999999999 1000000000000", "output": "0 0 999999999999" }, { "input": "1000000000000 5915587276", "output": "84 496909331184 503090668732" }, { "input": "1000000000000 1000000006", "output": "499 499000002994 500999996507" }, { "input": "549755813888 134217728", "output": "2047 274743689216 275012122625" }, { "input": "99999999999 3333333", "output": "14999 49996661667 50003323333" }, { "input": "9 1", "output": "2 2 5" }, { "input": "1000000000000 250000000001", "output": "1 250000000001 749999999998" }, { "input": "5 1", "output": "1 1 3" }, { "input": "3107038133 596040207", "output": "2 1192080414 1914957717" }, { "input": "1000000000000 73786977", "output": "6776 499980556152 500019437072" }, { "input": "1000000000000 73786976", "output": "6776 499980549376 500019443848" }, { "input": "1000000000000 25000000000", "output": "19 475000000000 524999999981" }, { "input": "216929598879 768233755932", "output": "0 0 216929598879" }, { "input": "1000000000000 250000000000", "output": "1 250000000000 749999999999" }, { "input": "1000000000000 100000000001", "output": "4 400000000004 599999999992" }, { "input": "100000000000 100000000001", "output": "0 0 100000000000" }, { "input": "900000000000 100281800001", "output": "4 401127200004 498872799992" }, { "input": "906028900004 109123020071", "output": "4 436492080284 469536819716" }, { "input": "1000000000000 1", "output": "250000000000 250000000000 500000000000" } ]
1,659,539,977
2,147,483,647
PyPy 3-64
OK
TESTS
44
62
0
import sys n, k = (int(el) for el in input().split()) if k > n: print(0, 0, n) sys.exit() '''a, b, c = 0, 0, 0 a = int((n // 2) ** 0.5) b = a * k c = n - (a + b) ''' '''for i in range((n // 2) ** 0.5, 0, -1): x = i y = n // 2 - i if max(y, x) / min(x, y) == k: a = min(x, y) b = max(x, y) break c = n - (a + b)''' a = (n // 2) // (k + 1) b = a * k c = n - (a + b) print(a, b, c)
Title: Diplomas and Certificates Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* students who have taken part in an olympiad. Now it's time to award the students. Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners. You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. Input Specification: The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. Output Specification: Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible. It's possible that there are no winners. Demo Input: ['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n'] Demo Output: ['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n'] Note: none
```python import sys n, k = (int(el) for el in input().split()) if k > n: print(0, 0, n) sys.exit() '''a, b, c = 0, 0, 0 a = int((n // 2) ** 0.5) b = a * k c = n - (a + b) ''' '''for i in range((n // 2) ** 0.5, 0, -1): x = i y = n // 2 - i if max(y, x) / min(x, y) == k: a = min(x, y) b = max(x, y) break c = n - (a + b)''' a = (n // 2) // (k + 1) b = a * k c = n - (a + b) print(a, b, c) ```
3
733
A
Grasshopper And the String
PROGRAMMING
1,000
[ "implementation" ]
null
null
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump. Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability. The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
[ "ABABBBACFEYUKOTT\n", "AAA\n" ]
[ "4", "1" ]
none
500
[ { "input": "ABABBBACFEYUKOTT", "output": "4" }, { "input": "AAA", "output": "1" }, { "input": "A", "output": "1" }, { "input": "B", "output": "2" }, { "input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU", "output": "39" }, { "input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI", "output": "1" }, { "input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC", "output": "85" }, { "input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ", "output": "18" }, { "input": "PKLKBWTXVJ", "output": "11" }, { "input": "CFHFPTGMOKXVLJJZJDQW", "output": "12" }, { "input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX", "output": "9" }, { "input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY", "output": "4" }, { "input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD", "output": "101" }, { "input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC", "output": "76" }, { "input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC", "output": "45" }, { "input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT", "output": "48" }, { "input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX", "output": "47" }, { "input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA", "output": "65" }, { "input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS", "output": "28" }, { "input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW", "output": "35" }, { "input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL", "output": "30" }, { "input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL", "output": "19" }, { "input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB", "output": "30" }, { "input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB", "output": "34" }, { "input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR", "output": "17" }, { "input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT", "output": "15" }, { "input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB", "output": "9" }, { "input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW", "output": "5" }, { "input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA", "output": "4" }, { "input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE", "output": "5" }, { "input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA", "output": "5" }, { "input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO", "output": "3" }, { "input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY", "output": "3" }, { "input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY", "output": "2" }, { "input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE", "output": "2" }, { "input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII", "output": "2" }, { "input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE", "output": "2" }, { "input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU", "output": "1" }, { "input": "ABBABBB", "output": "4" }, { "input": "ABCD", "output": "4" }, { "input": "XXYC", "output": "3" }, { "input": "YYY", "output": "1" }, { "input": "ABABBBBBBB", "output": "8" }, { "input": "YYYY", "output": "1" }, { "input": "YYYYY", "output": "1" }, { "input": "AXXX", "output": "4" }, { "input": "YYYYYYY", "output": "1" }, { "input": "BYYBBB", "output": "4" }, { "input": "YYYYYYYYY", "output": "1" }, { "input": "CAAAAA", "output": "2" }, { "input": "CCCACCCC", "output": "5" }, { "input": "ABABBBACFEYUKOTTTT", "output": "5" }, { "input": "AABBYYYYYYYY", "output": "3" }, { "input": "BYBACYC", "output": "2" }, { "input": "Y", "output": "1" }, { "input": "ABBBBBB", "output": "7" }, { "input": "BACDYDI", "output": "3" }, { "input": "XEXXXXXXXXXXXXXXX", "output": "16" }, { "input": "TTYTT", "output": "3" }, { "input": "AAYBC", "output": "3" }, { "input": "ABABBBACFEYUKOTTTTT", "output": "6" }, { "input": "YYAYY", "output": "1" }, { "input": "YZZY", "output": "3" }, { "input": "YY", "output": "1" }, { "input": "ZZYZZ", "output": "3" }, { "input": "YBBBY", "output": "4" }, { "input": "BBBACCCCCCC", "output": "8" }, { "input": "YBBBBY", "output": "5" }, { "input": "YYYYYYYYYY", "output": "1" }, { "input": "ABABBBBBBBBBBBB", "output": "13" } ]
1,625,394,863
2,147,483,647
Python 3
OK
TESTS
70
62
0
s = input()+' ' a = ['A', 'U', 'Y', 'E', 'O', 'I'] b = 0 c = 0 for i in s: c += 1 if i in a: if b < c: b = c c = 0 if b < c: b = c print(b)
Title: Grasshopper And the String Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump. Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability. The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'. Input Specification: The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100. Output Specification: Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels. Demo Input: ['ABABBBACFEYUKOTT\n', 'AAA\n'] Demo Output: ['4', '1'] Note: none
```python s = input()+' ' a = ['A', 'U', 'Y', 'E', 'O', 'I'] b = 0 c = 0 for i in s: c += 1 if i in a: if b < c: b = c c = 0 if b < c: b = c print(b) ```
3
978
B
File Name
PROGRAMMING
800
[ "greedy", "strings" ]
null
null
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed. Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx". You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii".
The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name. The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name.
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
[ "6\nxxxiii\n", "5\nxxoxx\n", "10\nxxxxxxxxxx\n" ]
[ "1\n", "0\n", "8\n" ]
In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
0
[ { "input": "6\nxxxiii", "output": "1" }, { "input": "5\nxxoxx", "output": "0" }, { "input": "10\nxxxxxxxxxx", "output": "8" }, { "input": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "98" }, { "input": "99\nxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxa", "output": "0" }, { "input": "3\nxxx", "output": "1" }, { "input": "77\naaabbbcccdddeeefffggghhhiiijjjkkklllmmmnnnooopppqqqrrrssstttuuuvvvwwwxxyyyzzz", "output": "0" }, { "input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxxxxmrx", "output": "41" }, { "input": "100\nxxxxxxxxxxxjtxxxxxxxxcxxxxxxcfxxxxzxxxxxxgxxxxxbxxxxbxxxxxxxxdycxxxxokixxxkizxxgcxxxxxxxxexxxxxfxxxc", "output": "49" }, { "input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxwxxxxx", "output": "41" }, { "input": "34\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "0" }, { "input": "5\nfcyju", "output": "0" }, { "input": "100\nihygyvdvyeifomhxhkhdkimquvgallbqharcyriyqkidnwykozuhvkwdldlztpabgyuflikychqpdenwzgtlzotyumjgdsrbxxxx", "output": "2" } ]
1,694,756,076
2,147,483,647
Python 3
OK
TESTS
13
46
0
size = int(input()) s = input() count, remove = 0, 0 for i in s: if i == 'x': count += 1 if count > 2: remove += 1 else: count = 0 if remove == 0: print(0) else: print(remove)
Title: File Name Time Limit: None seconds Memory Limit: None megabytes Problem Description: You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed. Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx". You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii". Input Specification: The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name. The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name. Output Specification: Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0. Demo Input: ['6\nxxxiii\n', '5\nxxoxx\n', '10\nxxxxxxxxxx\n'] Demo Output: ['1\n', '0\n', '8\n'] Note: In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
```python size = int(input()) s = input() count, remove = 0, 0 for i in s: if i == 'x': count += 1 if count > 2: remove += 1 else: count = 0 if remove == 0: print(0) else: print(remove) ```
3
463
B
Caisa and Pylons
PROGRAMMING
1,100
[ "brute force", "implementation", "math" ]
null
null
Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=&gt;<=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Print a single number representing the minimum number of dollars paid by Caisa.
[ "5\n3 4 3 2 4\n", "3\n4 4 4\n" ]
[ "4\n", "4\n" ]
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
1,000
[ { "input": "5\n3 4 3 2 4", "output": "4" }, { "input": "3\n4 4 4", "output": "4" }, { "input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730", "output": "4000" }, { "input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621", "output": "3993" }, { "input": "30\n30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "30" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n69", "output": "69" } ]
1,597,687,643
2,147,483,647
PyPy 3
OK
TESTS
49
187
10,956,800
n = int(input())+1 lst = list(map(int, input().split())) lst = [0] + lst money = 0 energy = 0 for i in range(n-1): if lst[i+1] == lst[i]: continue if lst[i+1] > lst[i]: if energy > lst[i+1] - lst[i]: energy -= lst[i+1] - lst[i] else: money += (lst[i+1] - lst[i]) - energy energy = 0 else: energy += lst[i] - lst[i+1] print(money)
Title: Caisa and Pylons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=&gt;<=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons. Output Specification: Print a single number representing the minimum number of dollars paid by Caisa. Demo Input: ['5\n3 4 3 2 4\n', '3\n4 4 4\n'] Demo Output: ['4\n', '4\n'] Note: In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
```python n = int(input())+1 lst = list(map(int, input().split())) lst = [0] + lst money = 0 energy = 0 for i in range(n-1): if lst[i+1] == lst[i]: continue if lst[i+1] > lst[i]: if energy > lst[i+1] - lst[i]: energy -= lst[i+1] - lst[i] else: money += (lst[i+1] - lst[i]) - energy energy = 0 else: energy += lst[i] - lst[i+1] print(money) ```
3
343
A
Rational Resistance
PROGRAMMING
1,600
[ "math", "number theory" ]
null
null
Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value. However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements: 1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel. With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected. Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.
The single input line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists.
Print a single number — the answer to the problem. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.
[ "1 1\n", "3 2\n", "199 200\n" ]
[ "1\n", "3\n", "200\n" ]
In the first sample, one resistor is enough. In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
500
[ { "input": "1 1", "output": "1" }, { "input": "3 2", "output": "3" }, { "input": "199 200", "output": "200" }, { "input": "1 1000000000000000000", "output": "1000000000000000000" }, { "input": "3 1", "output": "3" }, { "input": "21 8", "output": "7" }, { "input": "18 55", "output": "21" }, { "input": "1 2", "output": "2" }, { "input": "2 1", "output": "2" }, { "input": "1 3", "output": "3" }, { "input": "2 3", "output": "3" }, { "input": "1 4", "output": "4" }, { "input": "5 2", "output": "4" }, { "input": "2 5", "output": "4" }, { "input": "4 5", "output": "5" }, { "input": "3 5", "output": "4" }, { "input": "13 4", "output": "7" }, { "input": "21 17", "output": "9" }, { "input": "5 8", "output": "5" }, { "input": "13 21", "output": "7" }, { "input": "74 99", "output": "28" }, { "input": "2377 1055", "output": "33" }, { "input": "645597 134285", "output": "87" }, { "input": "29906716 35911991", "output": "92" }, { "input": "3052460231 856218974", "output": "82" }, { "input": "288565475053 662099878640", "output": "88" }, { "input": "11504415412768 12754036168327", "output": "163" }, { "input": "9958408561221547 4644682781404278", "output": "196" }, { "input": "60236007668635342 110624799949034113", "output": "179" }, { "input": "4 43470202936783249", "output": "10867550734195816" }, { "input": "16 310139055712567491", "output": "19383690982035476" }, { "input": "15 110897893734203629", "output": "7393192915613582" }, { "input": "439910263967866789 38", "output": "11576585893891241" }, { "input": "36 316049483082136289", "output": "8779152307837131" }, { "input": "752278442523506295 52", "output": "14466893125452056" }, { "input": "4052739537881 6557470319842", "output": "62" }, { "input": "44945570212853 72723460248141", "output": "67" }, { "input": "498454011879264 806515533049393", "output": "72" }, { "input": "8944394323791464 5527939700884757", "output": "77" }, { "input": "679891637638612258 420196140727489673", "output": "86" }, { "input": "1 923438", "output": "923438" }, { "input": "3945894354376 1", "output": "3945894354376" }, { "input": "999999999999999999 5", "output": "200000000000000004" }, { "input": "999999999999999999 1000000000000000000", "output": "1000000000000000000" }, { "input": "999999999999999991 1000000000000000000", "output": "111111111111111120" }, { "input": "999999999999999993 999999999999999991", "output": "499999999999999998" }, { "input": "3 1000000000000000000", "output": "333333333333333336" }, { "input": "1000000000000000000 3", "output": "333333333333333336" }, { "input": "10000000000 1000000001", "output": "100000019" }, { "input": "2 999999999999999999", "output": "500000000000000001" }, { "input": "999999999999999999 2", "output": "500000000000000001" }, { "input": "2 1000000001", "output": "500000002" }, { "input": "123 1000000000000000000", "output": "8130081300813023" } ]
1,408,675,941
2,147,483,647
Python 3
OK
TESTS
53
342
5,222,400
import fractions def resistance(): # print("Input the two values") _a, _b = input().split() a = int(_a) b = int(_b) answer = 1 f = fractions.Fraction(a,b) one = fractions.Fraction(1,1) while (f != one): n = f.numerator d = f.denominator # If either the num or denom are one, go straight up the line and finish if (n == 1) and (d == 1): print(answer) return if (n == 1 ): print(answer + d - 1) return if (d == 1): print(answer + n - 1) return if (f > 1): q = n//d n = n - q*d f = fractions.Fraction(n,d) answer = answer + q else: nbound = (1-f)/f n = nbound.numerator d = nbound.denominator then = n//d + 1 answer = answer + then f = f / (1-then*f) print(answer) resistance()
Title: Rational Resistance Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value. However, all Mike has is lots of identical resistors with unit resistance *R*0<==<=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements: 1. one resistor; 1. an element and one resistor plugged in sequence; 1. an element and one resistor plugged in parallel. With the consecutive connection the resistance of the new element equals *R*<==<=*R**e*<=+<=*R*0. With the parallel connection the resistance of the new element equals . In this case *R**e* equals the resistance of the element being connected. Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element. Input Specification: The single input line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=1018). It is guaranteed that the fraction is irreducible. It is guaranteed that a solution always exists. Output Specification: Print a single number — the answer to the problem. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier. Demo Input: ['1 1\n', '3 2\n', '199 200\n'] Demo Output: ['1\n', '3\n', '200\n'] Note: In the first sample, one resistor is enough. In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5305da389756aab6423d918a08ced468f05604df.png" style="max-width: 100.0%;max-height: 100.0%;"/>. We cannot make this element using two resistors.
```python import fractions def resistance(): # print("Input the two values") _a, _b = input().split() a = int(_a) b = int(_b) answer = 1 f = fractions.Fraction(a,b) one = fractions.Fraction(1,1) while (f != one): n = f.numerator d = f.denominator # If either the num or denom are one, go straight up the line and finish if (n == 1) and (d == 1): print(answer) return if (n == 1 ): print(answer + d - 1) return if (d == 1): print(answer + n - 1) return if (f > 1): q = n//d n = n - q*d f = fractions.Fraction(n,d) answer = answer + q else: nbound = (1-f)/f n = nbound.numerator d = nbound.denominator then = n//d + 1 answer = answer + then f = f / (1-then*f) print(answer) resistance() ```
3
837
A
Text Volume
PROGRAMMING
800
[ "implementation" ]
null
null
You are given a text of single-space separated words, consisting of small and capital Latin letters. Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text. Calculate the volume of the given text.
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text. The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters.
Print one integer number — volume of text.
[ "7\nNonZERO\n", "24\nthis is zero answer text\n", "24\nHarbour Space University\n" ]
[ "5\n", "0\n", "1\n" ]
In the first example there is only one word, there are 5 capital letters in it. In the second example all of the words contain 0 capital letters.
0
[ { "input": "7\nNonZERO", "output": "5" }, { "input": "24\nthis is zero answer text", "output": "0" }, { "input": "24\nHarbour Space University", "output": "1" }, { "input": "2\nWM", "output": "2" }, { "input": "200\nLBmJKQLCKUgtTxMoDsEerwvLOXsxASSydOqWyULsRcjMYDWdDCgaDvBfATIWPVSXlbcCLHPYahhxMEYUiaxoCebghJqvmRnaNHYTKLeOiaLDnATPZAOgSNfBzaxLymTGjfzvTegbXsAthTxyDTcmBUkqyGlVGZhoazQzVSoKbTFcCRvYsgSCwjGMxBfWEwMHuagTBxkz", "output": "105" }, { "input": "199\no A r v H e J q k J k v w Q F p O R y R Z o a K R L Z E H t X y X N y y p b x B m r R S q i A x V S u i c L y M n N X c C W Z m S j e w C w T r I S X T D F l w o k f t X u n W w p Z r A k I Y E h s g", "output": "1" }, { "input": "200\nhCyIdivIiISmmYIsCLbpKcTyHaOgTUQEwnQACXnrLdHAVFLtvliTEMlzBVzTesQbhXmcqvwPDeojglBMIjOXANfyQxCSjOJyO SIqOTnRzVzseGIDDYNtrwIusScWSuEhPyEmgQIVEzXofRptjeMzzhtUQxJgcUWILUhEaaRmYRBVsjoqgmyPIKwSajdlNPccOOtWrez", "output": "50" }, { "input": "1\ne", "output": "0" }, { "input": "1\nA", "output": "1" }, { "input": "200\nABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU VWXYZABCDE KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KZ", "output": "10" }, { "input": "200\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "200" }, { "input": "200\nffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff", "output": "0" }, { "input": "24\nHarbour Space UniversitY", "output": "2" }, { "input": "5\naA AA", "output": "2" }, { "input": "10\nas AS ASDA", "output": "4" }, { "input": "10\nas AS ASDZ", "output": "4" }, { "input": "3\na A", "output": "1" }, { "input": "24\nHarbour space UniversitY", "output": "2" }, { "input": "10\nas AS ASAa", "output": "3" }, { "input": "15\naAb ABCDFGRHTJS", "output": "11" }, { "input": "53\nsdfAZEZR AZE dfdf dsdRFGSDF ZZDZSD dfsd ERBGF dsfsdfR", "output": "6" }, { "input": "10\nABC ABc AB", "output": "3" }, { "input": "10\nA c de CDE", "output": "3" }, { "input": "4\nA AB", "output": "2" }, { "input": "18\nHARbour Space UNIV", "output": "4" }, { "input": "13\na b c d e f A", "output": "1" }, { "input": "6\nA B CA", "output": "2" }, { "input": "4\naa A", "output": "1" }, { "input": "3\nA a", "output": "1" } ]
1,504,031,389
2,147,483,647
Python 3
OK
TESTS
29
62
0
n = input() words = input() words = words.split() upper = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' maxWeight = 0 temp = 0 for word in words: for i in word: if i in upper: temp += 1 if temp > maxWeight: maxWeight = temp temp = 0 print(maxWeight)
Title: Text Volume Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a text of single-space separated words, consisting of small and capital Latin letters. Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text. Calculate the volume of the given text. Input Specification: The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text. The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters. Output Specification: Print one integer number — volume of text. Demo Input: ['7\nNonZERO\n', '24\nthis is zero answer text\n', '24\nHarbour Space University\n'] Demo Output: ['5\n', '0\n', '1\n'] Note: In the first example there is only one word, there are 5 capital letters in it. In the second example all of the words contain 0 capital letters.
```python n = input() words = input() words = words.split() upper = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' maxWeight = 0 temp = 0 for word in words: for i in word: if i in upper: temp += 1 if temp > maxWeight: maxWeight = temp temp = 0 print(maxWeight) ```
3
437
A
The Child and Homework
PROGRAMMING
1,300
[ "implementation" ]
null
null
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct. Fortunately the child knows how to solve such complicated test. The child will follow the algorithm: - If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice). You are given a multiple-choice questions, can you predict child's choose?
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length. Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
[ "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n", "A.ab\nB.abcde\nC.ab\nD.abc\n", "A.c\nB.cc\nC.c\nD.c\n" ]
[ "D\n", "C\n", "B\n" ]
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D. In the second sample, no choice is great, so the child will choose the luckiest choice C. In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.
500
[ { "input": "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute", "output": "D" }, { "input": "A.ab\nB.abcde\nC.ab\nD.abc", "output": "C" }, { "input": "A.c\nB.cc\nC.c\nD.c", "output": "B" }, { "input": "A.He_nan_de_yang_guang_zhao_yao_zhe_wo_men_mei_guo_ren_lian_shang_dou_xiao_kai_yan_wahaaaaaaaaaaaaaaaa\nB.Li_bai_li_bai_fei_liu_zhi_xia_san_qian_chi_yi_si_yin_he_luo_jiu_tian_li_bai_li_bai_li_bai_li_bai_shi\nC.Peng_yu_xiang_shi_zai_tai_shen_le_jian_zhi_jiu_shi_ye_jie_du_liu_a_si_mi_da_zhen_shi_tai_shen_le_a_a\nD.Wo_huo_le_si_shi_er_nian_zhen_de_shi_cong_lai_ye_mei_you_jian_guo_zhe_me_biao_zhun_de_yi_bai_ge_zi_a", "output": "C" }, { "input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____", "output": "C" }, { "input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_", "output": "D" }, { "input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____", "output": "C" }, { "input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_", "output": "D" }, { "input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h", "output": "A" }, { "input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_", "output": "C" }, { "input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg", "output": "B" }, { "input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___", "output": "D" }, { "input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS", "output": "C" }, { "input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_", "output": "C" }, { "input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi", "output": "C" }, { "input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__", "output": "C" }, { "input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h", "output": "A" }, { "input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_", "output": "C" }, { "input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg", "output": "B" }, { "input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___", "output": "D" }, { "input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS", "output": "C" }, { "input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_", "output": "C" }, { "input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi", "output": "C" }, { "input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__", "output": "C" }, { "input": "A.aaaaaa\nB.aaa\nC.aaa\nD.aaa", "output": "A" }, { "input": "A.aaa\nB.aaaaaa\nC.aaaaaa\nD.aaaaaa", "output": "A" }, { "input": "A.a\nB.b\nC.c\nD.d", "output": "C" }, { "input": "A._\nB.__\nC.____\nD.________", "output": "C" }, { "input": "A.____\nB.________\nC.________\nD._______", "output": "C" }, { "input": "A.h\nB.asdf\nC.asqw\nD.qwertasdfg", "output": "C" }, { "input": "A.aa\nB.aaaaa\nC.aaaaaa\nD.aaaaaaaaaaaaa", "output": "C" }, { "input": "A.ccc\nB.ccccccc\nC.ccc\nD.c", "output": "C" }, { "input": "A.c\nB.ccc\nC.cccccccccccccccccc\nD.cccccc", "output": "C" }, { "input": "A.aa\nB.bb\nC.cc\nD.ddd", "output": "C" }, { "input": "A.QW\nB.WERT\nC.QWER\nD.QWERTYUI", "output": "C" } ]
1,621,360,565
2,147,483,647
Python 3
OK
TESTS
35
62
0
choice = ['A', 'B', 'C', 'D'] size = [] min_, max_, great, index = float("inf"), float("-inf"), 0, 0 for i in range(4): ch = input()[2:] size.append((len(ch), i+1)) min_, max_ = min(len(ch), min_), max(len(ch), max_) size.sort() if min_ <= size[1][0]//2: great += 1 index = size[0][1] if max_ >= size[-2][0]*2: great += 1 index = size[-1][1] if great == 1: print(choice[index-1]) else: print("C")
Title: The Child and Homework Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct. Fortunately the child knows how to solve such complicated test. The child will follow the algorithm: - If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice). You are given a multiple-choice questions, can you predict child's choose? Input Specification: The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length. Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_". Output Specification: Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes). Demo Input: ['A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n', 'A.ab\nB.abcde\nC.ab\nD.abc\n', 'A.c\nB.cc\nC.c\nD.c\n'] Demo Output: ['D\n', 'C\n', 'B\n'] Note: In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D. In the second sample, no choice is great, so the child will choose the luckiest choice C. In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.
```python choice = ['A', 'B', 'C', 'D'] size = [] min_, max_, great, index = float("inf"), float("-inf"), 0, 0 for i in range(4): ch = input()[2:] size.append((len(ch), i+1)) min_, max_ = min(len(ch), min_), max(len(ch), max_) size.sort() if min_ <= size[1][0]//2: great += 1 index = size[0][1] if max_ >= size[-2][0]*2: great += 1 index = size[-1][1] if great == 1: print(choice[index-1]) else: print("C") ```
3
864
C
Bus
PROGRAMMING
1,500
[ "greedy", "implementation", "math" ]
null
null
A bus moves along the coordinate line *Ox* from the point *x*<==<=0 to the point *x*<==<=*a*. After starting from the point *x*<==<=0, it reaches the point *x*<==<=*a*, immediately turns back and then moves to the point *x*<==<=0. After returning to the point *x*<==<=0 it immediately goes back to the point *x*<==<=*a* and so on. Thus, the bus moves from *x*<==<=0 to *x*<==<=*a* and back. Moving from the point *x*<==<=0 to *x*<==<=*a* or from the point *x*<==<=*a* to *x*<==<=0 is called a bus journey. In total, the bus must make *k* journeys. The petrol tank of the bus can hold *b* liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point *x*<==<=*f*. This point is between points *x*<==<=0 and *x*<==<=*a*. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain *b* liters of gasoline. What is the minimum number of times the bus needs to refuel at the point *x*<==<=*f* to make *k* journeys? The first journey starts in the point *x*<==<=0.
The first line contains four integers *a*, *b*, *f*, *k* (0<=&lt;<=*f*<=&lt;<=*a*<=≤<=106, 1<=≤<=*b*<=≤<=109, 1<=≤<=*k*<=≤<=104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Print the minimum number of times the bus needs to refuel to make *k* journeys. If it is impossible for the bus to make *k* journeys, print -1.
[ "6 9 2 4\n", "6 10 2 4\n", "6 5 4 3\n" ]
[ "4\n", "2\n", "-1\n" ]
In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
1,500
[ { "input": "6 9 2 4", "output": "4" }, { "input": "6 10 2 4", "output": "2" }, { "input": "6 5 4 3", "output": "-1" }, { "input": "2 2 1 1", "output": "0" }, { "input": "10 4 6 10", "output": "-1" }, { "input": "3 1 1 1", "output": "-1" }, { "input": "2 1 1 1", "output": "1" }, { "input": "1000000 51923215 2302 10000", "output": "199" }, { "input": "10 11 3 2", "output": "-1" }, { "input": "20 50 10 25", "output": "11" }, { "input": "10 10 5 20", "output": "20" }, { "input": "15 65 5 50", "output": "12" }, { "input": "10 19 1 5", "output": "3" }, { "input": "10 19 9 5", "output": "3" }, { "input": "23 46 12 2", "output": "0" }, { "input": "23 46 12 3", "output": "1" }, { "input": "20 20 19 1", "output": "0" }, { "input": "20 23 17 2", "output": "1" }, { "input": "100 70 50 1", "output": "1" }, { "input": "100 70 70 2", "output": "2" }, { "input": "140 480 139 40", "output": "18" }, { "input": "1000000 1000000000 1 1000", "output": "0" }, { "input": "100000 1000000 50000 1000", "output": "100" }, { "input": "1000000 1000000 500000 1000", "output": "1000" }, { "input": "1000000 1000000 500000 10000", "output": "10000" }, { "input": "1000000 2500000 500000 9999", "output": "4998" }, { "input": "1000000 1500000 500000 9999", "output": "9997" }, { "input": "1000000 1500000 500000 10000", "output": "9998" }, { "input": "1000000 1 1 1", "output": "-1" }, { "input": "2 1000000000 1 1", "output": "0" }, { "input": "1000000 1000000000 1 1", "output": "0" }, { "input": "1000000 1 999999 1", "output": "-1" }, { "input": "1000000 1000000000 999999 1", "output": "0" }, { "input": "2 1 1 10000", "output": "-1" }, { "input": "1000000 1 1 10000", "output": "-1" }, { "input": "1000000 1000000000 1 10000", "output": "10" }, { "input": "1000000 1 999999 10000", "output": "-1" }, { "input": "2 1000000000 1 10000", "output": "0" }, { "input": "1000000 1000000000 999999 10000", "output": "10" }, { "input": "10000 78393 3000 9999", "output": "1428" }, { "input": "1000000 8839233 302200 9999", "output": "1249" }, { "input": "900005 3333333 210000 9999", "output": "3332" }, { "input": "6 7 4 2", "output": "2" }, { "input": "3 1 2 1", "output": "-1" }, { "input": "150 100 1 1", "output": "-1" }, { "input": "10 5 6 1", "output": "-1" }, { "input": "51 81 36 38", "output": "36" }, { "input": "100 159 80 2", "output": "1" } ]
1,616,295,812
2,147,483,647
Python 3
OK
TESTS
48
77
307,200
a,b,f,k = map(int,input().split()) tf = a-f gas = b-f if gas < 0: print(-1) exit() ans = 0 for i in range(k): if i%2 == 0: if i == k-1: if gas < tf: gas = b ans +=1 else: if gas < tf*2: gas = b ans +=1 if i == k-1: gas -= tf else: gas -= tf*2 if gas < 0: print(-1) exit() else: if i == k-1: if gas < f: gas = b ans +=1 else: if gas < f*2: gas = b ans +=1 if i == k-1: gas -= f else: gas -= f*2 if gas < 0: print(-1) exit() print(ans)
Title: Bus Time Limit: None seconds Memory Limit: None megabytes Problem Description: A bus moves along the coordinate line *Ox* from the point *x*<==<=0 to the point *x*<==<=*a*. After starting from the point *x*<==<=0, it reaches the point *x*<==<=*a*, immediately turns back and then moves to the point *x*<==<=0. After returning to the point *x*<==<=0 it immediately goes back to the point *x*<==<=*a* and so on. Thus, the bus moves from *x*<==<=0 to *x*<==<=*a* and back. Moving from the point *x*<==<=0 to *x*<==<=*a* or from the point *x*<==<=*a* to *x*<==<=0 is called a bus journey. In total, the bus must make *k* journeys. The petrol tank of the bus can hold *b* liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point *x*<==<=*f*. This point is between points *x*<==<=0 and *x*<==<=*a*. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain *b* liters of gasoline. What is the minimum number of times the bus needs to refuel at the point *x*<==<=*f* to make *k* journeys? The first journey starts in the point *x*<==<=0. Input Specification: The first line contains four integers *a*, *b*, *f*, *k* (0<=&lt;<=*f*<=&lt;<=*a*<=≤<=106, 1<=≤<=*b*<=≤<=109, 1<=≤<=*k*<=≤<=104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. Output Specification: Print the minimum number of times the bus needs to refuel to make *k* journeys. If it is impossible for the bus to make *k* journeys, print -1. Demo Input: ['6 9 2 4\n', '6 10 2 4\n', '6 5 4 3\n'] Demo Output: ['4\n', '2\n', '-1\n'] Note: In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
```python a,b,f,k = map(int,input().split()) tf = a-f gas = b-f if gas < 0: print(-1) exit() ans = 0 for i in range(k): if i%2 == 0: if i == k-1: if gas < tf: gas = b ans +=1 else: if gas < tf*2: gas = b ans +=1 if i == k-1: gas -= tf else: gas -= tf*2 if gas < 0: print(-1) exit() else: if i == k-1: if gas < f: gas = b ans +=1 else: if gas < f*2: gas = b ans +=1 if i == k-1: gas -= f else: gas -= f*2 if gas < 0: print(-1) exit() print(ans) ```
3
899
C
Dividing the numbers
PROGRAMMING
1,300
[ "constructive algorithms", "graphs", "math" ]
null
null
Petya has *n* integers: 1,<=2,<=3,<=...,<=*n*. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible. Help Petya to split the integers. Each of *n* integers should be exactly in one group.
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of integers Petya has.
Print the smallest possible absolute difference in the first line. In the second line print the size of the first group, followed by the integers in that group. You can print these integers in arbitrary order. If there are multiple answers, print any of them.
[ "4\n", "2\n" ]
[ "0\n2 1 4 \n", "1\n1 1 \n" ]
In the first example you have to put integers 1 and 4 in the first group, and 2 and 3 in the second. This way the sum in each group is 5, and the absolute difference is 0. In the second example there are only two integers, and since both groups should be non-empty, you have to put one integer in the first group and one in the second. This way the absolute difference of sums of integers in each group is 1.
1,500
[ { "input": "4", "output": "0\n2 1 4 " }, { "input": "2", "output": "1\n1 1 " }, { "input": "3", "output": "0\n1\n3 " }, { "input": "5", "output": "1\n3\n1 2 5 " }, { "input": "59998", "output": "1\n29999 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 149 152 153 156 157 160 161 164 165 168 169 172 173 176 177 180 181 184 185 188 189 192 193 196 197 200 201 204 205 208 209 212 213 216 217 220 221 224 225 228 229 232 233 236 237 240 241 244 245 248 249 252 253 256 257 260 261 264 265 268 269 272 273 276 277 ..." }, { "input": "60000", "output": "0\n30000 1 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37 40 41 44 45 48 49 52 53 56 57 60 61 64 65 68 69 72 73 76 77 80 81 84 85 88 89 92 93 96 97 100 101 104 105 108 109 112 113 116 117 120 121 124 125 128 129 132 133 136 137 140 141 144 145 148 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67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 1..." }, { "input": "18759", "output": "0\n9379\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 15..." } ]
1,666,845,193
2,147,483,647
PyPy 3-64
OK
TESTS
60
62
2,969,600
n = int(input()) if n%4 == 0 or n%4 ==3: print(0) else: print(1) res = [] i = n%4 if i != 0: res.append(i) i += 4 while i <= n: res.append(i) res.append(i-3) i += 4 print(len(res), end=" ") print(' '.join([str(rr) for rr in res]))
Title: Dividing the numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has *n* integers: 1,<=2,<=3,<=...,<=*n*. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible. Help Petya to split the integers. Each of *n* integers should be exactly in one group. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of integers Petya has. Output Specification: Print the smallest possible absolute difference in the first line. In the second line print the size of the first group, followed by the integers in that group. You can print these integers in arbitrary order. If there are multiple answers, print any of them. Demo Input: ['4\n', '2\n'] Demo Output: ['0\n2 1 4 \n', '1\n1 1 \n'] Note: In the first example you have to put integers 1 and 4 in the first group, and 2 and 3 in the second. This way the sum in each group is 5, and the absolute difference is 0. In the second example there are only two integers, and since both groups should be non-empty, you have to put one integer in the first group and one in the second. This way the absolute difference of sums of integers in each group is 1.
```python n = int(input()) if n%4 == 0 or n%4 ==3: print(0) else: print(1) res = [] i = n%4 if i != 0: res.append(i) i += 4 while i <= n: res.append(i) res.append(i-3) i += 4 print(len(res), end=" ") print(' '.join([str(rr) for rr in res])) ```
3
595
A
Vitaly and Night
PROGRAMMING
800
[ "constructive algorithms", "implementation" ]
null
null
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment. Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on. Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively. Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
[ "2 2\n0 0 0 1\n1 0 1 1\n", "1 3\n1 1 0 1 0 0\n" ]
[ "3\n", "2\n" ]
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off. In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
500
[ { "input": "2 2\n0 0 0 1\n1 0 1 1", "output": "3" }, { "input": "1 3\n1 1 0 1 0 0", "output": "2" }, { "input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1", "output": "8" }, { "input": "1 5\n1 0 1 1 1 0 1 1 1 1", "output": "5" }, { "input": "1 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99" }, { "input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "6" }, { "input": "1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "100" }, { "input": "100 1\n0 0\n0 0\n0 1\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\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\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 1\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\n1 0", "output": "8" }, { "input": "100 1\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": "1 1\n0 0", "output": "0" }, { "input": "1 1\n0 1", "output": "1" }, { "input": "1 1\n1 0", "output": "1" }, { "input": "1 1\n1 1", "output": "1" } ]
1,600,780,941
2,147,483,647
Python 3
OK
TESTS
36
155
819,200
from collections import Counter import string import math import sys def array_int(): return [int(i) for i in sys.stdin.readline().split()] def vary(number_of_variables): if number_of_variables==1: return int(sys.stdin.readline()) if number_of_variables>=2: return map(int,sys.stdin.readline().split()) def makedict(var): return dict(Counter(var)) mod=100000007 n,m=vary(2) count=0 for i in range(n): num=array_int() for i in range(0,2*m,2): if num[i]==1 or num[i+1]==1: count+=1 print(count)
Title: Vitaly and Night Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment. Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on. Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping. Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively. Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'. Output Specification: Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping. Demo Input: ['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n'] Demo Output: ['3\n', '2\n'] Note: In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off. In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
```python from collections import Counter import string import math import sys def array_int(): return [int(i) for i in sys.stdin.readline().split()] def vary(number_of_variables): if number_of_variables==1: return int(sys.stdin.readline()) if number_of_variables>=2: return map(int,sys.stdin.readline().split()) def makedict(var): return dict(Counter(var)) mod=100000007 n,m=vary(2) count=0 for i in range(n): num=array_int() for i in range(0,2*m,2): if num[i]==1 or num[i+1]==1: count+=1 print(count) ```
3
61
A
Ultra-Fast Mathematician
PROGRAMMING
800
[ "implementation" ]
A. Ultra-Fast Mathematician
2
256
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Write one line — the corresponding answer. Do not omit the leading 0s.
[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]
[ "1110001\n", "111\n", "0100\n", "00010\n" ]
none
500
[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { 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"00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010", "output": "00010000000110110101000011001000000100100110111010011111101010001010000" }, { "input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001", "output": "000100100000000110011100100001010110101001100101110010010011111001110111" }, { "input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000", "output": "1000111100010011010110011101000000101010101100011111100001101111001010010" }, { "input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011", "output": "01000010011110111001011011110000001011000111101101101010010110001010100100" }, { 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"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" } ]
1,559,027,606
2,147,483,647
Python 3
OK
TESTS
102
124
0
print(''.join(['0' if n == m else '1' for n, m in zip(input(), input())]))
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 print(''.join(['0' if n == m else '1' for n, m in zip(input(), input())])) ```
3.969
431
A
Black Square
PROGRAMMING
800
[ "implementation" ]
null
null
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone. In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip. You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104). The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Print a single integer — the total number of calories that Jury wastes.
[ "1 2 3 4\n123214\n", "1 5 3 2\n11221\n" ]
[ "13\n", "13\n" ]
none
500
[ { "input": "1 2 3 4\n123214", "output": "13" }, { "input": "1 5 3 2\n11221", "output": "13" }, { "input": "5 5 5 1\n3422", "output": "16" }, { "input": "4 3 2 1\n2", "output": "3" }, { "input": "5651 6882 6954 4733\n2442313421", "output": "60055" }, { "input": "0 0 0 0\n4132", "output": "0" }, { "input": "3163 5778 83 7640\n11141442444", "output": "64270" }, { "input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442", "output": "420780" }, { "input": "0 0 0 0\n1", "output": "0" }, { "input": "1 2 3 4\n4", "output": "4" }, { "input": "2343 7653 1242 5432\n1", "output": "2343" }, { "input": "2343 7653 1242 5432\n2", "output": "7653" }, { "input": "2343 7653 1242 5432\n3", "output": "1242" }, { "input": "2343 7653 1242 5432\n4", "output": "5432" }, { "input": "1 2 3 4\n123412", "output": "13" }, { "input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111", "output": "2650" }, { "input": "1 2 3 4\n11111111111111111111111111111111111111111111111111", "output": "50" }, { "input": "1 2 3 4\n23123231321231231231231231221232123121312321", "output": "87" }, { "input": "1 2 3 4\n1111111111111222222222233333333333444444444444444", "output": "126" }, { "input": "2 3 1 4\n121321232412342112312313213123123412131231231232", "output": "105" } ]
1,689,918,751
2,147,483,647
Python 3
OK
TESTS
49
77
2,355,200
calories = list(map(int, input().split())) stripes = [int(x) for x in input()] sumi = [] for stripe in stripes: if stripe == 1: sumi.append(calories[0]) elif stripe == 2: sumi.append(calories[1]) elif stripe == 3: sumi.append(calories[2]) else: sumi.append(calories[3]) print(sum(sumi))
Title: Black Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone. In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip. You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares? Input Specification: The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104). The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip. Output Specification: Print a single integer — the total number of calories that Jury wastes. Demo Input: ['1 2 3 4\n123214\n', '1 5 3 2\n11221\n'] Demo Output: ['13\n', '13\n'] Note: none
```python calories = list(map(int, input().split())) stripes = [int(x) for x in input()] sumi = [] for stripe in stripes: if stripe == 1: sumi.append(calories[0]) elif stripe == 2: sumi.append(calories[1]) elif stripe == 3: sumi.append(calories[2]) else: sumi.append(calories[3]) print(sum(sumi)) ```
3
453
A
Little Pony and Expected Maximum
PROGRAMMING
1,600
[ "probabilities" ]
null
null
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
[ "6 1\n", "6 3\n", "2 2\n" ]
[ "3.500000000000\n", "4.958333333333\n", "1.750000000000\n" ]
Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
500
[ { "input": "6 1", "output": "3.500000000000" }, { "input": "6 3", "output": "4.958333333333" }, { "input": "2 2", "output": "1.750000000000" }, { "input": "5 4", "output": "4.433600000000" }, { "input": "5 8", "output": "4.814773760000" }, { "input": "3 10", "output": "2.982641534996" }, { "input": "3 6", "output": "2.910836762689" }, { "input": "1 8", "output": "1.000000000000" }, { "input": "24438 9", "output": "21994.699969310015" }, { "input": "94444 9", "output": "85000.099992058866" }, { "input": "8 66716", "output": "8.000000000000" }, { "input": "4 25132", "output": "4.000000000000" }, { "input": "51520 73331", "output": "51519.682650242677" }, { "input": "54230 31747", "output": "54228.743352775018" }, { "input": "24236 90163", "output": "24235.975171545670" }, { "input": "26946 99523", "output": "26945.974480086279" }, { "input": "50323 7", "output": "44033.124988408454" }, { "input": "53033 3", "output": "39775.249995286234" }, { "input": "55743 5", "output": "46452.999992525307" }, { "input": "59964 79", "output": "59214.949890211828" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "3 1", "output": "2.000000000000" }, { "input": "1 2", "output": "1.000000000000" }, { "input": "53513 34040", "output": "53511.875329020870" }, { "input": "100000 100000", "output": "99999.418033254507" }, { "input": "1 100000", "output": "1.000000000000" }, { "input": "100000 1", "output": "50000.499999999935" }, { "input": "2 100000", "output": "2.000000000000" }, { "input": "100000 2", "output": "66667.166665000332" }, { "input": "50000 100000", "output": "49999.843487110789" }, { "input": "99999 1111", "output": "99909.571915885972" }, { "input": "99999 99999", "output": "99998.418033254609" }, { "input": "1000 1000", "output": "999.419018443269" }, { "input": "50000 50000", "output": "49999.418043215679" }, { "input": "88888 88888", "output": "88887.418034499773" }, { "input": "99999 100000", "output": "99998.418042461126" }, { "input": "100000 555", "output": "99820.643422392372" }, { "input": "10000 10000", "output": "9999.418122897887" }, { "input": "100000 5001", "output": "99980.503829474910" }, { "input": "100000 1000", "output": "99900.599066768002" } ]
1,442,911,306
2,147,483,647
PyPy 3
OK
TESTS
41
124
5,120,000
m, n = [int(x) for x in input().split()] print(sum([i *((i/m) ** n - ((i-1)/m) ** n) for i in range(1, m + 1)]))
Title: Little Pony and Expected Maximum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times. Input Specification: A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105). Output Specification: Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4. Demo Input: ['6 1\n', '6 3\n', '2 2\n'] Demo Output: ['3.500000000000\n', '4.958333333333\n', '1.750000000000\n'] Note: Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
```python m, n = [int(x) for x in input().split()] print(sum([i *((i/m) ** n - ((i-1)/m) ** n) for i in range(1, m + 1)])) ```
3
483
A
Counterexample
PROGRAMMING
1,100
[ "brute force", "implementation", "math", "number theory" ]
null
null
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one. Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime. You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=&lt;<=*b*<=&lt;<=*c*<=≤<=*r*. More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=&lt;<=*b*<=&lt;<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order. If the counterexample does not exist, print the single number -1.
[ "2 4\n", "10 11\n", "900000000000000009 900000000000000029\n" ]
[ "2 3 4\n", "-1\n", "900000000000000009 900000000000000010 900000000000000021\n" ]
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are. In the second sample you cannot form a group of three distinct integers, so the answer is -1. In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
500
[ { "input": "2 4", "output": "2 3 4" }, { "input": "10 11", "output": "-1" }, { "input": "900000000000000009 900000000000000029", "output": "900000000000000009 900000000000000010 900000000000000021" }, { "input": "640097987171091791 640097987171091835", "output": "640097987171091792 640097987171091793 640097987171091794" }, { "input": "19534350415104721 19534350415104725", "output": "19534350415104722 19534350415104723 19534350415104724" }, { "input": "933700505788726243 933700505788726280", "output": "933700505788726244 933700505788726245 933700505788726246" }, { "input": "1 3", "output": "-1" }, { "input": "1 4", "output": "2 3 4" }, { "input": "1 1", "output": "-1" }, { "input": "266540997167959130 266540997167959164", "output": "266540997167959130 266540997167959131 266540997167959132" }, { "input": "267367244641009850 267367244641009899", "output": "267367244641009850 267367244641009851 267367244641009852" }, { "input": "268193483524125978 268193483524125993", "output": "268193483524125978 268193483524125979 268193483524125980" }, { "input": "269019726702209402 269019726702209432", "output": "269019726702209402 269019726702209403 269019726702209404" }, { "input": "269845965585325530 269845965585325576", "output": "269845965585325530 269845965585325531 269845965585325532" }, { "input": "270672213058376250 270672213058376260", "output": "270672213058376250 270672213058376251 270672213058376252" }, { "input": "271498451941492378 271498451941492378", "output": "-1" }, { "input": "272324690824608506 272324690824608523", "output": "272324690824608506 272324690824608507 272324690824608508" }, { "input": "273150934002691930 273150934002691962", "output": "273150934002691930 273150934002691931 273150934002691932" }, { "input": "996517375802030516 996517375802030524", "output": "996517375802030516 996517375802030517 996517375802030518" }, { "input": "997343614685146644 997343614685146694", "output": "997343614685146644 997343614685146645 997343614685146646" }, { "input": "998169857863230068 998169857863230083", "output": "998169857863230068 998169857863230069 998169857863230070" }, { "input": "998996101041313492 998996101041313522", "output": "998996101041313492 998996101041313493 998996101041313494" }, { "input": "999822344219396916 999822344219396961", "output": "999822344219396916 999822344219396917 999822344219396918" }, { "input": "648583102513043 648583102513053", "output": "648583102513044 648583102513045 648583102513046" }, { "input": "266540997167959130 266540997167959131", "output": "-1" }, { "input": "267367244641009850 267367244641009850", "output": "-1" }, { "input": "268193483524125978 268193483524125979", "output": "-1" }, { "input": "269019726702209402 269019726702209402", "output": "-1" }, { "input": "269845965585325530 269845965585325530", "output": "-1" }, { "input": "270672213058376250 270672213058376254", "output": "270672213058376250 270672213058376251 270672213058376252" }, { "input": "271498451941492378 271498451941492379", "output": "-1" }, { "input": "272324690824608506 272324690824608508", "output": "272324690824608506 272324690824608507 272324690824608508" }, { "input": "273150934002691930 273150934002691931", "output": "-1" }, { "input": "996517375802030516 996517375802030518", "output": "996517375802030516 996517375802030517 996517375802030518" }, { "input": "997343614685146644 997343614685146644", "output": "-1" }, { "input": "2147483647 2147483649", "output": "-1" }, { "input": "3 5", "output": "-1" }, { "input": "1 7", "output": "2 3 4" }, { "input": "9 12", "output": "9 11 12" }, { "input": "4 4", "output": "-1" }, { "input": "11 13", "output": "-1" }, { "input": "2 2", "output": "-1" } ]
1,414,172,797
2,797
Python 3
OK
TESTS
42
62
0
def gcd(a, b): rem = 0 while b != 0: rem = a%b a = b b = rem return a a, b = map(int, input().split()) if b-a == 1: print (-1) else: found = False for i in range(a, b+1): for j in range(i+1, b+1): for k in range(j+1, b+1): if gcd(i, j) == 1 and gcd(j, k) == 1 and gcd(i, k) != 1: found = True print (i, j, k) break if found: break if found: break if not found: print (-1)
Title: Counterexample Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one. Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime. You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=&lt;<=*b*<=&lt;<=*c*<=≤<=*r*. More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=&lt;<=*b*<=&lt;<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime. Input Specification: The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50). Output Specification: Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order. If the counterexample does not exist, print the single number -1. Demo Input: ['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n'] Demo Output: ['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n'] Note: In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are. In the second sample you cannot form a group of three distinct integers, so the answer is -1. In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
```python def gcd(a, b): rem = 0 while b != 0: rem = a%b a = b b = rem return a a, b = map(int, input().split()) if b-a == 1: print (-1) else: found = False for i in range(a, b+1): for j in range(i+1, b+1): for k in range(j+1, b+1): if gcd(i, j) == 1 and gcd(j, k) == 1 and gcd(i, k) != 1: found = True print (i, j, k) break if found: break if found: break if not found: print (-1) ```
3
727
A
Transformation: from A to B
PROGRAMMING
1,000
[ "brute force", "dfs and similar", "math" ]
null
null
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations: - multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1). You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible. Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
If there is no way to get *b* from *a*, print "NO" (without quotes). Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where: - *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=&lt;<=*i*<=≤<=*k*). If there are multiple answers, print any of them.
[ "2 162\n", "4 42\n", "100 40021\n" ]
[ "YES\n5\n2 4 8 81 162 \n", "NO\n", "YES\n5\n100 200 2001 4002 40021 \n" ]
none
1,000
[ { "input": "2 162", "output": "YES\n5\n2 4 8 81 162 " }, { "input": "4 42", "output": "NO" }, { "input": "100 40021", "output": "YES\n5\n100 200 2001 4002 40021 " }, { "input": "1 111111111", "output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 " }, { "input": "1 1000000000", "output": "NO" }, { "input": "999999999 1000000000", "output": "NO" }, { "input": "1 2", "output": "YES\n2\n1 2 " }, { "input": "1 536870912", "output": "YES\n30\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 " }, { "input": "11111 11111111", "output": "YES\n4\n11111 111111 1111111 11111111 " }, { "input": "59139 946224", "output": "YES\n5\n59139 118278 236556 473112 946224 " }, { "input": "9859 19718", "output": "YES\n2\n9859 19718 " }, { "input": "25987 51974222", "output": "YES\n5\n25987 259871 2598711 25987111 51974222 " }, { "input": "9411 188222222", "output": "YES\n6\n9411 94111 941111 9411111 94111111 188222222 " }, { "input": "25539 510782222", "output": "YES\n6\n25539 255391 2553911 25539111 255391111 510782222 " }, { "input": "76259 610072", "output": "YES\n4\n76259 152518 305036 610072 " }, { "input": "92387 184774", "output": "YES\n2\n92387 184774 " }, { "input": "8515 85151111", "output": "YES\n5\n8515 85151 851511 8515111 85151111 " }, { "input": "91939 9193911", "output": "YES\n3\n91939 919391 9193911 " }, { "input": "30518 610361", "output": "YES\n3\n30518 61036 610361 " }, { "input": "46646 373168844", "output": "YES\n7\n46646 466461 932922 9329221 93292211 186584422 373168844 " }, { "input": "30070 300701", "output": "YES\n2\n30070 300701 " }, { "input": "13494 1079528", "output": "YES\n5\n13494 134941 269882 539764 1079528 " }, { "input": "96918 775344422", "output": "YES\n7\n96918 193836 1938361 3876722 38767221 387672211 775344422 " }, { "input": "13046 260921", "output": "YES\n3\n13046 26092 260921 " }, { "input": "29174 5834811", "output": "YES\n4\n29174 58348 583481 5834811 " }, { "input": "79894 319576421", "output": "YES\n6\n79894 798941 1597882 15978821 31957642 319576421 " }, { "input": "96022 1920442", "output": "YES\n3\n96022 960221 1920442 " }, { "input": "79446 6355681", "output": "YES\n5\n79446 158892 317784 635568 6355681 " }, { "input": "5440 27853056", "output": "YES\n11\n5440 10880 108801 217602 435204 870408 1740816 3481632 6963264 13926528 27853056 " }, { "input": "250000000 705032705", "output": "NO" }, { "input": "17 35", "output": "NO" }, { "input": "1 3", "output": "NO" }, { "input": "2 11", "output": "NO" } ]
1,603,421,718
2,147,483,647
Python 3
OK
TESTS
33
108
307,200
num_possuido, num_desejado = map(int, input().split()) aux = [] aux.append(num_desejado) while(num_desejado > num_possuido): if (num_desejado % 10) == 1: num_desejado = num_desejado // 10 aux.append(num_desejado) elif (num_desejado % 2 == 0): num_desejado = num_desejado // 2 aux.append(num_desejado) else: print("NO") exit() tamanho_da_seq = len(aux) seq = "" aux.reverse() for i in aux: seq += str(i) + " " if(num_desejado == num_possuido): print("YES") print(tamanho_da_seq) print(seq) else: print("NO")
Title: Transformation: from A to B Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations: - multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1). You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible. Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*. Input Specification: The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have. Output Specification: If there is no way to get *b* from *a*, print "NO" (without quotes). Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where: - *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=&lt;<=*i*<=≤<=*k*). If there are multiple answers, print any of them. Demo Input: ['2 162\n', '4 42\n', '100 40021\n'] Demo Output: ['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n'] Note: none
```python num_possuido, num_desejado = map(int, input().split()) aux = [] aux.append(num_desejado) while(num_desejado > num_possuido): if (num_desejado % 10) == 1: num_desejado = num_desejado // 10 aux.append(num_desejado) elif (num_desejado % 2 == 0): num_desejado = num_desejado // 2 aux.append(num_desejado) else: print("NO") exit() tamanho_da_seq = len(aux) seq = "" aux.reverse() for i in aux: seq += str(i) + " " if(num_desejado == num_possuido): print("YES") print(tamanho_da_seq) print(seq) else: print("NO") ```
3
453
A
Little Pony and Expected Maximum
PROGRAMMING
1,600
[ "probabilities" ]
null
null
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
[ "6 1\n", "6 3\n", "2 2\n" ]
[ "3.500000000000\n", "4.958333333333\n", "1.750000000000\n" ]
Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
500
[ { "input": "6 1", "output": "3.500000000000" }, { "input": "6 3", "output": "4.958333333333" }, { "input": "2 2", "output": "1.750000000000" }, { "input": "5 4", "output": "4.433600000000" }, { "input": "5 8", "output": "4.814773760000" }, { "input": "3 10", "output": "2.982641534996" }, { "input": "3 6", "output": "2.910836762689" }, { "input": "1 8", "output": "1.000000000000" }, { "input": "24438 9", "output": "21994.699969310015" }, { "input": "94444 9", "output": "85000.099992058866" }, { "input": "8 66716", "output": "8.000000000000" }, { "input": "4 25132", "output": "4.000000000000" }, { "input": "51520 73331", "output": "51519.682650242677" }, { "input": "54230 31747", "output": "54228.743352775018" }, { "input": "24236 90163", "output": "24235.975171545670" }, { "input": "26946 99523", "output": "26945.974480086279" }, { "input": "50323 7", "output": "44033.124988408454" }, { "input": "53033 3", "output": "39775.249995286234" }, { "input": "55743 5", "output": "46452.999992525307" }, { "input": "59964 79", "output": "59214.949890211828" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "1 1", "output": "1.000000000000" }, { "input": "3 1", "output": "2.000000000000" }, { "input": "1 2", "output": "1.000000000000" }, { "input": "53513 34040", "output": "53511.875329020870" }, { "input": "100000 100000", "output": "99999.418033254507" }, { "input": "1 100000", "output": "1.000000000000" }, { "input": "100000 1", "output": "50000.499999999935" }, { "input": "2 100000", "output": "2.000000000000" }, { "input": "100000 2", "output": "66667.166665000332" }, { "input": "50000 100000", "output": "49999.843487110789" }, { "input": "99999 1111", "output": "99909.571915885972" }, { "input": "99999 99999", "output": "99998.418033254609" }, { "input": "1000 1000", "output": "999.419018443269" }, { "input": "50000 50000", "output": "49999.418043215679" }, { "input": "88888 88888", "output": "88887.418034499773" }, { "input": "99999 100000", "output": "99998.418042461126" }, { "input": "100000 555", "output": "99820.643422392372" }, { "input": "10000 10000", "output": "9999.418122897887" }, { "input": "100000 5001", "output": "99980.503829474910" }, { "input": "100000 1000", "output": "99900.599066768002" } ]
1,579,862,214
2,147,483,647
PyPy 3
OK
TESTS
41
171
8,294,400
m, n = map(int, input().split()) powers = [] for j in range(m + 1): powers.append(pow(j / m, n)) print(sum(j * (powers[j] - powers[j - 1]) for j in range(1, m + 1)))
Title: Little Pony and Expected Maximum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game. The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times. Input Specification: A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105). Output Specification: Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4. Demo Input: ['6 1\n', '6 3\n', '2 2\n'] Demo Output: ['3.500000000000\n', '4.958333333333\n', '1.750000000000\n'] Note: Consider the third test example. If you've made two tosses: 1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. The probability of each outcome is 0.25, that is expectation equals to: You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
```python m, n = map(int, input().split()) powers = [] for j in range(m + 1): powers.append(pow(j / m, n)) print(sum(j * (powers[j] - powers[j - 1]) for j in range(1, m + 1))) ```
3
739
A
Alyona and mex
PROGRAMMING
1,700
[ "constructive algorithms", "greedy" ]
null
null
Alyona's mother wants to present an array of *n* non-negative integers to Alyona. The array should be special. Alyona is a capricious girl so after she gets the array, she inspects *m* of its subarrays. Subarray is a set of some subsequent elements of the array. The *i*-th subarray is described with two integers *l**i* and *r**i*, and its elements are *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*]. Alyona is going to find mex for each of the chosen subarrays. Among these *m* mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible. You are to find an array *a* of *n* elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible. The mex of a set *S* is a minimum possible non-negative integer that is not in *S*.
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *m* lines contain information about the subarrays chosen by Alyona. The *i*-th of these lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), that describe the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
In the first line print single integer — the maximum possible minimum mex. In the second line print *n* integers — the array *a*. All the elements in *a* should be between 0 and 109. It is guaranteed that there is an optimal answer in which all the elements in *a* are between 0 and 109. If there are multiple solutions, print any of them.
[ "5 3\n1 3\n2 5\n4 5\n", "4 2\n1 4\n2 4\n" ]
[ "2\n1 0 2 1 0\n", "3\n5 2 0 1" ]
The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.
500
[ { "input": "5 3\n1 3\n2 5\n4 5", "output": "2\n0 1 0 1 0" }, { "input": "4 2\n1 4\n2 4", "output": "3\n0 1 2 0" }, { "input": "1 1\n1 1", "output": "1\n0" }, { "input": "2 1\n2 2", "output": "1\n0 0" }, { "input": "5 6\n2 4\n2 3\n1 4\n3 4\n2 5\n1 3", "output": "2\n0 1 0 1 0" }, { "input": "8 3\n2 3\n2 8\n3 6", "output": "2\n0 1 0 1 0 1 0 1" }, { "input": "10 10\n1 9\n4 8\n4 8\n5 9\n1 9\n3 8\n1 6\n1 9\n1 6\n6 9", "output": "4\n0 1 2 3 0 1 2 3 0 1" }, { "input": "3 6\n1 3\n1 3\n1 1\n1 1\n3 3\n3 3", "output": "1\n0 0 0" }, { "input": "3 3\n1 3\n2 2\n1 3", "output": "1\n0 0 0" }, { "input": "6 8\n3 5\n3 6\n4 6\n2 5\n2 5\n1 3\n3 6\n3 5", "output": "3\n0 1 2 0 1 2" }, { "input": "10 4\n4 10\n4 6\n6 8\n1 10", "output": "3\n0 1 2 0 1 2 0 1 2 0" }, { "input": "9 1\n1 1", "output": "1\n0 0 0 0 0 0 0 0 0" }, { "input": "3 8\n2 3\n1 3\n1 2\n2 3\n1 3\n2 2\n1 2\n1 2", "output": "1\n0 0 0" }, { "input": "3 8\n1 2\n1 2\n1 1\n2 3\n2 3\n1 1\n1 3\n1 3", "output": "1\n0 0 0" }, { "input": "7 3\n7 7\n3 7\n5 7", "output": "1\n0 0 0 0 0 0 0" }, { "input": "9 9\n4 5\n5 8\n1 8\n4 8\n3 4\n7 8\n1 4\n7 8\n6 7", "output": "2\n0 1 0 1 0 1 0 1 0" }, { "input": "10 10\n1 5\n7 10\n2 10\n2 5\n2 9\n5 9\n3 10\n4 9\n6 9\n2 7", "output": "4\n0 1 2 3 0 1 2 3 0 1" }, { "input": "8 7\n5 8\n3 7\n1 8\n3 4\n2 8\n2 7\n4 6", "output": "2\n0 1 0 1 0 1 0 1" }, { "input": "10 3\n2 4\n8 10\n4 6", "output": "3\n0 1 2 0 1 2 0 1 2 0" }, { "input": "5 8\n3 4\n1 5\n3 4\n3 5\n3 4\n1 4\n1 5\n2 5", "output": "2\n0 1 0 1 0" }, { "input": "9 4\n5 9\n3 8\n2 8\n1 4", "output": "4\n0 1 2 3 0 1 2 3 0" }, { "input": "7 7\n2 4\n1 4\n3 6\n2 6\n5 7\n3 6\n1 4", "output": "3\n0 1 2 0 1 2 0" }, { "input": "6 2\n4 6\n2 5", "output": "3\n0 1 2 0 1 2" }, { "input": "7 9\n6 7\n1 2\n2 5\n4 7\n1 7\n5 6\n2 3\n6 7\n1 2", "output": "2\n0 1 0 1 0 1 0" } ]
1,591,777,633
2,147,483,647
PyPy 3
OK
TESTS
69
1,076
11,264,000
n,m=list(map(int,input().split())) a=[] for i in range(m): b=list(map(int,input().split())) a.append(b[1]-b[0]+1) c=min(a) b=[0]*n for i in range(n): b[i]=i%c print(c) print(*b)
Title: Alyona and mex Time Limit: None seconds Memory Limit: None megabytes Problem Description: Alyona's mother wants to present an array of *n* non-negative integers to Alyona. The array should be special. Alyona is a capricious girl so after she gets the array, she inspects *m* of its subarrays. Subarray is a set of some subsequent elements of the array. The *i*-th subarray is described with two integers *l**i* and *r**i*, and its elements are *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*]. Alyona is going to find mex for each of the chosen subarrays. Among these *m* mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible. You are to find an array *a* of *n* elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible. The mex of a set *S* is a minimum possible non-negative integer that is not in *S*. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *m* lines contain information about the subarrays chosen by Alyona. The *i*-th of these lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), that describe the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*]. Output Specification: In the first line print single integer — the maximum possible minimum mex. In the second line print *n* integers — the array *a*. All the elements in *a* should be between 0 and 109. It is guaranteed that there is an optimal answer in which all the elements in *a* are between 0 and 109. If there are multiple solutions, print any of them. Demo Input: ['5 3\n1 3\n2 5\n4 5\n', '4 2\n1 4\n2 4\n'] Demo Output: ['2\n1 0 2 1 0\n', '3\n5 2 0 1'] Note: The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.
```python n,m=list(map(int,input().split())) a=[] for i in range(m): b=list(map(int,input().split())) a.append(b[1]-b[0]+1) c=min(a) b=[0]*n for i in range(n): b[i]=i%c print(c) print(*b) ```
3
50
A
Domino piling
PROGRAMMING
800
[ "greedy", "math" ]
A. Domino piling
2
256
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output one number — the maximal number of dominoes, which can be placed.
[ "2 4\n", "3 3\n" ]
[ "4\n", "4\n" ]
none
500
[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]
1,662,904,462
2,147,483,647
PyPy 3-64
OK
TESTS
35
124
0
s=input().split(' ') m = int(s[0]) n = int(s[1]) print(m*n // 2)
Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none
```python s=input().split(' ') m = int(s[0]) n = int(s[1]) print(m*n // 2) ```
3.969
1,005
A
Tanya and Stairways
PROGRAMMING
800
[ "implementation" ]
null
null
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
[ "7\n1 2 3 1 2 3 4\n", "4\n1 1 1 1\n", "5\n1 2 3 4 5\n", "5\n1 2 1 2 1\n" ]
[ "2\n3 4 ", "4\n1 1 1 1 ", "1\n5 ", "3\n2 2 1 " ]
none
0
[ { "input": "7\n1 2 3 1 2 3 4", "output": "2\n3 4 " }, { "input": "4\n1 1 1 1", "output": "4\n1 1 1 1 " }, { "input": "5\n1 2 3 4 5", "output": "1\n5 " }, { "input": "5\n1 2 1 2 1", "output": "3\n2 2 1 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1", "output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 " }, { "input": "2\n1 2", "output": "1\n2 " }, { "input": "3\n1 1 2", "output": "2\n1 2 " }, { "input": "4\n1 1 2 3", "output": "2\n1 3 " }, { "input": "8\n1 2 3 1 2 3 4 5", "output": "2\n3 5 " }, { "input": "5\n1 1 1 2 3", "output": "3\n1 1 3 " } ]
1,693,767,591
2,147,483,647
Python 3
OK
TESTS
16
46
0
inp1 = int(input()) inp2 = input().split() all_count = 1 stairs_count = '' for i in range(inp1 - 1): if inp2[i + 1] == '1': all_count += 1 stairs_count += f"{inp2[i]} " stairs_count += inp2[-1] print(all_count) print(stairs_count) # Sun Sep 03 2023 21:59:50 GMT+0300 (Moscow Standard Time)
Title: Tanya and Stairways Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Input Specification: The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Output Specification: In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways. Demo Input: ['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n'] Demo Output: ['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 '] Note: none
```python inp1 = int(input()) inp2 = input().split() all_count = 1 stairs_count = '' for i in range(inp1 - 1): if inp2[i + 1] == '1': all_count += 1 stairs_count += f"{inp2[i]} " stairs_count += inp2[-1] print(all_count) print(stairs_count) # Sun Sep 03 2023 21:59:50 GMT+0300 (Moscow Standard Time) ```
3
265
A
Colorful Stones (Simplified Edition)
PROGRAMMING
800
[ "implementation" ]
null
null
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively. Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times. Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move. You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction. Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Print the final 1-based position of Liss in a single line.
[ "RGB\nRRR\n", "RRRBGBRBBB\nBBBRR\n", "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n" ]
[ "2\n", "3\n", "15\n" ]
none
500
[ { "input": "RGB\nRRR", "output": "2" }, { "input": "RRRBGBRBBB\nBBBRR", "output": "3" }, { "input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB", "output": "15" }, { "input": "G\nRRBBRBRRBR", "output": "1" }, { "input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB", "output": "1" }, { "input": "RRGGBRGRBG\nBRRGGBBGGR", "output": "7" }, { "input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG", "output": "15" }, { "input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR", "output": "8" }, { "input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR", "output": "16" }, { "input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR", "output": "18" }, { "input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG", "output": "23" }, { "input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR", "output": "14" }, { "input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG", "output": "12" }, { "input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB", "output": "12" }, { "input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR", "output": "2" }, { "input": "R\nB", "output": "1" }, { "input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "50" }, { "input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG", "output": "1" }, { "input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG", "output": "1" } ]
1,699,722,220
2,147,483,647
Python 3
OK
TESTS
19
62
0
s = input() c = 0 for v in input(): c += (v == s[c]) print(c + 1)
Title: Colorful Stones (Simplified Edition) Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively. Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times. Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move. You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction. Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. Input Specification: The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. Output Specification: Print the final 1-based position of Liss in a single line. Demo Input: ['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n'] Demo Output: ['2\n', '3\n', '15\n'] Note: none
```python s = input() c = 0 for v in input(): c += (v == s[c]) print(c + 1) ```
3
732
A
Buy a Shovel
PROGRAMMING
800
[ "brute force", "constructive algorithms", "implementation", "math" ]
null
null
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop. In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9). What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins". Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
[ "117 3\n", "237 7\n", "15 2\n" ]
[ "9\n", "1\n", "2\n" ]
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change. In the second example it is enough for Polycarp to buy one shovel. In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
500
[ { "input": "117 3", "output": "9" }, { "input": "237 7", "output": "1" }, { "input": "15 2", "output": "2" }, { "input": "1 1", "output": "1" }, { "input": "1 9", "output": "9" }, { "input": "1000 3", "output": "1" }, { "input": "1000 1", "output": "1" }, { "input": "1000 9", "output": "1" }, { "input": "1 2", "output": "2" }, { "input": "999 9", "output": "1" }, { "input": "999 8", "output": "2" }, { "input": "105 6", "output": "2" }, { "input": "403 9", "output": "3" }, { "input": "546 4", "output": "4" }, { "input": "228 9", "output": "5" }, { "input": "57 2", "output": "6" }, { "input": "437 9", "output": "7" }, { "input": "997 6", "output": "8" }, { "input": "109 1", "output": "9" }, { "input": "998 9", "output": "5" }, { "input": "4 2", "output": "3" }, { "input": "9 3", "output": "7" }, { "input": "8 2", "output": "4" }, { "input": "1 3", "output": "3" }, { "input": "1 4", "output": "4" }, { "input": "1 5", "output": "5" }, { "input": "1 6", "output": "6" }, { "input": "1 7", "output": "7" }, { "input": "1 8", "output": "8" }, { "input": "100 3", "output": "1" }, { "input": "1000 2", "output": "1" }, { "input": "1000 4", "output": "1" }, { "input": "1000 5", "output": "1" }, { "input": "1000 6", "output": "1" }, { "input": "1000 7", "output": "1" }, { "input": "1000 8", "output": "1" }, { "input": "23 4", "output": "8" }, { "input": "33 1", "output": "7" }, { "input": "33 2", "output": "4" }, { "input": "666 5", "output": "5" }, { "input": "2 3", "output": "5" }, { "input": "5 5", "output": "1" }, { "input": "3 6", "output": "2" }, { "input": "12 4", "output": "2" }, { "input": "15 5", "output": "1" }, { "input": "2 5", "output": "5" }, { "input": "25 5", "output": "1" }, { "input": "2 9", "output": "5" }, { "input": "6 7", "output": "5" }, { "input": "8 9", "output": "5" }, { "input": "2 7", "output": "5" }, { "input": "4 7", "output": "5" }, { "input": "2 1", "output": "5" }, { "input": "261 1", "output": "1" } ]
1,674,661,916
2,147,483,647
PyPy 3
OK
TESTS
54
93
0
k, r= map(int,input().split()) count = 1 sum=k for i in range(1000): if sum%10==0: break elif (sum-r)%10==0: break else: sum += k count += 1 print (count)
Title: Buy a Shovel Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop. In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9). What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. Input Specification: The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins". Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. Output Specification: Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. Demo Input: ['117 3\n', '237 7\n', '15 2\n'] Demo Output: ['9\n', '1\n', '2\n'] Note: In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change. In the second example it is enough for Polycarp to buy one shovel. In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
```python k, r= map(int,input().split()) count = 1 sum=k for i in range(1000): if sum%10==0: break elif (sum-r)%10==0: break else: sum += k count += 1 print (count) ```
3
38
A
Army
PROGRAMMING
800
[ "implementation" ]
A. Army
2
256
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
[ "3\n5 6\n1 2\n", "3\n5 6\n1 3\n" ]
[ "5\n", "11\n" ]
none
0
[ { "input": "3\n5 6\n1 2", "output": "5" }, { "input": "3\n5 6\n1 3", "output": "11" }, { "input": "2\n55\n1 2", "output": "55" }, { "input": "3\n85 78\n1 3", "output": "163" }, { "input": "4\n63 4 49\n2 3", "output": "4" }, { "input": "5\n93 83 42 56\n2 5", "output": "181" }, { "input": "6\n22 9 87 89 57\n1 6", "output": "264" }, { "input": "7\n52 36 31 23 74 78\n2 7", "output": "242" }, { "input": "8\n82 14 24 5 91 49 94\n3 8", "output": "263" }, { "input": "9\n12 40 69 39 59 21 59 5\n4 6", "output": "98" }, { "input": "10\n95 81 32 59 71 30 50 61 100\n1 6", "output": "338" }, { "input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14", "output": "617" }, { "input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17", "output": "399" }, { "input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23", "output": "846" }, { "input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16", "output": "730" }, { "input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35", "output": "1663" }, { "input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26", "output": "862" }, { "input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40", "output": "1061" }, { "input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28", "output": "344" }, { "input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60", "output": "1502" }, { "input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41", "output": "1767" }, { "input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24", "output": "219" }, { "input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44", "output": "1713" }, { "input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15", "output": "570" }, { "input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26", "output": "328" }, { "input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68", "output": "3385" }, { "input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77", "output": "3319" }, { "input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36", "output": "514" }, { "input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92", "output": "2599" }, { "input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27", "output": "618" }, { "input": "2\n3\n1 2", "output": "3" }, { "input": "5\n6 8 22 22\n2 3", "output": "8" }, { "input": "6\n3 12 27 28 28\n3 4", "output": "27" }, { "input": "9\n1 2 2 2 2 3 3 5\n3 7", "output": "9" }, { "input": "10\n1 1 1 1 1 1 1 1 1\n6 8", "output": "2" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17", "output": "23" }, { "input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8", "output": "23" }, { "input": "35\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\n30 31", "output": "2" }, { "input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45", "output": "30" }, { "input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17", "output": "285" }, { "input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58", "output": "173" }, { "input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65", "output": "126" }, { "input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65", "output": "326" }, { "input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48", "output": "784" }, { "input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93", "output": "1636" }, { "input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96", "output": "898" }, { "input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90", "output": "3016" }, { "input": "99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n66 95", "output": "29" }, { "input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52", "output": "82" }, { "input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44", "output": "2268" } ]
1,586,665,062
2,147,483,647
PyPy 3
OK
TESTS
50
280
0
n=int(input()) c=list(map(int,input().split())) a,b=map(int,input().split()) print(sum(c[a-1:b-1]))
Title: Army Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. Input Specification: The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated. Output Specification: Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. Demo Input: ['3\n5 6\n1 2\n', '3\n5 6\n1 3\n'] Demo Output: ['5\n', '11\n'] Note: none
```python n=int(input()) c=list(map(int,input().split())) a,b=map(int,input().split()) print(sum(c[a-1:b-1])) ```
3.93
339
A
Helpful Maths
PROGRAMMING
800
[ "greedy", "implementation", "sortings", "strings" ]
null
null
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation. The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3. You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Print the new sum that Xenia can count.
[ "3+2+1\n", "1+1+3+1+3\n", "2\n" ]
[ "1+2+3\n", "1+1+1+3+3\n", "2\n" ]
none
500
[ { "input": "3+2+1", "output": "1+2+3" }, { "input": "1+1+3+1+3", "output": "1+1+1+3+3" }, { "input": "2", "output": "2" }, { "input": "2+2+1+1+3", "output": "1+1+2+2+3" }, { "input": "2+1+2+2+2+3+1+3+1+2", "output": "1+1+1+2+2+2+2+2+3+3" }, { "input": "1+2+1+2+2+2+2+1+3+3", "output": "1+1+1+2+2+2+2+2+3+3" }, { "input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "1", "output": "1" }, { "input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3", "output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "3", "output": "3" }, { "input": "1+1", "output": "1+1" }, { "input": "1+2", "output": "1+2" }, { "input": "1+3", "output": "1+3" }, { "input": "2+1", "output": "1+2" }, { "input": "2+2", "output": "2+2" }, { "input": "2+3", "output": "2+3" }, { "input": "3+1", "output": "1+3" }, { "input": "3+2", "output": "2+3" }, { "input": "3+3", "output": "3+3" } ]
1,694,690,827
2,147,483,647
Python 3
OK
TESTS
21
92
0
s = input() count_1 = s.count('1') count_2 = s.count('2') count_3 = s.count('3') sorted_digits = [str(i) for i in [1, 2, 3] for _ in range(eval(f'count_{i}'))] rearranged_sum = '+'.join(sorted_digits) print(rearranged_sum)
Title: Helpful Maths Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation. The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3. You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. Input Specification: The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. Output Specification: Print the new sum that Xenia can count. Demo Input: ['3+2+1\n', '1+1+3+1+3\n', '2\n'] Demo Output: ['1+2+3\n', '1+1+1+3+3\n', '2\n'] Note: none
```python s = input() count_1 = s.count('1') count_2 = s.count('2') count_3 = s.count('3') sorted_digits = [str(i) for i in [1, 2, 3] for _ in range(eval(f'count_{i}'))] rearranged_sum = '+'.join(sorted_digits) print(rearranged_sum) ```
3
600
E
Lomsat gelral
PROGRAMMING
2,300
[ "data structures", "dfs and similar", "dsu", "trees" ]
null
null
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour. Let's call colour *c* dominating in the subtree of vertex *v* if there are no other colours that appear in the subtree of vertex *v* more times than colour *c*. So it's possible that two or more colours will be dominating in the subtree of some vertex. The subtree of vertex *v* is the vertex *v* and all other vertices that contains vertex *v* in each path to the root. For each vertex *v* find the sum of all dominating colours in the subtree of vertex *v*.
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of vertices in the tree. The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=*n*), *c**i* — the colour of the *i*-th vertex. Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the edge of the tree. The first vertex is the root of the tree.
Print *n* integers — the sums of dominating colours for each vertex.
[ "4\n1 2 3 4\n1 2\n2 3\n2 4\n", "15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13\n" ]
[ "10 9 3 4\n", "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\n" ]
none
0
[ { "input": "4\n1 2 3 4\n1 2\n2 3\n2 4", "output": "10 9 3 4" }, { "input": "15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13", "output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3" } ]
1,663,856,147
2,147,483,647
Python 3
OK
TESTS
28
1,559
26,009,600
import sys readline = sys.stdin.readline write = sys.stdout.write def dsu_on_tree(N, G, Prop): order = [] prt = [-1]*N que = [0] pn = [-1]*N used = [0]*N used[0] = 1 while que: v = que.pop() for w in G[v]: if used[w]: continue used[w] = 1 que.append(w) prt[w] = v pn[v] = len(order) order.append(v) sz = [0]*N tmp_ps = [[i] for i in range(N)] h_ps = [] for v in reversed(order): p = prt[v] m_sz = 0 h = -1 s = 1 for w in G[v]: if p == w: continue if m_sz < sz[w]: m_sz = sz[w] h = w s += sz[w] if h != -1: tmp_ps[v] = tmp_ps[h] tmp_ps[v].append(v) for w in G[v]: if w == p or w == h: continue h_ps.append(tmp_ps[w]) sz[v] = s h_ps.append(tmp_ps[0]) ans = [-1]*N ph = Prop() for path in h_ps: h = -1 for v in path: p = prt[v] for w in G[v]: if w == p or w == h: continue for c in order[pn[w]:pn[w] + sz[w]]: ph.add(c) ph.add(v) ans[v] = ph.get(v) h = v for c in order[pn[h]:pn[h] + sz[h]]: ph.remove(c) return ans def solve(): N = int(readline()) *C, = map(int, readline().split()) G = [[] for i in range(N)] for i in range(N-1): x, y = map(int, readline().split()) G[x-1].append(y-1) G[y-1].append(x-1) class Prop: def __init__(self): self.ts = [0]*(N+1) self.cs = [0]*(N+1) self.mc = 0 def add(self, v): c = C[v] r = self.cs[c] if r != 0: self.ts[r] -= c self.ts[r+1] += c if self.mc < r+1: self.mc = r+1 self.cs[c] += 1 def remove(self, v): c = C[v] r = self.cs[c] #assert r > 0 self.ts[r] -= c if r != 1: self.ts[r-1] += c self.cs[c] -= 1 if self.ts[r] == 0: self.mc -= 1 def get(self, i): return self.ts[self.mc] ans = dsu_on_tree(N, G, Prop) write(" ".join(map(str, ans))) write("\n") solve()
Title: Lomsat gelral Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour. Let's call colour *c* dominating in the subtree of vertex *v* if there are no other colours that appear in the subtree of vertex *v* more times than colour *c*. So it's possible that two or more colours will be dominating in the subtree of some vertex. The subtree of vertex *v* is the vertex *v* and all other vertices that contains vertex *v* in each path to the root. For each vertex *v* find the sum of all dominating colours in the subtree of vertex *v*. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of vertices in the tree. The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=*n*), *c**i* — the colour of the *i*-th vertex. Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the edge of the tree. The first vertex is the root of the tree. Output Specification: Print *n* integers — the sums of dominating colours for each vertex. Demo Input: ['4\n1 2 3 4\n1 2\n2 3\n2 4\n', '15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13\n'] Demo Output: ['10 9 3 4\n', '6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\n'] Note: none
```python import sys readline = sys.stdin.readline write = sys.stdout.write def dsu_on_tree(N, G, Prop): order = [] prt = [-1]*N que = [0] pn = [-1]*N used = [0]*N used[0] = 1 while que: v = que.pop() for w in G[v]: if used[w]: continue used[w] = 1 que.append(w) prt[w] = v pn[v] = len(order) order.append(v) sz = [0]*N tmp_ps = [[i] for i in range(N)] h_ps = [] for v in reversed(order): p = prt[v] m_sz = 0 h = -1 s = 1 for w in G[v]: if p == w: continue if m_sz < sz[w]: m_sz = sz[w] h = w s += sz[w] if h != -1: tmp_ps[v] = tmp_ps[h] tmp_ps[v].append(v) for w in G[v]: if w == p or w == h: continue h_ps.append(tmp_ps[w]) sz[v] = s h_ps.append(tmp_ps[0]) ans = [-1]*N ph = Prop() for path in h_ps: h = -1 for v in path: p = prt[v] for w in G[v]: if w == p or w == h: continue for c in order[pn[w]:pn[w] + sz[w]]: ph.add(c) ph.add(v) ans[v] = ph.get(v) h = v for c in order[pn[h]:pn[h] + sz[h]]: ph.remove(c) return ans def solve(): N = int(readline()) *C, = map(int, readline().split()) G = [[] for i in range(N)] for i in range(N-1): x, y = map(int, readline().split()) G[x-1].append(y-1) G[y-1].append(x-1) class Prop: def __init__(self): self.ts = [0]*(N+1) self.cs = [0]*(N+1) self.mc = 0 def add(self, v): c = C[v] r = self.cs[c] if r != 0: self.ts[r] -= c self.ts[r+1] += c if self.mc < r+1: self.mc = r+1 self.cs[c] += 1 def remove(self, v): c = C[v] r = self.cs[c] #assert r > 0 self.ts[r] -= c if r != 1: self.ts[r-1] += c self.cs[c] -= 1 if self.ts[r] == 0: self.mc -= 1 def get(self, i): return self.ts[self.mc] ans = dsu_on_tree(N, G, Prop) write(" ".join(map(str, ans))) write("\n") solve() ```
3
35
A
Shell Game
PROGRAMMING
1,000
[ "implementation" ]
A. Shell Game
2
64
Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too?
The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3.
In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles.
[ "1\n1 2\n2 1\n2 1\n", "1\n2 1\n3 1\n1 3\n" ]
[ "2\n", "2\n" ]
none
500
[ { "input": "1\n1 2\n2 1\n2 1", "output": "2" }, { "input": "1\n2 1\n3 1\n1 3", "output": "2" }, { "input": "3\n3 1\n2 1\n1 2", "output": "1" }, { "input": "1\n1 3\n1 2\n2 3", "output": "2" }, { "input": "3\n3 2\n3 1\n3 1", "output": "2" }, { "input": "1\n2 1\n1 3\n1 3", "output": "2" }, { "input": "3\n3 1\n2 3\n3 2", "output": "1" }, { "input": "2\n1 3\n1 2\n2 1", "output": "2" }, { "input": "1\n1 3\n3 2\n1 2", "output": "1" }, { "input": "1\n1 3\n1 3\n2 3", "output": "1" }, { "input": "2\n1 2\n2 3\n2 1", "output": "2" }, { "input": "3\n1 3\n3 2\n2 1", "output": "2" }, { "input": "1\n1 2\n2 1\n2 3", "output": "1" }, { "input": "1\n2 3\n1 3\n1 2", "output": "3" }, { "input": "2\n3 1\n3 2\n2 3", "output": "2" }, { "input": "2\n1 3\n3 1\n3 1", "output": "2" }, { "input": "1\n3 2\n1 3\n3 1", "output": "1" }, { "input": "3\n1 3\n1 2\n1 3", "output": "2" }, { "input": "1\n3 2\n3 1\n1 2", "output": "3" }, { "input": "2\n2 3\n1 3\n1 3", "output": "3" } ]
1,502,784,278
2,147,483,647
Python 3
OK
TESTS
20
124
4,608,000
fo = open('input.txt', 'r') idx = fo.readline() idx = int(idx) for i in range(3): s = fo.readline() a = s.split() for j in range(len(a)): a[j] = int(a[j]) if idx in a: a.remove(idx) idx = a[0] fo.close() fi = open('output.txt', 'w') fi.write(str(idx)) fi.close()
Title: Shell Game Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too? Input Specification: The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3. Output Specification: In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles. Demo Input: ['1\n1 2\n2 1\n2 1\n', '1\n2 1\n3 1\n1 3\n'] Demo Output: ['2\n', '2\n'] Note: none
```python fo = open('input.txt', 'r') idx = fo.readline() idx = int(idx) for i in range(3): s = fo.readline() a = s.split() for j in range(len(a)): a[j] = int(a[j]) if idx in a: a.remove(idx) idx = a[0] fo.close() fi = open('output.txt', 'w') fi.write(str(idx)) fi.close() ```
3.934668
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" } ]
1,619,179,310
2,147,483,647
Python 3
OK
TESTS
81
124
0
n = int(input()) x_s = 0 y_s = 0 z_s = 0 while n > 0: x,y,z = map(int, input().split()) x_s += x y_s += y z_s += z n -= 1 if x_s == 0 and y_s == 0 and z_s == 0: print("YES") else: print("NO")
Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none
```python n = int(input()) x_s = 0 y_s = 0 z_s = 0 while n > 0: x,y,z = map(int, input().split()) x_s += x y_s += y z_s += z n -= 1 if x_s == 0 and y_s == 0 and z_s == 0: print("YES") else: print("NO") ```
3.969
496
A
Minimum Difficulty
PROGRAMMING
900
[ "brute force", "implementation", "math" ]
null
null
Mike is trying rock climbing but he is awful at it. There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=&lt;<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height. Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions. Help Mike determine the minimum difficulty of the track after removing one hold.
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds. The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Print a single number — the minimum difficulty of the track after removing a single hold.
[ "3\n1 4 6\n", "5\n1 2 3 4 5\n", "5\n1 2 3 7 8\n" ]
[ "5\n", "2\n", "4\n" ]
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5. In the second test after removing every hold the difficulty equals 2. In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
500
[ { "input": "3\n1 4 6", "output": "5" }, { "input": "5\n1 2 3 4 5", "output": "2" }, { "input": "5\n1 2 3 7 8", "output": "4" }, { "input": "3\n1 500 1000", "output": "999" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "2" }, { "input": "10\n1 4 9 16 25 36 49 64 81 100", "output": "19" }, { "input": "10\n300 315 325 338 350 365 379 391 404 416", "output": "23" }, { "input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112", "output": "2" }, { "input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500", "output": "310" }, { "input": "3\n159 282 405", "output": "246" }, { "input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741", "output": "241" }, { "input": "10\n218 300 388 448 535 629 680 740 836 925", "output": "111" }, { "input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996", "output": "20" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000", "output": "901" }, { "input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978", "output": "605" }, { "input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999", "output": "803" }, { "input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932", "output": "17" }, { "input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000", "output": "54" }, { "input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000", "output": "74" } ]
1,418,834,647
847
Python 3
OK
TESTS
19
61
0
holds= int(input()) heights= [int(i) for i in input().split()] ans= 999999999 real= heights for i in range(1, len(heights)-1): #print(i, heights, real) app= heights[i] heights.remove(app) k= 0 for j in range(len(heights)-1): k= max(heights[j+1]-heights[j], k) ans= min(ans, k) heights.append(app) heights= sorted(heights) print(ans)
Title: Minimum Difficulty Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mike is trying rock climbing but he is awful at it. There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=&lt;<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height. Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions. Help Mike determine the minimum difficulty of the track after removing one hold. Input Specification: The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds. The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one). Output Specification: Print a single number — the minimum difficulty of the track after removing a single hold. Demo Input: ['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n'] Demo Output: ['5\n', '2\n', '4\n'] Note: In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5. In the second test after removing every hold the difficulty equals 2. In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
```python holds= int(input()) heights= [int(i) for i in input().split()] ans= 999999999 real= heights for i in range(1, len(heights)-1): #print(i, heights, real) app= heights[i] heights.remove(app) k= 0 for j in range(len(heights)-1): k= max(heights[j+1]-heights[j], k) ans= min(ans, k) heights.append(app) heights= sorted(heights) print(ans) ```
3
475
B
Strongly Connected City
PROGRAMMING
1,400
[ "brute force", "dfs and similar", "graphs", "implementation" ]
null
null
Imagine a city with *n* horizontal streets crossing *m* vertical streets, forming an (*n*<=-<=1)<=×<=(*m*<=-<=1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection. The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
The first line of input contains two integers *n* and *m*, (2<=≤<=*n*,<=*m*<=≤<=20), denoting the number of horizontal streets and the number of vertical streets. The second line contains a string of length *n*, made of characters '&lt;' and '&gt;', denoting direction of each horizontal street. If the *i*-th character is equal to '&lt;', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south. The third line contains a string of length *m*, made of characters '^' and 'v', denoting direction of each vertical street. If the *i*-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
[ "3 3\n&gt;&lt;&gt;\nv^v\n", "4 6\n&lt;&gt;&lt;&gt;\nv^v^v^\n" ]
[ "NO\n", "YES\n" ]
The figure above shows street directions in the second sample test case.
1,000
[ { "input": "3 3\n><>\nv^v", "output": "NO" }, { "input": "4 6\n<><>\nv^v^v^", "output": "YES" }, { "input": "2 2\n<>\nv^", "output": "YES" }, { "input": "2 2\n>>\n^v", "output": "NO" }, { "input": "3 3\n>><\n^^v", "output": "YES" }, { "input": "3 4\n>><\n^v^v", "output": "YES" }, { "input": "3 8\n>><\nv^^^^^^^", "output": "NO" }, { "input": "7 2\n<><<<<>\n^^", "output": "NO" }, { "input": "4 5\n><<<\n^^^^v", "output": "YES" }, { "input": "2 20\n><\n^v^^v^^v^^^v^vv^vv^^", "output": "NO" }, { "input": "2 20\n<>\nv^vv^v^^vvv^^^v^vvv^", "output": "YES" }, { "input": "20 2\n<><<><<>><<<>><><<<<\n^^", "output": "NO" }, { "input": "20 2\n><>><>><>><<<><<><><\n^v", "output": "YES" }, { "input": "11 12\n><<<><><<>>\nvv^^^^vvvvv^", "output": "NO" }, { "input": "4 18\n<<>>\nv^v^v^^vvvv^v^^vv^", "output": "YES" }, { "input": "16 11\n<<<<>><><<<<<><<\nvv^v^vvvv^v", "output": "NO" }, { "input": "14 7\n><<<<>>>>>>><<\nvv^^^vv", "output": "NO" }, { "input": "5 14\n<<><>\nv^vv^^vv^v^^^v", "output": "NO" }, { "input": "8 18\n>>>><>>>\nv^vv^v^^^^^vvv^^vv", "output": "NO" }, { "input": "18 18\n<<><>><<>><>><><<<\n^^v^v^vvvv^v^vv^vv", "output": "NO" }, { "input": "4 18\n<<<>\n^^^^^vv^vv^^vv^v^v", "output": "NO" }, { "input": "19 18\n><><>>><<<<<>>><<<>\n^^v^^v^^v^vv^v^vvv", "output": "NO" }, { "input": "14 20\n<<<><><<>><><<\nvvvvvvv^v^vvvv^^^vv^", "output": "NO" }, { "input": "18 18\n><>>><<<>><><>>>><\nvv^^^^v^v^^^^v^v^^", "output": "NO" }, { "input": "8 18\n<><<<>>>\n^^^^^^v^^^vv^^vvvv", "output": "NO" }, { "input": "11 12\n><><><<><><\n^^v^^^^^^^^v", "output": "YES" }, { "input": "4 18\n<<>>\nv^v^v^^vvvv^v^^vv^", "output": "YES" }, { "input": "16 11\n>><<><<<<>>><><<\n^^^^vvvv^vv", "output": "YES" }, { "input": "14 7\n<><><<<>>>><>>\nvv^^v^^", "output": "YES" }, { "input": "5 14\n>>>><\n^v^v^^^vv^vv^v", "output": "YES" }, { "input": "8 18\n<<<><>>>\nv^^vvv^^v^v^vvvv^^", "output": "YES" }, { "input": "18 18\n><><<><><>>><>>>><\n^^vvv^v^^^v^vv^^^v", "output": "YES" }, { "input": "4 18\n<<>>\nv^v^v^^vvvv^v^^vv^", "output": "YES" }, { "input": "19 18\n>>>><><<>>><<<><<<<\n^v^^^^vv^^v^^^^v^v", "output": "YES" }, { "input": "14 20\n<>><<<><<>>>>>\nvv^^v^^^^v^^vv^^vvv^", "output": "YES" }, { "input": "18 18\n><><<><><>>><>>>><\n^^vvv^v^^^v^vv^^^v", "output": "YES" }, { "input": "8 18\n<<<><>>>\nv^^vvv^^v^v^vvvv^^", "output": "YES" }, { "input": "20 19\n<><>>>>><<<<<><<>>>>\nv^vv^^vvvvvv^vvvv^v", "output": "NO" }, { "input": "20 19\n<<<><<<>><<<>><><><>\nv^v^vvv^vvv^^^vvv^^", "output": "YES" }, { "input": "19 20\n<><<<><><><<<<<<<<>\n^v^^^^v^^vvvv^^^^vvv", "output": "NO" }, { "input": "19 20\n>>>>>>>><>>><><<<><\n^v^v^^^vvv^^^v^^vvvv", "output": "YES" }, { "input": "20 20\n<<<>>>><>><<>><<>>>>\n^vvv^^^^vv^^^^^v^^vv", "output": "NO" }, { "input": "20 20\n>>><><<><<<<<<<><<><\nvv^vv^vv^^^^^vv^^^^^", "output": "NO" }, { "input": "20 20\n><<><<<<<<<>>><>>><<\n^^^^^^^^vvvv^vv^vvvv", "output": "YES" }, { "input": "20 20\n<>>>>>>>><>>><>><<<>\nvv^^vv^^^^v^vv^v^^^^", "output": "YES" }, { "input": "20 20\n><>><<>><>>>>>>>><<>\n^^v^vv^^^vvv^v^^^vv^", "output": "NO" }, { "input": "20 20\n<<<<><<>><><<<>><<><\nv^^^^vvv^^^vvvv^v^vv", "output": "NO" }, { "input": "20 20\n><<<><<><>>><><<<<<<\nvv^^vvv^^v^^v^vv^vvv", "output": "NO" }, { "input": "20 20\n<<>>><>>>><<<<>>><<>\nv^vv^^^^^vvv^^v^^v^v", "output": "NO" }, { "input": "20 20\n><<><<><<<<<<>><><>>\nv^^^v^vv^^v^^vvvv^vv", "output": "NO" }, { "input": "20 20\n<<<<<<<<><>><><>><<<\n^vvv^^^v^^^vvv^^^^^v", "output": "NO" }, { "input": "20 20\n>>><<<<<>>><><><<><<\n^^^vvv^^^v^^v^^v^vvv", "output": "YES" }, { "input": "20 20\n<><<<><><>><><><<<<>\n^^^vvvv^vv^v^^^^v^vv", "output": "NO" }, { "input": "20 20\n>>>>>>>>>><>>><>><>>\n^vvv^^^vv^^^^^^vvv^v", "output": "NO" }, { "input": "20 20\n<><>><><<<<<>><<>>><\nv^^^v^v^v^vvvv^^^vv^", "output": "NO" }, { "input": "20 20\n><<<><<<><<<><>>>><<\nvvvv^^^^^vv^v^^vv^v^", "output": "NO" }, { "input": "20 20\n<<><<<<<<>>>>><<<>>>\nvvvvvv^v^vvv^^^^^^^^", "output": "YES" }, { "input": "20 20\n><<><<>>>>><><>><>>>\nv^^^^vvv^^^^^v^v^vv^", "output": "NO" }, { "input": "20 20\n<<>>><>><<>>>><<<><<\n^^vvv^^vvvv^vv^^v^v^", "output": "NO" }, { "input": "20 20\n><<>><>>>><<><>><><<\n^v^^^^^^vvvv^v^v^v^^", "output": "NO" }, { "input": "20 20\n<<><<<<><><<>>><>>>>\n^^vvvvv^v^^^^^^^vvv^", "output": "NO" }, { "input": "20 20\n>><<<<<<><>>>><>>><>\n^^^v^v^vv^^vv^vvv^^^", "output": "NO" }, { "input": "20 20\n>>>>>>>>>>>>>>>>>>>>\nvvvvvvvvvvvvvvvvvvvv", "output": "NO" }, { "input": "2 2\n><\nv^", "output": "NO" }, { "input": "2 2\n<>\n^v", "output": "NO" }, { "input": "3 3\n>><\nvvv", "output": "NO" }, { "input": "2 3\n<>\nv^^", "output": "YES" }, { "input": "4 4\n>>><\nvvv^", "output": "NO" }, { "input": "20 20\n<><><><><><><><><><>\nvvvvvvvvvvvvvvvvvvvv", "output": "NO" }, { "input": "4 4\n<>>>\nv^^^", "output": "YES" }, { "input": "20 20\n<><><><><><><><><><>\nv^v^v^v^v^v^v^v^v^v^", "output": "YES" }, { "input": "2 3\n<>\n^v^", "output": "NO" }, { "input": "4 3\n<><>\n^vv", "output": "NO" }, { "input": "3 3\n<<>\nvv^", "output": "YES" }, { "input": "2 3\n><\nvv^", "output": "NO" }, { "input": "7 6\n>>><>><\n^vv^vv", "output": "YES" }, { "input": "2 2\n<<\nv^", "output": "NO" }, { "input": "3 3\n>><\n^^^", "output": "NO" }, { "input": "3 3\n<><\nv^v", "output": "NO" }, { "input": "20 20\n><><><><><><><><><><\n^v^v^v^v^v^v^v^v^v^v", "output": "YES" }, { "input": "4 4\n<>>>\nvvv^", "output": "YES" } ]
1,683,560,376
2,147,483,647
PyPy 3
OK
TESTS
81
155
8,396,800
n,m=list(map(int,input().split())) di=input() dj=input() def canmove(i,j): return i>=0 and j>=0 and i<n and j<m def dfs(i,j,seen): seen[i][j]=True pi=i pj=j if di[i]=='>': pj+=1 else: pj-=1 if dj[j]=='^': pi-=1 else: pi+=1 if canmove(pi,j) and not seen[pi][j]: dfs(pi,j,seen) if canmove(i,pj) and not seen[i][pj]: dfs(i,pj,seen) def solve(): for i in range(n): for j in range(m): seen = [[False] * m for k in range(n)] if not seen[i][j]: dfs(i,j,seen) for row in seen: for b in row: if b==False: return False return True if solve(): print("YES") else: print("NO")
Title: Strongly Connected City Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imagine a city with *n* horizontal streets crossing *m* vertical streets, forming an (*n*<=-<=1)<=×<=(*m*<=-<=1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection. The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern. Input Specification: The first line of input contains two integers *n* and *m*, (2<=≤<=*n*,<=*m*<=≤<=20), denoting the number of horizontal streets and the number of vertical streets. The second line contains a string of length *n*, made of characters '&lt;' and '&gt;', denoting direction of each horizontal street. If the *i*-th character is equal to '&lt;', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south. The third line contains a string of length *m*, made of characters '^' and 'v', denoting direction of each vertical street. If the *i*-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east. Output Specification: If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO". Demo Input: ['3 3\n&gt;&lt;&gt;\nv^v\n', '4 6\n&lt;&gt;&lt;&gt;\nv^v^v^\n'] Demo Output: ['NO\n', 'YES\n'] Note: The figure above shows street directions in the second sample test case.
```python n,m=list(map(int,input().split())) di=input() dj=input() def canmove(i,j): return i>=0 and j>=0 and i<n and j<m def dfs(i,j,seen): seen[i][j]=True pi=i pj=j if di[i]=='>': pj+=1 else: pj-=1 if dj[j]=='^': pi-=1 else: pi+=1 if canmove(pi,j) and not seen[pi][j]: dfs(pi,j,seen) if canmove(i,pj) and not seen[i][pj]: dfs(i,pj,seen) def solve(): for i in range(n): for j in range(m): seen = [[False] * m for k in range(n)] if not seen[i][j]: dfs(i,j,seen) for row in seen: for b in row: if b==False: return False return True if solve(): print("YES") else: print("NO") ```
3
116
A
Tram
PROGRAMMING
800
[ "implementation" ]
null
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
[ { "input": "4\n0 3\n2 5\n4 2\n4 0", "output": "6" }, { "input": "5\n0 4\n4 6\n6 5\n5 4\n4 0", "output": "6" }, { "input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0", "output": "18" }, { "input": "3\n0 1\n1 1\n1 0", "output": "1" }, { "input": "4\n0 1\n0 1\n1 0\n1 0", "output": "2" }, { "input": "3\n0 0\n0 0\n0 0", "output": "0" }, { "input": "3\n0 1000\n1000 1000\n1000 0", "output": "1000" }, { "input": "5\n0 73\n73 189\n189 766\n766 0\n0 0", "output": "766" }, { "input": "5\n0 0\n0 0\n0 0\n0 1\n1 0", "output": "1" }, { "input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0", "output": "1011" }, { "input": "5\n0 1\n1 2\n2 1\n1 2\n2 0", "output": "2" }, { "input": "5\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0", "output": "22" }, { "input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0", "output": "1000" }, { "input": "10\n0 592\n258 598\n389 203\n249 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" } ]
1,694,962,601
2,147,483,647
Python 3
OK
TESTS
71
92
0
tc=int(input()) persons=0 max_list=[] for x in range(tc): exit_,enter=map(int,input().split()) persons=(persons+enter)-exit_ max_list.append(persons) print(max(max_list))
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 tc=int(input()) persons=0 max_list=[] for x in range(tc): exit_,enter=map(int,input().split()) persons=(persons+enter)-exit_ max_list.append(persons) print(max(max_list)) ```
3
549
A
Face Detection
PROGRAMMING
900
[ "implementation", "strings" ]
null
null
The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them. In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=×<=2 square, such that from the four letters of this square you can make word "face". You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.
The first line contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the height and the width of the image, respectively. Next *n* lines define the image. Each line contains *m* lowercase Latin letters.
In the single line print the number of faces on the image.
[ "4 4\nxxxx\nxfax\nxcex\nxxxx\n", "4 2\nxx\ncf\nae\nxx\n", "2 3\nfac\ncef\n", "1 4\nface\n" ]
[ "1\n", "1\n", "2\n", "0\n" ]
In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column: In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column. In the third sample two faces are shown: In the fourth sample the image has no faces on it.
250
[ { "input": "4 4\nxxxx\nxfax\nxcex\nxxxx", "output": "1" }, { "input": "4 2\nxx\ncf\nae\nxx", "output": "1" }, { "input": "2 3\nfac\ncef", "output": "2" }, { "input": "1 4\nface", "output": "0" }, { "input": "5 5\nwmmwn\nlurcm\nkeetd\nfokon\ncxxgx", "output": "0" }, { "input": "5 5\nkjxbw\neacra\nxefhx\nucmcz\npgtjk", "output": "1" }, { "input": "1 1\np", "output": "0" }, { "input": "2 5\nacdmw\nefazb", "output": "1" }, { "input": "5 2\ndz\nda\nsx\nyu\nzz", "output": "0" }, { "input": "5 5\nxeljd\nwriac\nveief\nlcacf\nbqefn", "output": "2" }, { "input": "5 5\nacnbx\nefacp\nlrefa\norqce\nzvbay", "output": "3" }, { "input": "5 5\nbyjvu\nkmaca\nalefe\nwcacg\nrefez", "output": "5" }, { "input": "5 5\npuxac\nbbaef\naccfa\nefaec\nligsr", "output": "5" }, { "input": "37 4\nacjo\nefac\nacef\nefac\nwpef\nicac\naefe\ncfac\naece\ncfaf\nyqce\nmiaf\nirce\nycaf\naefc\ncfae\nrsnc\nbacz\nqefb\npdhs\nffac\nfaef\nacfd\nacmi\nefvm\nacaz\nefpn\nacao\nefer\nacap\nefec\nacaf\nefef\nacbj\nefac\nacef\nefoz", "output": "49" }, { "input": "7 3\njac\naef\ncfa\naec\ncfq\ndig\nxyq", "output": "5" }, { "input": "35 1\ny\na\nk\ng\ni\nd\nv\nn\nl\nx\nu\nx\nu\no\nd\nf\nk\nj\nr\nm\nq\ns\nc\nd\nc\nm\nv\nh\nn\ne\nl\nt\nz\ny\no", "output": "0" }, { "input": "9 46\nuuexbaacesjclggslacermcbkxlcxhdgqtacdwfryxzuxc\naclrsaefakndbnzlkefenuphgcgoedhkaxefjtnkgfeaca\nefuqunpmfxdyyffyhvracozzrxlpekhtsrfhlilfmyhefg\numyacfzffvicqtdpiulefnwcojuwtfbvlxkfsiapdnzpqo\nactefvuxqptremlqjhdbdwacjxdxitxjktecvefacamjcz\neflarseklqrkayhosverpfefzirqigzlxezabhzeferkwm\nztpypwxyohtacsnnvoigqwcachtaceyxlkhfhgwbnfmklb\nfzpqeymhljiefctvqtbgcdjefkxefarbciyosmktmoziac\nbssxizyrzyhacszttwhrzspbfkgfcejybdjcpuczhrkqef", "output": "20" }, { "input": "7 30\nmjfracgaacacctacrreyrlkacuacay\nrlacefacefeftaeftkacacaefcefev\nacefacefacraccfaeaefefecaeacaf\nefacefacefaefaecfcfacacaecfefa\nncefacefacecacfaeaecefefcaeace\nfafaceacuafaefadcfcafacaefcfea\nzsvefafukcecfarkaeaecefecailgu", "output": "95" }, { "input": "2 2\nff\nee", "output": "0" }, { "input": "2 2\nfa\ncc", "output": "0" }, { "input": "2 2\nfa\naa", "output": "0" }, { "input": "2 2\nfa\nea", "output": "0" }, { "input": "2 2\nfa\ndd", "output": "0" }, { "input": "2 2\nee\nff", "output": "0" }, { "input": "2 2\ncc\ncf", "output": "0" }, { "input": "2 2\naa\nae", "output": "0" }, { "input": "2 2\nef\nac", "output": "1" }, { "input": "2 2\nfe\nca", "output": "1" } ]
1,435,226,434
2,147,483,647
Python 3
OK
TESTS
37
62
0
n, m = map(int, input().split()) a = [input() for i in range(n)] print(sum([{a[i][j], a[i + 1][j], a[i][j + 1], a[i + 1][j + 1]} == set('face') for j in range(m - 1) for i in range(n - 1)]))
Title: Face Detection Time Limit: None seconds Memory Limit: None megabytes Problem Description: The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them. In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=×<=2 square, such that from the four letters of this square you can make word "face". You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap. Input Specification: The first line contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the height and the width of the image, respectively. Next *n* lines define the image. Each line contains *m* lowercase Latin letters. Output Specification: In the single line print the number of faces on the image. Demo Input: ['4 4\nxxxx\nxfax\nxcex\nxxxx\n', '4 2\nxx\ncf\nae\nxx\n', '2 3\nfac\ncef\n', '1 4\nface\n'] Demo Output: ['1\n', '1\n', '2\n', '0\n'] Note: In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column: In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column. In the third sample two faces are shown: In the fourth sample the image has no faces on it.
```python n, m = map(int, input().split()) a = [input() for i in range(n)] print(sum([{a[i][j], a[i + 1][j], a[i][j + 1], a[i + 1][j + 1]} == set('face') for j in range(m - 1) for i in range(n - 1)])) ```
3
75
A
Life Without Zeros
PROGRAMMING
1,000
[ "implementation" ]
A. Life Without Zeros
2
256
Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
[ "101\n102\n", "105\n106\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "101\n102", "output": "YES" }, { "input": "105\n106", "output": "NO" }, { "input": "544\n397", "output": "YES" }, { "input": "822\n280", "output": "NO" }, { "input": "101\n413", "output": "NO" }, { "input": "309\n139", "output": "NO" }, { "input": "693\n970", "output": "NO" }, { "input": "981\n1", "output": "YES" }, { "input": "352\n276", "output": "YES" }, { "input": "164\n691", "output": "YES" }, { "input": "110036\n43", "output": "YES" }, { "input": "100\n1000", "output": "NO" }, { "input": "1000000000\n1000000000", "output": "YES" }, { "input": "999999999\n999999999", "output": "YES" }, { "input": "6\n4", "output": "NO" }, { "input": "123456\n876543", "output": "YES" }, { "input": "1234567\n9876543", "output": "NO" }, { "input": "1111111\n1119111", "output": "NO" }, { "input": "10502405\n40403403", "output": "YES" }, { "input": "1\n1", "output": "YES" }, { "input": "5\n4", "output": "YES" }, { "input": "1\n6", "output": "YES" }, { "input": "40\n17", "output": "NO" }, { "input": "35\n81", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "2081\n7450", "output": "NO" }, { "input": "84355694\n373629503", "output": "NO" }, { "input": "885992048\n510468672", "output": "NO" }, { "input": "540144752\n647307841", "output": "NO" }, { "input": "341781108\n784147010", "output": "NO" }, { "input": "960440938\n572334653", "output": "NO" }, { "input": "657414510\n768771903", "output": "NO" }, { "input": "501871728\n725074574", "output": "NO" }, { "input": "968\n851", "output": "YES" }, { "input": "943\n939", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "611\n41", "output": "YES" }, { "input": "737\n289", "output": "NO" }, { "input": "168\n571", "output": "YES" }, { "input": "868\n463", "output": "YES" }, { "input": "815732007\n913882008", "output": "NO" }, { "input": "459039901\n453033850", "output": "NO" }, { "input": "975305404\n484601256", "output": "NO" } ]
1,634,489,552
2,147,483,647
Python 3
OK
TESTS
43
92
0
#n, m, k = map(int,input().split()) #a, b = map(int,input().split()) a = input() b = input() c = str(int(a)+int(b)) s_a = "" s_b = "" s_c = "" for i in range(0, len(a)): if(a[i]!='0'): s_a += a[i] for i in range(0, len(b)): if(b[i] != '0'): s_b+=b[i] for i in range(0, len(c)): if(c[i] != '0'): s_c+=c[i] if(int(s_c)==int(s_a)+int(s_b)): print("YES") else: print("NO")
Title: Life Without Zeros Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation. Input Specification: The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*. Output Specification: The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise. Demo Input: ['101\n102\n', '105\n106\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python #n, m, k = map(int,input().split()) #a, b = map(int,input().split()) a = input() b = input() c = str(int(a)+int(b)) s_a = "" s_b = "" s_c = "" for i in range(0, len(a)): if(a[i]!='0'): s_a += a[i] for i in range(0, len(b)): if(b[i] != '0'): s_b+=b[i] for i in range(0, len(c)): if(c[i] != '0'): s_c+=c[i] if(int(s_c)==int(s_a)+int(s_b)): print("YES") else: print("NO") ```
3.977
501
A
Contest
PROGRAMMING
900
[ "implementation" ]
null
null
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points.
[ "500 1000 20 30\n", "1000 1000 1 1\n", "1500 1000 176 177\n" ]
[ "Vasya\n", "Tie\n", "Misha\n" ]
none
500
[ { "input": "500 1000 20 30", "output": "Vasya" }, { "input": "1000 1000 1 1", "output": "Tie" }, { "input": "1500 1000 176 177", "output": "Misha" }, { "input": "1500 1000 74 177", "output": "Misha" }, { "input": "750 2500 175 178", "output": "Vasya" }, { "input": "750 1000 54 103", "output": "Tie" }, { "input": "2000 1250 176 130", "output": "Tie" }, { "input": "1250 1750 145 179", "output": "Tie" }, { "input": "2000 2000 176 179", "output": "Tie" }, { "input": "1500 1500 148 148", "output": "Tie" }, { "input": "2750 1750 134 147", "output": "Misha" }, { "input": "3250 250 175 173", "output": "Misha" }, { "input": "500 500 170 176", "output": "Misha" }, { "input": "250 1000 179 178", "output": "Vasya" }, { "input": "3250 1000 160 138", "output": "Misha" }, { "input": "3000 2000 162 118", "output": "Tie" }, { "input": "1500 1250 180 160", "output": "Tie" }, { "input": "1250 2500 100 176", "output": "Tie" }, { "input": "3500 3500 177 178", "output": "Tie" }, { "input": "3000 3250 16 34", "output": "Tie" }, { "input": "1750 3000 137 49", "output": "Vasya" }, { "input": "500 1500 179 71", "output": "Vasya" }, { "input": "1250 2000 101 180", "output": "Misha" }, { "input": "250 750 180 176", "output": "Vasya" }, { "input": "2250 2250 163 145", "output": "Vasya" }, { "input": "3000 3000 176 78", "output": "Vasya" }, { "input": "250 3500 8 178", "output": "Vasya" }, { "input": "1750 1250 179 180", "output": "Misha" }, { "input": "2750 1750 13 164", "output": "Misha" }, { "input": "1750 2250 178 53", "output": "Vasya" }, { "input": "2500 2750 73 179", "output": "Misha" }, { "input": "1000 3500 178 175", "output": "Vasya" }, { "input": "1000 500 7 162", "output": "Misha" }, { "input": "1000 250 175 48", "output": "Misha" }, { "input": "1750 500 166 177", "output": "Misha" }, { "input": "250 250 0 0", "output": "Tie" }, { "input": "250 3500 0 0", "output": "Vasya" }, { "input": "250 3500 0 180", "output": "Vasya" }, { "input": "3500 3500 180 180", "output": "Tie" }, { "input": "3500 250 0 180", "output": "Misha" } ]
1,641,721,141
2,147,483,647
PyPy 3-64
OK
TESTS
40
108
0
a, b, c, d = map(int, input().split()) vasya = max(3 * b // 10, b - b // 250 * d) misha = max(3 * a // 10, a - a // 250 * c) if vasya > misha: print("Vasya") elif misha > vasya: print("Misha") else: print("Tie")
Title: Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. Input Specification: The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). Output Specification: Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points. Demo Input: ['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n'] Demo Output: ['Vasya\n', 'Tie\n', 'Misha\n'] Note: none
```python a, b, c, d = map(int, input().split()) vasya = max(3 * b // 10, b - b // 250 * d) misha = max(3 * a // 10, a - a // 250 * c) if vasya > misha: print("Vasya") elif misha > vasya: print("Misha") else: print("Tie") ```
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" } ]
1,593,018,134
2,147,483,647
PyPy 3
OK
TESTS
81
280
20,172,800
n = int(input()) data = [] for i in range(n): data.append(list(map(int, input().split()))) sum1 = sum(x[0] for x in data) sum2 = sum(x[1] for x in data) sum3 = sum(x[2] for x in data) if sum1 == 0 and sum2 == 0 and sum3 == 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()) data = [] for i in range(n): data.append(list(map(int, input().split()))) sum1 = sum(x[0] for x in data) sum2 = sum(x[1] for x in data) sum3 = sum(x[2] for x in data) if sum1 == 0 and sum2 == 0 and sum3 == 0: print("YES") else: print("NO") ```
3.892425
312
B
Archer
PROGRAMMING
1,300
[ "math", "probabilities" ]
null
null
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner. Output the probability that SmallR will win the match.
A single line contains four integers .
Print a single real number, the probability that SmallR will win the match. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
[ "1 2 1 2\n" ]
[ "0.666666666667" ]
none
1,000
[ { "input": "1 2 1 2", "output": "0.666666666667" }, { "input": "1 3 1 3", "output": "0.600000000000" }, { "input": "1 3 2 3", "output": "0.428571428571" }, { "input": "3 4 3 4", "output": "0.800000000000" }, { "input": "1 2 10 11", "output": "0.523809523810" }, { "input": "4 5 4 5", "output": "0.833333333333" }, { "input": "466 701 95 721", "output": "0.937693791148" }, { "input": "268 470 444 885", "output": "0.725614009325" }, { "input": "632 916 713 821", "output": "0.719292895126" }, { "input": "269 656 918 992", "output": "0.428937461623" }, { "input": "71 657 187 695", "output": "0.310488463257" }, { "input": "435 852 973 978", "output": "0.511844133157" }, { "input": "518 816 243 359", "output": "0.719734031025" }, { "input": "882 962 311 811", "output": "0.966386645447" }, { "input": "684 774 580 736", "output": "0.906051574446" }, { "input": "486 868 929 999", "output": "0.577723252958" }, { "input": "132 359 996 998", "output": "0.368154532345" }, { "input": "933 977 266 450", "output": "0.972879407907" }, { "input": "298 833 615 872", "output": "0.441270817024" }, { "input": "34 554 14 958", "output": "0.817324099167" }, { "input": "836 934 800 905", "output": "0.906105535462" }, { "input": "482 815 69 509", "output": "0.914365577772" }, { "input": "284 423 137 521", "output": "0.885974839378" }, { "input": "648 881 486 703", "output": "0.800911421248" }, { "input": "450 885 755 836", "output": "0.533901011176" }, { "input": "533 773 823 998", "output": "0.729222130525" }, { "input": "897 957 92 898", "output": "0.993193806364" }, { "input": "699 925 441 928", "output": "0.866816866175" }, { "input": "64 704 148 603", "output": "0.289486317811" }, { "input": "719 735 626 990", "output": "0.986124079764" }, { "input": "1 1000 1 1000", "output": "0.500250125063" } ]
1,602,868,270
2,147,483,647
Python 3
OK
TESTS
31
109
0
a, b, c, d = [int(i) for i in input().split()] p1 = a/b p2 = c/d awns = p1 / (1 - (1-p1) * (1 - p2)) print(awns)
Title: Archer Time Limit: None seconds Memory Limit: None megabytes Problem Description: SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner. Output the probability that SmallR will win the match. Input Specification: A single line contains four integers . Output Specification: Print a single real number, the probability that SmallR will win the match. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6. Demo Input: ['1 2 1 2\n'] Demo Output: ['0.666666666667'] Note: none
```python a, b, c, d = [int(i) for i in input().split()] p1 = a/b p2 = c/d awns = p1 / (1 - (1-p1) * (1 - p2)) print(awns) ```
3
1,000
B
Light It Up
PROGRAMMING
1,500
[ "greedy" ]
null
null
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$.
Print the only integer — maximum possible total time when the lamp is lit.
[ "3 10\n4 6 7\n", "2 12\n1 10\n", "2 7\n3 4\n" ]
[ "8\n", "9\n", "6\n" ]
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
0
[ { "input": "3 10\n4 6 7", "output": "8" }, { "input": "2 12\n1 10", "output": "9" }, { "input": "2 7\n3 4", "output": "6" }, { "input": "1 2\n1", "output": "1" }, { "input": "5 10\n1 3 5 6 8", "output": "6" }, { "input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "1 10\n1", "output": "9" }, { "input": "1 10000000\n1", "output": "9999999" }, { "input": "1 8\n1", "output": "7" }, { "input": "7 17\n1 5 9 10 11 14 16", "output": "9" }, { "input": "4 17\n1 5 9 10", "output": "12" }, { "input": "5 12\n1 2 3 4 5", "output": "9" }, { "input": "2 1000000000\n100 111", "output": "999999989" }, { "input": "3 90591\n90579 90580 90581", "output": "90589" }, { "input": "5 16\n1 2 3 4 5", "output": "13" }, { "input": "6 981231233\n1 6 15 30 130 1875", "output": "981229468" }, { "input": "1 3\n1", "output": "2" }, { "input": "3 12\n4 7 11", "output": "8" }, { "input": "1 100000\n1", "output": "99999" }, { "input": "2 5\n1 3", "output": "3" }, { "input": "1 1000000000\n1", "output": "999999999" }, { "input": "1 1000000000\n999999999", "output": "999999999" }, { "input": "3 4\n1 2 3", "output": "2" }, { "input": "3 5\n1 2 3", "output": "3" }, { "input": "1 4\n2", "output": "3" }, { "input": "3 5\n1 3 4", "output": "3" }, { "input": "9 20\n5 9 11 12 14 15 16 17 19", "output": "12" }, { "input": "2 4\n1 2", "output": "3" }, { "input": "1 10\n2", "output": "9" }, { "input": "2 6\n2 4", "output": "4" }, { "input": "2 4\n1 3", "output": "2" }, { "input": "3 6\n1 2 4", "output": "4" }, { "input": "7 9863\n65 96 97 98 101 112 1115", "output": "9819" } ]
1,635,650,294
2,147,483,647
Python 3
OK
TESTS
39
217
12,902,400
n, M = map(int, input().split()) op = list(map(int, input().split())) switch = 1 time = [0] * (n+2) op.append(0) op.sort() for i in range(1, n+1): time[i] += time[i-1] + switch*(op[i]-op[i-1]) switch = 1 - switch time[n+1] = time[n] + switch * (M-op[n]) maxHour = time[n+1] for i in range(1, n+1): maxHour = max(maxHour, time[i]+M-op[i]-(time[n+1]-time[i])-1) print(maxHour)
Title: Light It Up Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up. Input Specification: First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$. Output Specification: Print the only integer — maximum possible total time when the lamp is lit. Demo Input: ['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n'] Demo Output: ['8\n', '9\n', '6\n'] Note: In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
```python n, M = map(int, input().split()) op = list(map(int, input().split())) switch = 1 time = [0] * (n+2) op.append(0) op.sort() for i in range(1, n+1): time[i] += time[i-1] + switch*(op[i]-op[i-1]) switch = 1 - switch time[n+1] = time[n] + switch * (M-op[n]) maxHour = time[n+1] for i in range(1, n+1): maxHour = max(maxHour, time[i]+M-op[i]-(time[n+1]-time[i])-1) print(maxHour) ```
3
255
A
Greg's Workout
PROGRAMMING
800
[ "implementation" ]
null
null
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous.
[ "2\n2 8\n", "3\n5 1 10\n", "7\n3 3 2 7 9 6 8\n" ]
[ "biceps\n", "back\n", "chest\n" ]
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
500
[ { "input": "2\n2 8", "output": "biceps" }, { "input": "3\n5 1 10", "output": "back" }, { "input": "7\n3 3 2 7 9 6 8", "output": "chest" }, { "input": "4\n5 6 6 2", "output": "chest" }, { "input": "5\n8 2 2 6 3", "output": "chest" }, { "input": "6\n8 7 2 5 3 4", "output": "chest" }, { "input": "8\n7 2 9 10 3 8 10 6", "output": "chest" }, { "input": "9\n5 4 2 3 4 4 5 2 2", "output": "chest" }, { "input": "10\n4 9 8 5 3 8 8 10 4 2", "output": "biceps" }, { "input": "11\n10 9 7 6 1 3 9 7 1 3 5", "output": "chest" }, { "input": "12\n24 22 6 16 5 21 1 7 2 19 24 5", "output": "chest" }, { "input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24", "output": "chest" }, { "input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7", "output": "back" }, { "input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12", "output": "chest" }, { "input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8", "output": "biceps" }, { "input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19", "output": "chest" }, { "input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21", "output": "back" }, { "input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24", "output": "chest" }, { "input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20", "output": "chest" }, { "input": "1\n10", "output": "chest" }, { "input": "2\n15 3", "output": "chest" }, { "input": "3\n21 11 19", "output": "chest" }, { "input": "4\n19 24 13 15", "output": "chest" }, { "input": "5\n4 24 1 9 19", "output": "biceps" }, { "input": "6\n6 22 24 7 15 24", "output": "back" }, { "input": "7\n10 8 23 23 14 18 14", "output": "chest" }, { "input": "8\n5 16 8 9 17 16 14 7", "output": "biceps" }, { "input": "9\n12 3 10 23 6 4 22 13 12", "output": "chest" }, { "input": "10\n1 9 20 18 20 17 7 24 23 2", "output": "back" }, { "input": "11\n22 25 8 2 18 15 1 13 1 11 4", "output": "biceps" }, { "input": "12\n20 12 14 2 15 6 24 3 11 8 11 14", "output": "chest" }, { "input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18", "output": "back" }, { "input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22", "output": "biceps" }, { "input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6", "output": "back" }, { "input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14", "output": "chest" }, { "input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10", "output": "biceps" }, { "input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17", "output": "chest" }, { "input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14", "output": "back" }, { "input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6", "output": "back" }, { "input": "1\n19", "output": "chest" }, { "input": "2\n1 7", "output": "biceps" }, { "input": "3\n18 18 23", "output": "back" }, { "input": "4\n12 15 1 13", "output": "chest" }, { "input": "5\n11 14 25 21 21", "output": "biceps" }, { "input": "6\n11 9 12 11 22 18", "output": "biceps" }, { "input": "7\n11 1 16 20 21 25 20", "output": "chest" }, { "input": "8\n1 2 20 9 3 22 17 4", "output": "back" }, { "input": "9\n19 2 10 19 15 20 3 1 13", "output": "back" }, { "input": "10\n11 2 11 8 21 16 2 3 19 9", "output": "back" }, { "input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24", "output": "chest" }, { "input": "12\n4 24 21 3 13 24 22 13 12 21 1 15", "output": "back" }, { "input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3", "output": "biceps" }, { "input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11", "output": "biceps" }, { "input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17", "output": "back" }, { "input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17", "output": "chest" }, { "input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8", "output": "chest" }, { "input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22", "output": "chest" }, { "input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24", "output": "chest" }, { "input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23", "output": "biceps" }, { "input": "1\n19", "output": "chest" }, { "input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22", "output": "biceps" } ]
1,691,486,648
2,147,483,647
PyPy 3-64
OK
TESTS
61
124
0
n = int(input()) j = 0 a = [0] * 3 data = input().split() for el in data: a[j] += int(el) j += 1 if j == 3: j = 0 if a.index(max(a)) == 0: print('chest') elif a.index(max(a)) == 1: print('biceps') else: print('back')
Title: Greg's Workout Time Limit: None seconds Memory Limit: None megabytes Problem Description: Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. Output Specification: Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous. Demo Input: ['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n'] Demo Output: ['biceps\n', 'back\n', 'chest\n'] Note: In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
```python n = int(input()) j = 0 a = [0] * 3 data = input().split() for el in data: a[j] += int(el) j += 1 if j == 3: j = 0 if a.index(max(a)) == 0: print('chest') elif a.index(max(a)) == 1: print('biceps') else: print('back') ```
3
369
A
Valera and Plates
PROGRAMMING
900
[ "greedy", "implementation" ]
null
null
Valera is a lazy student. He has *m* clean bowls and *k* clean plates. Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates. When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.
The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish.
Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.
[ "3 1 1\n1 2 1\n", "4 3 1\n1 1 1 1\n", "3 1 2\n2 2 2\n", "8 2 2\n1 2 1 2 1 2 1 2\n" ]
[ "1\n", "1\n", "0\n", "4\n" ]
In the first sample Valera will wash a bowl only on the third day, so the answer is one. In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once. In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.
500
[ { "input": "3 1 1\n1 2 1", "output": "1" }, { "input": "4 3 1\n1 1 1 1", "output": "1" }, { "input": "3 1 2\n2 2 2", "output": "0" }, { "input": "8 2 2\n1 2 1 2 1 2 1 2", "output": "4" }, { "input": "2 100 100\n2 2", "output": "0" }, { "input": "1 1 1\n2", "output": "0" }, { "input": "233 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 1 2 1 2 2 2 2 2 2 2 2 1 1 2 1 2 1 2 2", "output": "132" }, { "input": "123 100 1\n2 2 2 1 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 1 2 2 2 1 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 2 2 1 1 1 1 2 1 2 2 1 2 2 2 1 1 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1 2 1 2 1 2 1 1 1", "output": "22" }, { "input": "188 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1", "output": "87" }, { "input": "3 1 2\n1 1 1", "output": "2" }, { "input": "3 2 2\n1 1 1", "output": "1" }, { "input": "3 2 1\n1 1 1", "output": "1" }, { "input": "3 1 1\n1 1 1", "output": "2" }, { "input": "5 1 2\n2 2 2 2 2", "output": "2" }, { "input": "5 2 2\n2 2 2 2 2", "output": "1" }, { "input": "5 2 1\n2 2 2 2 2", "output": "2" }, { "input": "5 1 1\n2 2 2 2 2", "output": "3" }, { "input": "1 1 2\n2", "output": "0" }, { "input": "1 2 2\n2", "output": "0" }, { "input": "1 2 1\n2", "output": "0" }, { "input": "1 1 1\n2", "output": "0" }, { "input": "6 3 1\n1 1 2 2 2 2", "output": "2" }, { "input": "100 40 20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "40" }, { "input": "7 5 2\n2 2 1 1 1 1 1", "output": "0" }, { "input": "10 4 4\n2 2 2 2 2 2 1 1 1 1", "output": "2" }, { "input": "3 2 1\n2 1 1", "output": "0" }, { "input": "7 6 1\n2 1 1 1 1 1 1", "output": "0" }, { "input": "7 5 1\n1 1 1 2 2 2 2", "output": "1" }, { "input": "5 3 1\n1 1 2 2 2", "output": "1" }, { "input": "3 1 1\n2 2 2", "output": "1" }, { "input": "5 2 2\n2 2 2 2 2", "output": "1" }, { "input": "3 1 3\n1 1 1", "output": "2" }, { "input": "5 2 1\n1 1 2 2 2", "output": "2" }, { "input": "4 3 2\n2 1 1 1", "output": "0" }, { "input": "4 2 1\n1 2 2 2", "output": "1" }, { "input": "14 4 7\n1 1 1 2 2 2 2 2 2 2 2 2 2 2", "output": "3" }, { "input": "12 10 4\n2 2 2 2 2 2 1 1 1 1 1 1", "output": "0" }, { "input": "5 3 2\n2 2 1 1 1", "output": "0" } ]
1,582,659,141
2,147,483,647
Python 3
OK
TESTS
63
109
0
n,m,k=map(int,input().split()) a=list(map(int,input().split())) s=0 for i in range(n): if m<1 and k<1: s+=1 elif a[i]==1: if m<1: s+=1 else: m-=1 else: if k>0: k-=1 elif m>0: m-=1 else: s+=1 print(s)
Title: Valera and Plates Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera is a lazy student. He has *m* clean bowls and *k* clean plates. Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates. When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally. Input Specification: The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish. Output Specification: Print a single integer — the minimum number of times Valera will need to wash a plate/bowl. Demo Input: ['3 1 1\n1 2 1\n', '4 3 1\n1 1 1 1\n', '3 1 2\n2 2 2\n', '8 2 2\n1 2 1 2 1 2 1 2\n'] Demo Output: ['1\n', '1\n', '0\n', '4\n'] Note: In the first sample Valera will wash a bowl only on the third day, so the answer is one. In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once. In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.
```python n,m,k=map(int,input().split()) a=list(map(int,input().split())) s=0 for i in range(n): if m<1 and k<1: s+=1 elif a[i]==1: if m<1: s+=1 else: m-=1 else: if k>0: k-=1 elif m>0: m-=1 else: s+=1 print(s) ```
3
546
A
Soldier and Bananas
PROGRAMMING
800
[ "brute force", "implementation", "math" ]
null
null
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana). He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
[ "3 17 4\n" ]
[ "13" ]
none
500
[ { "input": "3 17 4", "output": "13" }, { "input": "1 2 1", "output": "0" }, { "input": "1 1 1", "output": "0" }, { "input": "1 5 6", "output": "16" }, { "input": "1 1000000000 1", "output": "0" }, { "input": "1000 0 1000", "output": "500500000" }, { "input": "859 453892 543", "output": "126416972" }, { "input": "1000 1000000000 1000", "output": "0" }, { "input": "1000 500500000 1000", "output": "0" }, { "input": "1000 500500001 1000", "output": "0" }, { "input": "1000 500499999 1000", "output": "1" }, { "input": "634 87973 214", "output": "14497197" }, { "input": "432 10000 241", "output": "12587552" }, { "input": "111 111111111 111", "output": "0" }, { "input": "20 43 3", "output": "77" } ]
1,695,757,272
2,147,483,647
Python 3
OK
TESTS
15
46
5,632,000
a,b,c = list(map(int,input().split())) if 1<=a<=1000 and 1<=c<=1000 and 0<=b<=pow(10,9): sum = 0 for i in range(1, c + 1): sum += i tk = sum * a print(tk - b) if b < tk else print(0)
Title: Soldier and Bananas Time Limit: None seconds Memory Limit: None megabytes Problem Description: A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana). He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas? Input Specification: The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants. Output Specification: Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0. Demo Input: ['3 17 4\n'] Demo Output: ['13'] Note: none
```python a,b,c = list(map(int,input().split())) if 1<=a<=1000 and 1<=c<=1000 and 0<=b<=pow(10,9): sum = 0 for i in range(1, c + 1): sum += i tk = sum * a print(tk - b) if b < tk else print(0) ```
3
682
B
Alyona and Mex
PROGRAMMING
1,200
[ "sortings" ]
null
null
Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all. Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array. Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array. The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.
[ "5\n1 3 3 3 6\n", "2\n2 1\n" ]
[ "5\n", "3\n" ]
In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5. To reach the answer to the second sample case one must not decrease any of the array elements.
1,000
[ { "input": "5\n1 3 3 3 6", "output": "5" }, { "input": "2\n2 1", "output": "3" }, { "input": "1\n1", "output": "2" }, { "input": "1\n1000000000", "output": "2" }, { "input": "1\n2", "output": "2" }, { "input": "2\n1 1", "output": "2" }, { "input": "2\n1 3", "output": "3" }, { "input": "2\n2 2", "output": "3" }, { "input": "2\n2 3", "output": "3" }, { "input": "2\n3 3", "output": "3" }, { "input": "3\n1 1 1", "output": "2" }, { "input": "3\n2 1 1", "output": "3" }, { "input": "3\n3 1 1", "output": "3" }, { "input": "3\n1 1 4", "output": "3" }, { "input": "3\n2 1 2", "output": "3" }, { "input": "3\n3 2 1", "output": "4" }, { "input": "3\n2 4 1", "output": "4" }, { "input": "3\n3 3 1", "output": "4" }, { "input": "3\n1 3 4", "output": "4" }, { "input": "3\n4 1 4", "output": "4" }, { "input": "3\n2 2 2", "output": "3" }, { "input": "3\n3 2 2", "output": "4" }, { "input": "3\n4 2 2", "output": "4" }, { "input": "3\n2 3 3", "output": "4" }, { "input": "3\n4 2 3", "output": "4" }, { "input": "3\n4 4 2", "output": "4" }, { "input": "3\n3 3 3", "output": "4" }, { "input": "3\n4 3 3", "output": "4" }, { "input": "3\n4 3 4", "output": "4" }, { "input": "3\n4 4 4", "output": "4" }, { "input": "4\n1 1 1 1", "output": "2" }, { "input": "4\n1 1 2 1", "output": "3" }, { "input": "4\n1 1 3 1", "output": "3" }, { "input": "4\n1 4 1 1", "output": "3" }, { "input": "4\n1 2 1 2", "output": "3" }, { "input": "4\n1 3 2 1", "output": "4" }, { "input": "4\n2 1 4 1", "output": "4" }, { "input": "4\n3 3 1 1", "output": "4" }, { "input": "4\n1 3 4 1", "output": "4" }, { "input": "4\n1 1 4 4", "output": "4" }, { "input": "4\n2 2 2 1", "output": "3" }, { "input": "4\n1 2 2 3", "output": "4" }, { "input": "4\n2 4 1 2", "output": "4" }, { "input": "4\n3 3 1 2", "output": "4" }, { "input": "4\n2 3 4 1", "output": "5" }, { "input": "4\n1 4 2 4", "output": "5" }, { "input": "4\n3 1 3 3", "output": "4" }, { "input": "4\n3 4 3 1", "output": "5" }, { "input": "4\n1 4 4 3", "output": "5" }, { "input": "4\n4 1 4 4", "output": "5" }, { "input": "4\n2 2 2 2", "output": "3" }, { "input": "4\n2 2 3 2", "output": "4" }, { "input": "4\n2 2 2 4", "output": "4" }, { "input": "4\n2 2 3 3", "output": "4" }, { "input": "4\n2 2 3 4", "output": "5" }, { "input": "4\n2 4 4 2", "output": "5" }, { "input": "4\n2 3 3 3", "output": "4" }, { "input": "4\n2 4 3 3", "output": "5" }, { "input": "4\n4 4 2 3", "output": "5" }, { "input": "4\n4 4 4 2", "output": "5" }, { "input": "4\n3 3 3 3", "output": "4" }, { "input": "4\n3 3 3 4", "output": "5" }, { "input": "4\n4 3 3 4", "output": "5" }, { "input": "4\n4 4 3 4", "output": "5" }, { "input": "4\n4 4 4 4", "output": "5" }, { "input": "11\n1 1 1 1 1 1 1 1 1 3 3", "output": "4" }, { "input": "20\n1 1 1 1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8", "output": "9" }, { "input": "4\n2 2 2 3", "output": "4" }, { "input": "3\n1 1 2", "output": "3" }, { "input": "15\n1 2 2 20 23 25 28 60 66 71 76 77 79 99 100", "output": "15" }, { "input": "7\n1 2 2 2 5 5 1", "output": "5" }, { "input": "4\n1 1 1 2", "output": "3" }, { "input": "5\n1 1 1 1 10000", "output": "3" }, { "input": "5\n1 1 1 1 2", "output": "3" }, { "input": "7\n1 3 3 3 3 3 6", "output": "5" }, { "input": "4\n1 1 1 3", "output": "3" }, { "input": "10\n1 1 1 1 1 1 1 1 1 100", "output": "3" }, { "input": "4\n1 1 2 2", "output": "3" }, { "input": "5\n1 1 1 3 4", "output": "4" }, { "input": "8\n1 1 1 1 2 2 3 40", "output": "5" }, { "input": "5\n1 1 1 1 1", "output": "2" }, { "input": "7\n1 2 2 2 2 2 4", "output": "4" }, { "input": "10\n1 1 1 10000000 10000000 10000000 10000000 10000000 10000000 10000000", "output": "9" }, { "input": "10\n1 1 1 1 1 1 1 1 2 3", "output": "4" }, { "input": "4\n8 8 8 8", "output": "5" }, { "input": "5\n5 6 6 6 7", "output": "6" } ]
1,520,951,509
2,147,483,647
Python 3
OK
TESTS
127
171
14,438,400
n=int(input()) a=sorted(list(map(int,input().split())));m=1 for i in range(n): if a[i]>=m: a[i]=m m+=1 print(m)
Title: Alyona and Mex Time Limit: None seconds Memory Limit: None megabytes Problem Description: Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all. Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array. Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array. The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. Output Specification: Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations. Demo Input: ['5\n1 3 3 3 6\n', '2\n2 1\n'] Demo Output: ['5\n', '3\n'] Note: In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5. To reach the answer to the second sample case one must not decrease any of the array elements.
```python n=int(input()) a=sorted(list(map(int,input().split())));m=1 for i in range(n): if a[i]>=m: a[i]=m m+=1 print(m) ```
3
9
C
Hexadecimal's Numbers
PROGRAMMING
1,200
[ "brute force", "implementation", "math" ]
C. Hexadecimal's Numbers
1
64
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy. But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).
Output the only number — answer to the problem.
[ "10\n" ]
[ "2" ]
For *n* = 10 the answer includes numbers 1 and 10.
0
[ { "input": "10", "output": "2" }, { "input": "20", "output": "3" }, { "input": "72", "output": "3" }, { "input": "99", "output": "3" }, { "input": "100", "output": "4" }, { "input": "101", "output": "5" }, { "input": "102", "output": "5" }, { "input": "111", "output": "7" }, { "input": "112", "output": "7" }, { "input": "745", "output": "7" }, { "input": "23536", "output": "31" }, { "input": "1", "output": "1" }, { "input": "1010011", "output": "83" }, { "input": "312410141", "output": "511" }, { "input": "1000000000", "output": "512" }, { "input": "999999999", "output": "511" }, { "input": "111111111", "output": "511" }, { "input": "101010101", "output": "341" }, { "input": "121212121", "output": "511" }, { "input": "106341103", "output": "383" }, { "input": "901556123", "output": "511" }, { "input": "832513432", "output": "511" }, { "input": "3", "output": "1" }, { "input": "732875234", "output": "511" }, { "input": "7", "output": "1" }, { "input": "9", "output": "1" }, { "input": "2", "output": "1" }, { "input": "11", "output": "3" }, { "input": "12", "output": "3" }, { "input": "13", "output": "3" }, { "input": "101020101", "output": "351" }, { "input": "111100100", "output": "484" }, { "input": "110110101", "output": "437" }, { "input": "100111001", "output": "313" }, { "input": "100100", "output": "36" }, { "input": "110100102", "output": "421" } ]
1,595,098,502
2,147,483,647
PyPy 3
OK
TESTS
36
312
22,016,000
from itertools import product n = input() count = 0 for i in range(1, len(n)+1): for comb in product("10", repeat=i): a = "".join(comb) if a[0]=="0": continue a = int(a) if a <= int(n) and a > 0: #print(a) count += 1 print(count)
Title: Hexadecimal's Numbers Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy. But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully. Input Specification: Input data contains the only number *n* (1<=≤<=*n*<=≤<=109). Output Specification: Output the only number — answer to the problem. Demo Input: ['10\n'] Demo Output: ['2'] Note: For *n* = 10 the answer includes numbers 1 and 10.
```python from itertools import product n = input() count = 0 for i in range(1, len(n)+1): for comb in product("10", repeat=i): a = "".join(comb) if a[0]=="0": continue a = int(a) if a <= int(n) and a > 0: #print(a) count += 1 print(count) ```
3.679968
999
A
Mishka and Contest
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve?
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Print one integer — the maximum number of problems Mishka can solve.
[ "8 4\n4 2 3 1 5 1 6 4\n", "5 2\n3 1 2 1 3\n", "5 100\n12 34 55 43 21\n" ]
[ "5\n", "0\n", "5\n" ]
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
0
[ { "input": "8 4\n4 2 3 1 5 1 6 4", "output": "5" }, { "input": "5 2\n3 1 2 1 3", "output": "0" }, { "input": "5 100\n12 34 55 43 21", "output": "5" }, { "input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62", "output": "100" }, { "input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98", "output": "98" }, { "input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2", "output": "39" }, { "input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16", "output": "99" }, { "input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1", "output": "1" }, { "input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "1 5\n4", "output": "1" }, { "input": "3 2\n1 4 1", "output": "2" }, { "input": "1 2\n100", "output": "0" }, { "input": "7 4\n4 2 3 4 4 2 3", "output": "7" }, { "input": "1 2\n1", "output": "1" }, { "input": "1 2\n15", "output": "0" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "5 3\n3 4 3 2 1", "output": "4" }, { "input": "1 1\n2", "output": "0" }, { "input": "1 5\n1", "output": "1" }, { "input": "6 6\n7 1 1 1 1 1", "output": "5" }, { "input": "5 5\n6 5 5 5 5", "output": "4" }, { "input": "1 4\n2", "output": "1" }, { "input": "9 4\n1 2 1 2 4 2 1 2 1", "output": "9" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 10\n5", "output": "1" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4", "output": "71" }, { "input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90", "output": "63" }, { "input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8", "output": "61" }, { "input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22", "output": "72" }, { "input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2", "output": "52" }, { "input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90", "output": "60" }, { "input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10", "output": "56" }, { "input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 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48 30 13 48 48 48 17 9 48", "output": "99" }, { "input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57", "output": "99" }, { "input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93", "output": "99" }, { "input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33", "output": "3" }, { "input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51", "output": "3" }, { "input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49", "output": "6" }, { "input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8", "output": "66" }, { "input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95", "output": "0" }, { "input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66", "output": "0" }, { "input": "2 8\n8 8", "output": "2" }, { "input": "1 6\n3", "output": "1" }, { "input": "1 5\n5", "output": "1" } ]
1,626,773,176
2,147,483,647
PyPy 3
OK
TESTS
48
109
20,172,800
n,k=map(int,input().split()) l=list(map(int,input().split())) i=0 j=n-1 ans=0 while i<=j: if l[i]>k and l[j]>k: break if l[i]<=k: ans+=1 i+=1 if l[j]<=k and j>i: ans+=1 j-=1 print(ans)
Title: Mishka and Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve? Input Specification: The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list. Output Specification: Print one integer — the maximum number of problems Mishka can solve. Demo Input: ['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n'] Demo Output: ['5\n', '0\n', '5\n'] Note: In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
```python n,k=map(int,input().split()) l=list(map(int,input().split())) i=0 j=n-1 ans=0 while i<=j: if l[i]>k and l[j]>k: break if l[i]<=k: ans+=1 i+=1 if l[j]<=k and j>i: ans+=1 j-=1 print(ans) ```
3
609
A
USB Flash Drives
PROGRAMMING
800
[ "greedy", "implementation", "sortings" ]
null
null
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes. Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives. The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file. Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes. It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
[ "3\n5\n2\n1\n3\n", "3\n6\n2\n3\n2\n", "2\n5\n5\n10\n" ]
[ "2\n", "3\n", "1\n" ]
In the first example Sean needs only two USB flash drives — the first and the third. In the second example Sean needs all three USB flash drives. In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
0
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"100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460", "output": "78" }, { "input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2", "output": "74" }, { "input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433", "output": "42" }, { "input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63", "output": "2" }, { "input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440", "output": "1" }, { "input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8", "output": "3" }, { "input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68", "output": "7" }, { "input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116", "output": "6" }, { "input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36", "output": "30" }, { "input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718", "output": "52" }, { "input": "1\n1\n1", "output": "1" } ]
1,507,831,605
2,147,483,647
Python 3
OK
TESTS
34
62
0
# -*- coding: utf-8 -*- """ Created on Thu Oct 12 18:44:49 2017 @author: ASUS """ n = int(input ()) m = int (input ()) L = [] for i in range (n): L.append(int (input ())) L.sort () L = L[::-1] s = 0 t=0 for i in L : if i >= m : t = 1 break elif i < m : if s < m : t+=1 s+= i else : break print (t)
Title: USB Flash Drives Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes. Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives. Input Specification: The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives. The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file. Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes. It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*. Output Specification: Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives. Demo Input: ['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n'] Demo Output: ['2\n', '3\n', '1\n'] Note: In the first example Sean needs only two USB flash drives — the first and the third. In the second example Sean needs all three USB flash drives. In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
```python # -*- coding: utf-8 -*- """ Created on Thu Oct 12 18:44:49 2017 @author: ASUS """ n = int(input ()) m = int (input ()) L = [] for i in range (n): L.append(int (input ())) L.sort () L = L[::-1] s = 0 t=0 for i in L : if i >= m : t = 1 break elif i < m : if s < m : t+=1 s+= i else : break print (t) ```
3
427
A
Police Recruits
PROGRAMMING
800
[ "implementation" ]
null
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" } ]
1,692,425,898
2,147,483,647
Python 3
OK
TESTS
63
62
9,420,800
n = int(input("")) l=[int(x) for x in input().split()] crime = 0 police = 0 crime_unsolved = 0 for i in l: if i==-1: if police<=0: crime_unsolved+=1 else: police-=1 if i>=1: police+=i print(crime_unsolved)
Title: Police Recruits Time Limit: None seconds Memory Limit: None megabytes Problem Description: The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups. Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime. If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated. Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. Input Specification: The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers. If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. Output Specification: Print a single integer, the number of crimes which will go untreated. Demo Input: ['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n'] Demo Output: ['2\n', '1\n', '8\n'] Note: Lets consider the second example: 1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired. The answer is one, as one crime (on step 5) will go untreated.
```python n = int(input("")) l=[int(x) for x in input().split()] crime = 0 police = 0 crime_unsolved = 0 for i in l: if i==-1: if police<=0: crime_unsolved+=1 else: police-=1 if i>=1: police+=i print(crime_unsolved) ```
3
931
A
Friends Meeting
PROGRAMMING
800
[ "brute force", "greedy", "implementation", "math" ]
null
null
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*. Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6. The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend. The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend. It is guaranteed that *a*<=≠<=*b*.
Print the minimum possible total tiredness if the friends meet in the same point.
[ "3\n4\n", "101\n99\n", "5\n10\n" ]
[ "1\n", "2\n", "9\n" ]
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1. In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2. In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
500
[ { "input": "3\n4", "output": "1" }, { "input": "101\n99", "output": "2" }, { "input": "5\n10", "output": "9" }, { "input": "1\n2", "output": "1" }, { "input": "1\n1000", "output": "250000" }, { "input": "999\n1000", "output": "1" }, { "input": "1000\n999", "output": "1" }, { "input": "1000\n1", "output": "250000" }, { "input": "2\n1", "output": "1" }, { "input": "2\n999", "output": "249001" }, { "input": "2\n998", "output": "248502" }, { "input": "999\n2", "output": "249001" }, { "input": "998\n2", "output": "248502" }, { "input": "2\n1000", "output": "249500" }, { "input": "1000\n2", "output": "249500" }, { "input": "1\n999", "output": "249500" }, { "input": "999\n1", "output": "249500" }, { "input": "188\n762", "output": "82656" }, { "input": "596\n777", "output": "8281" }, { "input": "773\n70", "output": "123904" }, { "input": "825\n729", "output": "2352" }, { "input": "944\n348", "output": "89102" }, { "input": "352\n445", "output": "2209" }, { "input": "529\n656", "output": "4096" }, { "input": "19\n315", "output": "22052" }, { "input": "138\n370", "output": "13572" }, { "input": "546\n593", "output": "576" }, { "input": "285\n242", "output": "484" }, { "input": "773\n901", "output": "4160" }, { "input": "892\n520", "output": "34782" }, { "input": "864\n179", "output": "117649" }, { "input": "479\n470", "output": "25" }, { "input": "967\n487", "output": "57840" }, { "input": "648\n106", "output": "73712" }, { "input": "58\n765", "output": "125316" }, { "input": "235\n56", "output": "8100" }, { "input": "285\n153", "output": "4422" }, { "input": "943\n13", "output": "216690" }, { "input": "675\n541", "output": "4556" }, { "input": "4\n912", "output": "206570" } ]
1,520,183,148
5,448
Python 3
OK
TESTS
40
77
5,632,000
def suma(q): sump = 0 for i in range(1,q+1): sump = sump + i return(sump) a = int(input()) b = int(input()) c = (a+b)//2 c1 = abs(c - a) c2 = abs(c - b) print(suma(c1)+suma(c2))
Title: Friends Meeting Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*. Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6. The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point. Input Specification: The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend. The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend. It is guaranteed that *a*<=≠<=*b*. Output Specification: Print the minimum possible total tiredness if the friends meet in the same point. Demo Input: ['3\n4\n', '101\n99\n', '5\n10\n'] Demo Output: ['1\n', '2\n', '9\n'] Note: In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1. In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2. In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
```python def suma(q): sump = 0 for i in range(1,q+1): sump = sump + i return(sump) a = int(input()) b = int(input()) c = (a+b)//2 c1 = abs(c - a) c2 = abs(c - b) print(suma(c1)+suma(c2)) ```
3
456
A
Laptops
PROGRAMMING
1,100
[ "sortings" ]
null
null
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop. Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops. Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality). All *a**i* are distinct. All *b**i* are distinct.
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
[ "2\n1 2\n2 1\n" ]
[ "Happy Alex\n" ]
none
500
[ { "input": "2\n1 2\n2 1", "output": "Happy Alex" }, { "input": "2\n1 1\n2 2", "output": "Poor Alex" }, { "input": "3\n2 2\n3 3\n1 1", "output": "Poor Alex" }, { "input": "3\n3 3\n1 2\n2 1", "output": "Happy Alex" }, { "input": "1\n1 1", "output": "Poor Alex" }, { "input": "3\n2 3\n1 1\n3 2", "output": "Happy Alex" }, { "input": "4\n4 1\n3 2\n2 3\n1 4", "output": "Happy Alex" }, { "input": "2\n2 1\n1 2", "output": "Happy Alex" }, { "input": "3\n3 2\n1 1\n2 3", "output": "Happy Alex" }, { "input": "3\n1 2\n2 3\n3 1", "output": "Happy Alex" }, { "input": "3\n1 2\n2 1\n3 3", "output": "Happy Alex" }, { "input": "3\n1 2\n2 3\n3 1", "output": "Happy Alex" } ]
1,673,354,589
2,147,483,647
Python 3
OK
TESTS
46
358
9,728,000
q = [[int(x) for x in input().split()] for _ in range(int(input()))] q.sort(key=lambda x: x[1], reverse=True) found = False for i in range(len(q)-1): if ( q[i][1] > q[i+1][1] and q[i][0] < q[i+1][0] ): found = True break print(["Poor Alex", "Happy Alex"][found])
Title: Laptops Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop. Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops. Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality). All *a**i* are distinct. All *b**i* are distinct. Output Specification: If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). Demo Input: ['2\n1 2\n2 1\n'] Demo Output: ['Happy Alex\n'] Note: none
```python q = [[int(x) for x in input().split()] for _ in range(int(input()))] q.sort(key=lambda x: x[1], reverse=True) found = False for i in range(len(q)-1): if ( q[i][1] > q[i+1][1] and q[i][0] < q[i+1][0] ): found = True break print(["Poor Alex", "Happy Alex"][found]) ```
3
821
A
Okabe and Future Gadget Laboratory
PROGRAMMING
800
[ "implementation" ]
null
null
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column. Help Okabe determine whether a given lab is good!
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab. The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105).
Print "Yes" if the given lab is good and "No" otherwise. You can output each letter in upper or lower case.
[ "3\n1 1 2\n2 3 1\n6 4 1\n", "3\n1 5 2\n1 1 1\n1 2 3\n" ]
[ "Yes\n", "No\n" ]
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes". In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
500
[ { "input": "3\n1 1 2\n2 3 1\n6 4 1", "output": "Yes" }, { "input": "3\n1 5 2\n1 1 1\n1 2 3", "output": "No" }, { "input": "1\n1", "output": "Yes" }, { "input": "4\n1 1 1 1\n1 11 1 2\n2 5 1 4\n3 9 4 1", "output": "Yes" }, { "input": "4\n1 1 1 1\n1 7 1 1\n1 3 1 2\n2 6 3 1", "output": "Yes" }, { "input": "4\n1 1 1 1\n1 12 1 2\n4 4 1 3\n5 10 6 1", "output": "Yes" }, { "input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n5 11 6 1", "output": "Yes" }, { "input": "4\n1 1 1 1\n1 13 1 2\n4 5 1 3\n7 11 6 1", "output": "No" }, { "input": "4\n1 1 1 1\n1 14 1 6\n4 5 1 5\n3 8 2 1", "output": "Yes" }, { "input": "10\n3 2 6 5 1 2 1 1 2 1\n1 2 2 3 1 4 1 1 1 1\n1 1 3 1 2 2 2 1 1 1\n1 3 1 2 1 2 1 1 1 1\n12 1 4 3 3 1 1 1 2 1\n8 1 1 1 1 3 6 5 1 3\n1 1 1 2 4 1 3 2 1 1\n1 1 1 1 1 1 1 2 1 1\n1 2 1 1 1 1 1 1 1 1\n2 1 1 1 1 1 1 1 1 2", "output": "Yes" }, { "input": "4\n1 2 3 5\n1 1 1 1\n1 1 1 1\n1 1 1 1", "output": "No" }, { "input": "1\n2", "output": "No" }, { "input": "3\n1 1 4\n1 1 2\n1 1 2", "output": "No" }, { "input": "1\n5", "output": "No" }, { "input": "3\n1 2 1\n2 2 3\n1 3 1", "output": "No" }, { "input": "3\n2 1 1\n2 1 1\n2 1 1", "output": "No" }, { "input": "1\n60000", "output": "No" } ]
1,616,882,841
2,147,483,647
PyPy 3
OK
TESTS
31
140
2,252,800
cases = int(input()) n = cases matrix = [] while cases: cases -= 1 row = list(map(int, input().split())) matrix.append(row) for i in range(n): for j in range(n): if matrix[i][j] == 1: continue broken = False for k in range(n): for m in range(n): if matrix[i][k] + matrix[m][j] == matrix[i][j]: broken = True if not broken: print("No") exit() print("Yes")
Title: Okabe and Future Gadget Laboratory Time Limit: None seconds Memory Limit: None megabytes Problem Description: Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an *n* by *n* square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every *x*,<=*y* such that 1<=≤<=*x*,<=*y*<=≤<=*n* and *a**x*,<=*y*<=≠<=1, there should exist two indices *s* and *t* so that *a**x*,<=*y*<==<=*a**x*,<=*s*<=+<=*a**t*,<=*y*, where *a**i*,<=*j* denotes the integer in *i*-th row and *j*-th column. Help Okabe determine whether a given lab is good! Input Specification: The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=50) — the size of the lab. The next *n* lines contain *n* space-separated integers denoting a row of the grid. The *j*-th integer in the *i*-th row is *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=105). Output Specification: Print "Yes" if the given lab is good and "No" otherwise. You can output each letter in upper or lower case. Demo Input: ['3\n1 1 2\n2 3 1\n6 4 1\n', '3\n1 5 2\n1 1 1\n1 2 3\n'] Demo Output: ['Yes\n', 'No\n'] Note: In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes". In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
```python cases = int(input()) n = cases matrix = [] while cases: cases -= 1 row = list(map(int, input().split())) matrix.append(row) for i in range(n): for j in range(n): if matrix[i][j] == 1: continue broken = False for k in range(n): for m in range(n): if matrix[i][k] + matrix[m][j] == matrix[i][j]: broken = True if not broken: print("No") exit() print("Yes") ```
3
50
A
Domino piling
PROGRAMMING
800
[ "greedy", "math" ]
A. Domino piling
2
256
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output one number — the maximal number of dominoes, which can be placed.
[ "2 4\n", "3 3\n" ]
[ "4\n", "4\n" ]
none
500
[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]
1,662,742,934
2,147,483,647
Python 3
OK
TESTS
35
92
0
num = input().split() m = num[0] n = num[1] sum = int(n)*int(m) / 2 print(int(sum))
Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none
```python num = input().split() m = num[0] n = num[1] sum = int(n)*int(m) / 2 print(int(sum)) ```
3.977
660
A
Co-prime Array
PROGRAMMING
1,200
[ "greedy", "implementation", "math", "number theory" ]
null
null
You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible. In each move you can insert any positive integral number you want not greater than 109 in any place in the array. An array is co-prime if any two adjacent numbers of it are co-prime. In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1.
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime. The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it. If there are multiple answers you can print any one of them.
[ "3\n2 7 28\n" ]
[ "1\n2 7 9 28\n" ]
none
0
[ { "input": "3\n2 7 28", "output": "1\n2 7 1 28" }, { "input": "1\n1", "output": "0\n1" }, { "input": "1\n548", "output": "0\n548" }, { "input": "1\n963837006", "output": "0\n963837006" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "0\n1 1 1 1 1 1 1 1 1 1" }, { "input": "10\n26 723 970 13 422 968 875 329 234 983", "output": "2\n26 723 970 13 422 1 968 875 1 329 234 983" }, { "input": "10\n319645572 758298525 812547177 459359946 355467212 304450522 807957797 916787906 239781206 242840396", "output": "7\n319645572 1 758298525 1 812547177 1 459359946 1 355467212 1 304450522 807957797 916787906 1 239781206 1 242840396" }, { "input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1", "output": "19\n1 1 1 1 2 1 1 1 1 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1" }, { "input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917", "output": "38\n591 1 417 1 888 251 792 1 847 685 3 182 461 102 1 348 1 555 956 771 901 712 1 878 1 580 631 342 1 333 1 285 899 525 1 725 537 718 929 653 84 1 788 1 104 355 624 803 1 253 853 201 995 536 1 184 65 1 205 1 540 1 652 549 1 777 248 405 677 950 431 580 1 600 1 846 1 328 429 134 983 526 103 500 963 400 23 1 276 1 704 1 570 757 410 1 658 507 620 1 984 1 244 1 486 1 454 1 802 411 985 303 635 283 96 1 597 1 855 1 775 139 839 1 839 61 219 986 1 776 1 72 1 729 1 69 20 917" }, { "input": "5\n472882027 472882027 472882027 472882027 472882027", "output": "4\n472882027 1 472882027 1 472882027 1 472882027 1 472882027" }, { "input": "2\n1000000000 1000000000", "output": "1\n1000000000 1 1000000000" }, { "input": "2\n8 6", "output": "1\n8 1 6" }, { "input": "3\n100000000 1000000000 1000000000", "output": "2\n100000000 1 1000000000 1 1000000000" }, { "input": "5\n1 2 3 4 5", "output": "0\n1 2 3 4 5" }, { "input": "20\n2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000", "output": "19\n2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000" }, { "input": "2\n223092870 23", "output": "1\n223092870 1 23" }, { "input": "2\n100000003 100000003", "output": "1\n100000003 1 100000003" }, { "input": "2\n999999937 999999937", "output": "1\n999999937 1 999999937" }, { "input": "4\n999 999999937 999999937 999", "output": "1\n999 999999937 1 999999937 999" }, { "input": "2\n999999929 999999929", "output": "1\n999999929 1 999999929" }, { "input": "2\n1049459 2098918", "output": "1\n1049459 1 2098918" }, { "input": "2\n352229 704458", "output": "1\n352229 1 704458" }, { "input": "2\n7293 4011", "output": "1\n7293 1 4011" }, { "input": "2\n5565651 3999930", "output": "1\n5565651 1 3999930" }, { "input": "2\n997 997", "output": "1\n997 1 997" }, { "input": "3\n9994223 9994223 9994223", "output": "2\n9994223 1 9994223 1 9994223" }, { "input": "2\n99999998 1000000000", "output": "1\n99999998 1 1000000000" }, { "input": "3\n1000000000 1000000000 1000000000", "output": "2\n1000000000 1 1000000000 1 1000000000" }, { "input": "2\n130471 130471", "output": "1\n130471 1 130471" }, { "input": "3\n1000000000 2 2", "output": "2\n1000000000 1 2 1 2" }, { "input": "2\n223092870 66526", "output": "1\n223092870 1 66526" }, { "input": "14\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491 436077930 570018449", "output": "10\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491 436077930 1 570018449" }, { "input": "2\n3996017 3996017", "output": "1\n3996017 1 3996017" }, { "input": "2\n999983 999983", "output": "1\n999983 1 999983" }, { "input": "2\n618575685 773990454", "output": "1\n618575685 1 773990454" }, { "input": "3\n9699690 3 7", "output": "1\n9699690 1 3 7" }, { "input": "2\n999999999 999999996", "output": "1\n999999999 1 999999996" }, { "input": "2\n99999910 99999910", "output": "1\n99999910 1 99999910" }, { "input": "12\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491", "output": "9\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491" }, { "input": "3\n999999937 999999937 999999937", "output": "2\n999999937 1 999999937 1 999999937" }, { "input": "2\n99839 99839", "output": "1\n99839 1 99839" }, { "input": "3\n19999909 19999909 19999909", "output": "2\n19999909 1 19999909 1 19999909" }, { "input": "4\n1 1000000000 1 1000000000", "output": "0\n1 1000000000 1 1000000000" }, { "input": "2\n64006 64006", "output": "1\n64006 1 64006" }, { "input": "2\n1956955 1956955", "output": "1\n1956955 1 1956955" }, { "input": "3\n1 1000000000 1000000000", "output": "1\n1 1000000000 1 1000000000" }, { "input": "2\n982451707 982451707", "output": "1\n982451707 1 982451707" }, { "input": "2\n999999733 999999733", "output": "1\n999999733 1 999999733" }, { "input": "3\n999999733 999999733 999999733", "output": "2\n999999733 1 999999733 1 999999733" }, { "input": "2\n3257 3257", "output": "1\n3257 1 3257" }, { "input": "2\n223092870 181598", "output": "1\n223092870 1 181598" }, { "input": "3\n959919409 105935 105935", "output": "2\n959919409 1 105935 1 105935" }, { "input": "2\n510510 510510", "output": "1\n510510 1 510510" }, { "input": "3\n223092870 1000000000 1000000000", "output": "2\n223092870 1 1000000000 1 1000000000" }, { "input": "14\n1000000000 2 1000000000 3 1000000000 6 1000000000 1000000000 15 1000000000 1000000000 1000000000 100000000 1000", "output": "11\n1000000000 1 2 1 1000000000 3 1000000000 1 6 1 1000000000 1 1000000000 1 15 1 1000000000 1 1000000000 1 1000000000 1 100000000 1 1000" }, { "input": "7\n1 982451653 982451653 1 982451653 982451653 982451653", "output": "3\n1 982451653 1 982451653 1 982451653 1 982451653 1 982451653" }, { "input": "2\n100000007 100000007", "output": "1\n100000007 1 100000007" }, { "input": "3\n999999757 999999757 999999757", "output": "2\n999999757 1 999999757 1 999999757" }, { "input": "3\n99999989 99999989 99999989", "output": "2\n99999989 1 99999989 1 99999989" }, { "input": "5\n2 4 982451707 982451707 3", "output": "2\n2 1 4 982451707 1 982451707 3" }, { "input": "2\n20000014 20000014", "output": "1\n20000014 1 20000014" }, { "input": "2\n99999989 99999989", "output": "1\n99999989 1 99999989" }, { "input": "2\n111546435 111546435", "output": "1\n111546435 1 111546435" }, { "input": "2\n55288874 33538046", "output": "1\n55288874 1 33538046" }, { "input": "5\n179424673 179424673 179424673 179424673 179424673", "output": "4\n179424673 1 179424673 1 179424673 1 179424673 1 179424673" }, { "input": "2\n199999978 199999978", "output": "1\n199999978 1 199999978" }, { "input": "2\n1000000000 2", "output": "1\n1000000000 1 2" }, { "input": "3\n19999897 19999897 19999897", "output": "2\n19999897 1 19999897 1 19999897" }, { "input": "2\n19999982 19999982", "output": "1\n19999982 1 19999982" }, { "input": "2\n10000007 10000007", "output": "1\n10000007 1 10000007" }, { "input": "3\n999999937 999999937 2", "output": "1\n999999937 1 999999937 2" }, { "input": "5\n2017 2017 2017 2017 2017", "output": "4\n2017 1 2017 1 2017 1 2017 1 2017" }, { "input": "2\n19999909 39999818", "output": "1\n19999909 1 39999818" }, { "input": "2\n62615533 7919", "output": "1\n62615533 1 7919" }, { "input": "5\n39989 39989 33 31 29", "output": "1\n39989 1 39989 33 31 29" }, { "input": "2\n1000000000 100000", "output": "1\n1000000000 1 100000" }, { "input": "2\n1938 10010", "output": "1\n1938 1 10010" }, { "input": "2\n199999 199999", "output": "1\n199999 1 199999" }, { "input": "2\n107273 107273", "output": "1\n107273 1 107273" }, { "input": "3\n49999 49999 49999", "output": "2\n49999 1 49999 1 49999" }, { "input": "2\n1999966 1999958", "output": "1\n1999966 1 1999958" }, { "input": "2\n86020 300846", "output": "1\n86020 1 300846" }, { "input": "2\n999999997 213", "output": "1\n999999997 1 213" }, { "input": "2\n200000014 200000434", "output": "1\n200000014 1 200000434" } ]
1,629,872,043
243
PyPy 3
OK
TESTS
93
124
22,528,000
import math import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) s = set() for i in range(n - 1): if math.gcd(a[i], a[i + 1]) ^ 1: s.add(i) k = len(s) ans = [] for i in range(n): ans.append(a[i]) if i in s: ans.append(1) print(k) print(*ans)
Title: Co-prime Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible. In each move you can insert any positive integral number you want not greater than 109 in any place in the array. An array is co-prime if any two adjacent numbers of it are co-prime. In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. Output Specification: Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime. The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it. If there are multiple answers you can print any one of them. Demo Input: ['3\n2 7 28\n'] Demo Output: ['1\n2 7 9 28\n'] Note: none
```python import math import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) s = set() for i in range(n - 1): if math.gcd(a[i], a[i + 1]) ^ 1: s.add(i) k = len(s) ans = [] for i in range(n): ans.append(a[i]) if i in s: ans.append(1) print(k) print(*ans) ```
3
801
B
Valued Keys
PROGRAMMING
900
[ "constructive algorithms", "greedy", "strings" ]
null
null
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length. The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2. For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel". You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
The first line of input contains the string *x*. The second line of input contains the string *y*. Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1. Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
[ "ab\naa\n", "nzwzl\nniwel\n", "ab\nba\n" ]
[ "ba\n", "xiyez\n", "-1\n" ]
The first case is from the statement. Another solution for the second case is "zizez" There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) =  "ba".
1,000
[ { "input": "ab\naa", "output": "ba" }, { "input": "nzwzl\nniwel", "output": "xiyez" }, { "input": "ab\nba", "output": "-1" }, { "input": "r\nl", "output": "l" }, { "input": "d\ny", "output": "-1" }, { "input": "yvowz\ncajav", "output": "cajav" }, { "input": "lwzjp\ninjit", "output": "-1" }, { "input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm", "output": "eodnlemiicdedmkcgavqbnqmm" }, { "input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs", "output": "-1" }, { "input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf", "output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf" }, { "input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen", "output": "-1" }, { "input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb", "output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb" }, { "input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym", "output": "-1" }, { "input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib", "output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib" }, { "input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew", "output": "-1" }, { "input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv", "output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv" }, { "input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns", "output": "-1" }, { "input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf", "output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf" }, { "input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp", "output": "-1" }, { "input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp", "output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp" }, { "input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej", "output": "-1" }, { "input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe", "output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe" }, { "input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang", "output": "-1" }, { "input": "ftfr\nftfr", "output": "ftfr" }, { "input": "ftr\nftr", "output": "ftr" }, { "input": "shftr\nshftr", "output": "shftr" }, { "input": "vkvkkv\nvkvkkv", "output": "vkvkkv" }, { "input": "ftrd\nftrd", "output": "ftrd" }, { "input": "fztr\nfztr", "output": "fztr" }, { "input": "frtr\nfrtr", "output": "frtr" }, { "input": "shdftr\nshdftr", "output": "shdftr" }, { "input": "zzz\nzzz", "output": "zzz" }, { "input": "shtr\nshtr", "output": "shtr" }, { "input": "aaaaa\nzzzzz", "output": "-1" }, { "input": "efr\nefr", "output": "efr" } ]
1,594,895,567
2,147,483,647
PyPy 3
OK
TESTS
35
155
20,172,800
x,y=input(),input() print('-1' if any(y[i]>x[i] for i in range(len(x))) else y)
Title: Valued Keys Time Limit: None seconds Memory Limit: None megabytes Problem Description: You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length. The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2. For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel". You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists. Input Specification: The first line of input contains the string *x*. The second line of input contains the string *y*. Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100. Output Specification: If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1. Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters. Demo Input: ['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n'] Demo Output: ['ba\n', 'xiyez\n', '-1\n'] Note: The first case is from the statement. Another solution for the second case is "zizez" There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) =  "ba".
```python x,y=input(),input() print('-1' if any(y[i]>x[i] for i in range(len(x))) else y) ```
3
515
C
Drazil and Factorial
PROGRAMMING
1,400
[ "greedy", "math", "sortings" ]
null
null
Drazil is playing a math game with Varda. Let's define for positive integer *x* as a product of factorials of its digits. For example, . First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions: 1. *x* doesn't contain neither digit 0 nor digit 1. 2. = . Help friends find such number.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*. The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
[ "4\n1234\n", "3\n555\n" ]
[ "33222\n", "555\n" ]
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>
1,000
[ { "input": "4\n1234", "output": "33222" }, { "input": "3\n555", "output": "555" }, { "input": "15\n012345781234578", "output": "7777553333222222222222" }, { "input": "1\n8", "output": "7222" }, { "input": "10\n1413472614", "output": "75333332222222" }, { "input": "8\n68931246", "output": "77553333332222222" }, { "input": "7\n4424368", "output": "75333332222222222" }, { "input": "6\n576825", "output": "7755532222" }, { "input": "5\n97715", "output": "7775332" }, { "input": "3\n915", "output": "75332" }, { "input": "2\n26", "output": "532" }, { "input": "1\n4", "output": "322" }, { "input": "15\n028745260720699", "output": "7777755533333332222222222" }, { "input": "13\n5761790121605", "output": "7775555333322" }, { "input": "10\n3312667105", "output": "755533332" }, { "input": "1\n7", "output": "7" }, { "input": "15\n989898989898989", "output": "777777777777777333333333333333322222222222222222222222222222" }, { "input": "15\n000000000000007", "output": "7" }, { "input": "15\n999999999999990", "output": "77777777777777333333333333333333333333333322222222222222" }, { "input": "1\n2", "output": "2" }, { "input": "1\n3", "output": "3" }, { "input": "1\n4", "output": "322" }, { "input": "1\n5", "output": "5" }, { "input": "1\n6", "output": "53" }, { "input": "1\n7", "output": "7" }, { "input": "1\n9", "output": "7332" }, { "input": "2\n09", "output": "7332" }, { "input": "13\n1337251172966", "output": "777555333333222" }, { "input": "15\n987654329876543", "output": "777777555533333333332222222222222" }, { "input": "9\n234567899", "output": "777755333333322222222" }, { "input": "2\n99", "output": "77333322" }, { "input": "2\n66", "output": "5533" }, { "input": "3\n999", "output": "777333333222" }, { "input": "5\n99999", "output": "77777333333333322222" }, { "input": "9\n123456789", "output": "77755333332222222" }, { "input": "9\n987654321", "output": "77755333332222222" }, { "input": "3\n666", "output": "555333" }, { "input": "2\n26", "output": "532" }, { "input": "6\n555777", "output": "777555" }, { "input": "10\n1234567899", "output": "777755333333322222222" }, { "input": "4\n6666", "output": "55553333" }, { "input": "4\n9754", "output": "775333222" }, { "input": "2\n95", "output": "75332" }, { "input": "14\n11122233344455", "output": "55333333222222222" }, { "input": "12\n836544897832", "output": "77777553333333222222222222222" } ]
1,590,175,707
2,147,483,647
PyPy 3
OK
TESTS
45
155
0
def f_app(arr, number): if number == '2': arr.append('2') if number == '3': arr.append('3') if number == '4': arr.extend(['3','2','2']) if number == '5': arr.append('5') if number == '6': arr.extend(['5','3']) if number == '7': arr.append('7') if number == '8': arr.extend(['7','2','2','2']) if number == '9': arr.extend(['7','3','3','2']) return arr def solve(s): arr = [] for letter in s: arr = f_app(arr, letter) return ''.join(sorted(arr, reverse=True)) n = int(input()) s = input() print(solve(s))
Title: Drazil and Factorial Time Limit: None seconds Memory Limit: None megabytes Problem Description: Drazil is playing a math game with Varda. Let's define for positive integer *x* as a product of factorials of its digits. For example, . First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions: 1. *x* doesn't contain neither digit 0 nor digit 1. 2. = . Help friends find such number. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*. The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. Output Specification: Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Demo Input: ['4\n1234\n', '3\n555\n'] Demo Output: ['33222\n', '555\n'] Note: In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>
```python def f_app(arr, number): if number == '2': arr.append('2') if number == '3': arr.append('3') if number == '4': arr.extend(['3','2','2']) if number == '5': arr.append('5') if number == '6': arr.extend(['5','3']) if number == '7': arr.append('7') if number == '8': arr.extend(['7','2','2','2']) if number == '9': arr.extend(['7','3','3','2']) return arr def solve(s): arr = [] for letter in s: arr = f_app(arr, letter) return ''.join(sorted(arr, reverse=True)) n = int(input()) s = input() print(solve(s)) ```
3
38
B
Chess
PROGRAMMING
1,200
[ "brute force", "implementation", "math" ]
B. Chess
2
256
Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one. Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Print a single number which is the required number of ways.
[ "a1\nb2\n", "a8\nd4\n" ]
[ "44\n", "38\n" ]
none
0
[ { "input": "a1\nb2", "output": "44" }, { "input": "a8\nd4", "output": "38" }, { "input": "a8\nf1", "output": "42" }, { "input": "f8\nh3", "output": "42" }, { "input": "g8\nb7", "output": "42" }, { "input": "h1\ng5", "output": "42" }, { "input": "c6\nb5", "output": "39" }, { "input": "c1\nd2", "output": "42" }, { "input": "g3\nh4", "output": "42" }, { "input": "e3\ng5", "output": "38" }, { "input": "f8\na3", "output": "40" }, { "input": "a2\nh8", "output": "43" }, { "input": "a3\nc5", "output": "40" }, { "input": "g1\ne6", "output": "39" }, { "input": "e1\na7", "output": "41" }, { "input": "b5\nc1", "output": "39" }, { "input": "b2\ne1", "output": "43" }, { "input": "h8\ng2", "output": "43" }, { "input": "a3\nd6", "output": "38" }, { "input": "g6\nb7", "output": "39" }, { "input": "c8\ne6", "output": "40" }, { "input": "e6\nf2", "output": "35" }, { "input": "b6\nd8", "output": "41" }, { "input": "a4\nd1", "output": "42" }, { "input": "b5\nh8", "output": "40" }, { "input": "h6\na1", "output": "42" }, { "input": "c3\na8", "output": "39" }, { "input": "g5\nd2", "output": "38" }, { "input": "b6\ng7", "output": "39" }, { "input": "h6\na8", "output": "43" }, { "input": "a8\nb7", "output": "44" }, { "input": "c8\nb2", "output": "41" }, { "input": "e4\nc1", "output": "37" }, { "input": "f1\nc3", "output": "38" }, { "input": "a3\nc8", "output": "41" }, { "input": "e8\nb6", "output": "40" }, { "input": "a1\nb7", "output": "43" }, { "input": "g2\nb7", "output": "40" }, { "input": "e1\nd6", "output": "38" }, { "input": "e5\nh6", "output": "39" } ]
1,550,574,200
2,147,483,647
Python 3
OK
TESTS
40
218
307,200
rock=input() knight=input() busyDots=set() busyDots.add(knight) rockColumn=rock[0] rockRow=int(rock[1]) for code in range(ord('a'), ord('i')): busyDots.add(str(chr(code))+str(rockRow)) for i in range(1,9): busyDots.add(str(rockColumn)+str(i)) knightColumn=knight[0] knightRow=int(knight[1]) #up up left if(knightRow+2<=8 and chr(ord(knightColumn)-1)>='a'): busyDots.add(str(chr(ord(knightColumn)-1))+str((knightRow+2))) #up left left if(knightRow+1<=8 and chr(ord(knightColumn)-2)>='a'): busyDots.add( str(chr(ord(knightColumn)-2)) + str((knightRow+1)) ) #up up right if(knightRow+2<=8 and chr(ord(knightColumn)+1)<='h'): busyDots.add( str(chr(ord(knightColumn)+1)) + str((knightRow+2)) ) #up right right if(knightRow+1<=8 and chr(ord(knightColumn)+2)<='h'): busyDots.add( str(chr(ord(knightColumn)+2)) + str((knightRow+1)) ) #down down left if(knightRow-2>=1 and chr(ord(knightColumn)-1)>='a'): busyDots.add( str(chr(ord(knightColumn)-1)) + str((knightRow-2)) ) #down left left if(knightRow-1>=1 and chr(ord(knightColumn)-2)>='a'): busyDots.add( str(chr(ord(knightColumn)-2)) + str((knightRow-1)) ) #down down right if(knightRow-2>=1 and chr(ord(knightColumn)+1)<='h'): busyDots.add( str(chr(ord(knightColumn)+1)) + str((knightRow-2)) ) #down right right if(knightRow-1>=1 and chr(ord(knightColumn)+2)<='h'): busyDots.add( str(chr(ord(knightColumn)+2)) + str((knightRow-1)) ) #rock research #up up left if(rockRow+2<=8 and chr(ord(rockColumn)-1)>='a'): busyDots.add (str(chr(ord(rockColumn)-1))+str((rockRow+2))) #up left left if(rockRow+1<=8 and chr(ord(rockColumn)-2)>='a'): busyDots.add( str(chr(ord(rockColumn)-2)) + str((rockRow+1)) ) #up up right if(rockRow+2<=8 and chr(ord(rockColumn)+1)<='h'): busyDots.add( str(chr(ord(rockColumn)+1)) + str((rockRow+2)) ) #up right right if(rockRow+1<=8 and chr(ord(rockColumn)+2)<='h'): busyDots.add( str(chr(ord(rockColumn)+2)) + str((rockRow+1)) ) #down down left if(rockRow-2>=1 and chr(ord(rockColumn)-1)>='a'): busyDots.add( str(chr(ord(rockColumn)-1)) + str((rockRow-2)) ) #down left left if(rockRow-1>=1 and chr(ord(rockColumn)-2)>='a'): busyDots.add( str(chr(ord(rockColumn)-2)) + str((rockRow-1)) ) #down down right if(rockRow-2>=1 and chr(ord(rockColumn)+1)<='h'): busyDots.add( str(chr(ord(rockColumn)+1)) + str((rockRow-2)) ) #down right right if(rockRow-1>=1 and chr(ord(rockColumn)+2)<='h'): busyDots.add( str(chr(ord(rockColumn)+2)) + str((rockRow-1)) ) print(64-len(busyDots))
Title: Chess Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one. Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square. Input Specification: The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide. Output Specification: Print a single number which is the required number of ways. Demo Input: ['a1\nb2\n', 'a8\nd4\n'] Demo Output: ['44\n', '38\n'] Note: none
```python rock=input() knight=input() busyDots=set() busyDots.add(knight) rockColumn=rock[0] rockRow=int(rock[1]) for code in range(ord('a'), ord('i')): busyDots.add(str(chr(code))+str(rockRow)) for i in range(1,9): busyDots.add(str(rockColumn)+str(i)) knightColumn=knight[0] knightRow=int(knight[1]) #up up left if(knightRow+2<=8 and chr(ord(knightColumn)-1)>='a'): busyDots.add(str(chr(ord(knightColumn)-1))+str((knightRow+2))) #up left left if(knightRow+1<=8 and chr(ord(knightColumn)-2)>='a'): busyDots.add( str(chr(ord(knightColumn)-2)) + str((knightRow+1)) ) #up up right if(knightRow+2<=8 and chr(ord(knightColumn)+1)<='h'): busyDots.add( str(chr(ord(knightColumn)+1)) + str((knightRow+2)) ) #up right right if(knightRow+1<=8 and chr(ord(knightColumn)+2)<='h'): busyDots.add( str(chr(ord(knightColumn)+2)) + str((knightRow+1)) ) #down down left if(knightRow-2>=1 and chr(ord(knightColumn)-1)>='a'): busyDots.add( str(chr(ord(knightColumn)-1)) + str((knightRow-2)) ) #down left left if(knightRow-1>=1 and chr(ord(knightColumn)-2)>='a'): busyDots.add( str(chr(ord(knightColumn)-2)) + str((knightRow-1)) ) #down down right if(knightRow-2>=1 and chr(ord(knightColumn)+1)<='h'): busyDots.add( str(chr(ord(knightColumn)+1)) + str((knightRow-2)) ) #down right right if(knightRow-1>=1 and chr(ord(knightColumn)+2)<='h'): busyDots.add( str(chr(ord(knightColumn)+2)) + str((knightRow-1)) ) #rock research #up up left if(rockRow+2<=8 and chr(ord(rockColumn)-1)>='a'): busyDots.add (str(chr(ord(rockColumn)-1))+str((rockRow+2))) #up left left if(rockRow+1<=8 and chr(ord(rockColumn)-2)>='a'): busyDots.add( str(chr(ord(rockColumn)-2)) + str((rockRow+1)) ) #up up right if(rockRow+2<=8 and chr(ord(rockColumn)+1)<='h'): busyDots.add( str(chr(ord(rockColumn)+1)) + str((rockRow+2)) ) #up right right if(rockRow+1<=8 and chr(ord(rockColumn)+2)<='h'): busyDots.add( str(chr(ord(rockColumn)+2)) + str((rockRow+1)) ) #down down left if(rockRow-2>=1 and chr(ord(rockColumn)-1)>='a'): busyDots.add( str(chr(ord(rockColumn)-1)) + str((rockRow-2)) ) #down left left if(rockRow-1>=1 and chr(ord(rockColumn)-2)>='a'): busyDots.add( str(chr(ord(rockColumn)-2)) + str((rockRow-1)) ) #down down right if(rockRow-2>=1 and chr(ord(rockColumn)+1)<='h'): busyDots.add( str(chr(ord(rockColumn)+1)) + str((rockRow-2)) ) #down right right if(rockRow-1>=1 and chr(ord(rockColumn)+2)<='h'): busyDots.add( str(chr(ord(rockColumn)+2)) + str((rockRow-1)) ) print(64-len(busyDots)) ```
3.944928
709
A
Juicer
PROGRAMMING
900
[ "implementation" ]
null
null
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one. The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Print one integer — the number of times Kolya will have to empty the waste section.
[ "2 7 10\n5 6\n", "1 5 10\n7\n", "3 10 10\n5 7 7\n", "1 1 1\n1\n" ]
[ "1\n", "0\n", "1\n", "0\n" ]
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards. In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
500
[ { "input": "2 7 10\n5 6", "output": "1" }, { "input": "1 5 10\n7", "output": "0" }, { "input": "3 10 10\n5 7 7", "output": "1" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "2 951637 951638\n44069 951637", "output": "1" }, { "input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126", "output": "12" }, { "input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850", "output": "36" }, { "input": "2 6 8\n2 1", "output": "0" }, { "input": "5 15 16\n7 11 5 12 8", "output": "2" }, { "input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574", "output": "4" }, { "input": "5 234613 716125\n642626 494941 234613 234613 234613", "output": "0" }, { "input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547", "output": "0" }, { "input": "5 10 20\n10 10 10 10 1", "output": "1" }, { "input": "5 10 11\n10 10 10 10 1", "output": "2" }, { "input": "3 10 10\n4 3 3", "output": "0" }, { "input": "3 5 5\n5 5 5", "output": "1" }, { "input": "3 4 14\n5 5 5", "output": "0" }, { "input": "2 7 10\n1234 1234", "output": "0" }, { "input": "1 5 6\n10", "output": "0" }, { "input": "3 4 6\n1 2 3", "output": "0" }, { "input": "5 10 12\n13 13 13 13 13", "output": "0" }, { "input": "3 4 5\n5 7 9", "output": "0" }, { "input": "3 10 100\n5 5 5", "output": "0" }, { "input": "5 1 2\n2 2 2 2 2", "output": "0" }, { "input": "5 5 5\n5 5 5 5 5", "output": "2" }, { "input": "2 2 5\n5 5", "output": "0" }, { "input": "3 1 4\n2 2 2", "output": "0" } ]
1,621,119,821
2,147,483,647
Python 3
OK
TESTS
58
124
7,270,400
N,MaxSize,WasteSize = [int(x) for x in input().split()] Oranges = [int(x) for x in input().split()] CurrWaste = 0 TimesEmptied = 0 for i in Oranges: if CurrWaste > WasteSize: TimesEmptied += 1 CurrWaste = 0 if i <= MaxSize: CurrWaste += i if CurrWaste > WasteSize: TimesEmptied += 1 print(TimesEmptied)
Title: Juicer Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one. The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? Input Specification: The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. Output Specification: Print one integer — the number of times Kolya will have to empty the waste section. Demo Input: ['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n'] Demo Output: ['1\n', '0\n', '1\n', '0\n'] Note: In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards. In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
```python N,MaxSize,WasteSize = [int(x) for x in input().split()] Oranges = [int(x) for x in input().split()] CurrWaste = 0 TimesEmptied = 0 for i in Oranges: if CurrWaste > WasteSize: TimesEmptied += 1 CurrWaste = 0 if i <= MaxSize: CurrWaste += i if CurrWaste > WasteSize: TimesEmptied += 1 print(TimesEmptied) ```
3
920
B
Tea Queue
PROGRAMMING
1,200
[ "implementation" ]
null
null
Recently *n* students from city S moved to city P to attend a programming camp. They moved there by train. In the evening, all students in the train decided that they want to drink some tea. Of course, no two people can use the same teapot simultaneously, so the students had to form a queue to get their tea. *i*-th student comes to the end of the queue at the beginning of *l**i*-th second. If there are multiple students coming to the queue in the same moment, then the student with greater index comes after the student with lesser index. Students in the queue behave as follows: if there is nobody in the queue before the student, then he uses the teapot for exactly one second and leaves the queue with his tea; otherwise the student waits for the people before him to get their tea. If at the beginning of *r**i*-th second student *i* still cannot get his tea (there is someone before him in the queue), then he leaves the queue without getting any tea. For each student determine the second he will use the teapot and get his tea (if he actually gets it).
The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=1000). Then *t* test cases follow. The first line of each test case contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of students. Then *n* lines follow. Each line contains two integer *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=5000) — the second *i*-th student comes to the end of the queue, and the second he leaves the queue if he still cannot get his tea. It is guaranteed that for every condition *l**i*<=-<=1<=≤<=*l**i* holds. The sum of *n* over all test cases doesn't exceed 1000. Note that in hacks you have to set *t*<==<=1.
For each test case print *n* integers. *i*-th of them must be equal to the second when *i*-th student gets his tea, or 0 if he leaves without tea.
[ "2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3\n" ]
[ "1 2 \n1 0 2 \n" ]
The example contains 2 tests: 1. During 1-st second, students 1 and 2 come to the queue, and student 1 gets his tea. Student 2 gets his tea during 2-nd second. 1. During 1-st second, students 1 and 2 come to the queue, student 1 gets his tea, and student 2 leaves without tea. During 2-nd second, student 3 comes and gets his tea.
0
[ { "input": "2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3", "output": "1 2 \n1 0 2 " }, { "input": "19\n1\n1 1\n1\n1 2\n1\n1 1000\n1\n1 2000\n1\n2 2\n1\n2 3\n1\n2 1000\n1\n2 2000\n1\n1999 1999\n1\n1999 2000\n1\n2000 2000\n2\n1 1\n1 1\n2\n1 1\n1 2\n2\n1 2\n1 1\n2\n1 2000\n1 1\n2\n1 1\n1 2000\n2\n1 2000\n2 2\n2\n2 2000\n2 2\n2\n2 2\n2 2000", "output": "1 \n1 \n1 \n1 \n2 \n2 \n2 \n2 \n1999 \n1999 \n2000 \n1 0 \n1 2 \n1 0 \n1 0 \n1 2 \n1 2 \n2 0 \n2 3 " }, { "input": "1\n11\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "1 0 0 0 0 0 0 0 0 0 0 " }, { "input": "1\n5\n1 1\n1 2\n1 5\n1 1\n1 1", "output": "1 2 3 0 0 " } ]
1,517,587,423
5,323
Python 3
OK
TESTS
8
62
5,734,400
t = int(input()) while t: t -= 1 n = int(input()) wait = [] time = 1 leave = [] for i in range(n): a, b = input().split(" ") wait.append((int(a), int(b))) for i in wait: if i[1] < time: leave.append(0) else: if i[0] <= time: leave.append(time) time += 1 else: leave.append(i[0]) time = i[0] + 1 for i in leave: print(i, end=" ") print("")
Title: Tea Queue Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently *n* students from city S moved to city P to attend a programming camp. They moved there by train. In the evening, all students in the train decided that they want to drink some tea. Of course, no two people can use the same teapot simultaneously, so the students had to form a queue to get their tea. *i*-th student comes to the end of the queue at the beginning of *l**i*-th second. If there are multiple students coming to the queue in the same moment, then the student with greater index comes after the student with lesser index. Students in the queue behave as follows: if there is nobody in the queue before the student, then he uses the teapot for exactly one second and leaves the queue with his tea; otherwise the student waits for the people before him to get their tea. If at the beginning of *r**i*-th second student *i* still cannot get his tea (there is someone before him in the queue), then he leaves the queue without getting any tea. For each student determine the second he will use the teapot and get his tea (if he actually gets it). Input Specification: The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=1000). Then *t* test cases follow. The first line of each test case contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of students. Then *n* lines follow. Each line contains two integer *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=5000) — the second *i*-th student comes to the end of the queue, and the second he leaves the queue if he still cannot get his tea. It is guaranteed that for every condition *l**i*<=-<=1<=≤<=*l**i* holds. The sum of *n* over all test cases doesn't exceed 1000. Note that in hacks you have to set *t*<==<=1. Output Specification: For each test case print *n* integers. *i*-th of them must be equal to the second when *i*-th student gets his tea, or 0 if he leaves without tea. Demo Input: ['2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3\n'] Demo Output: ['1 2 \n1 0 2 \n'] Note: The example contains 2 tests: 1. During 1-st second, students 1 and 2 come to the queue, and student 1 gets his tea. Student 2 gets his tea during 2-nd second. 1. During 1-st second, students 1 and 2 come to the queue, student 1 gets his tea, and student 2 leaves without tea. During 2-nd second, student 3 comes and gets his tea.
```python t = int(input()) while t: t -= 1 n = int(input()) wait = [] time = 1 leave = [] for i in range(n): a, b = input().split(" ") wait.append((int(a), int(b))) for i in wait: if i[1] < time: leave.append(0) else: if i[0] <= time: leave.append(time) time += 1 else: leave.append(i[0]) time = i[0] + 1 for i in leave: print(i, end=" ") print("") ```
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" } ]
1,639,708,469
2,147,483,647
Python 3
OK
TESTS
40
122
0
x = input() y = input() reversedString = "" if(len(x) == len(y)): reversedString = x[::-1] if(reversedString == y): 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 x = input() y = input() reversedString = "" if(len(x) == len(y)): reversedString = x[::-1] if(reversedString == y): print("YES") else: print("NO") ```
3.9695
59
A
Word
PROGRAMMING
800
[ "implementation", "strings" ]
A. Word
2
256
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
[ "HoUse\n", "ViP\n", "maTRIx\n" ]
[ "house\n", "VIP\n", "matrix\n" ]
none
500
[ { "input": "HoUse", "output": "house" }, { "input": "ViP", "output": "VIP" }, { "input": "maTRIx", "output": "matrix" }, { "input": "BNHWpnpawg", "output": "bnhwpnpawg" }, { "input": "VTYGP", "output": "VTYGP" }, { "input": "CHNenu", "output": "chnenu" }, { "input": "ERPZGrodyu", "output": "erpzgrodyu" }, { "input": "KSXBXWpebh", "output": "KSXBXWPEBH" }, { "input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv", "output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv" }, { "input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd", "output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd" }, { "input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv", "output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv" }, { "input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg", "output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg" }, { "input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak", "output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK" }, { "input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw", "output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW" }, { "input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB", "output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB" }, { "input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge", "output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge" }, { "input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw", "output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw" }, { "input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks", "output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks" }, { "input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz", "output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ" }, { "input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg", "output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG" }, { "input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO", "output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO" }, { "input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd", "output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD" }, { "input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas", "output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas" }, { "input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm", "output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM" }, { "input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn", "output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn" }, { "input": "t", "output": "t" }, { "input": "N", "output": "N" }, { "input": "kv", "output": "kv" }, { "input": "Ur", "output": "ur" }, { "input": "CN", "output": "CN" } ]
1,638,455,450
2,147,483,647
Python 3
OK
TESTS
30
92
0
#59A Words #check how many upper and lowercase letters then print all as lowercase or uppercase s = input() uppercase = int(sum(1 for c in s if c.isupper())) lowercase = int(sum(1 for c in s if c.islower())) if (uppercase > lowercase): print(s.upper()) else: print(s.lower())
Title: Word Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. Input Specification: The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. Output Specification: Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. Demo Input: ['HoUse\n', 'ViP\n', 'maTRIx\n'] Demo Output: ['house\n', 'VIP\n', 'matrix\n'] Note: none
```python #59A Words #check how many upper and lowercase letters then print all as lowercase or uppercase s = input() uppercase = int(sum(1 for c in s if c.isupper())) lowercase = int(sum(1 for c in s if c.islower())) if (uppercase > lowercase): print(s.upper()) else: print(s.lower()) ```
3.977
270
A
Fancy Fence
PROGRAMMING
1,100
[ "geometry", "implementation", "math" ]
null
null
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
The first line of input contains an integer *t* (0<=&lt;<=*t*<=&lt;<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=&lt;<=*a*<=&lt;<=180) — the angle the robot can make corners at measured in degrees.
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
[ "3\n30\n60\n90\n" ]
[ "NO\nYES\nYES\n" ]
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.
500
[ { "input": "3\n30\n60\n90", "output": "NO\nYES\nYES" }, { "input": "6\n1\n2\n3\n170\n179\n25", "output": "NO\nNO\nNO\nYES\nYES\nNO" } ]
1,694,584,074
2,147,483,647
Python 3
OK
TESTS
3
92
0
# -*- coding: utf-8 -*- """ Created on Wed Sep 13 13:41:09 2023 @author: mac """ t = int(input()) for i in range(t): a = int(input()) ans = False for n in range(3,361): if 360 % n == 0 and a == 180 - 360 // n: ans = True break if ans: print("YES") else: print("NO")
Title: Fancy Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? Input Specification: The first line of input contains an integer *t* (0<=&lt;<=*t*<=&lt;<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=&lt;<=*a*<=&lt;<=180) — the angle the robot can make corners at measured in degrees. Output Specification: For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. Demo Input: ['3\n30\n60\n90\n'] Demo Output: ['NO\nYES\nYES\n'] Note: In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.
```python # -*- coding: utf-8 -*- """ Created on Wed Sep 13 13:41:09 2023 @author: mac """ t = int(input()) for i in range(t): a = int(input()) ans = False for n in range(3,361): if 360 % n == 0 and a == 180 - 360 // n: ans = True break if ans: print("YES") else: print("NO") ```
3
242
C
King's Path
PROGRAMMING
1,800
[ "dfs and similar", "graphs", "hashing", "shortest paths" ]
null
null
The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*). You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way. Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.
The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily. It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.
If there is no path between the initial and final position along allowed cells, print -1. Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one.
[ "5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n", "3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n", "1 1 2 10\n2\n1 1 3\n2 6 10\n" ]
[ "4\n", "6\n", "-1\n" ]
none
1,500
[ { "input": "5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5", "output": "4" }, { "input": "3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10", "output": "6" }, { "input": "1 1 2 10\n2\n1 1 3\n2 6 10", "output": "-1" }, { "input": "9 8 7 8\n9\n10 6 6\n10 6 6\n7 7 8\n9 5 6\n8 9 9\n9 5 5\n9 8 8\n8 5 6\n9 10 10", "output": "2" }, { "input": "6 15 7 15\n9\n6 15 15\n7 14 14\n6 15 15\n9 14 14\n7 14 16\n6 15 15\n6 15 15\n7 14 14\n8 15 15", "output": "1" }, { "input": "13 16 20 10\n18\n13 16 16\n20 10 10\n19 10 10\n12 15 15\n20 10 10\n18 11 11\n19 10 10\n19 10 10\n20 10 10\n19 10 10\n20 10 10\n20 10 10\n19 10 10\n18 11 11\n13 16 16\n12 15 15\n19 10 10\n19 10 10", "output": "-1" }, { "input": "89 29 88 30\n16\n87 31 31\n14 95 95\n98 88 89\n96 88 88\n14 97 97\n13 97 98\n100 88 88\n88 32 32\n99 88 89\n90 29 29\n87 31 31\n15 94 96\n89 29 29\n88 32 32\n97 89 89\n88 29 30", "output": "1" }, { "input": "30 14 39 19\n31\n35 7 11\n37 11 12\n32 13 13\n37 5 6\n46 13 13\n37 14 14\n31 13 13\n43 13 19\n45 15 19\n46 13 13\n32 17 17\n41 14 19\n30 14 14\n43 13 17\n34 16 18\n44 11 19\n38 13 13\n40 12 20\n37 16 18\n46 16 18\n34 10 14\n36 9 10\n36 15 19\n38 15 19\n42 13 19\n33 14 15\n35 15 19\n33 17 18\n39 12 20\n36 5 7\n45 12 12", "output": "9" }, { "input": "2 1 1 1\n2\n1 1 2\n2 1 2", "output": "1" }, { "input": "1 1 1 2\n5\n1000000000 1 10000\n19920401 1188 5566\n1000000000 1 10000\n1 1 10000\n5 100 200", "output": "1" }, { "input": "1 1 1000000000 2\n5\n1000000000 1 10000\n19920401 1188 5566\n1000000000 1 10000\n1 1 10000\n5 100 200", "output": "-1" } ]
1,611,688,803
2,147,483,647
PyPy 3
OK
TESTS
43
1,122
16,896,000
###### ### ####### ####### ## # ##### ### ##### # # # # # # # # # # # # # ### # # # # # # # # # # # # # ### ###### ######### # # # # # # ######### # ###### ######### # # # # # # ######### # # # # # # # # # # # #### # # # # # # # # # # ## # # # # # ###### # # ####### ####### # # ##### # # # # from __future__ import print_function # for PyPy2 from collections import Counter, OrderedDict from itertools import permutations as perm from fractions import Fraction from collections import deque from sys import stdin from bisect import * from heapq import * from math import * g = lambda : stdin.readline().strip() gl = lambda : g().split() gil = lambda : [int(var) for var in gl()] gfl = lambda : [float(var) for var in gl()] gcl = lambda : list(g()) gbs = lambda : [int(var) for var in g()] mod = int(1e9)+7 inf = float("inf") sx, sy, ex, ey = gil() d, h = {}, [] for _ in range(gil()[0]): r, L, R = gil() for c in range(L, R+1): d[r, c] = inf if (sx, sy) not in d or (ex, ey) not in d : print(0 if (sx, sy) == (ex, ey) else -1) exit() d[sx, sy] = 0 h.append((0, sx, sy)) perm = [(-1, -1), (1, 0), (0, 1), (1, 1), (0, -1), (-1, 0), (-1, 1), (1, -1)] while h: di, x, y = heappop(h) if d[x, y] != di:continue for dx, dy in perm: p = (x+dx, y+dy) if p in d and d[p] > di+1: d[p] = di + 1 heappush(h, (di+1, p[0], p[1])) print(-1 if d[ex, ey] == inf else d[ex, ey])
Title: King's Path Time Limit: None seconds Memory Limit: None megabytes Problem Description: The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*). You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way. Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point. Input Specification: The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily. It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105. Output Specification: If there is no path between the initial and final position along allowed cells, print -1. Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one. Demo Input: ['5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n', '3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n', '1 1 2 10\n2\n1 1 3\n2 6 10\n'] Demo Output: ['4\n', '6\n', '-1\n'] Note: none
```python ###### ### ####### ####### ## # ##### ### ##### # # # # # # # # # # # # # ### # # # # # # # # # # # # # ### ###### ######### # # # # # # ######### # ###### ######### # # # # # # ######### # # # # # # # # # # # #### # # # # # # # # # # ## # # # # # ###### # # ####### ####### # # ##### # # # # from __future__ import print_function # for PyPy2 from collections import Counter, OrderedDict from itertools import permutations as perm from fractions import Fraction from collections import deque from sys import stdin from bisect import * from heapq import * from math import * g = lambda : stdin.readline().strip() gl = lambda : g().split() gil = lambda : [int(var) for var in gl()] gfl = lambda : [float(var) for var in gl()] gcl = lambda : list(g()) gbs = lambda : [int(var) for var in g()] mod = int(1e9)+7 inf = float("inf") sx, sy, ex, ey = gil() d, h = {}, [] for _ in range(gil()[0]): r, L, R = gil() for c in range(L, R+1): d[r, c] = inf if (sx, sy) not in d or (ex, ey) not in d : print(0 if (sx, sy) == (ex, ey) else -1) exit() d[sx, sy] = 0 h.append((0, sx, sy)) perm = [(-1, -1), (1, 0), (0, 1), (1, 1), (0, -1), (-1, 0), (-1, 1), (1, -1)] while h: di, x, y = heappop(h) if d[x, y] != di:continue for dx, dy in perm: p = (x+dx, y+dy) if p in d and d[p] > di+1: d[p] = di + 1 heappush(h, (di+1, p[0], p[1])) print(-1 if d[ex, ey] == inf else d[ex, ey]) ```
3
136
A
Presents
PROGRAMMING
800
[ "implementation" ]
null
null
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]
[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]
none
500
[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71", "output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90" }, { "input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9", "output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9" }, { "input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44", "output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20" }, { "input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17", "output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28" }, { "input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7", "output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5" }, { "input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16", "output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43" }, { "input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65", "output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43" }, { "input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16", "output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45" }, { "input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 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43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12" }, { "input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87", "output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82" }, { "input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25", "output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97" }, { "input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61", "output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96" }, { "input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92", "output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96" }, { "input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6", "output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36" }, { "input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1" } ]
1,683,036,512
2,147,483,647
PyPy 3-64
OK
TESTS
101
124
0
n = int(input()) l = list(map(int, input().split(' '))) ans = [0] * n for x in range(n): y = l[x] y -= 1 ans[y] = (x + 1) print(*ans)
Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none
```python n = int(input()) l = list(map(int, input().split(' '))) ans = [0] * n for x in range(n): y = l[x] y -= 1 ans[y] = (x + 1) print(*ans) ```
3
868
A
Bark to Unlock
PROGRAMMING
900
[ "brute force", "implementation", "strings" ]
null
null
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters. Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
The first line contains two lowercase English letters — the password on the phone. The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows. The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise. You can print each letter in arbitrary case (upper or lower).
[ "ya\n4\nah\noy\nto\nha\n", "hp\n2\nht\ntp\n", "ah\n1\nha\n" ]
[ "YES\n", "NO\n", "YES\n" ]
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES". In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring. In the third example the string "hahahaha" contains "ah" as a substring.
250
[ { "input": "ya\n4\nah\noy\nto\nha", "output": "YES" }, { "input": "hp\n2\nht\ntp", "output": "NO" }, { "input": "ah\n1\nha", "output": "YES" }, { "input": "bb\n4\nba\nab\naa\nbb", "output": "YES" }, { "input": "bc\n4\nca\nba\nbb\ncc", "output": "YES" }, { "input": "ba\n4\ncd\nad\ncc\ncb", "output": "YES" }, { "input": "pg\n4\nzl\nxs\ndi\nxn", "output": "NO" }, { "input": "bn\n100\ndf\nyb\nze\nml\nyr\nof\nnw\nfm\ndw\nlv\nzr\nhu\nzt\nlw\nld\nmo\nxz\ntp\nmr\nou\nme\npx\nvp\nes\nxi\nnr\nbx\nqc\ngm\njs\nkn\ntw\nrq\nkz\nuc\nvc\nqr\nab\nna\nro\nya\nqy\ngu\nvk\nqk\ngs\nyq\nop\nhw\nrj\neo\nlz\nbh\nkr\nkb\nma\nrd\nza\nuf\nhq\nmc\nmn\nti\nwn\nsh\nax\nsi\nnd\ntz\ndu\nfj\nkl\nws\now\nnf\nvr\nye\nzc\niw\nfv\nkv\noo\nsm\nbc\nrs\nau\nuz\nuv\ngh\nsu\njn\ndz\nrl\nwj\nbk\nzl\nas\nms\nit\nwu", "output": "YES" }, { "input": "bb\n1\naa", "output": "NO" }, { "input": "qm\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm", "output": "NO" }, { "input": "mq\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm", "output": "YES" }, { "input": "aa\n1\naa", "output": "YES" }, { "input": "bb\n1\nbb", "output": "YES" }, { "input": "ba\n1\ncc", "output": "NO" }, { "input": "ha\n1\nha", "output": "YES" }, { "input": "aa\n1\naa", "output": "YES" }, { "input": "ez\n1\njl", "output": "NO" }, { "input": "aa\n2\nab\nba", "output": "YES" }, { "input": "aa\n2\nca\ncc", "output": "NO" }, { "input": "dd\n2\nac\ndc", "output": "NO" }, { "input": "qc\n2\nyc\nkr", "output": "NO" }, { "input": "aa\n3\nba\nbb\nab", "output": "YES" }, { "input": "ca\n3\naa\nbb\nab", "output": "NO" }, { "input": "ca\n3\nbc\nbd\nca", "output": "YES" }, { "input": "dd\n3\nmt\nrg\nxl", "output": "NO" }, { "input": "be\n20\nad\ncd\ncb\ndb\ndd\naa\nab\nca\nae\ned\ndc\nbb\nba\nda\nee\nea\ncc\nac\nec\neb", "output": "YES" }, { "input": "fc\n20\nca\nbb\nce\nfd\nde\nfa\ncc\nec\nfb\nfc\nff\nbe\ncf\nba\ndb\ned\naf\nae\nda\nef", "output": "YES" }, { "input": "ca\n20\ndc\naf\ndf\neg\naa\nbc\nea\nbd\nab\ndb\ngc\nfb\nba\nbe\nee\ngf\ncf\nag\nga\nca", "output": "YES" }, { "input": "ke\n20\nzk\nra\nbq\nqz\nwt\nzg\nmz\nuk\nge\nuv\nud\nfd\neh\ndm\nsk\nki\nfv\ntp\nat\nfb", "output": "YES" }, { "input": "hh\n50\nag\nhg\ndg\nfh\neg\ngh\ngd\nda\nbh\nab\nhf\ndc\nhb\nfe\nad\nec\nac\nfd\nca\naf\ncg\nhd\neb\nce\nhe\nha\ngb\nea\nae\nfb\nff\nbe\nch\nhh\nee\nde\nge\ngf\naa\ngg\neh\ned\nbf\nfc\nah\nga\nbd\ncb\nbg\nbc", "output": "YES" }, { "input": "id\n50\nhi\ndc\nfg\nee\ngi\nhc\nac\nih\ndg\nfc\nde\ned\nie\neb\nic\ncf\nib\nfa\ngc\nba\nbe\nga\nha\nhg\nia\ndf\nab\nei\neh\nad\nii\nci\ndh\nec\nif\ndi\nbg\nag\nhe\neg\nca\nae\ndb\naa\nid\nfh\nhh\ncc\nfb\ngb", "output": "YES" }, { "input": "fe\n50\nje\nbi\nbg\ngc\nfb\nig\ndf\nji\ndg\nfe\nfc\ncf\ngf\nai\nhe\nac\nch\nja\ngh\njf\nge\ncb\nij\ngb\ncg\naf\neh\nee\nhd\njd\njb\nii\nca\nci\nga\nab\nhi\nag\nfj\nej\nfi\nie\ndj\nfg\nef\njc\njg\njh\nhf\nha", "output": "YES" }, { "input": "rn\n50\nba\nec\nwg\nao\nlk\nmz\njj\ncf\nfa\njk\ndy\nsz\njs\nzr\nqv\ntx\nwv\nrd\nqw\nls\nrr\nvt\nrx\nkc\neh\nnj\niq\nyi\nkh\nue\nnv\nkz\nrn\nes\nua\nzf\nvu\nll\neg\nmj\ncz\nzj\nxz\net\neb\nci\nih\nig\nam\nvd", "output": "YES" }, { "input": "ee\n100\nah\nfb\ncd\nbi\nii\nai\nid\nag\nie\nha\ndi\nec\nae\nce\njb\ndg\njg\ngd\ngf\nda\nih\nbd\nhj\ngg\nhb\ndf\ned\nfh\naf\nja\nci\nfc\nic\nji\nac\nhi\nfj\nch\nbc\njd\naa\nff\nad\ngj\nej\nde\nee\nhe\ncf\nga\nia\ncg\nbb\nhc\nbe\ngi\njf\nbg\naj\njj\nbh\nfe\ndj\nef\ngb\nge\ndb\nig\ncj\ndc\nij\njh\nei\ndd\nib\nhf\neg\nbf\nfg\nab\ngc\nfd\nhd\ngh\neh\njc\neb\nhh\nca\nje\nbj\nif\nea\nhg\nfa\ncc\nba\ndh\ncb\nfi", "output": "YES" }, { "input": "if\n100\njd\nbc\nje\nhi\nga\nde\nkb\nfc\ncd\ngd\naj\ncb\nei\nbf\ncf\ndk\ndb\ncg\nki\ngg\nkg\nfa\nkj\nii\njf\njg\ngb\nbh\nbg\neh\nhj\nhb\ndg\ndj\njc\njb\nce\ndi\nig\nci\ndf\nji\nhc\nfk\naf\nac\ngk\nhd\nae\nkd\nec\nkc\neb\nfh\nij\nie\nca\nhh\nkf\nha\ndd\nif\nef\nih\nhg\nej\nfe\njk\nea\nib\nck\nhf\nak\ngi\nch\ndc\nba\nke\nad\nka\neg\njh\nja\ngc\nfd\ncc\nab\ngj\nik\nfg\nbj\nhe\nfj\nge\ngh\nhk\nbk\ned\nid\nfi", "output": "YES" }, { "input": "kd\n100\nek\nea\nha\nkf\nkj\ngh\ndl\nfj\nal\nga\nlj\nik\ngd\nid\ncb\nfh\ndk\nif\nbh\nkb\nhc\nej\nhk\ngc\ngb\nef\nkk\nll\nlf\nkh\ncl\nlh\njj\nil\nhh\nci\ndb\ndf\ngk\njg\nch\nbd\ncg\nfg\nda\neb\nlg\ndg\nbk\nje\nbg\nbl\njl\ncj\nhb\nei\naa\ngl\nka\nfa\nfi\naf\nkc\nla\ngi\nij\nib\nle\ndi\nck\nag\nlc\nca\nge\nie\nlb\nke\nii\nae\nig\nic\nhe\ncf\nhd\nak\nfb\nhi\ngf\nad\nba\nhg\nbi\nkl\nac\ngg\ngj\nbe\nlk\nld\naj", "output": "YES" }, { "input": "ab\n1\nab", "output": "YES" }, { "input": "ya\n1\nya", "output": "YES" }, { "input": "ay\n1\nyb", "output": "NO" }, { "input": "ax\n2\nii\nxa", "output": "YES" }, { "input": "hi\n1\nhi", "output": "YES" }, { "input": "ag\n1\nag", "output": "YES" }, { "input": "th\n1\nth", "output": "YES" }, { "input": "sb\n1\nsb", "output": "YES" }, { "input": "hp\n1\nhp", "output": "YES" }, { "input": "ah\n1\nah", "output": "YES" }, { "input": "ta\n1\nta", "output": "YES" }, { "input": "tb\n1\ntb", "output": "YES" }, { "input": "ab\n5\nca\nda\nea\nfa\nka", "output": "NO" }, { "input": "ac\n1\nac", "output": "YES" }, { "input": "ha\n2\nha\nzz", "output": "YES" }, { "input": "ok\n1\nok", "output": "YES" }, { "input": "bc\n1\nbc", "output": "YES" }, { "input": "az\n1\nzz", "output": "NO" }, { "input": "ab\n2\nba\ntt", "output": "YES" }, { "input": "ah\n2\nap\nhp", "output": "NO" }, { "input": "sh\n1\nsh", "output": "YES" }, { "input": "az\n1\nby", "output": "NO" }, { "input": "as\n1\nas", "output": "YES" }, { "input": "ab\n2\nab\ncd", "output": "YES" }, { "input": "ab\n2\nxa\nza", "output": "NO" }, { "input": "ab\n2\net\nab", "output": "YES" }, { "input": "ab\n1\naa", "output": "NO" }, { "input": "ab\n2\nab\nde", "output": "YES" }, { "input": "ah\n2\nba\nha", "output": "YES" }, { "input": "ha\n3\ndd\ncc\nha", "output": "YES" }, { "input": "oo\n1\nox", "output": "NO" }, { "input": "ab\n2\nax\nbx", "output": "NO" }, { "input": "ww\n4\nuw\now\npo\nko", "output": "NO" }, { "input": "ay\n1\nay", "output": "YES" }, { "input": "yo\n1\nyo", "output": "YES" }, { "input": "ba\n1\nba", "output": "YES" }, { "input": "qw\n1\nqw", "output": "YES" }, { "input": "la\n1\nla", "output": "YES" }, { "input": "ab\n2\nbb\nbc", "output": "NO" }, { "input": "aa\n2\nab\nac", "output": "NO" }, { "input": "ah\n2\nbb\nha", "output": "YES" }, { "input": "ya\n42\nab\nac\nad\nae\naf\nag\nah\nai\nak\naj\nba\nbc\nbd\nbe\nbf\nbg\nbh\nbi\nbk\nbj\ncb\nca\ncd\nce\ncf\ncg\nch\nci\nck\ncj\ndb\ndc\nda\nde\ndf\ndg\ndh\ndi\ndk\ndj\nef\nek", "output": "NO" }, { "input": "ab\n3\nab\nxx\nyy", "output": "YES" }, { "input": "ab\n2\nab\ncc", "output": "YES" }, { "input": "sa\n2\nxx\nas", "output": "YES" }, { "input": "ma\n1\nma", "output": "YES" }, { "input": "ba\n1\nbb", "output": "NO" }, { "input": "bc\n1\nab", "output": "NO" }, { "input": "fa\n1\nfa", "output": "YES" }, { "input": "ap\n1\nap", "output": "YES" }, { "input": "ab\n1\nbb", "output": "NO" }, { "input": "bk\n1\nbk", "output": "YES" }, { "input": "xy\n2\nxy\naa", "output": "YES" }, { "input": "ab\n2\nza\nbz", "output": "YES" } ]
1,631,633,143
2,147,483,647
PyPy 3
OK
TESTS
90
108
21,401,600
pwd = input() chk = [] for _ in range(int(input())): chk.append(input()) for a in chk: for b in chk: if pwd in a+b: print("Yes"); exit() print("No")
Title: Bark to Unlock Time Limit: None seconds Memory Limit: None megabytes Problem Description: As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters. Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not. Input Specification: The first line contains two lowercase English letters — the password on the phone. The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows. The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct. Output Specification: Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise. You can print each letter in arbitrary case (upper or lower). Demo Input: ['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES". In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring. In the third example the string "hahahaha" contains "ah" as a substring.
```python pwd = input() chk = [] for _ in range(int(input())): chk.append(input()) for a in chk: for b in chk: if pwd in a+b: print("Yes"); exit() print("No") ```
3
50
A
Domino piling
PROGRAMMING
800
[ "greedy", "math" ]
A. Domino piling
2
256
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output one number — the maximal number of dominoes, which can be placed.
[ "2 4\n", "3 3\n" ]
[ "4\n", "4\n" ]
none
500
[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]
1,611,809,397
2,147,483,647
Python 3
OK
TESTS
35
124
0
dom=2 num=list(map(int,input().split(' '))) ar=num[0]*num[1] sum=0 c=0 while True: sum=sum+dom if sum<=ar: c+=1 else: break print(c)
Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none
```python dom=2 num=list(map(int,input().split(' '))) ar=num[0]*num[1] sum=0 c=0 while True: sum=sum+dom if sum<=ar: c+=1 else: break print(c) ```
3.969
115
A
Party
PROGRAMMING
900
[ "dfs and similar", "graphs", "trees" ]
null
null
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed?
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Print a single integer denoting the minimum number of groups that will be formed in the party.
[ "5\n-1\n1\n2\n1\n-1\n" ]
[ "3\n" ]
For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5
500
[ { "input": "5\n-1\n1\n2\n1\n-1", "output": "3" }, { "input": "4\n-1\n1\n2\n3", "output": "4" }, { "input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11", "output": "4" }, { "input": "6\n-1\n-1\n2\n3\n1\n1", "output": "3" }, { "input": "3\n-1\n1\n1", "output": "2" }, { "input": "1\n-1", "output": "1" }, { "input": "2\n2\n-1", "output": "2" }, { "input": "2\n-1\n-1", "output": "1" }, { "input": "3\n2\n-1\n1", "output": "3" }, { "input": "3\n-1\n-1\n-1", "output": "1" }, { "input": "5\n4\n5\n1\n-1\n4", "output": "3" }, { "input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7", "output": "4" }, { "input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4", "output": "5" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1", "output": "2" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1", "output": "1" }, { "input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11", "output": "12" }, { "input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2", "output": "11" }, { "input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6", "output": "6" }, { "input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1", "output": "3" }, { "input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1", "output": "2" }, { "input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1", "output": "3" }, { "input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8", "output": "5" }, { "input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10", "output": "6" }, { "input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1", "output": "4" }, { "input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9", "output": "4" }, { "input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8", "output": "3" }, { "input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11", "output": "7" }, { "input": "7\n7\n6\n-1\n-1\n4\n4\n-1", "output": "3" }, { "input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34", "output": "6" }, { "input": "5\n2\n3\n4\n5\n-1", "output": "5" } ]
1,698,963,370
2,147,483,647
PyPy 3
OK
TESTS
106
310
3,276,800
n = int(input()) ranks = [None for i in range(n)] boss = [int(input()) for i in range(n)] for i in range(0, n): if ranks[i] != None: continue stack = [i] while stack: last = stack[-1] b = boss[last] if b == -1: ranks[last] = 1 stack.pop() elif ranks[b - 1] != None: ranks[last] = ranks[b - 1] + 1 stack.pop() else: stack.append(b - 1) ans = max(ranks) print(ans)
Title: Party Time Limit: None seconds Memory Limit: None megabytes Problem Description: A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. Output Specification: Print a single integer denoting the minimum number of groups that will be formed in the party. Demo Input: ['5\n-1\n1\n2\n1\n-1\n'] Demo Output: ['3\n'] Note: For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5
```python n = int(input()) ranks = [None for i in range(n)] boss = [int(input()) for i in range(n)] for i in range(0, n): if ranks[i] != None: continue stack = [i] while stack: last = stack[-1] b = boss[last] if b == -1: ranks[last] = 1 stack.pop() elif ranks[b - 1] != None: ranks[last] = ranks[b - 1] + 1 stack.pop() else: stack.append(b - 1) ans = max(ranks) print(ans) ```
3
208
A
Dubstep
PROGRAMMING
900
[ "strings" ]
null
null
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
[ "WUBWUBABCWUB\n", "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n" ]
[ "ABC ", "WE ARE THE CHAMPIONS MY FRIEND " ]
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
500
[ { "input": "WUBWUBABCWUB", "output": "ABC " }, { "input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB", "output": "WE ARE THE CHAMPIONS MY FRIEND " }, { "input": "WUBWUBWUBSR", "output": "SR " }, { "input": "RWUBWUBWUBLWUB", "output": "R L " }, { "input": "ZJWUBWUBWUBJWUBWUBWUBL", "output": "ZJ J L " }, { "input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB", "output": "C B E Q " }, { "input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB", "output": "JKD WBIRAQKF YE WV " }, { "input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB", "output": "KSDHEMIXUJ R S H " }, { "input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB", "output": "OG X I KO " }, { "input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH", "output": "Q QQ I WW JOPJPBRH " }, { "input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB", "output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C " }, { "input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV", "output": "E IQMJNIQ GZZBQZAUHYP PMR DCV " }, { "input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB", "output": "FV BPS RXNETCJ JDMBH B V B " }, { "input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL", "output": "FBQ IDFSY CTWDM SXO QI L " }, { "input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL", "output": "I QLHD YIIKZDFQ CX U K NL " }, { "input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE", "output": "K UPDYXGOKU AGOAH IZD IY V P E " }, { "input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB", "output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ " }, { "input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB", "output": "PAMJGY XGPQM TKGSXUY E N H E " }, { "input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB", "output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB " }, { "input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM", "output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M " }, { "input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW", "output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W " }, { "input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG", "output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G " }, { "input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN", "output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N " }, { "input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG", "output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG " }, { "input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB", "output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L " }, { "input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB", "output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U " }, { "input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB", "output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ " }, { "input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB", "output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J " }, { "input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO", "output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O " }, { "input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR", "output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R " }, { "input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB", "output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM " }, { "input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB", "output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H " }, { "input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB", "output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S " }, { "input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB", "output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J " }, { "input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF", "output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F " }, { "input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY", "output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y " }, { "input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB", "output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO " }, { "input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW", "output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W " }, { "input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD", "output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D " }, { "input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB", "output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A " }, { "input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB", "output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y " }, { "input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB", "output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R " }, { "input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY", "output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y " }, { "input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE", "output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E " }, { "input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS", "output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S " }, { "input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP", "output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP " }, { "input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB", "output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y " }, { "input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL", "output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL " }, { "input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO", "output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O " }, { "input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU", "output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU " }, { "input": "A", "output": "A " }, { "input": "WUBA", "output": "A " }, { "input": "WUBWUBA", "output": "A " }, { "input": "AWUBWUBWUB", "output": "A " }, { "input": "AWUBBWUBCWUBD", "output": "A B C D " }, { "input": "WUBWWUBWUBWUBUWUBWUBBWUB", "output": "W U B " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "WUWUBBWWUBUB", "output": "WU BW UB " }, { "input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB", "output": "WUAB " }, { "input": "ZWUB", "output": "Z " }, { "input": "WU", "output": "WU " }, { "input": "UB", "output": "UB " }, { "input": "U", "output": "U " }, { "input": "WUBW", "output": "W " }, { "input": "WUBWU", "output": "WU " }, { "input": "WUWUB", "output": "WU " }, { "input": "UBWUB", "output": "UB " }, { "input": "WUWUBUBWUBUWUB", "output": "WU UB U " }, { "input": "WUBWWUBAWUB", "output": "W A " }, { "input": "WUUUUU", "output": "WUUUUU " } ]
1,693,481,810
2,147,483,647
Python 3
OK
TESTS
71
92
0
s=input() c=0 a="" if 'WUB' in s: s=s.replace('WUB',' ') print(s.strip())
Title: Dubstep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. Input Specification: The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. Output Specification: Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. Demo Input: ['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n'] Demo Output: ['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND '] Note: In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
```python s=input() c=0 a="" if 'WUB' in s: s=s.replace('WUB',' ') print(s.strip()) ```
3
144
A
Arrival of the General
PROGRAMMING
800
[ "implementation" ]
null
null
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground. By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important. For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong. Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
[ "4\n33 44 11 22\n", "7\n10 10 58 31 63 40 76\n" ]
[ "2\n", "10\n" ]
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11). In the second sample the colonel may swap the soldiers in the following sequence: 1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
500
[ { "input": "4\n33 44 11 22", "output": "2" }, { "input": "7\n10 10 58 31 63 40 76", "output": "10" }, { "input": "2\n88 89", "output": "1" }, { "input": "5\n100 95 100 100 88", "output": "0" }, { "input": "7\n48 48 48 48 45 45 45", "output": "0" }, { "input": "10\n68 47 67 29 63 71 71 65 54 56", "output": "10" }, { "input": "15\n77 68 96 60 92 75 61 60 66 79 80 65 60 95 92", "output": "4" }, { "input": "3\n1 2 1", "output": "1" }, { "input": "20\n30 30 30 14 30 14 30 30 30 14 30 14 14 30 14 14 30 14 14 14", "output": "0" }, { "input": "35\n37 41 46 39 47 39 44 47 44 42 44 43 47 39 46 39 38 42 39 37 40 44 41 42 41 42 39 42 36 36 42 36 42 42 42", "output": "7" }, { "input": "40\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 98 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99", "output": "47" }, { "input": "50\n48 52 44 54 53 56 62 49 39 41 53 39 40 64 53 50 62 48 40 52 51 48 40 52 61 62 62 61 48 64 55 57 56 40 48 58 41 60 60 56 64 50 64 45 48 45 46 63 59 57", "output": "50" }, { "input": "57\n7 24 17 19 6 19 10 11 12 22 14 5 5 11 13 10 24 19 24 24 24 11 21 20 4 14 24 24 18 13 24 3 20 3 3 3 3 9 3 9 22 22 16 3 3 3 15 11 3 3 8 17 10 13 3 14 13", "output": "3" }, { "input": "65\n58 50 35 44 35 37 36 58 38 36 58 56 56 49 48 56 58 43 40 44 52 44 58 58 57 50 43 35 55 39 38 49 53 56 50 42 41 56 34 57 49 38 34 51 56 38 58 40 53 46 48 34 38 43 49 49 58 56 41 43 44 34 38 48 36", "output": "3" }, { "input": "69\n70 48 49 48 49 71 48 53 55 69 48 53 54 58 53 63 48 48 69 67 72 75 71 75 74 74 57 63 65 60 48 48 65 48 48 51 50 49 62 53 76 68 76 56 76 76 64 76 76 57 61 76 73 51 59 76 65 50 69 50 76 67 76 63 62 74 74 58 73", "output": "73" }, { "input": "75\n70 65 64 71 71 64 71 64 68 71 65 64 65 68 71 66 66 69 68 63 69 65 71 69 68 68 71 67 71 65 65 65 71 71 65 69 63 66 62 67 64 63 62 64 67 65 62 69 62 64 69 62 67 64 67 70 64 63 64 64 69 62 62 64 70 62 62 68 67 69 62 64 66 70 68", "output": "7" }, { "input": "84\n92 95 84 85 94 80 90 86 80 92 95 84 86 83 86 83 93 91 95 92 84 88 82 84 84 84 80 94 93 80 94 80 95 83 85 80 95 95 80 84 86 92 83 81 90 87 81 89 92 93 80 87 90 85 93 85 93 94 93 89 94 83 93 91 80 83 90 94 95 80 95 92 85 84 93 94 94 82 91 95 95 89 85 94", "output": "15" }, { "input": "90\n86 87 72 77 82 71 75 78 61 67 79 90 64 94 94 74 85 87 73 76 71 71 60 69 77 73 76 80 82 57 62 57 57 83 76 72 75 87 72 94 77 85 59 82 86 69 62 80 95 73 83 94 79 85 91 68 85 74 93 95 68 75 89 93 83 78 95 78 83 77 81 85 66 92 63 65 75 78 67 91 77 74 59 86 77 76 90 67 70 64", "output": "104" }, { "input": "91\n94 98 96 94 95 98 98 95 98 94 94 98 95 95 99 97 97 94 95 98 94 98 96 98 96 98 97 95 94 94 94 97 94 96 98 98 98 94 96 95 94 95 97 97 97 98 94 98 96 95 98 96 96 98 94 97 96 98 97 95 97 98 94 95 94 94 97 94 96 97 97 93 94 95 95 94 96 98 97 96 94 98 98 96 96 96 96 96 94 96 97", "output": "33" }, { "input": "92\n44 28 32 29 41 41 36 39 40 39 41 35 41 28 35 27 41 34 28 38 43 43 41 38 27 26 28 36 30 29 39 32 35 35 32 30 39 30 37 27 41 41 28 30 43 31 35 33 36 28 44 40 41 35 31 42 37 38 37 34 39 40 27 40 33 33 44 43 34 33 34 34 35 38 38 37 30 39 35 41 45 42 41 32 33 33 31 30 43 41 43 43", "output": "145" }, { "input": "93\n46 32 52 36 39 30 57 63 63 30 32 44 27 59 46 38 40 45 44 62 35 36 51 48 39 58 36 51 51 51 48 58 59 36 29 35 31 49 64 60 34 38 42 56 33 42 52 31 63 34 45 51 35 45 33 53 33 62 31 38 66 29 51 54 28 61 32 45 57 41 36 34 47 36 31 28 67 48 52 46 32 40 64 58 27 53 43 57 34 66 43 39 26", "output": "76" }, { "input": "94\n56 55 54 31 32 42 46 29 24 54 40 40 20 45 35 56 32 33 51 39 26 56 21 56 51 27 29 39 56 52 54 43 43 55 48 51 44 49 52 49 23 19 19 28 20 26 45 33 35 51 42 36 25 25 38 23 21 35 54 50 41 20 37 28 42 20 22 43 37 34 55 21 24 38 19 41 45 34 19 33 44 54 38 31 23 53 35 32 47 40 39 31 20 34", "output": "15" }, { "input": "95\n57 71 70 77 64 64 76 81 81 58 63 75 81 77 71 71 71 60 70 70 69 67 62 64 78 64 69 62 76 76 57 70 68 77 70 68 73 77 79 73 60 57 69 60 74 65 58 75 75 74 73 73 65 75 72 57 81 62 62 70 67 58 76 57 79 81 68 64 58 77 70 59 79 64 80 58 71 59 81 71 80 64 78 80 78 65 70 68 78 80 57 63 64 76 81", "output": "11" }, { "input": "96\n96 95 95 95 96 97 95 97 96 95 98 96 97 95 98 96 98 96 98 96 98 95 96 95 95 95 97 97 95 95 98 98 95 96 96 95 97 96 98 96 95 97 97 95 97 97 95 94 96 96 97 96 97 97 96 94 94 97 95 95 95 96 95 96 95 97 97 95 97 96 95 94 97 97 97 96 97 95 96 94 94 95 97 94 94 97 97 97 95 97 97 95 94 96 95 95", "output": "13" }, { "input": "97\n14 15 12 12 13 15 12 15 12 12 12 12 12 14 15 15 13 12 15 15 12 12 12 13 14 15 15 13 14 15 14 14 14 14 12 13 12 13 13 12 15 12 13 13 15 12 15 13 12 13 13 13 14 13 12 15 14 13 14 15 13 14 14 13 14 12 15 12 14 12 13 14 15 14 13 15 13 12 15 15 15 13 15 15 13 14 16 16 16 13 15 13 15 14 15 15 15", "output": "104" }, { "input": "98\n37 69 35 70 58 69 36 47 41 63 60 54 49 35 55 50 35 53 52 43 35 41 40 49 38 35 48 70 42 35 35 65 56 54 44 59 59 48 51 49 59 67 35 60 69 35 58 50 35 44 48 69 41 58 44 45 35 47 70 61 49 47 37 39 35 51 44 70 72 65 36 41 63 63 48 66 45 50 50 71 37 52 72 67 72 39 72 39 36 64 48 72 69 49 45 72 72 67", "output": "100" }, { "input": "99\n31 31 16 15 19 31 19 22 29 27 12 22 28 30 25 33 26 25 19 22 34 21 17 33 31 22 16 26 22 30 31 17 13 33 13 17 28 25 18 33 27 22 31 22 13 27 20 22 23 15 24 32 29 13 16 20 32 33 14 33 19 27 16 28 25 17 17 28 18 26 32 33 19 23 30 13 14 23 24 28 14 28 22 20 30 14 24 23 17 29 18 28 29 21 28 18 16 24 32", "output": "107" }, { "input": "100\n37 54 39 29 32 49 21 13 34 21 16 42 34 27 16 26 7 34 51 9 11 27 16 40 36 7 48 52 30 42 42 52 51 11 32 26 6 7 28 54 48 51 6 54 42 20 51 48 46 4 4 31 47 6 9 16 8 23 36 50 49 30 47 37 45 24 48 26 8 50 18 24 27 13 39 7 37 34 5 43 42 19 34 23 52 38 35 6 29 11 53 49 21 29 36 22 45 33 51 22", "output": "50" }, { "input": "100\n41 5 20 23 5 47 28 28 64 5 12 14 34 56 22 48 6 52 39 55 62 40 24 54 39 28 56 27 35 51 14 24 38 10 55 20 39 62 52 2 30 55 49 48 15 23 59 44 31 28 66 29 62 59 31 13 28 12 50 45 41 47 29 27 64 66 14 39 62 22 44 63 24 5 63 17 42 7 25 41 49 33 31 51 63 3 37 66 19 9 2 40 16 45 56 21 34 29 39 61", "output": "59" }, { "input": "100\n66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95", "output": "99" }, { "input": "100\n81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10", "output": "0" }, { "input": "3\n2 1 2", "output": "1" }, { "input": "2\n100 99", "output": "0" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100", "output": "197" }, { "input": "100\n87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87", "output": "0" }, { "input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "0" }, { "input": "2\n1 1", "output": "0" }, { "input": "3\n2 2 2", "output": "0" }, { "input": "3\n1 3 2", "output": "2" } ]
1,695,321,183
2,147,483,647
Python 3
OK
TESTS
39
92
0
n = int(input()) a = [int(i) for i in input().split()] maxn = max(a) maxn = a.index(maxn) for i in range(maxn, 0, -1): a[i], a[i-1] = a[i-1], a[i] b = [a[len(a)-1]] for i in range(len(a)-1, 0, -1): b.append(a[i-1]) minn = min(b) minn = b.index(minn) steps = maxn+minn print(steps)
Title: Arrival of the General Time Limit: None seconds Memory Limit: None megabytes Problem Description: A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground. By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important. For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong. Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct. Input Specification: The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different. Output Specification: Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like. Demo Input: ['4\n33 44 11 22\n', '7\n10 10 58 31 63 40 76\n'] Demo Output: ['2\n', '10\n'] Note: In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11). In the second sample the colonel may swap the soldiers in the following sequence: 1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
```python n = int(input()) a = [int(i) for i in input().split()] maxn = max(a) maxn = a.index(maxn) for i in range(maxn, 0, -1): a[i], a[i-1] = a[i-1], a[i] b = [a[len(a)-1]] for i in range(len(a)-1, 0, -1): b.append(a[i-1]) minn = min(b) minn = b.index(minn) steps = maxn+minn print(steps) ```
3
758
A
Holiday Of Equality
PROGRAMMING
800
[ "implementation", "math" ]
null
null
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
In the only line print the integer *S* — the minimum number of burles which are had to spend.
[ "5\n0 1 2 3 4\n", "5\n1 1 0 1 1\n", "3\n1 3 1\n", "1\n12\n" ]
[ "10", "1", "4", "0" ]
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
500
[ { "input": "5\n0 1 2 3 4", "output": "10" }, { "input": "5\n1 1 0 1 1", "output": "1" }, { "input": "3\n1 3 1", "output": "4" }, { "input": "1\n12", "output": "0" }, { "input": "3\n1 2 3", "output": "3" }, { "input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281", "output": "5464380" }, { "input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353", "output": "9535765" }, { "input": "3\n1 3 2", "output": "3" }, { "input": "3\n2 1 3", "output": "3" }, { "input": "3\n2 3 1", "output": "3" }, { "input": "3\n3 1 2", "output": "3" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n228503", "output": "0" }, { "input": "2\n32576 550340", "output": "517764" }, { "input": "3\n910648 542843 537125", "output": "741328" }, { "input": "4\n751720 572344 569387 893618", "output": "787403" }, { "input": "6\n433864 631347 597596 794426 713555 231193", "output": "1364575" }, { "input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107", "output": "4647430" }, { "input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025", "output": "13488674" }, { "input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763", "output": "21993384" }, { "input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351", "output": "26984185" }, { "input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642", "output": "32816391" }, { "input": "1\n0", "output": "0" }, { "input": "1\n1000000", "output": "0" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000", "output": "99000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99000000" }, { "input": "3\n0 0 0", "output": "0" }, { "input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "3\n5 0 0", "output": "10" }, { "input": "5\n2 10 0 0 0", "output": "38" } ]
1,680,241,952
2,147,483,647
Python 3
OK
TESTS
41
46
0
n=input() s=list(map(int,input().split())) print(max(s)*len(s)-sum(s))
Title: Holiday Of Equality Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. Input Specification: The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. Output Specification: In the only line print the integer *S* — the minimum number of burles which are had to spend. Demo Input: ['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n'] Demo Output: ['10', '1', '4', '0'] Note: In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
```python n=input() s=list(map(int,input().split())) print(max(s)*len(s)-sum(s)) ```
3
61
A
Ultra-Fast Mathematician
PROGRAMMING
800
[ "implementation" ]
A. Ultra-Fast Mathematician
2
256
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Write one line — the corresponding answer. Do not omit the leading 0s.
[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]
[ "1110001\n", "111\n", "0100\n", "00010\n" ]
none
500
[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": 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"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": 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"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101", "output": "101101110110110010011100001011111100100001110000101100110000100011011100110" }, { "input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001", "output": "1110111111110010111000001100101010101011010100101010100101100011001001111111" }, { "input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111", "output": "10111011000111000101110100101000100111011011100011110110000101010001111010111" }, { "input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110", "output": 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"0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]
1,632,566,696
2,147,483,647
Python 3
OK
TESTS
102
78
6,758,400
n1 = input() n2 = input() req= "" for i in range(len(n1)): if n1[i]!=n2[i]: req+="1" else: req+="0" print(req)
Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none
```python n1 = input() n2 = input() req= "" for i in range(len(n1)): if n1[i]!=n2[i]: req+="1" else: req+="0" print(req) ```
3.967911
251
A
Points on Line
PROGRAMMING
1,300
[ "binary search", "combinatorics", "two pointers" ]
null
null
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*. Note that the order of the points inside the group of three chosen points doesn't matter.
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got. It is guaranteed that the coordinates of the points in the input strictly increase.
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
[ "4 3\n1 2 3 4\n", "4 2\n-3 -2 -1 0\n", "5 19\n1 10 20 30 50\n" ]
[ "4\n", "2\n", "1\n" ]
In the first sample any group of three points meets our conditions. In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}. In the third sample only one group does: {1, 10, 20}.
500
[ { "input": "4 3\n1 2 3 4", "output": "4" }, { "input": "4 2\n-3 -2 -1 0", "output": "2" }, { "input": "5 19\n1 10 20 30 50", "output": "1" }, { "input": "10 5\n31 36 43 47 48 50 56 69 71 86", "output": "2" }, { "input": "10 50\n1 4 20 27 65 79 82 83 99 100", "output": "25" }, { "input": "10 90\n24 27 40 41 61 69 73 87 95 97", "output": "120" }, { "input": "100 100\n-98 -97 -96 -93 -92 -91 -90 -87 -86 -84 -81 -80 -79 -78 -76 -75 -73 -71 -69 -67 -65 -64 -63 -62 -61 -54 -51 -50 -49 -48 -46 -45 -44 -37 -36 -33 -30 -28 -27 -16 -15 -13 -12 -10 -9 -7 -6 -5 -4 2 3 5 8 9 10 11 13 14 15 16 17 19 22 24 25 26 27 28 30 31 32 36 40 43 45 46 47 50 51 52 53 58 60 63 69 70 73 78 80 81 82 85 88 89 90 91 95 96 97 99", "output": "79351" }, { "input": "1 14751211\n847188590", "output": "0" }, { "input": "2 1000000000\n-907894512 -289906312", "output": "0" }, { "input": "2 1000000000\n-14348867 1760823", "output": "0" }, { "input": "3 1000000000\n-5 -1 1", "output": "1" } ]
1,681,706,480
2,147,483,647
PyPy 3-64
OK
TESTS
39
280
12,288,000
n, d = map(int, input().split()) L = [int(i) for i in input().split()] c = 0 for i in range(n): l = i - 1 r = n while r - l > 1: e = (l + r) // 2 if L[e] <= L[i] + d: l = e else: r = e if l - i > 1: c += (l - i) * (l - i - 1) // 2 print(c)
Title: Points on Line Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*. Note that the order of the points inside the group of three chosen points doesn't matter. Input Specification: The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got. It is guaranteed that the coordinates of the points in the input strictly increase. Output Specification: Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*. Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n'] Demo Output: ['4\n', '2\n', '1\n'] Note: In the first sample any group of three points meets our conditions. In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}. In the third sample only one group does: {1, 10, 20}.
```python n, d = map(int, input().split()) L = [int(i) for i in input().split()] c = 0 for i in range(n): l = i - 1 r = n while r - l > 1: e = (l + r) // 2 if L[e] <= L[i] + d: l = e else: r = e if l - i > 1: c += (l - i) * (l - i - 1) // 2 print(c) ```
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" } ]
1,469,192,333
2,147,483,647
Python 3
OK
TESTS
25
154
0
n,m=[int(i) for i in input().split()] d=n*(n+1)//2 dk=m//d c=m-dk*d i=1 while c>=i: c=c-i i=i+1 print(c)
Title: Chips Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*. The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. Output Specification: Print the number of chips the presenter ended up with. Demo Input: ['4 11\n', '17 107\n', '3 8\n'] Demo Output: ['0\n', '2\n', '1\n'] Note: In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes. In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
```python n,m=[int(i) for i in input().split()] d=n*(n+1)//2 dk=m//d c=m-dk*d i=1 while c>=i: c=c-i i=i+1 print(c) ```
3.9615
0
none
none
none
0
[ "none" ]
null
null
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
[ "fixprefixsuffix\n", "abcdabc\n" ]
[ "fix", "Just a legend" ]
none
0
[ { "input": "fixprefixsuffix", "output": "fix" }, { "input": "abcdabc", "output": "Just a legend" }, { "input": "qwertyqwertyqwerty", "output": "qwerty" }, { "input": "papapapap", "output": "papap" }, { "input": "aaaaaaaaaa", "output": "aaaaaaaa" }, { "input": "ghbdtn", "output": "Just a legend" }, { "input": "a", "output": "Just a legend" }, { "input": "aa", "output": "Just a legend" }, { "input": "ab", "output": "Just a legend" }, { "input": "aaa", "output": "a" }, { "input": "aba", "output": "Just a legend" }, { "input": "aab", "output": "Just a legend" }, { "input": "abb", "output": "Just a legend" }, { "input": "abc", "output": "Just a legend" }, { "input": "aaabaabaaaaab", "output": "Just a legend" }, { "input": "aabaaabaaaaab", "output": "aab" }, { "input": "aaabaaaabab", "output": "Just a legend" }, { "input": "abcabcabcabcabc", "output": "abcabcabc" }, { "input": "aaaaabaaaa", "output": "aaaa" }, { "input": "aaaabaaaaaaa", "output": "aaaa" }, { "input": "ghghghgxghghghg", "output": "ghghg" }, { "input": "kincenvizh", "output": "Just a legend" }, { "input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw", "output": "Just a legend" }, { "input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx", "output": "Just a legend" }, { "input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb", "output": "nfbdzgdlbjhrlvfryyjb" }, { "input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp", "output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp" }, { "input": "abcazabca", "output": "a" } ]
1,588,263,914
2,147,483,647
PyPy 3
OK
TESTS
97
590
16,486,400
def LPS(pat): M = len(pat) j = 0 lps = [0 for i in range(M)] i = 1 while i < M: if pat[i]== pat[j]: j += 1 lps[i] = j i += 1 else: if j != 0: j = lps[j-1] else: lps[i] = 0 i += 1 return lps s = input() lps = LPS(s) ans = lps[-1] ct = 0 for i in range(len(s)): if(lps[i] == ans): ct += 1 if(ct < 2): ans = lps[ans-1] if(ans > 0): print(s[:ans]) else: print("Just a legend")
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. Input Specification: You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. Output Specification: Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. Demo Input: ['fixprefixsuffix\n', 'abcdabc\n'] Demo Output: ['fix', 'Just a legend'] Note: none
```python def LPS(pat): M = len(pat) j = 0 lps = [0 for i in range(M)] i = 1 while i < M: if pat[i]== pat[j]: j += 1 lps[i] = j i += 1 else: if j != 0: j = lps[j-1] else: lps[i] = 0 i += 1 return lps s = input() lps = LPS(s) ans = lps[-1] ct = 0 for i in range(len(s)): if(lps[i] == ans): ct += 1 if(ct < 2): ans = lps[ans-1] if(ans > 0): print(s[:ans]) else: print("Just a legend") ```
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
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1,564,468,681
2,147,483,647
Python 3
OK
TESTS
102
124
0
a1 = input() a = int(a1,2) b = int(input(), 2) c = bin(a^b) c = c[2:] if len(c) < len(str(a1)): c = "0"*(len(a1) - len(c)) + c print(c)
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 a1 = input() a = int(a1,2) b = int(input(), 2) c = bin(a^b) c = c[2:] if len(c) < len(str(a1)): c = "0"*(len(a1) - len(c)) + c print(c) ```
3.969
606
B
Testing Robots
PROGRAMMING
1,600
[ "implementation" ]
null
null
The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (*x*0,<=*y*0) of a rectangular squared field of size *x*<=×<=*y*, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly *x*·*y* tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same. After placing the objects on the field the robot will have to run a sequence of commands given by string *s*, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code. Moving to the left decreases coordinate *y*, and moving to the right increases it. Similarly, moving up decreases the *x* coordinate, and moving down increases it. The tests can go on for very long, so your task is to predict their results. For each *k* from 0 to *length*(*s*) your task is to find in how many tests the robot will run exactly *k* commands before it blows up.
The first line of the input contains four integers *x*, *y*, *x*0, *y*0 (1<=≤<=*x*,<=*y*<=≤<=500,<=1<=≤<=*x*0<=≤<=*x*,<=1<=≤<=*y*0<=≤<=*y*) — the sizes of the field and the starting coordinates of the robot. The coordinate axis *X* is directed downwards and axis *Y* is directed to the right. The second line contains a sequence of commands *s*, which should be fulfilled by the robot. It has length from 1 to 100<=000 characters and only consists of characters 'L', 'R', 'U', 'D'.
Print the sequence consisting of (*length*(*s*)<=+<=1) numbers. On the *k*-th position, starting with zero, print the number of tests where the robot will run exactly *k* commands before it blows up.
[ "3 4 2 2\nUURDRDRL\n", "2 2 2 2\nULD\n" ]
[ "1 1 0 1 1 1 1 0 6\n", "1 1 1 1\n" ]
In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16bfda1e4f41cc00665c31f0a1d754d68cd9b4ab.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
1,000
[ { "input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6" }, { "input": "2 2 2 2\nULD", "output": "1 1 1 1" }, { "input": "1 1 1 1\nURDLUURRDDLLURDL", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0" }, { "input": "15 17 8 9\nURRDLUULLDD", "output": "1 1 1 1 1 1 0 1 1 1 1 245" }, { "input": "15 17 8 9\nURRDLUULLDDDRRUR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241" }, { "input": "15 17 8 9\nURRDLUULLDDDRRURR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241" }, { "input": "1 2 1 1\nR", "output": "1 1" }, { "input": "2 1 1 1\nD", "output": "1 1" }, { "input": "1 2 1 2\nLR", "output": "1 1 0" }, { "input": "2 1 2 1\nUD", "output": "1 1 0" }, { "input": "4 4 2 2\nDRUL", "output": "1 1 1 1 12" }, { "input": "4 4 3 3\nLUDRUL", "output": "1 1 1 0 0 1 12" }, { "input": "15 17 8 9\nURRDLU", "output": "1 1 1 1 1 1 249" }, { "input": "15 17 8 9\nURRDLUULLDDR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244" }, { "input": "15 17 8 9\nURRDLUULLDDRR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244" }, { "input": "15 17 8 9\nURRDLUULLDDRRR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244" }, { "input": "15 17 8 9\nURRDLUULLDDRRRR", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244" }, { "input": "15 17 8 9\nURRDLUULLDDRRRRU", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243" } ]
1,594,964,613
2,147,483,647
Python 3
OK
TESTS
68
327
9,011,200
x, y, x0, y0 = list(map(int, input().split())) s = input() board = [[0] * y for _ in range(x)] ans = [0] * (len(s) + 1) ans[0] = 1 board[x0 - 1][y0 - 1] = 1 for i in range(len(s)): ch = s[i] if ch == 'U' and x0 > 1: x0 -= 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'D' and x0 < x: x0 += 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'R' and y0 < y: y0 += 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'L' and y0 > 1: y0 -= 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 cnt0 = sum([row.count(0) for row in board]) ans[-1] += cnt0 print(*ans)
Title: Testing Robots Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (*x*0,<=*y*0) of a rectangular squared field of size *x*<=×<=*y*, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly *x*·*y* tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same. After placing the objects on the field the robot will have to run a sequence of commands given by string *s*, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code. Moving to the left decreases coordinate *y*, and moving to the right increases it. Similarly, moving up decreases the *x* coordinate, and moving down increases it. The tests can go on for very long, so your task is to predict their results. For each *k* from 0 to *length*(*s*) your task is to find in how many tests the robot will run exactly *k* commands before it blows up. Input Specification: The first line of the input contains four integers *x*, *y*, *x*0, *y*0 (1<=≤<=*x*,<=*y*<=≤<=500,<=1<=≤<=*x*0<=≤<=*x*,<=1<=≤<=*y*0<=≤<=*y*) — the sizes of the field and the starting coordinates of the robot. The coordinate axis *X* is directed downwards and axis *Y* is directed to the right. The second line contains a sequence of commands *s*, which should be fulfilled by the robot. It has length from 1 to 100<=000 characters and only consists of characters 'L', 'R', 'U', 'D'. Output Specification: Print the sequence consisting of (*length*(*s*)<=+<=1) numbers. On the *k*-th position, starting with zero, print the number of tests where the robot will run exactly *k* commands before it blows up. Demo Input: ['3 4 2 2\nUURDRDRL\n', '2 2 2 2\nULD\n'] Demo Output: ['1 1 0 1 1 1 1 0 6\n', '1 1 1 1\n'] Note: In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16bfda1e4f41cc00665c31f0a1d754d68cd9b4ab.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
```python x, y, x0, y0 = list(map(int, input().split())) s = input() board = [[0] * y for _ in range(x)] ans = [0] * (len(s) + 1) ans[0] = 1 board[x0 - 1][y0 - 1] = 1 for i in range(len(s)): ch = s[i] if ch == 'U' and x0 > 1: x0 -= 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'D' and x0 < x: x0 += 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'R' and y0 < y: y0 += 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 if ch == 'L' and y0 > 1: y0 -= 1 if board[x0 - 1][y0 - 1] == 0: ans[i + 1] = 1 board[x0 - 1][y0 - 1] = 1 cnt0 = sum([row.count(0) for row in board]) ans[-1] += cnt0 print(*ans) ```
3
629
A
Far Relative’s Birthday Cake
PROGRAMMING
800
[ "brute force", "combinatorics", "constructive algorithms", "implementation" ]
null
null
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be? Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake. Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
[ "3\n.CC\nC..\nC.C\n", "4\nCC..\nC..C\n.CC.\n.CC.\n" ]
[ "4\n", "9\n" ]
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are: 1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
500
[ { "input": "3\n.CC\nC..\nC.C", "output": "4" }, { "input": "4\nCC..\nC..C\n.CC.\n.CC.", "output": "9" }, { "input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C", "output": "46" }, { "input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.", "output": "84" }, { "input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.", "output": "80" }, { "input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C", "output": "144" }, { "input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC", "output": "190" }, { "input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..", "output": "228" }, { "input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..", "output": "2103" }, { "input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC", "output": "2071" }, { "input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.", "output": "1160" }, { "input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C", "output": "789" }, { "input": "1\n.", "output": "0" }, { "input": "3\n.CC\nC..\nC.C", "output": "4" }, { "input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C", "output": "529" }, { "input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C", "output": "874" }, { "input": "2\nCC\nCC", "output": "4" }, { "input": "3\nC..\nC..\nC..", "output": "3" } ]
1,593,448,876
2,147,483,647
PyPy 3
OK
TESTS
48
155
21,606,400
from sys import stdin ##################################################################### def iinput(): return int(stdin.readline()) def minput(): return map(int, stdin.readline().split()) def linput(): return list(map(int, stdin.readline().split())) ##################################################################### n = iinput() mat = [] for _ in range(n): mat.append(input()) pair = 0 row, col = [0]*n, [0]*n for i in range(n): for j in range(n): if mat[i][j]=='C': col[j] += 1 row[i] += 1 for e in row: pair += (e*(e-1))//2 for e in col: pair += (e*(e-1))//2 print(pair)
Title: Far Relative’s Birthday Cake Time Limit: None seconds Memory Limit: None megabytes Problem Description: Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be? Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. Input Specification: In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake. Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. Output Specification: Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. Demo Input: ['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n'] Demo Output: ['4\n', '9\n'] Note: If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are: 1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)
```python from sys import stdin ##################################################################### def iinput(): return int(stdin.readline()) def minput(): return map(int, stdin.readline().split()) def linput(): return list(map(int, stdin.readline().split())) ##################################################################### n = iinput() mat = [] for _ in range(n): mat.append(input()) pair = 0 row, col = [0]*n, [0]*n for i in range(n): for j in range(n): if mat[i][j]=='C': col[j] += 1 row[i] += 1 for e in row: pair += (e*(e-1))//2 for e in col: pair += (e*(e-1))//2 print(pair) ```
3
152
A
Marks
PROGRAMMING
900
[ "implementation" ]
null
null
Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group.
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.
Print the single number — the number of successful students in the given group.
[ "3 3\n223\n232\n112\n", "3 5\n91728\n11828\n11111\n" ]
[ "2\n", "3\n" ]
In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.
500
[ { "input": "3 3\n223\n232\n112", "output": "2" }, { "input": "3 5\n91728\n11828\n11111", "output": "3" }, { "input": "2 2\n48\n27", "output": "1" }, { "input": "2 1\n4\n6", "output": "1" }, { "input": "1 2\n57", "output": "1" }, { "input": "1 1\n5", "output": "1" }, { "input": "3 4\n2553\n6856\n5133", "output": "2" }, { "input": "8 7\n6264676\n7854895\n3244128\n2465944\n8958761\n1378945\n3859353\n6615285", "output": "6" }, { "input": "9 8\n61531121\n43529859\n18841327\n88683622\n98995641\n62741632\n57441743\n49396792\n63381994", "output": "4" }, { "input": "10 20\n26855662887514171367\n48525577498621511535\n47683778377545341138\n47331616748732562762\n44876938191354974293\n24577238399664382695\n42724955594463126746\n79187344479926159359\n48349683283914388185\n82157191115518781898", "output": "9" }, { "input": "20 15\n471187383859588\n652657222494199\n245695867594992\n726154672861295\n614617827782772\n862889444974692\n373977167653235\n645434268565473\n785993468314573\n722176861496755\n518276853323939\n723712762593348\n728935312568886\n373898548522463\n769777587165681\n247592995114377\n182375946483965\n497496542536127\n988239919677856\n859844339819143", "output": "18" }, { "input": "13 9\n514562255\n322655246\n135162979\n733845982\n473117129\n513967187\n965649829\n799122777\n661249521\n298618978\n659352422\n747778378\n723261619", "output": "11" }, { "input": "75 1\n2\n3\n8\n3\n2\n1\n3\n1\n5\n1\n5\n4\n8\n8\n4\n2\n5\n1\n7\n6\n3\n2\n2\n3\n5\n5\n2\n4\n7\n7\n9\n2\n9\n5\n1\n4\n9\n5\n2\n4\n6\n6\n3\n3\n9\n3\n3\n2\n3\n4\n2\n6\n9\n1\n1\n1\n1\n7\n2\n3\n2\n9\n7\n4\n9\n1\n7\n5\n6\n8\n3\n4\n3\n4\n6", "output": "7" }, { "input": "92 3\n418\n665\n861\n766\n529\n416\n476\n676\n561\n995\n415\n185\n291\n176\n776\n631\n556\n488\n118\n188\n437\n496\n466\n131\n914\n118\n766\n365\n113\n897\n386\n639\n276\n946\n759\n169\n494\n837\n338\n351\n783\n311\n261\n862\n598\n132\n246\n982\n575\n364\n615\n347\n374\n368\n523\n132\n774\n161\n552\n492\n598\n474\n639\n681\n635\n342\n516\n483\n141\n197\n571\n336\n175\n596\n481\n327\n841\n133\n142\n146\n246\n396\n287\n582\n556\n996\n479\n814\n497\n363\n963\n162", "output": "23" }, { "input": "100 1\n1\n6\n9\n1\n1\n5\n5\n4\n6\n9\n6\n1\n7\n8\n7\n3\n8\n8\n7\n6\n2\n1\n5\n8\n7\n3\n5\n4\n9\n7\n1\n2\n4\n1\n6\n5\n1\n3\n9\n4\n5\n8\n1\n2\n1\n9\n7\n3\n7\n1\n2\n2\n2\n2\n3\n9\n7\n2\n4\n7\n1\n6\n8\n1\n5\n6\n1\n1\n2\n9\n7\n4\n9\n1\n9\n4\n1\n3\n5\n2\n4\n4\n6\n5\n1\n4\n5\n8\n4\n7\n6\n5\n6\n9\n5\n8\n1\n5\n1\n6", "output": "10" }, { "input": "100 2\n71\n87\n99\n47\n22\n87\n49\n73\n21\n12\n77\n43\n18\n41\n78\n62\n61\n16\n64\n89\n81\n54\n53\n92\n93\n94\n68\n93\n15\n68\n42\n93\n28\n19\n86\n16\n97\n17\n11\n43\n72\n76\n54\n95\n58\n53\n48\n45\n85\n85\n74\n21\n44\n51\n89\n75\n76\n17\n38\n62\n81\n22\n66\n59\n89\n85\n91\n87\n12\n97\n52\n87\n43\n89\n51\n58\n57\n98\n78\n68\n82\n41\n87\n29\n75\n72\n48\n14\n35\n71\n74\n91\n66\n67\n42\n98\n52\n54\n22\n41", "output": "21" }, { "input": "5 20\n11111111111111111111\n11111111111111111111\n11111111111111111111\n99999999999999999999\n11111111111111111119", "output": "2" }, { "input": "3 3\n111\n111\n999", "output": "1" }, { "input": "3 3\n119\n181\n711", "output": "3" }, { "input": "15 5\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111", "output": "15" }, { "input": "2 20\n22222222222222222222\n11111111111111111111", "output": "1" }, { "input": "3 3\n233\n222\n111", "output": "2" }, { "input": "4 15\n222222222222222\n111111111111119\n111111111111119\n111111111111111", "output": "3" }, { "input": "4 1\n1\n9\n9\n9", "output": "3" }, { "input": "3 3\n123\n321\n132", "output": "3" }, { "input": "3 3\n113\n332\n322", "output": "3" }, { "input": "2 100\n2222222222222222222222222222222222222222222222222222222222222222222222221222222222222222222222222222\n1111111111111111111111111111111111111111111111111111111111111111111111119111111111111111111111111111", "output": "2" }, { "input": "3 3\n321\n231\n123", "output": "3" }, { "input": "2 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222", "output": "1" }, { "input": "3 3\n221\n111\n111", "output": "3" } ]
1,501,874,593
2,147,483,647
Python 3
OK
TESTS
44
62
4,608,000
n, m = map(int, input().split()) a = [] for i in range(n): a.append(input()) b = [] for j in range(m): b.append(max(a[i][j] for i in range(n))) #print(b) cnt = 0 for i in range(n): for j in range(m): if a[i][j] == b[j]: cnt += 1 break print(cnt)
Title: Marks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group. Input Specification: The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces. Output Specification: Print the single number — the number of successful students in the given group. Demo Input: ['3 3\n223\n232\n112\n', '3 5\n91728\n11828\n11111\n'] Demo Output: ['2\n', '3\n'] Note: In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.
```python n, m = map(int, input().split()) a = [] for i in range(n): a.append(input()) b = [] for j in range(m): b.append(max(a[i][j] for i in range(n))) #print(b) cnt = 0 for i in range(n): for j in range(m): if a[i][j] == b[j]: cnt += 1 break print(cnt) ```
3
845
C
Two TVs
PROGRAMMING
1,500
[ "data structures", "greedy", "sortings" ]
null
null
Polycarp is a great fan of television. He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*. Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV. Polycarp wants to check out all *n* shows. Are two TVs enough to do so?
The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows. Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=&lt;<=*r**i*<=≤<=109) — starting and ending time of *i*-th show.
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
[ "3\n1 2\n2 3\n4 5\n", "4\n1 2\n2 3\n2 3\n1 2\n" ]
[ "YES\n", "NO\n" ]
none
0
[ { "input": "3\n1 2\n2 3\n4 5", "output": "YES" }, { "input": "4\n1 2\n2 3\n2 3\n1 2", "output": "NO" }, { "input": "4\n0 1\n1 2\n2 3\n3 4", "output": "YES" }, { "input": "3\n1 2\n2 3\n2 4", "output": "NO" }, { "input": "3\n0 100\n0 100\n0 100", "output": "NO" }, { "input": "1\n0 1000000000", "output": "YES" }, { "input": "2\n0 1\n0 1", "output": "YES" }, { "input": "3\n2 3\n4 5\n1 6", "output": "YES" }, { "input": "5\n1 3\n1 4\n4 10\n5 8\n9 11", "output": "YES" }, { "input": "3\n1 2\n1 2\n2 3", "output": "NO" }, { "input": "4\n1 100\n10 15\n20 25\n30 35", "output": "YES" }, { "input": "3\n1 8\n6 7\n8 11", "output": "YES" }, { "input": "5\n1 2\n3 5\n4 7\n8 9\n5 10", "output": "NO" }, { "input": "4\n1 7\n2 3\n4 5\n6 7", "output": "YES" }, { "input": "4\n1 100\n50 51\n60 90\n51 52", "output": "NO" }, { "input": "3\n1 10\n2 9\n3 8", "output": "NO" }, { "input": "2\n0 4\n0 4", "output": "YES" }, { "input": "2\n0 2\n0 6", "output": "YES" }, { "input": "5\n3 4\n21 26\n12 17\n9 14\n15 16", "output": "YES" }, { "input": "5\n1 4\n13 15\n11 12\n9 15\n2 5", "output": "YES" }, { "input": "4\n16 19\n9 14\n14 15\n15 19", "output": "YES" }, { "input": "5\n16 19\n23 29\n3 8\n23 26\n22 23", "output": "NO" }, { "input": "5\n19 23\n12 17\n16 21\n20 23\n8 10", "output": "NO" }, { "input": "5\n8 10\n4 10\n3 4\n14 15\n17 19", "output": "YES" }, { "input": "3\n2 8\n5 7\n6 7", "output": "NO" }, { "input": "5\n10 12\n4 6\n21 24\n9 12\n7 13", "output": "NO" }, { "input": "5\n0 3\n14 16\n6 8\n5 9\n9 15", "output": "YES" }, { "input": "5\n6 12\n23 25\n6 7\n19 25\n10 11", "output": "YES" }, { "input": "5\n15 18\n23 24\n23 28\n22 24\n15 19", "output": "NO" }, { "input": "4\n1 8\n8 9\n5 7\n1 4", "output": "YES" }, { "input": "3\n6 10\n1 9\n2 5", "output": "YES" }, { "input": "3\n1 8\n5 6\n6 9", "output": "NO" }, { "input": "4\n2 3\n5 9\n8 10\n9 10", "output": "NO" }, { "input": "4\n0 8\n6 7\n5 9\n1 4", "output": "NO" }, { "input": "3\n6 9\n0 1\n0 2", "output": "YES" }, { "input": "5\n0 6\n21 25\n18 19\n0 3\n6 12", "output": "YES" }, { "input": "4\n1 5\n6 9\n4 8\n1 3", "output": "YES" }, { "input": "2\n2 5\n0 5", "output": "YES" }, { "input": "4\n5 8\n11 15\n3 7\n10 14", "output": "YES" }, { "input": "3\n12 14\n0 4\n2 3", "output": "YES" }, { "input": "4\n4 10\n0 1\n2 10\n0 5", "output": "NO" }, { "input": "4\n0 3\n0 1\n2 4\n2 5", "output": "NO" } ]
1,566,867,529
2,147,483,647
Python 3
OK
TESTS
58
904
22,118,400
# https://vjudge.net/contest/321560#problem/E n = int(input()) shows = [] MAX_TVS = 2 for i in range(n): show_info = [ int(x) for x in input().split() ] shows.append(show_info) shows.sort(key=lambda x : x[0]) tvs = [] i = 0 while len(tvs) <= MAX_TVS and i < n: if len(tvs) == MAX_TVS: if shows[i][0] > tvs[0][1]: tvs.pop(0) elif shows[i][0] > tvs[1][1]: tvs.pop(1) tvs.append(shows[i]) i += 1 result = 'YES' if len(tvs) > MAX_TVS: result = 'NO' print(result)
Title: Two TVs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is a great fan of television. He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*. Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV. Polycarp wants to check out all *n* shows. Are two TVs enough to do so? Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows. Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=&lt;<=*r**i*<=≤<=109) — starting and ending time of *i*-th show. Output Specification: If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes). Demo Input: ['3\n1 2\n2 3\n4 5\n', '4\n1 2\n2 3\n2 3\n1 2\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python # https://vjudge.net/contest/321560#problem/E n = int(input()) shows = [] MAX_TVS = 2 for i in range(n): show_info = [ int(x) for x in input().split() ] shows.append(show_info) shows.sort(key=lambda x : x[0]) tvs = [] i = 0 while len(tvs) <= MAX_TVS and i < n: if len(tvs) == MAX_TVS: if shows[i][0] > tvs[0][1]: tvs.pop(0) elif shows[i][0] > tvs[1][1]: tvs.pop(1) tvs.append(shows[i]) i += 1 result = 'YES' if len(tvs) > MAX_TVS: result = 'NO' print(result) ```
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" } ]
1,684,865,793
2,147,483,647
Python 3
OK
TESTS
102
46
0
n1 = input() n2 = int(input()) n3 = [] size = len(n1) n1 = int(n1) while n1 != 0 or n2 != 0: if n1%10 != n2%10: n3.append(1) if n1%10 == n2%10: n3.append(0) n1 = n1//10 n2 = n2//10 for j in range(size-len(n3)): print(0, end = "") i = len(n3) - 1 while i >= 0: print(n3[i], end = "") i -=1
Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none
```python n1 = input() n2 = int(input()) n3 = [] size = len(n1) n1 = int(n1) while n1 != 0 or n2 != 0: if n1%10 != n2%10: n3.append(1) if n1%10 == n2%10: n3.append(0) n1 = n1//10 n2 = n2//10 for j in range(size-len(n3)): print(0, end = "") i = len(n3) - 1 while i >= 0: print(n3[i], end = "") i -=1 ```
3.9885
617
A
Elephant
PROGRAMMING
800
[ "math" ]
null
null
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
[ "5\n", "12\n" ]
[ "1\n", "3\n" ]
In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
500
[ { "input": "5", "output": "1" }, { "input": "12", "output": "3" }, { "input": "999999", "output": "200000" }, { "input": "41", "output": "9" }, { "input": "1000000", "output": "200000" }, { "input": "1", "output": "1" }, { "input": "2", "output": "1" }, { "input": "3", "output": "1" }, { "input": "4", "output": "1" }, { "input": "534204", "output": "106841" }, { "input": "469569", "output": "93914" }, { "input": "502877", "output": "100576" }, { "input": "942212", "output": "188443" }, { "input": "97", "output": "20" }, { "input": "53", "output": "11" }, { "input": "89", "output": "18" }, { "input": "574", "output": "115" }, { "input": "716", "output": "144" }, { "input": "729", "output": "146" }, { "input": "8901", "output": "1781" }, { "input": "3645", "output": "729" }, { "input": "4426", "output": "886" }, { "input": "46573", "output": "9315" }, { "input": "86380", "output": "17276" }, { "input": "94190", "output": "18838" }, { "input": "999990", "output": "199998" }, { "input": "999991", "output": "199999" }, { "input": "999992", "output": "199999" }, { "input": "999993", "output": "199999" }, { "input": "999994", "output": "199999" }, { "input": "999995", "output": "199999" }, { "input": "999996", "output": "200000" }, { "input": "999997", "output": "200000" }, { "input": "999998", "output": "200000" } ]
1,696,237,346
2,147,483,647
Python 3
OK
TESTS
34
46
0
x = int(input()) complete_groups = x // 5 if x % 5 != 0: steps_needed = complete_groups + 1 else: steps_needed = complete_groups print(steps_needed)
Title: Elephant Time Limit: None seconds Memory Limit: None megabytes Problem Description: An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. Input Specification: The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. Output Specification: Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. Demo Input: ['5\n', '12\n'] Demo Output: ['1\n', '3\n'] Note: In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
```python x = int(input()) complete_groups = x // 5 if x % 5 != 0: steps_needed = complete_groups + 1 else: steps_needed = complete_groups print(steps_needed) ```
3
81
A
Plug-in
PROGRAMMING
1,400
[ "implementation" ]
A. Plug-in
1
256
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu". Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something! Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
[ "hhoowaaaareyyoouu\n", "reallazy\n", "abacabaabacabaa\n" ]
[ "wre", "rezy", "a" ]
none
500
[ { "input": "hhoowaaaareyyoouu", "output": "wre" }, { "input": "reallazy", "output": "rezy" }, { "input": "abacabaabacabaa", "output": "a" }, { "input": "xraccabccbry", "output": "xy" }, { "input": "a", "output": "a" }, { "input": "b", "output": "b" }, { "input": "il", "output": "il" }, { "input": "gfj", "output": "gfj" }, { "input": "babbbbbababa", "output": "babababa" }, { "input": "babbbbabbabbbababbabbbbbbabaabaababaaabbbbbabbbbaaaaabbaaabbaabaabbbbabbbababbabaaabbababaaababbbaaa", "output": "babababababababababababa" }, { "input": "aab", "output": "b" }, { "input": "abb", "output": "a" }, { "input": "aba", "output": "aba" }, { "input": "ab", "output": "ab" } ]
1,623,963,615
2,147,483,647
Python 3
OK
TESTS
32
218
2,867,200
import sys class Solution: def plugin(self,s:str)->str: size = len(s) current =1 stack = [s[0]] while current<size: if stack and stack[-1] == s[current]: stack.pop() else: stack.append(s[current]) current+=1 return "".join(stack) if __name__ == '__main__': s = Solution() for text in sys.stdin: print(s.plugin(text))
Title: Plug-in Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu". Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something! Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result. Input Specification: The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters. Output Specification: Print the given string after it is processed. It is guaranteed that the result will contain at least one character. Demo Input: ['hhoowaaaareyyoouu\n', 'reallazy\n', 'abacabaabacabaa\n'] Demo Output: ['wre', 'rezy', 'a'] Note: none
```python import sys class Solution: def plugin(self,s:str)->str: size = len(s) current =1 stack = [s[0]] while current<size: if stack and stack[-1] == s[current]: stack.pop() else: stack.append(s[current]) current+=1 return "".join(stack) if __name__ == '__main__': s = Solution() for text in sys.stdin: print(s.plugin(text)) ```
3.885659
550
A
Two Substrings
PROGRAMMING
1,500
[ "brute force", "dp", "greedy", "implementation", "strings" ]
null
null
You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters.
Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
[ "ABA\n", "BACFAB\n", "AXBYBXA\n" ]
[ "NO\n", "YES\n", "NO\n" ]
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".
1,000
[ { "input": "ABA", "output": "NO" }, { "input": "BACFAB", "output": "YES" }, { "input": "AXBYBXA", "output": "NO" }, { "input": "ABABAB", "output": "YES" }, { "input": "BBBBBBBBBB", "output": "NO" }, { "input": "ABBA", "output": "YES" }, { "input": "ABAXXXAB", "output": "YES" }, { "input": "TESTABAXXABTEST", "output": "YES" }, { "input": "A", "output": "NO" }, { "input": "B", "output": "NO" }, { "input": "X", "output": "NO" }, { "input": "BA", "output": "NO" }, { "input": "AB", "output": "NO" }, { "input": "AA", "output": "NO" }, { "input": "BB", "output": "NO" }, { "input": "BAB", "output": "NO" }, { "input": "AAB", "output": "NO" }, { "input": "BAA", "output": "NO" }, { "input": "ABB", "output": "NO" }, { "input": "BBA", "output": "NO" }, { "input": "AAA", "output": "NO" }, { "input": "BBB", "output": "NO" }, { "input": "AXBXBXA", "output": "NO" }, { "input": "SKDSKDJABSDBADKFJDK", "output": "YES" }, { "input": "ABAXXBBXXAA", "output": "NO" }, { "input": "ABAB", "output": "NO" }, { "input": "BABA", "output": "NO" }, { "input": "AAAB", "output": "NO" }, { "input": "AAAA", "output": "NO" }, { "input": "AABA", "output": "NO" }, { "input": "ABAA", "output": "NO" }, { "input": "BAAA", "output": "NO" }, { "input": "AABB", "output": "NO" }, { "input": "BAAB", "output": "YES" }, { "input": "BBAA", "output": "NO" }, { "input": "BBBA", "output": "NO" }, { "input": "BBAB", "output": "NO" }, { "input": "BABB", "output": "NO" }, { "input": "ABBB", "output": "NO" }, { "input": "BBBB", "output": "NO" }, { "input": "BABAB", "output": "YES" }, { "input": "ABABA", "output": "YES" }, { "input": "AAABAAACBBBC", "output": "NO" }, { "input": "AABBBACBBBBBBAACBCCACBBAABBBBBCAACABAACABCACCBCBCCCBCBCABCBBCCBCBBAACBACAABACBBCACCBCCACCABBCBABBAAC", "output": "YES" }, { "input": "CBBABDDBBADAC", "output": "YES" }, { "input": "ABYFAB", "output": "NO" }, { "input": "BABXXAB", "output": "YES" }, { "input": "ABAXAB", "output": "YES" }, { "input": "ABABXAB", "output": "YES" }, { "input": "ABXABXABXABXABXBAXBAXBAXBA", "output": "YES" }, { "input": "QQQQQQQQQABABQQQQQQQQ", "output": "NO" } ]
1,684,870,340
2,147,483,647
PyPy 3-64
OK
TESTS
85
62
4,915,200
s=str(input()) n=len(s) l1,l2=[i for i in range(n-1)if s[i:i+2]=="AB"],[i for i in range(n-1)if s[i:i+2]=="BA"] print("YES" if any(abs(i-j)>=2 for i in l1 for j in l2) else "NO")
Title: Two Substrings Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). Input Specification: The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters. Output Specification: Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. Demo Input: ['ABA\n', 'BACFAB\n', 'AXBYBXA\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n'] Note: In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".
```python s=str(input()) n=len(s) l1,l2=[i for i in range(n-1)if s[i:i+2]=="AB"],[i for i in range(n-1)if s[i:i+2]=="BA"] print("YES" if any(abs(i-j)>=2 for i in l1 for j in l2) else "NO") ```
3
450
A
Jzzhu and Children
PROGRAMMING
1,000
[ "implementation" ]
null
null
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies. Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm: 1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty. Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output a single integer, representing the number of the last child.
[ "5 2\n1 3 1 4 2\n", "6 4\n1 1 2 2 3 3\n" ]
[ "4\n", "6\n" ]
Let's consider the first sample. Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home. Child 4 is the last one who goes home.
500
[ { "input": "5 2\n1 3 1 4 2", "output": "4" }, { "input": "6 4\n1 1 2 2 3 3", "output": "6" }, { "input": "7 3\n6 1 5 4 2 3 1", "output": "4" }, { "input": "10 5\n2 7 3 6 2 5 1 3 4 5", "output": "4" }, { "input": "100 1\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": "100" }, { "input": "9 3\n9 5 2 3 7 1 8 4 6", "output": "7" }, { "input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58", "output": "16" }, { "input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54", "output": "48" }, { "input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1", "output": "44" }, { "input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3", "output": "46" }, { "input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78", "output": "40" }, { "input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69", "output": "11" }, { "input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95", "output": "50" }, { "input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9", "output": "86" }, { "input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7", "output": "93" }, { "input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42", "output": "77" }, { "input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48", "output": "97" }, { "input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78", "output": "20" }, { "input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41", "output": "97" }, { "input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82", "output": "51" }, { "input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58", "output": "97" }, { "input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88", "output": "47" }, { "input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51", "output": "98" }, { "input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67", "output": "94" }, { "input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19", "output": "95" }, { "input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49", "output": "88" }, { "input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1", "output": "99" }, { "input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95", "output": "100" }, { "input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15", "output": "98" }, { "input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16", "output": "100" }, { "input": "1 3\n5", "output": "1" }, { "input": "1 1\n100", "output": "1" }, { "input": "2 3\n4 2", "output": "1" }, { "input": "2 5\n99 97", "output": "2" }, { "input": "3 4\n7 5 2", "output": "2" }, { "input": "3 50\n47 86 51", "output": "3" }, { "input": "5 100\n82 100 85 1 37", "output": "5" }, { "input": "5 20\n40 39 21 5 20", "output": "3" }, { "input": "1 27\n81", "output": "1" }, { "input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1", "output": "7" }, { "input": "2 1\n100 2", "output": "1" }, { "input": "2 2\n6 4", "output": "1" }, { "input": "5 2\n6 4 4 1 1", "output": "1" }, { "input": "1 4\n3", "output": "1" }, { "input": "3 2\n1 5 3", "output": "2" }, { "input": "3 1\n3 2 2", "output": "1" }, { "input": "3 1\n2 3 2", "output": "2" }, { "input": "5 1\n5 1 1 1 1", "output": "1" }, { "input": "2 3\n7 4", "output": "1" } ]
1,699,187,328
2,147,483,647
Python 3
OK
TESTS
49
46
0
n,m = map(int,input().split()) l= list(map(int,input().split())) l1 = [i for i in range(n)] while len(l)>1: l[0] -= m if l[0]>0: l.append(l[0]) l1.append(l1[0]) l.pop(0) l1.pop(0) else: l.pop(0) l1.pop(0) print(l1[0]+1)
Title: Jzzhu and Children Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies. Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm: 1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty. Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? Input Specification: The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). Output Specification: Output a single integer, representing the number of the last child. Demo Input: ['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n'] Demo Output: ['4\n', '6\n'] Note: Let's consider the first sample. Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home. Child 4 is the last one who goes home.
```python n,m = map(int,input().split()) l= list(map(int,input().split())) l1 = [i for i in range(n)] while len(l)>1: l[0] -= m if l[0]>0: l.append(l[0]) l1.append(l1[0]) l.pop(0) l1.pop(0) else: l.pop(0) l1.pop(0) print(l1[0]+1) ```
3
38
A
Army
PROGRAMMING
800
[ "implementation" ]
A. Army
2
256
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
[ "3\n5 6\n1 2\n", "3\n5 6\n1 3\n" ]
[ "5\n", "11\n" ]
none
0
[ { "input": "3\n5 6\n1 2", "output": "5" }, { "input": "3\n5 6\n1 3", "output": "11" }, { "input": "2\n55\n1 2", "output": "55" }, { "input": "3\n85 78\n1 3", "output": "163" }, { "input": "4\n63 4 49\n2 3", "output": "4" }, { "input": "5\n93 83 42 56\n2 5", "output": "181" }, { "input": "6\n22 9 87 89 57\n1 6", "output": "264" }, { "input": "7\n52 36 31 23 74 78\n2 7", "output": "242" }, { "input": "8\n82 14 24 5 91 49 94\n3 8", "output": "263" }, { "input": "9\n12 40 69 39 59 21 59 5\n4 6", "output": "98" }, { "input": "10\n95 81 32 59 71 30 50 61 100\n1 6", "output": "338" }, { "input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14", "output": "617" }, { "input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17", "output": "399" }, { "input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23", "output": "846" }, { "input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16", "output": "730" }, { "input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35", "output": "1663" }, { "input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26", "output": "862" }, { "input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40", "output": "1061" }, { "input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28", "output": "344" }, { "input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60", "output": "1502" }, { "input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41", "output": "1767" }, { "input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24", "output": "219" }, { "input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44", "output": "1713" }, { "input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15", "output": "570" }, { "input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26", "output": "328" }, { "input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68", "output": "3385" }, { "input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77", "output": "3319" }, { "input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36", "output": "514" }, { "input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92", "output": "2599" }, { "input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27", "output": "618" }, { "input": "2\n3\n1 2", "output": "3" }, { "input": "5\n6 8 22 22\n2 3", "output": "8" }, { "input": "6\n3 12 27 28 28\n3 4", "output": "27" }, { "input": "9\n1 2 2 2 2 3 3 5\n3 7", "output": "9" }, { "input": "10\n1 1 1 1 1 1 1 1 1\n6 8", "output": "2" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17", "output": "23" }, { "input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8", "output": "23" }, { "input": "35\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\n30 31", "output": "2" }, { "input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45", "output": "30" }, { "input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17", "output": "285" }, { "input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58", "output": "173" }, { "input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65", "output": "126" }, { "input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65", "output": "326" }, { "input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48", "output": "784" }, { "input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93", "output": "1636" }, { "input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96", "output": "898" }, { "input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90", "output": "3016" }, { "input": "99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n66 95", "output": "29" }, { "input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52", "output": "82" }, { "input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44", "output": "2268" } ]
1,659,487,723
2,147,483,647
PyPy 3-64
OK
TESTS
50
124
0
import sys input = sys.stdin.readline n = int(input()) d = list(map(int, input().split())) a, b = map(int, input().split()) ans = sum(d[a - 1:b - 1]) print(ans)
Title: Army Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. Input Specification: The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated. Output Specification: Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. Demo Input: ['3\n5 6\n1 2\n', '3\n5 6\n1 3\n'] Demo Output: ['5\n', '11\n'] Note: none
```python import sys input = sys.stdin.readline n = int(input()) d = list(map(int, input().split())) a, b = map(int, input().split()) ans = sum(d[a - 1:b - 1]) print(ans) ```
3.969
102
B
Sum of Digits
PROGRAMMING
1,000
[ "implementation" ]
B. Sum of Digits
2
265
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
[ "0\n", "10\n", "991\n" ]
[ "0\n", "1\n", "3\n" ]
In the first sample the number already is one-digit — Herald can't cast a spell. The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once. The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
1,000
[ { "input": "0", "output": "0" }, { "input": "10", "output": "1" }, { "input": "991", "output": "3" }, { "input": "99", "output": "2" }, { "input": "100", "output": "1" }, { "input": "123456789", "output": "2" }, { "input": "32", "output": "1" }, { "input": "86", "output": "2" }, { "input": "2", "output": "0" }, { "input": "8", "output": "0" }, { "input": "34", "output": "1" }, { "input": "13", "output": "1" }, { "input": "28", "output": "2" }, { "input": "23", "output": "1" }, { "input": "57", "output": "2" }, { "input": "29", "output": "2" }, { "input": "353869285", "output": "3" }, { "input": "549548646", "output": "2" }, { "input": "858893424", "output": "2" }, { "input": "644818852", "output": "3" }, { "input": "360322525", "output": "3" }, { "input": "122937520", "output": "2" }, { "input": "288403032", "output": "2" }, { "input": "677257481", "output": "3" }, { "input": "58059561", "output": "3" }, { "input": "211288847", "output": "2" }, { "input": "339900034079539584", "output": "2" }, { "input": "784084029310269952", "output": "3" }, { "input": "774730557264864000", "output": "3" }, { "input": "478233807148352256", "output": "3" }, { "input": "165646874056812544", "output": "3" }, { "input": "477533739511673792", "output": "3" }, { "input": "660119944377002240", "output": "3" }, { "input": "448375218506078080", "output": "3" }, { "input": "77213334785795616", "output": "3" }, { "input": "165767221702271872", "output": "3" }, { "input": "1", "output": "0" }, { "input": "5", "output": "0" }, { "input": "8", "output": "0" }, { "input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561", "output": "3" }, { "input": "9", "output": "0" } ]
1,611,650,873
2,147,483,647
Python 3
OK
TESTS
51
592
307,200
def add(k): sum = 0 for i in k: sum += int(i) return sum n = input() spell = 0 if int(n) > 9: while add(n) > 9: sum_of_digits = add(n) n = str(sum_of_digits) spell += 1 print(spell+1) else: print(0)
Title: Sum of Digits Time Limit: 2 seconds Memory Limit: 265 megabytes Problem Description: Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit? Input Specification: The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes. Output Specification: Print the number of times a number can be replaced by the sum of its digits until it only contains one digit. Demo Input: ['0\n', '10\n', '991\n'] Demo Output: ['0\n', '1\n', '3\n'] Note: In the first sample the number already is one-digit — Herald can't cast a spell. The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once. The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
```python def add(k): sum = 0 for i in k: sum += int(i) return sum n = input() spell = 0 if int(n) > 9: while add(n) > 9: sum_of_digits = add(n) n = str(sum_of_digits) spell += 1 print(spell+1) else: print(0) ```
3.851447
755
A
PolandBall and Hypothesis
PROGRAMMING
800
[ "brute force", "graphs", "math", "number theory" ]
null
null
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number". Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
[ "3\n", "4\n" ]
[ "1", "2" ]
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. For the first sample testcase, 3·1 + 1 = 4. We can output 1. In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
500
[ { "input": "3", "output": "1" }, { "input": "4", "output": "2" }, { "input": "10", "output": "2" }, { "input": "153", "output": "1" }, { "input": "1000", "output": "1" }, { "input": "1", "output": "3" }, { "input": "2", "output": "4" }, { "input": "5", "output": "1" }, { "input": "6", "output": "4" }, { "input": "7", "output": "1" }, { "input": "8", "output": "1" }, { "input": "9", "output": "1" }, { "input": "11", "output": "1" }, { "input": "998", "output": "1" }, { "input": "996", "output": "3" }, { "input": "36", "output": "4" }, { "input": "210", "output": "4" }, { "input": "270", "output": "4" }, { "input": "306", "output": "4" }, { "input": "330", "output": "5" }, { "input": "336", "output": "4" }, { "input": "600", "output": "4" }, { "input": "726", "output": "4" }, { "input": "988", "output": "1" }, { "input": "12", "output": "2" }, { "input": "987", "output": "1" }, { "input": "13", "output": "1" }, { "input": "986", "output": "1" }, { "input": "14", "output": "1" }, { "input": "985", "output": "1" }, { "input": "15", "output": "1" }, { "input": "984", "output": "1" }, { "input": "16", "output": "2" }, { "input": "983", "output": "1" }, { "input": "17", "output": "1" }, { "input": "982", "output": "2" }, { "input": "18", "output": "3" }, { "input": "981", "output": "1" }, { "input": "19", "output": "1" }, { "input": "980", "output": "1" }, { "input": "20", "output": "1" }, { "input": "979", "output": "1" }, { "input": "21", "output": "1" }, { "input": "978", "output": "1" }, { "input": "22", "output": "2" }, { "input": "977", "output": "1" }, { "input": "23", "output": "1" }, { "input": "976", "output": "2" }, { "input": "24", "output": "1" }, { "input": "975", "output": "1" }, { "input": "25", "output": "1" }, { "input": "2", "output": "4" }, { "input": "6", "output": "4" }, { "input": "996", "output": "3" }, { "input": "8", "output": "1" }, { "input": "998", "output": "1" }, { "input": "330", "output": "5" }, { "input": "18", "output": "3" }, { "input": "12", "output": "2" }, { "input": "24", "output": "1" }, { "input": "999", "output": "1" }, { "input": "726", "output": "4" }, { "input": "876", "output": "3" } ]
1,607,713,606
2,147,483,647
PyPy 3
OK
TESTS
63
140
1,228,800
def cp(m): for i in range(2,m//2): if m%i == 0: return False return True n=int(input()) for j in range(1,1000): x=cp(n*j+1) if not x: print(j) break
Title: PolandBall and Hypothesis Time Limit: None seconds Memory Limit: None megabytes Problem Description: PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number". Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*. Input Specification: The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis. Output Specification: Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists. Demo Input: ['3\n', '4\n'] Demo Output: ['1', '2'] Note: A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. For the first sample testcase, 3·1 + 1 = 4. We can output 1. In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
```python def cp(m): for i in range(2,m//2): if m%i == 0: return False return True n=int(input()) for j in range(1,1000): x=cp(n*j+1) if not x: print(j) break ```
3
526
B
Om Nom and Dark Park
PROGRAMMING
1,400
[ "dfs and similar", "greedy", "implementation" ]
null
null
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him. The park consists of 2*n*<=+<=1<=-<=1 squares connected by roads so that the scheme of the park is a full binary tree of depth *n*. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2*n*,<=2*n*<=+<=1,<=...,<=2*n*<=+<=1<=-<=1 and these exits lead straight to the Om Nom friends' houses. From each square *i* (2<=≤<=*i*<=&lt;<=2*n*<=+<=1) there is a road to the square . Thus, it is possible to go from the park entrance to each of the exits by walking along exactly *n* roads. Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe. He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
The first line contains integer *n* (1<=≤<=*n*<=≤<=10) — the number of roads on the path from the entrance to any exit. The next line contains 2*n*<=+<=1<=-<=2 numbers *a*2,<=*a*3,<=... *a*2*n*<=+<=1<=-<=1 — the initial numbers of street lights on each road of the park. Here *a**i* is the number of street lights on the road between squares *i* and . All numbers *a**i* are positive integers, not exceeding 100.
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
[ "2\n1 2 3 4 5 6\n" ]
[ "5\n" ]
Picture for the sample test. Green color denotes the additional street lights.
500
[ { "input": "2\n1 2 3 4 5 6", "output": "5" }, { "input": "2\n1 2 3 3 2 2", "output": "0" }, { "input": "1\n39 52", "output": "13" }, { "input": "2\n59 96 34 48 8 72", "output": "139" }, { "input": "3\n87 37 91 29 58 45 51 74 70 71 47 38 91 89", "output": "210" }, { "input": "5\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82 91 20 64 52 70 6 88 53 47 30 47 34 14 11 22 42 15 28 54 37 48 29 3 14 13 18 77 90 58 54 38 94 49 45 66 13 74 11 14 64 72 95 54 73 79 41 35", "output": "974" }, { "input": "1\n49 36", "output": "13" }, { "input": "1\n77 88", "output": "11" }, { "input": "1\n1 33", "output": "32" }, { "input": "2\n72 22 81 23 14 75", "output": "175" }, { "input": "2\n100 70 27 1 68 52", "output": "53" }, { "input": "2\n24 19 89 82 22 21", "output": "80" }, { "input": "3\n86 12 92 91 3 68 57 56 76 27 33 62 71 84", "output": "286" }, { "input": "3\n14 56 53 61 57 45 40 44 31 9 73 2 61 26", "output": "236" }, { "input": "3\n35 96 7 43 10 14 16 36 95 92 16 50 59 55", "output": "173" }, { "input": "4\n1 97 18 48 96 65 24 91 17 45 36 27 74 93 78 86 39 55 53 21 26 68 31 33 79 63 80 92 1 26", "output": "511" }, { "input": "4\n25 42 71 29 50 30 99 79 77 24 76 66 68 23 97 99 65 17 75 62 66 46 48 4 40 71 98 57 21 92", "output": "603" }, { "input": "4\n49 86 17 7 3 6 86 71 36 10 27 10 58 64 12 16 88 67 93 3 15 20 58 87 97 91 11 6 34 62", "output": "470" }, { "input": "5\n16 87 36 16 81 53 87 35 63 56 47 91 81 95 80 96 91 7 58 99 25 28 47 60 7 69 49 14 51 52 29 30 83 23 21 52 100 26 91 14 23 94 72 70 40 12 50 32 54 52 18 74 5 15 62 3 48 41 24 25 56 43", "output": "1060" }, { "input": "5\n40 27 82 94 38 22 66 23 18 34 87 31 71 28 95 5 14 61 76 52 66 6 60 40 68 77 70 63 64 18 47 13 82 55 34 64 30 1 29 24 24 9 65 17 29 96 61 76 72 23 32 26 90 39 54 41 35 66 71 29 75 48", "output": "1063" }, { "input": "5\n64 72 35 68 92 95 45 15 77 16 26 74 61 65 18 22 32 19 98 97 14 84 70 23 29 1 87 28 88 89 73 79 69 88 43 60 64 64 66 39 17 27 46 71 18 83 73 20 90 77 49 70 84 63 50 72 26 87 26 37 78 65", "output": "987" }, { "input": "6\n35 61 54 77 70 50 53 70 4 66 58 47 76 100 78 5 43 50 55 93 13 93 59 92 30 74 22 23 98 70 19 56 90 92 19 7 28 53 45 77 42 91 71 56 19 83 100 53 13 93 37 13 70 60 16 13 76 3 12 22 17 26 50 6 63 7 25 41 92 29 36 80 11 4 10 14 77 75 53 82 46 24 56 46 82 36 80 75 8 45 24 22 90 34 45 76 18 38 86 43 7 49 80 56 90 53 12 51 98 47 44 58 32 4 2 6 3 60 38 72 74 46 30 86 1 98", "output": "2499" }, { "input": "6\n63 13 100 54 31 15 29 58 59 44 2 99 70 33 97 14 70 12 73 42 65 71 68 67 87 83 43 84 18 41 37 22 81 24 27 11 57 28 83 92 39 1 56 15 16 67 16 97 31 52 50 65 63 89 8 52 55 20 71 27 28 35 86 92 94 60 10 65 83 63 89 71 34 20 78 40 34 62 2 86 100 81 87 69 25 4 52 17 57 71 62 38 1 3 54 71 34 85 20 60 80 23 82 47 4 19 7 18 14 18 28 27 4 55 26 71 45 9 2 40 67 28 32 19 81 92", "output": "2465" }, { "input": "6\n87 62 58 32 81 92 12 50 23 27 38 39 64 74 16 35 84 59 91 87 14 48 90 47 44 95 64 45 31 11 67 5 80 60 36 15 91 3 21 2 40 24 37 69 5 50 23 37 49 19 68 21 49 9 100 94 45 41 22 31 31 48 25 70 25 25 95 88 82 1 37 53 49 31 57 74 94 45 55 93 43 37 13 85 59 72 15 68 3 90 96 55 100 64 63 69 43 33 66 84 57 97 87 34 23 89 97 77 39 89 8 92 68 13 50 36 95 61 71 96 73 13 30 49 57 89", "output": "2513" } ]
1,499,946,387
2,147,483,647
Python 3
OK
TESTS
38
62
5,529,600
n = int(input()) ar = [int(i) for i in input().split()] tree = [] it = 0 for i in range(n): tr = [] for j in range(it,it+2**(i+1)): tr.append(ar[j]) tree.append(tr) it += 2**(i+1) add = 0 for i in range(n-1,0,-1): for j in range(len(tree[i-1])): diff = tree[i][2*j] - tree[i][2*j+1] if diff >= 0: add += diff tree[i-1][j] += tree[i][2*j] else: add -= diff tree[i-1][j] += tree[i][2*j+1] del tree[i] diff = tree[0][0] - tree[0][1] if diff >= 0: add += diff else: add -= diff print(add)
Title: Om Nom and Dark Park Time Limit: None seconds Memory Limit: None megabytes Problem Description: Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him. The park consists of 2*n*<=+<=1<=-<=1 squares connected by roads so that the scheme of the park is a full binary tree of depth *n*. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2*n*,<=2*n*<=+<=1,<=...,<=2*n*<=+<=1<=-<=1 and these exits lead straight to the Om Nom friends' houses. From each square *i* (2<=≤<=*i*<=&lt;<=2*n*<=+<=1) there is a road to the square . Thus, it is possible to go from the park entrance to each of the exits by walking along exactly *n* roads. Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe. He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=10) — the number of roads on the path from the entrance to any exit. The next line contains 2*n*<=+<=1<=-<=2 numbers *a*2,<=*a*3,<=... *a*2*n*<=+<=1<=-<=1 — the initial numbers of street lights on each road of the park. Here *a**i* is the number of street lights on the road between squares *i* and . All numbers *a**i* are positive integers, not exceeding 100. Output Specification: Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe. Demo Input: ['2\n1 2 3 4 5 6\n'] Demo Output: ['5\n'] Note: Picture for the sample test. Green color denotes the additional street lights.
```python n = int(input()) ar = [int(i) for i in input().split()] tree = [] it = 0 for i in range(n): tr = [] for j in range(it,it+2**(i+1)): tr.append(ar[j]) tree.append(tr) it += 2**(i+1) add = 0 for i in range(n-1,0,-1): for j in range(len(tree[i-1])): diff = tree[i][2*j] - tree[i][2*j+1] if diff >= 0: add += diff tree[i-1][j] += tree[i][2*j] else: add -= diff tree[i-1][j] += tree[i][2*j+1] del tree[i] diff = tree[0][0] - tree[0][1] if diff >= 0: add += diff else: add -= diff print(add) ```
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" } ]
1,547,484,945
2,147,483,647
Python 3
OK
TESTS
81
248
0
n = int(input()) v1 = 0 v2 = 0 v3 = 0 i = 0 for i in range (n): x , y , z = map(int, input().split()) v1 = v1 + x; v2 = v2 + y; v3 = v3 + z; if v1 == 0 and v2 == 0 and v3 == 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()) v1 = 0 v2 = 0 v3 = 0 i = 0 for i in range (n): x , y , z = map(int, input().split()) v1 = v1 + x; v2 = v2 + y; v3 = v3 + z; if v1 == 0 and v2 == 0 and v3 == 0 : print("YES") else : print("NO") ```
3.938
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" } ]
1,604,212,531
2,147,483,647
Python 3
OK
TESTS
40
218
0
print((lambda x,y: "YES" if(x[::-1] == y) else "NO")(input(), input()))
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 print((lambda x,y: "YES" if(x[::-1] == y) else "NO")(input(), input())) ```
3.9455
96
A
Football
PROGRAMMING
900
[ "implementation", "strings" ]
A. Football
2
256
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Print "YES" if the situation is dangerous. Otherwise, print "NO".
[ "001001\n", "1000000001\n" ]
[ "NO\n", "YES\n" ]
none
500
[ { "input": "001001", "output": "NO" }, { "input": "1000000001", "output": "YES" }, { "input": "00100110111111101", "output": "YES" }, { "input": "11110111111111111", "output": "YES" }, { "input": "01", "output": "NO" }, { "input": "10100101", "output": "NO" }, { "input": "1010010100000000010", "output": "YES" }, { "input": "101010101", "output": "NO" }, { "input": "000000000100000000000110101100000", "output": "YES" }, { "input": "100001000000110101100000", "output": "NO" }, { "input": "100001000011010110000", "output": "NO" }, { "input": "010", "output": "NO" }, { "input": "10101011111111111111111111111100", "output": "YES" }, { "input": "1001101100", "output": "NO" }, { "input": "1001101010", "output": "NO" }, { "input": "1111100111", "output": "NO" }, { "input": "00110110001110001111", "output": "NO" }, { "input": "11110001001111110001", "output": "NO" }, { "input": "10001111001011111101", "output": "NO" }, { "input": "10000010100000001000110001010100001001001010011", "output": "YES" }, { "input": "01111011111010111100101100001011001010111110000010", "output": "NO" }, { "input": "00100000100100101110011001011011101110110110010100", "output": "NO" }, { "input": "10110100110001001011110101110010100010000000000100101010111110111110100011", "output": "YES" }, { "input": "00011101010101111001011011001101101011111101000010100000111000011100101011", "output": "NO" }, { "input": "01110000110100110101110100111000101101011101011110110100100111100001110111", "output": "NO" }, { "input": "11110110011000100111100111101101011111110100010101011011111101110110110111", "output": "YES" }, { "input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100", "output": "NO" }, { "input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010", "output": "NO" }, { "input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110", "output": "NO" }, { "input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000", "output": "NO" }, { "input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110", "output": "YES" }, { "input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000", "output": "NO" }, { "input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101", "output": "YES" }, { "input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101", "output": "YES" }, { "input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000", "output": "NO" }, { "input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000", "output": "YES" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "YES" }, { "input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111", "output": "YES" }, { "input": "10100101000", "output": "NO" }, { "input": "11110111011101", "output": "NO" }, { "input": "10000000", "output": "YES" }, { "input": "00000001", "output": "YES" }, { "input": "01111111", "output": "YES" }, { "input": "11111110", "output": "YES" } ]
1,665,387,165
2,147,483,647
Python 3
OK
TESTS
44
62
0
p=input() z=0 o=0 flag=0 for i in p: if i == "0": z+=1 o=0 if z >= 7 and flag==0: print("YES") flag=1 if i == "1": o+=1 z=0 if o >= 7 and flag==0: print("YES") flag=1 if flag==0: print("NO")
Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. Input Specification: The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. Output Specification: Print "YES" if the situation is dangerous. Otherwise, print "NO". Demo Input: ['001001\n', '1000000001\n'] Demo Output: ['NO\n', 'YES\n'] Note: none
```python p=input() z=0 o=0 flag=0 for i in p: if i == "0": z+=1 o=0 if z >= 7 and flag==0: print("YES") flag=1 if i == "1": o+=1 z=0 if o >= 7 and flag==0: print("YES") flag=1 if flag==0: print("NO") ```
3.9845
977
A
Wrong Subtraction
PROGRAMMING
800
[ "implementation" ]
null
null
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm: - if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit). You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions. It is guaranteed that the result will be positive integer number.
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Print one integer number — the result of the decreasing $n$ by one $k$ times. It is guaranteed that the result will be positive integer number.
[ "512 4\n", "1000000000 9\n" ]
[ "50\n", "1\n" ]
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
0
[ { "input": "512 4", "output": "50" }, { "input": "1000000000 9", "output": "1" }, { "input": "131203 11", "output": "12" }, { "input": "999999999 50", "output": "9999" }, { "input": "999999999 49", "output": "99990" }, { "input": "131203 9", "output": "130" }, { "input": "900000000 16", "output": "1" }, { "input": "909090909 50", "output": "3" }, { "input": "1001 2", "output": "100" }, { "input": "5 2", "output": "3" }, { "input": "2 1", "output": "1" } ]
1,697,692,017
2,147,483,647
Python 3
OK
TESTS
11
46
0
nk=list(map(int, input().split())) for i in range(nk[1]): if nk[0]%10==0: nk[0]/=10 else: nk[0]-=1 print(int(nk[0]))
Title: Wrong Subtraction Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm: - if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit). You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions. It is guaranteed that the result will be positive integer number. Input Specification: The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly. Output Specification: Print one integer number — the result of the decreasing $n$ by one $k$ times. It is guaranteed that the result will be positive integer number. Demo Input: ['512 4\n', '1000000000 9\n'] Demo Output: ['50\n', '1\n'] Note: The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
```python nk=list(map(int, input().split())) for i in range(nk[1]): if nk[0]%10==0: nk[0]/=10 else: nk[0]-=1 print(int(nk[0])) ```
3
723
A
The New Year: Meeting Friends
PROGRAMMING
800
[ "implementation", "math", "sortings" ]
null
null
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer.
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Print one integer — the minimum total distance the friends need to travel in order to meet together.
[ "7 1 4\n", "30 20 10\n" ]
[ "6\n", "20\n" ]
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
500
[ { "input": "7 1 4", "output": "6" }, { "input": "30 20 10", "output": "20" }, { "input": "1 4 100", "output": "99" }, { "input": "100 1 91", "output": "99" }, { "input": "1 45 100", "output": "99" }, { "input": "1 2 3", "output": "2" }, { "input": "71 85 88", "output": "17" }, { "input": "30 38 99", "output": "69" }, { "input": "23 82 95", "output": "72" }, { "input": "22 41 47", "output": "25" }, { "input": "9 94 77", "output": "85" }, { "input": "1 53 51", "output": "52" }, { "input": "25 97 93", "output": "72" }, { "input": "42 53 51", "output": "11" }, { "input": "81 96 94", "output": "15" }, { "input": "21 5 93", "output": "88" }, { "input": "50 13 75", "output": "62" }, { "input": "41 28 98", "output": "70" }, { "input": "69 46 82", "output": "36" }, { "input": "87 28 89", "output": "61" }, { "input": "44 45 40", "output": "5" }, { "input": "86 97 68", "output": "29" }, { "input": "43 92 30", "output": "62" }, { "input": "16 70 1", "output": "69" }, { "input": "40 46 19", "output": "27" }, { "input": "71 38 56", "output": "33" }, { "input": "82 21 80", "output": "61" }, { "input": "75 8 35", "output": "67" }, { "input": "75 24 28", "output": "51" }, { "input": "78 23 56", "output": "55" }, { "input": "85 31 10", "output": "75" }, { "input": "76 50 9", "output": "67" }, { "input": "95 37 34", "output": "61" }, { "input": "84 61 35", "output": "49" }, { "input": "87 85 37", "output": "50" }, { "input": "1 3 2", "output": "2" }, { "input": "4 2 6", "output": "4" }, { "input": "6 9 3", "output": "6" }, { "input": "12 4 8", "output": "8" }, { "input": "15 10 5", "output": "10" }, { "input": "1 50 17", "output": "49" }, { "input": "10 5 15", "output": "10" }, { "input": "8 1 9", "output": "8" }, { "input": "3 5 4", "output": "2" }, { "input": "2 1 3", "output": "2" }, { "input": "1 8 2", "output": "7" }, { "input": "1 100 2", "output": "99" }, { "input": "1 4 6", "output": "5" } ]
1,698,610,266
2,147,483,647
Python 3
OK
TESTS
48
46
0
x,y,z=sorted(map(int,input().split())) print(abs(x-y)+abs(z-y))
Title: The New Year: Meeting Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year? It's guaranteed that the optimal answer is always integer. Input Specification: The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. Output Specification: Print one integer — the minimum total distance the friends need to travel in order to meet together. Demo Input: ['7 1 4\n', '30 20 10\n'] Demo Output: ['6\n', '20\n'] Note: In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
```python x,y,z=sorted(map(int,input().split())) print(abs(x-y)+abs(z-y)) ```
3
902
B
Coloring a Tree
PROGRAMMING
1,200
[ "dfs and similar", "dsu", "greedy" ]
null
null
You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1. Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0. You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*. It is guaranteed that you have to color each vertex in a color different from 0. You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)).
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree. The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=&lt;<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*. The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into. It is guaranteed that the given graph is a tree.
Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.
[ "6\n1 2 2 1 5\n2 1 1 1 1 1\n", "7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n" ]
[ "3\n", "5\n" ]
The tree from the first sample is shown on the picture (numbers are vetices' indices): <img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/> On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors): <img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/> On seond step we color all vertices in the subtree of vertex 5 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/> On third step we color all vertices in the subtree of vertex 2 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/> The tree from the second sample is shown on the picture (numbers are vetices' indices): <img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/> On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors): <img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/> On second step we color all vertices in the subtree of vertex 3 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/> On third step we color all vertices in the subtree of vertex 6 into color 2: <img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/> On fourth step we color all vertices in the subtree of vertex 4 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/> On fith step we color all vertices in the subtree of vertex 7 into color 3: <img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
1,000
[ { "input": "6\n1 2 2 1 5\n2 1 1 1 1 1", "output": "3" }, { "input": "7\n1 1 2 3 1 4\n3 3 1 1 1 2 3", "output": "5" }, { "input": "2\n1\n2 2", "output": "1" }, { "input": "3\n1 1\n2 2 2", "output": "1" }, { "input": "4\n1 2 1\n1 2 3 4", "output": "4" }, { "input": "4\n1 2 3\n4 1 2 4", "output": "4" }, { "input": "5\n1 2 1 4\n1 1 1 2 2", "output": "2" }, { "input": "3\n1 2\n2 1 1", "output": "2" }, { "input": "4\n1 1 1\n3 1 3 1", "output": "3" }, { "input": "4\n1 1 2\n4 1 4 1", "output": "2" }, { "input": "4\n1 2 2\n3 1 2 3", "output": "4" }, { "input": "3\n1 1\n1 2 2", "output": "3" } ]
1,584,396,342
2,147,483,647
Python 3
OK
TESTS
50
249
3,276,800
from queue import * n=int(input()) parent={} child={} for i in range(1,n+1): parent[i]=0 child[i]=[] l=list(map(int,input().split())) for i in range(2,n+1): parent[i]=l[i-2] child[l[i-2]].append(i) l=list(map(int,input().split())) color={} for i in range(1,n+1): color[i]=l[i-1] q=Queue() q.put(1) ans=0 while(not(q.empty())): e=q.get() c=0 if(e==1): c=0 else: c=color[parent[e]] if(c!=color[e]): ans+=1 for i in child[e]: q.put(i) print(ans)
Title: Coloring a Tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1. Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0. You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*. It is guaranteed that you have to color each vertex in a color different from 0. You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)). Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree. The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=&lt;<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*. The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into. It is guaranteed that the given graph is a tree. Output Specification: Print a single integer — the minimum number of steps you have to perform to color the tree into given colors. Demo Input: ['6\n1 2 2 1 5\n2 1 1 1 1 1\n', '7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n'] Demo Output: ['3\n', '5\n'] Note: The tree from the first sample is shown on the picture (numbers are vetices' indices): <img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/> On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors): <img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/> On seond step we color all vertices in the subtree of vertex 5 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/> On third step we color all vertices in the subtree of vertex 2 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/> The tree from the second sample is shown on the picture (numbers are vetices' indices): <img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/> On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors): <img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/> On second step we color all vertices in the subtree of vertex 3 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/> On third step we color all vertices in the subtree of vertex 6 into color 2: <img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/> On fourth step we color all vertices in the subtree of vertex 4 into color 1: <img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/> On fith step we color all vertices in the subtree of vertex 7 into color 3: <img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
```python from queue import * n=int(input()) parent={} child={} for i in range(1,n+1): parent[i]=0 child[i]=[] l=list(map(int,input().split())) for i in range(2,n+1): parent[i]=l[i-2] child[l[i-2]].append(i) l=list(map(int,input().split())) color={} for i in range(1,n+1): color[i]=l[i-1] q=Queue() q.put(1) ans=0 while(not(q.empty())): e=q.get() c=0 if(e==1): c=0 else: c=color[parent[e]] if(c!=color[e]): ans+=1 for i in child[e]: q.put(i) print(ans) ```
3
559
B
Equivalent Strings
PROGRAMMING
1,700
[ "divide and conquer", "hashing", "sortings", "strings" ]
null
null
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases: 1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1 As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent. Gerald has already completed this home task. Now it's your turn!
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
[ "aaba\nabaa\n", "aabb\nabab\n" ]
[ "YES\n", "NO\n" ]
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a". In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
1,000
[ { "input": "aaba\nabaa", "output": "YES" }, { "input": "aabb\nabab", "output": "NO" }, { "input": "a\na", "output": "YES" }, { "input": "a\nb", "output": "NO" }, { "input": "ab\nab", "output": "YES" }, { "input": "ab\nba", "output": "YES" }, { "input": "ab\nbb", "output": "NO" }, { "input": "zzaa\naazz", "output": "YES" }, { "input": "azza\nzaaz", "output": "YES" }, { "input": "abc\nabc", "output": "YES" }, { "input": "abc\nacb", "output": "NO" }, { "input": "azzz\nzzaz", "output": "YES" }, { "input": "abcd\ndcab", "output": "YES" }, { "input": "abcd\ncdab", "output": "YES" }, { "input": "abcd\ndcba", "output": "YES" }, { "input": "abcd\nacbd", "output": "NO" }, { "input": "oloaxgddgujq\noloaxgujqddg", "output": "YES" }, { "input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta", "output": "YES" }, { "input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz", "output": "YES" }, { "input": "snyaydaeobufdg\nsnyaydaeobufdg", "output": "YES" }, { "input": "baaaaa\nabaaaa", "output": "NO" }, { "input": "hhiisug\nmzdjwju", "output": "NO" }, { "input": "bbbabbabaaab\naaaabbabbbbb", "output": "NO" }, { "input": "bbaaab\naababb", "output": "NO" }, { "input": "aabbaaaa\naaaaabab", "output": "NO" }, { "input": "aab\naba", "output": "NO" }, { "input": "abcddd\nbacddd", "output": "NO" }, { "input": "qgiufelsfhanx\naaaaaaaaaaaaa", "output": "NO" }, { "input": "aabaababaaba\naababaaababa", "output": "NO" }, { "input": "nocdqzdriyyil\naaaaaaaaaaaaa", "output": "NO" }, { "input": "zdmctxl\nkojqhgw", "output": "NO" }, { "input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw", "output": "NO" }, { "input": "abc\nbac", "output": "NO" }, { "input": "ottceez\npcstdvz", "output": "NO" } ]
1,646,619,148
2,147,483,647
Python 3
OK
TESTS
104
249
921,600
def solve(s): if len(s) % 2: return s a = solve(s[:len(s) // 2]) b = solve(s[len(s) // 2:]) return a + b if a < b else b + a print('YES' if solve(input()) == solve(input()) else 'NO')
Title: Equivalent Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases: 1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1 As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent. Gerald has already completed this home task. Now it's your turn! Input Specification: The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. Output Specification: Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. Demo Input: ['aaba\nabaa\n', 'aabb\nabab\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a". In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
```python def solve(s): if len(s) % 2: return s a = solve(s[:len(s) // 2]) b = solve(s[len(s) // 2:]) return a + b if a < b else b + a print('YES' if solve(input()) == solve(input()) else 'NO') ```
3
394
C
Dominoes
PROGRAMMING
0
[ "constructive algorithms", "greedy" ]
null
null
During the break, we decided to relax and play dominoes. Our box with Domino was empty, so we decided to borrow the teacher's dominoes. The teacher responded instantly at our request. He put *nm* dominoes on the table as an *n*<=×<=2*m* rectangle so that each of the *n* rows contained *m* dominoes arranged horizontally. Each half of each domino contained number (0 or 1). We were taken aback, and the teacher smiled and said: "Consider some arrangement of dominoes in an *n*<=×<=2*m* matrix. Let's count for each column of the matrix the sum of numbers in this column. Then among all such sums find the maximum one. Can you rearrange the dominoes in the matrix in such a way that the maximum sum will be minimum possible? Note that it is prohibited to change the orientation of the dominoes, they all need to stay horizontal, nevertheless dominoes are allowed to rotate by 180 degrees. As a reward I will give you all my dominoes". We got even more taken aback. And while we are wondering what was going on, help us make an optimal matrix of dominoes.
The first line contains integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=103). In the next lines there is a description of the teachers' matrix. Each of next *n* lines contains *m* dominoes. The description of one domino is two integers (0 or 1), written without a space — the digits on the left and right half of the domino.
Print the resulting matrix of dominoes in the format: *n* lines, each of them contains *m* space-separated dominoes. If there are multiple optimal solutions, print any of them.
[ "2 3\n01 11 00\n00 01 11\n", "4 1\n11\n10\n01\n00\n" ]
[ "11 11 10\n00 00 01\n", "11\n10\n01\n00\n" ]
Consider the answer for the first sample. There, the maximum sum among all columns equals 1 (the number of columns is 6, and not 3). Obviously, this maximum can't be less than 1, then such matrix is optimal. Note that the dominoes can be rotated by 180 degrees.
1,500
[ { "input": "2 3\n01 11 00\n00 01 11", "output": "11 11 10\n00 00 01" }, { "input": "4 1\n11\n10\n01\n00", "output": "11\n10\n01\n00" }, { "input": "1 1\n00", "output": "00" }, { "input": "1 1\n01", "output": "10" }, { "input": "1 1\n11", "output": "11" }, { "input": "9 9\n01 00 00 01 00 01 11 11 11\n10 10 10 01 10 01 11 01 10\n10 00 10 00 11 01 00 10 00\n01 00 01 01 11 00 00 11 11\n11 00 10 11 01 01 11 00 01\n01 10 00 00 11 10 01 01 10\n11 10 11 00 11 11 01 10 10\n10 00 01 00 00 00 11 01 01\n00 11 01 00 10 01 10 00 01", "output": "11 11 11 11 11 11 11 11 11\n11 11 11 11 11 11 11 11 11\n10 10 10 10 10 10 10 10 10\n10 10 10 10 10 10 10 10 10\n10 10 10 01 01 01 01 01 01\n01 01 01 01 01 01 01 01 01\n01 01 01 00 00 00 00 01 01\n00 00 00 00 00 00 00 00 00\n00 00 00 00 00 00 00 00 00" }, { "input": "9 9\n10 10 10 01 10 11 11 01 10\n11 00 10 10 11 10 01 00 00\n10 00 11 01 00 01 01 11 10\n10 11 10 00 01 11 11 10 11\n01 11 11 01 11 00 10 00 01\n01 00 00 10 01 01 10 00 01\n11 10 11 10 01 00 00 11 00\n10 11 10 10 01 10 10 10 01\n10 10 10 10 11 11 01 00 11", "output": "11 11 11 11 11 11 11 11 11\n11 11 11 11 11 11 11 11 11\n11 11 10 10 10 10 10 10 10\n10 10 10 10 10 10 10 10 10\n10 10 10 10 10 10 10 01 01\n01 01 01 01 01 01 01 01 01\n01 01 01 01 01 01 01 01 01\n00 00 00 00 01 01 01 00 00\n00 00 00 00 00 00 00 00 00" }, { "input": "9 1\n01\n00\n01\n01\n00\n00\n00\n01\n11", "output": "11\n10\n10\n01\n01\n00\n00\n00\n00" }, { "input": "2 9\n11 10 11 10 10 11 00 10 00\n10 00 00 10 10 00 11 01 01", "output": "11 11 11 11 10 10 10 10 10\n00 00 00 00 00 01 01 01 01" }, { "input": "2 8\n10 01 01 11 10 10 01 10\n01 11 01 01 11 10 01 01", "output": "11 11 11 10 10 10 10 10\n10 10 01 01 01 01 01 01" }, { "input": "3 5\n00 10 10 11 01\n11 01 11 11 10\n10 11 00 00 00", "output": "11 11 11 11 11\n10 10 10 01 01\n00 00 01 00 00" }, { "input": "2 3\n00 10 01\n01 01 00", "output": "10 10 01\n00 01 00" }, { "input": "2 5\n01 00 01 01 00\n11 01 11 11 10", "output": "11 11 11 10 10\n10 00 00 01 01" } ]
1,548,375,756
2,147,483,647
Python 3
OK
TESTS
27
951
45,977,600
a,b=map(int,input().split()) c=list() x00=0 x01=0 x11=0 for i in range(a): c.append(list(input().split())) x11+=c[-1].count('11') x01+=c[-1].count('01')+c[-1].count('10') x00=a*b-x11-x01 new=[[]for i in range(b)] i=0 while x11>0: x11-=1 new[i].append('11') i+=1 i%=b step=1 reverse=False while x01>0: x01-=1 if reverse: new[i].append('01') else: new[i].append('10') i+=step if i==-1 or i==b: step=-step i+=step reverse=1-reverse for r in new: r+=['00']*(a-len(r)) for i in range(a): print(' '.join([new[k][i]for k in range(b)]))
Title: Dominoes Time Limit: None seconds Memory Limit: None megabytes Problem Description: During the break, we decided to relax and play dominoes. Our box with Domino was empty, so we decided to borrow the teacher's dominoes. The teacher responded instantly at our request. He put *nm* dominoes on the table as an *n*<=×<=2*m* rectangle so that each of the *n* rows contained *m* dominoes arranged horizontally. Each half of each domino contained number (0 or 1). We were taken aback, and the teacher smiled and said: "Consider some arrangement of dominoes in an *n*<=×<=2*m* matrix. Let's count for each column of the matrix the sum of numbers in this column. Then among all such sums find the maximum one. Can you rearrange the dominoes in the matrix in such a way that the maximum sum will be minimum possible? Note that it is prohibited to change the orientation of the dominoes, they all need to stay horizontal, nevertheless dominoes are allowed to rotate by 180 degrees. As a reward I will give you all my dominoes". We got even more taken aback. And while we are wondering what was going on, help us make an optimal matrix of dominoes. Input Specification: The first line contains integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=103). In the next lines there is a description of the teachers' matrix. Each of next *n* lines contains *m* dominoes. The description of one domino is two integers (0 or 1), written without a space — the digits on the left and right half of the domino. Output Specification: Print the resulting matrix of dominoes in the format: *n* lines, each of them contains *m* space-separated dominoes. If there are multiple optimal solutions, print any of them. Demo Input: ['2 3\n01 11 00\n00 01 11\n', '4 1\n11\n10\n01\n00\n'] Demo Output: ['11 11 10\n00 00 01\n', '11\n10\n01\n00\n'] Note: Consider the answer for the first sample. There, the maximum sum among all columns equals 1 (the number of columns is 6, and not 3). Obviously, this maximum can't be less than 1, then such matrix is optimal. Note that the dominoes can be rotated by 180 degrees.
```python a,b=map(int,input().split()) c=list() x00=0 x01=0 x11=0 for i in range(a): c.append(list(input().split())) x11+=c[-1].count('11') x01+=c[-1].count('01')+c[-1].count('10') x00=a*b-x11-x01 new=[[]for i in range(b)] i=0 while x11>0: x11-=1 new[i].append('11') i+=1 i%=b step=1 reverse=False while x01>0: x01-=1 if reverse: new[i].append('01') else: new[i].append('10') i+=step if i==-1 or i==b: step=-step i+=step reverse=1-reverse for r in new: r+=['00']*(a-len(r)) for i in range(a): print(' '.join([new[k][i]for k in range(b)])) ```
3
359
B
Permutation
PROGRAMMING
1,400
[ "constructive algorithms", "dp", "math" ]
null
null
A permutation *p* is an ordered group of numbers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers, each is no more than *n*. We'll define number *n* as the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*. Simon has a positive integer *n* and a non-negative integer *k*, such that 2*k*<=≤<=*n*. Help him find permutation *a* of length 2*n*, such that it meets this equation: .
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 0<=≤<=2*k*<=≤<=*n*).
Print 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* — the required permutation *a*. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
[ "1 0\n", "2 1\n", "4 0\n" ]
[ "1 2", "3 2 1 4\n", "2 7 4 6 1 3 5 8\n" ]
Record |*x*| represents the absolute value of number *x*. In the first sample |1 - 2| - |1 - 2| = 0. In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2. In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
1,000
[ { "input": "1 0", "output": "1 2" }, { "input": "2 1", "output": "3 2 1 4" }, { "input": "4 0", "output": "2 7 4 6 1 3 5 8" }, { "input": "50000 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "50000 25000", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "50000 24999", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "49999 24999", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "49999 3", "output": "2 1 4 3 6 5 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "1333 156", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "7563 3781", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "7563 3780", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "2 0", "output": "1 2 3 4" }, { "input": "3 0", "output": "1 2 3 4 5 6" }, { "input": "4 1", "output": "2 1 3 4 5 6 7 8" }, { "input": "4 2", "output": "2 1 4 3 5 6 7 8" }, { "input": "6 3", "output": "2 1 4 3 6 5 7 8 9 10 11 12" }, { "input": "48888 24444", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "50000 1", "output": "2 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "50000 1000", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 156..." }, { "input": "34 17", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 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" }, { "input": "43244 1233", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 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57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100" }, { "input": "55 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110" }, { "input": "5000 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "3 1", "output": "2 1 3 4 5 6" }, { "input": "7563 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "7563 1", "output": "2 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "7563 2", "output": "2 1 4 3 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "6 0", "output": "1 2 3 4 5 6 7 8 9 10 11 12" } ]
1,663,546,432
2,147,483,647
PyPy 3-64
OK
TESTS
30
124
8,806,400
s=input().split();print(int(s[1])+1,end=' ') for i in range(1,1+2*int(s[0])): if i!=int(s[1])+1: print(i,end=' ')
Title: Permutation Time Limit: None seconds Memory Limit: None megabytes Problem Description: A permutation *p* is an ordered group of numbers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers, each is no more than *n*. We'll define number *n* as the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*. Simon has a positive integer *n* and a non-negative integer *k*, such that 2*k*<=≤<=*n*. Help him find permutation *a* of length 2*n*, such that it meets this equation: . Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 0<=≤<=2*k*<=≤<=*n*). Output Specification: Print 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* — the required permutation *a*. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them. Demo Input: ['1 0\n', '2 1\n', '4 0\n'] Demo Output: ['1 2', '3 2 1 4\n', '2 7 4 6 1 3 5 8\n'] Note: Record |*x*| represents the absolute value of number *x*. In the first sample |1 - 2| - |1 - 2| = 0. In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2. In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
```python s=input().split();print(int(s[1])+1,end=' ') for i in range(1,1+2*int(s[0])): if i!=int(s[1])+1: print(i,end=' ') ```
3
0
none
none
none
0
[ "none" ]
null
null
A subsequence of length |*x*| of string *s*<==<=*s*1*s*2... *s*|*s*| (where |*s*| is the length of string *s*) is a string *x*<==<=*s**k*1*s**k*2... *s**k*|*x*| (1<=≤<=*k*1<=&lt;<=*k*2<=&lt;<=...<=&lt;<=*k*|*x*|<=≤<=|*s*|). You've got two strings — *s* and *t*. Let's consider all subsequences of string *s*, coinciding with string *t*. Is it true that each character of string *s* occurs in at least one of these subsequences? In other words, is it true that for all *i* (1<=≤<=*i*<=≤<=|*s*|), there is such subsequence *x*<==<=*s**k*1*s**k*2... *s**k*|*x*| of string *s*, that *x*<==<=*t* and for some *j* (1<=≤<=*j*<=≤<=|*x*|) *k**j*<==<=*i*.
The first line contains string *s*, the second line contains string *t*. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Print "Yes" (without the quotes), if each character of the string *s* occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
[ "abab\nab\n", "abacaba\naba\n", "abc\nba\n" ]
[ "Yes\n", "No\n", "No\n" ]
In the first sample string *t* can occur in the string *s* as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string *s* occurs at least once. In the second sample the 4-th character of the string *s* doesn't occur in any occurrence of string *t*. In the third sample there is no occurrence of string *t* in string *s*.
0
[ { "input": "abab\nab", "output": "Yes" }, { "input": "abacaba\naba", "output": "No" }, { "input": "abc\nba", "output": "No" }, { "input": "babbbbbaba\nab", "output": "No" }, { "input": "accbacabaa\nbada", "output": "No" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "Yes" }, { "input": "hqxibotrjfqrgwrydtlpkzeqrkfgzdylfokmaguwafdgwltgvqobnouttrrfuavfkqcoqxkfwsuseomraigoljzzbjukwaxcftvlhfzdypuinnxbluzfxajkabirvyawtxzcrxpoghucjypwinspnnbptsuamkbjqgttooxwcsqxrukwwtgrkxdujioqywqlugkjngfxrybvxjmlwszszljfgyouvgdilzseekxlsiujhod\nnghetuvcotztgttmr", "output": "No" }, { "input": "bacbbcbcacaacbabacbcbacaaaabbabaaccccacbcbbbabcacbacacabaabacacbaaacacbbccbcccbabccaacccccbbcabacbaacabaccccccacbbaccbabaaabaaccabcaaabcccccbbabccccccabacbaaababcbbbccbbabcabbbbaaabbccccbacbaacbcacbbaaccbaabcaaacbccccbcbababccbcccabbbabbba\nbacbbcbcacaacbabacbcbacaaaabbabaaccccacbcbbbabcacbacacabaabacacbaaacacbbccbcccbabccaacccccbbcabacbaacabaccccccacbbaccbabaaabaaccabcaaabcccccbbabccccccabacbaaababcbbbccbbabcabbbbaaabbccccbacbaacbcacbbaaccbaabcaaacbccccbcbababccbcccabbbabbba", "output": "Yes" }, { "input": "adbecbeaddbbebdaa\nadbecbeaddbbebdaa", "output": "Yes" }, { "input": "iiiiiiqqqqqqqqqqaaaaffffllllleeeeeeeekkkkkkkhhhhhhhhhhooooooddddddddlllllllliiiaaaaaaaaaaaaaaaaaooggggggggggllllllffffffcccccccpppppppdddddddddddccccbbbbbbbbbbkkkkfffffiiiiiiipppppppppccccnnnnnnnnnnnnnnkkkkkkkkkkqqqqppppppeeeeeeeeemmmmmmmmbbbbbbbaaaaaaffffllllljjjj\niqaflekhodliaaoglfcpdcbbkfipcnnkqpeembaflj", "output": "Yes" }, { "input": "cccbbiiiiiqqvvgggwwwwxxxxxxxxoooondddkkkpvvvdddddooqqxxxxxqqqqllllkkkkkkggggfwwwwwkkkfffeeeemmmmmmmqwwwwwwxxxxxxxdddddqqqqqqq\ncbiqvgwxxondkpvdoqxqlkgfwkfemqwxdq", "output": "Yes" }, { "input": "babaabaabb\nbbccb", "output": "No" }, { "input": "ctkhagrifaztmnxhmqztzibnmzzkojiztvrkfeoqegvwtbxlvvjhebshqaicsovtkcdovytimjggglyxlvglgunbohnkxargymbqvzgsnvjzgxivdgnaesgxqcruaopjuqsyyorrobnelehjnxcetveehlbmeskptivsuhuqupbieumycwczxyqjtwfofehfkpqmjngygwxkaviuyouiippgvlxjgtkxmhcwtzacbllsybgiujyryngapfwjkkyapfgxtcdpc\nctkhagrifaztmnxhmqztzibnmzzkojiztvrkfeoqegvwtbxlvvjhebshqaicsovtkcdovytimjggglyxlvglgunbohnkxargymbqvzgsnvjzgxivdgnaesgxcetveehlbmeskptivsuhuqupbieumycwczxyqjtwfofehfkpqmjngygwxkaviuyouiippgvlxjgtkxmhcwtzacbllsybgiujyryngapfwjkkyapfgxtcdpc", "output": "No" }, { "input": "adedadcababceeeaddadedddaeaccccbcccdaeeecaaeaebccebddddeedabbddeaaccdacebaeeccdeeddbecbdecddebe\nadedcceecebdccdbe", "output": "No" }, { "input": "cctckkhatkgrhktihcgififfgfctctkrgiakrifazzggfzczfkkahhafhcfgacccfakkarcatkfiktczkficahgiriakccfiztkhkgrfkrimgamighhtamrhxftaadwxgfggytwjccgkdpyyatctfdygxggkyycpjyfxyfdwtgytcacawjddjdctyfgddkfkypyxftxxtaddcxxpgfgxgdfggfdggdcddtgpxpctpddcdcpc\nctkhagrifaztmnxhmqztzibnmzzkojiztvrkfeoqegvwtbxlvvjhebshqaicsovtkcdovytimjggglyxlvglgunbohnkxargymbqvzgsnvjzgxivdgnaesgxcetveehlbmeskptivsuhuqupbieumycwczxyqjtwfofehfkpqmjngygwxkaviuyouiippgvlxjgtkxmhcwtzacbllsybgiujyryngapfwjkkyapfgxtcdpc", "output": "No" }, { "input": "iqqiaiiffiqlqfaaflfieflfillkkhqfolhehedqdqqfddlheifeoqeohhoadqkfiqeleeqdekhhahkaqqqiaqliiqlelkhdfodeafqfhogihlgoqafdiffkaekhqhgqfkcqiaaoodkkfeqkciqfeihkifeodhahdhddghaihkhahghlkcckicehechocfhfcdfeldelgaqhqfepipegklqiafhqglcdfgkfpoggldgfcglfbbpkkkfipipcnnkqpeembaflj\niqaflekhodliaaoglfcpdcbbkfipcnnkqpeembaflj", "output": "No" }, { "input": "iqqiaiiffiqlqfaaflfieflfillkkhqfolhehedqdqqfddlheifeoqeohhoadqkfiqeleeqdekhhahkaqqqiaqliiqlelkhdfodeafqfhogihlgoqafdiffkaekhqhgqfkcqiaaoodkkfeqkciqfeihkifeodhahdhddghaihkhahghlkcckicehechocfhfcdfeldelgaqhqfepipegklqiafhqglcdfaflekhodliaaoglfcpdcbbkfipcnnkqpeembaflj\niqaflekhodliaaoglfcpdcbbkfipcnnkqpeembaflj", "output": "Yes" }, { "input": "abaaaba\nabba", "output": "No" }, { "input": "abaca\nabca", "output": "No" }, { "input": "aaa\naaaa", "output": "No" }, { "input": "aa\naaa", "output": "No" }, { "input": "aaaa\naaa", "output": "Yes" }, { "input": "aa\naaaaaaaa", "output": "No" }, { "input": "aaaaaa\naaaaaaa", "output": "No" }, { "input": "aaaaaaaa\naaaaa", "output": "Yes" }, { "input": "abebea\nabeba", "output": "No" }, { "input": "ab\nabcd", "output": "No" }, { "input": "abcdad\nabcd", "output": "No" }, { "input": "abaaaaba\nabba", "output": "No" }, { "input": "ababa\nab", "output": "No" }, { "input": "cabcbac\ncabac", "output": "No" }, { "input": "ababcab\nabbcab", "output": "No" }, { "input": "abaaaaaaba\nabba", "output": "No" }, { "input": "abcbab\nabcab", "output": "No" }, { "input": "abc\nbac", "output": "No" }, { "input": "abcdadbcd\nabcd", "output": "Yes" } ]
1,692,035,053
2,147,483,647
PyPy 3-64
OK
TESTS
130
154
6,963,200
s = input() t = input() N = len(s) M = len(t) latest = [0] * 200 MaxL = [0] * (N + 1) MaxR = [0] * (N + 1) Cnt = 0 for i in range(1, N + 1): if Cnt < M and s[i - 1] == t[Cnt]: Cnt += 1 MaxL[i] = Cnt latest[ord(s[i - 1])] = Cnt else: MaxL[i] = latest[ord(s[i - 1])] t = t[::-1] latest = [0] * 200 Cnt = 0 for i in range(N, 0, -1): if Cnt < M and s[i - 1] == t[Cnt]: Cnt += 1 MaxR[i] = Cnt latest[ord(s[i - 1])] = Cnt else: MaxR[i] = latest[ord(s[i - 1])] flag = all(MaxL[i] + MaxR[i] >= M + 1 for i in range(1, N + 1)) if flag: print("Yes") else: print("No")# 1692035052.9421067
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: A subsequence of length |*x*| of string *s*<==<=*s*1*s*2... *s*|*s*| (where |*s*| is the length of string *s*) is a string *x*<==<=*s**k*1*s**k*2... *s**k*|*x*| (1<=≤<=*k*1<=&lt;<=*k*2<=&lt;<=...<=&lt;<=*k*|*x*|<=≤<=|*s*|). You've got two strings — *s* and *t*. Let's consider all subsequences of string *s*, coinciding with string *t*. Is it true that each character of string *s* occurs in at least one of these subsequences? In other words, is it true that for all *i* (1<=≤<=*i*<=≤<=|*s*|), there is such subsequence *x*<==<=*s**k*1*s**k*2... *s**k*|*x*| of string *s*, that *x*<==<=*t* and for some *j* (1<=≤<=*j*<=≤<=|*x*|) *k**j*<==<=*i*. Input Specification: The first line contains string *s*, the second line contains string *t*. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105. Output Specification: Print "Yes" (without the quotes), if each character of the string *s* occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise. Demo Input: ['abab\nab\n', 'abacaba\naba\n', 'abc\nba\n'] Demo Output: ['Yes\n', 'No\n', 'No\n'] Note: In the first sample string *t* can occur in the string *s* as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string *s* occurs at least once. In the second sample the 4-th character of the string *s* doesn't occur in any occurrence of string *t*. In the third sample there is no occurrence of string *t* in string *s*.
```python s = input() t = input() N = len(s) M = len(t) latest = [0] * 200 MaxL = [0] * (N + 1) MaxR = [0] * (N + 1) Cnt = 0 for i in range(1, N + 1): if Cnt < M and s[i - 1] == t[Cnt]: Cnt += 1 MaxL[i] = Cnt latest[ord(s[i - 1])] = Cnt else: MaxL[i] = latest[ord(s[i - 1])] t = t[::-1] latest = [0] * 200 Cnt = 0 for i in range(N, 0, -1): if Cnt < M and s[i - 1] == t[Cnt]: Cnt += 1 MaxR[i] = Cnt latest[ord(s[i - 1])] = Cnt else: MaxR[i] = latest[ord(s[i - 1])] flag = all(MaxL[i] + MaxR[i] >= M + 1 for i in range(1, N + 1)) if flag: print("Yes") else: print("No")# 1692035052.9421067 ```
3
804
A
Find Amir
PROGRAMMING
1,000
[ "constructive algorithms", "greedy", "math" ]
null
null
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends. There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of schools.
Print single integer: the minimum cost of tickets needed to visit all schools.
[ "2\n", "10\n" ]
[ "0\n", "4\n" ]
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
500
[ { "input": "2", "output": "0" }, { "input": "10", "output": "4" }, { "input": "43670", "output": "21834" }, { "input": "4217", "output": "2108" }, { "input": "17879", "output": "8939" }, { "input": "31809", "output": "15904" }, { "input": "40873", "output": "20436" }, { "input": "77859", "output": "38929" }, { "input": "53022", "output": "26510" }, { "input": "79227", "output": "39613" }, { "input": "100000", "output": "49999" }, { "input": "82801", "output": "41400" }, { "input": "5188", "output": "2593" }, { "input": "86539", "output": "43269" }, { "input": "12802", "output": "6400" }, { "input": "20289", "output": "10144" }, { "input": "32866", "output": "16432" }, { "input": "33377", "output": "16688" }, { "input": "31775", "output": "15887" }, { "input": "60397", "output": "30198" }, { "input": "100000", "output": "49999" }, { "input": "99999", "output": "49999" }, { "input": "99998", "output": "49998" }, { "input": "99997", "output": "49998" }, { "input": "99996", "output": "49997" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "3", "output": "1" }, { "input": "4", "output": "1" }, { "input": "1", "output": "0" }, { "input": "3", "output": "1" } ]
1,593,243,004
2,147,483,647
PyPy 3
OK
TESTS
31
140
20,172,800
n = int(input()) if n & 1 == 1: n /= 2 print(int(n)) else: n /= 2 n -= 1 print(int(n))
Title: Find Amir Time Limit: None seconds Memory Limit: None megabytes Problem Description: A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends. There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of schools. Output Specification: Print single integer: the minimum cost of tickets needed to visit all schools. Demo Input: ['2\n', '10\n'] Demo Output: ['0\n', '4\n'] Note: In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
```python n = int(input()) if n & 1 == 1: n /= 2 print(int(n)) else: n /= 2 n -= 1 print(int(n)) ```
3
215
A
Bicycle Chain
PROGRAMMING
900
[ "brute force", "implementation" ]
null
null
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation. We know that the *i*-th star on the pedal axle has *a**i* (0<=&lt;<=*a*1<=&lt;<=*a*2<=&lt;<=...<=&lt;<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=&lt;<=*b*1<=&lt;<=*b*2<=&lt;<=...<=&lt;<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value . Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears. In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing. The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing. It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
[ "2\n4 5\n3\n12 13 15\n", "4\n1 2 3 4\n5\n10 11 12 13 14\n" ]
[ "2\n", "1\n" ]
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
500
[ { "input": "2\n4 5\n3\n12 13 15", "output": "2" }, { "input": "4\n1 2 3 4\n5\n10 11 12 13 14", "output": "1" }, { "input": "1\n1\n1\n1", "output": "1" }, { "input": "2\n1 2\n1\n1", "output": "1" }, { "input": "1\n1\n2\n1 2", "output": "1" }, { "input": "4\n3 7 11 13\n4\n51 119 187 221", "output": "4" }, { "input": "4\n2 3 4 5\n3\n1 2 3", "output": "2" }, { "input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97", "output": "1" }, { "input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28", "output": "1" }, { "input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958", "output": "1" }, { "input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20", "output": "1" }, { "input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50", "output": "1" }, { "input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38", "output": "4" }, { "input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100", "output": "1" }, { "input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149", "output": "1" }, { "input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193", "output": "1" }, { "input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245", "output": "1" }, { "input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298", "output": "1" }, { "input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350", "output": "1" }, { "input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388", "output": "1" }, { "input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999", "output": "8" }, { "input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813", "output": "3" }, { "input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217", "output": "3" }, { "input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555", "output": "8" }, { "input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345", "output": "20" }, { "input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735", "output": "23" }, { "input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782", "output": "3" }, { "input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428", "output": "4" }, { "input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959", "output": "5" }, { "input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664", "output": "24" }, { "input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782", "output": "3" }, { "input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428", "output": "4" }, { "input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959", "output": "5" }, { "input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664", "output": "24" }, { "input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968", "output": "12" }, { "input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706", "output": "1" }, { "input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394", "output": "1" }, { "input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284", "output": "1" }, { "input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790", "output": "4" }, { "input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902", "output": "17" }, { "input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991", "output": "28" }, { "input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862", "output": "15" }, { "input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691", "output": "8" }, { "input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673", "output": "38" }, { "input": "3\n3 4 5\n3\n6 20 25", "output": "2" }, { "input": "4\n2 3 5 8\n4\n2 6 8 10", "output": "1" }, { "input": "4\n3 5 7 11\n4\n3 5 7 22", "output": "1" }, { "input": "2\n2 3\n3\n20 30 50", "output": "1" }, { "input": "3\n1 2 3\n4\n2 4 6 49", "output": "1" }, { "input": "2\n4 5\n3\n12 15 20", "output": "1" }, { "input": "3\n2 5 7\n3\n4 5 7", "output": "1" }, { "input": "3\n3 5 8\n3\n6 8 10", "output": "2" }, { "input": "2\n2 3\n4\n4 6 9 33", "output": "1" }, { "input": "2\n2 3\n4\n4 6 21 40", "output": "1" }, { "input": "3\n4 9 10\n3\n8 9 10", "output": "1" }, { "input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000", "output": "1" }, { "input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14", "output": "1" } ]
1,545,828,960
2,147,483,647
Python 3
OK
TESTS
57
218
0
n = int(input()) a = [int(i) for i in input().split()] m = int(input()) b = [int(i) for i in input().split()] l = [j // i for j in b for i in a if not j % i] print(l.count(max(l)))
Title: Bicycle Chain Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation. We know that the *i*-th star on the pedal axle has *a**i* (0<=&lt;<=*a*1<=&lt;<=*a*2<=&lt;<=...<=&lt;<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=&lt;<=*b*1<=&lt;<=*b*2<=&lt;<=...<=&lt;<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value . Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears. In the problem, fraction denotes division in real numbers, that is, no rounding is performed. Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing. The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing. It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces. Output Specification: Print the number of "integer" gears with the maximum ratio among all "integer" gears. Demo Input: ['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n'] Demo Output: ['2\n', '1\n'] Note: In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
```python n = int(input()) a = [int(i) for i in input().split()] m = int(input()) b = [int(i) for i in input().split()] l = [j // i for j in b for i in a if not j % i] print(l.count(max(l))) ```
3
242
B
Big Segment
PROGRAMMING
1,100
[ "implementation", "sortings" ]
null
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" } ]
1,613,992,817
2,147,483,647
Python 3
OK
TESTS
45
778
12,595,200
n=int(input()) arr=[list(map(int,input().split())) for i in range(n)] front=sorted(arr,key=lambda x:x[1]-x[0],reverse=True) # print(front) min1=min(arr,key=lambda x:x[0])[0] max2=max(arr,key=lambda x:x[1])[1] # print(min1,max2) if min1>=front[0][0] and max2<=front[0][1]: print(arr.index(front[0])+1) else: print(-1)
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 n=int(input()) arr=[list(map(int,input().split())) for i in range(n)] front=sorted(arr,key=lambda x:x[1]-x[0],reverse=True) # print(front) min1=min(arr,key=lambda x:x[0])[0] max2=max(arr,key=lambda x:x[1])[1] # print(min1,max2) if min1>=front[0][0] and max2<=front[0][1]: print(arr.index(front[0])+1) else: print(-1) ```
3
964
A
Splits
PROGRAMMING
800
[ "math" ]
null
null
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$. For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$. The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$. The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$. For a given $n$, find out the number of different weights of its splits.
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
Output one integer — the answer to the problem.
[ "7\n", "8\n", "9\n" ]
[ "4\n", "5\n", "5\n" ]
In the first sample, there are following possible weights of splits of $7$: Weight 1: [$\textbf 7$] Weight 2: [$\textbf 3$, $\textbf 3$, 1] Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1] Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$]
500
[ { "input": "7", "output": "4" }, { "input": "8", "output": "5" }, { "input": "9", "output": "5" }, { "input": "1", "output": "1" }, { "input": "286", "output": "144" }, { "input": "48", "output": "25" }, { "input": "941", "output": "471" }, { "input": "45154", "output": "22578" }, { "input": "60324", "output": "30163" }, { "input": "91840", "output": "45921" }, { "input": "41909", "output": "20955" }, { "input": "58288", "output": "29145" }, { "input": "91641", "output": "45821" }, { "input": "62258", "output": "31130" }, { "input": "79811", "output": "39906" }, { "input": "88740", "output": "44371" }, { "input": "12351", "output": "6176" }, { "input": "1960", "output": "981" }, { "input": "29239", "output": "14620" }, { "input": "85801", "output": "42901" }, { "input": "43255", "output": "21628" }, { "input": "13439", "output": "6720" }, { "input": "35668", "output": "17835" }, { "input": "19122", "output": "9562" }, { "input": "60169", "output": "30085" }, { "input": "50588", "output": "25295" }, { "input": "2467", "output": "1234" }, { "input": "39315", "output": "19658" }, { "input": "29950", "output": "14976" }, { "input": "17286", "output": "8644" }, { "input": "7359066", "output": "3679534" }, { "input": "1016391", "output": "508196" }, { "input": "7928871", "output": "3964436" }, { "input": "3968891", "output": "1984446" }, { "input": "2636452", "output": "1318227" }, { "input": "5076901", "output": "2538451" }, { "input": "9870265", "output": "4935133" }, { "input": "2453786", "output": "1226894" }, { "input": "7263670", "output": "3631836" }, { "input": "1890845", "output": "945423" }, { "input": "574128507", "output": "287064254" }, { "input": "648476655", "output": "324238328" }, { "input": "97349542", "output": "48674772" }, { "input": "716489761", "output": "358244881" }, { "input": "858771038", "output": "429385520" }, { "input": "520778784", "output": "260389393" }, { "input": "439004204", "output": "219502103" }, { "input": "589992198", "output": "294996100" }, { "input": "371106544", "output": "185553273" }, { "input": "894241590", "output": "447120796" }, { "input": "123957268", "output": "61978635" }, { "input": "234149297", "output": "117074649" }, { "input": "789954052", "output": "394977027" }, { "input": "667978920", "output": "333989461" }, { "input": "154647261", "output": "77323631" }, { "input": "751453521", "output": "375726761" }, { "input": "848862308", "output": "424431155" }, { "input": "323926781", "output": "161963391" }, { "input": "576768825", "output": "288384413" }, { "input": "31293802", "output": "15646902" }, { "input": "2", "output": "2" }, { "input": "1000000000", "output": "500000001" }, { "input": "3", "output": "2" } ]
1,609,163,693
2,147,483,647
Python 3
OK
TESTS
63
109
0
import math num=int(input()) k=math.ceil(num/2) if num%2==0: print(k+1) else: print(k)
Title: Splits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$. For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$. The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$. The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$. For a given $n$, find out the number of different weights of its splits. Input Specification: The first line contains one integer $n$ ($1 \leq n \leq 10^9$). Output Specification: Output one integer — the answer to the problem. Demo Input: ['7\n', '8\n', '9\n'] Demo Output: ['4\n', '5\n', '5\n'] Note: In the first sample, there are following possible weights of splits of $7$: Weight 1: [$\textbf 7$] Weight 2: [$\textbf 3$, $\textbf 3$, 1] Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1] Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$]
```python import math num=int(input()) k=math.ceil(num/2) if num%2==0: print(k+1) else: print(k) ```
3