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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" } ]
1,672,825,751
2,147,483,647
Python 3
OK
TESTS
45
92
0
import math t = input().split(" ") a, b, n, c = int(t[0]), int(t[1]), int(t[2]), 0 while n >= 0: n -= math.gcd(int(t[c % 2]), n) c = 1-c if c == 0:print(0) else:print(1)
Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 &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 import math t = input().split(" ") a, b, n, c = int(t[0]), int(t[1]), int(t[2]), 0 while n >= 0: n -= math.gcd(int(t[c % 2]), n) c = 1-c if c == 0:print(0) else:print(1) ```
3
272
A
Dima and Friends
PROGRAMMING
1,000
[ "implementation", "math" ]
null
null
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place. To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment. For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place. Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show. The numbers in the lines are separated by a single space.
In a single line print the answer to the problem.
[ "1\n1\n", "1\n2\n", "2\n3 5\n" ]
[ "3\n", "2\n", "3\n" ]
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend. In the second sample Dima can show 2 or 4 fingers.
500
[ { "input": "1\n1", "output": "3" }, { "input": "1\n2", "output": "2" }, { "input": "2\n3 5", "output": "3" }, { "input": "2\n3 5", "output": "3" }, { "input": "1\n5", "output": "3" }, { "input": "5\n4 4 3 5 1", "output": "4" }, { "input": "6\n2 3 2 2 1 3", "output": "4" }, { "input": "8\n2 2 5 3 4 3 3 2", "output": "4" }, { "input": "7\n4 1 3 2 2 4 5", "output": "4" }, { "input": "3\n3 5 1", "output": "4" }, { "input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2", "output": "5" }, { "input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1", "output": "4" }, { "input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4", "output": "5" }, { "input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3", "output": "5" }, { "input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3", "output": "4" }, { "input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2", "output": "4" }, { "input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3", "output": "5" }, { "input": "6\n4 2 3 1 3 5", "output": "4" }, { "input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4", "output": "5" }, { "input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2", "output": "5" }, { "input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3", "output": "5" }, { "input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1", "output": "5" }, { "input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5", "output": "5" }, { "input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3", "output": "5" }, { "input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2", "output": "5" }, { "input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1", "output": "5" }, { "input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5", "output": "5" }, { "input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3", "output": "5" }, { "input": "100\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 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 4", "output": "4" }, { "input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "5" } ]
1,687,932,770
2,147,483,647
PyPy 3-64
OK
TESTS
30
122
0
n=int(input('')) f=input('') b=f.split(' ') c=0 w=0 for i in b: c+=int(i) for j in range(1,6): if (c+j)%(n+1)!=1: w+=1 else: continue print(w)
Title: Dima and Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place. To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment. For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place. Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show. The numbers in the lines are separated by a single space. Output Specification: In a single line print the answer to the problem. Demo Input: ['1\n1\n', '1\n2\n', '2\n3 5\n'] Demo Output: ['3\n', '2\n', '3\n'] Note: In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend. In the second sample Dima can show 2 or 4 fingers.
```python n=int(input('')) f=input('') b=f.split(' ') c=0 w=0 for i in b: c+=int(i) for j in range(1,6): if (c+j)%(n+1)!=1: w+=1 else: continue print(w) ```
3
289
A
Polo the Penguin and Segments
PROGRAMMING
1,100
[ "brute force", "implementation" ]
null
null
Little penguin Polo adores integer segments, that is, pairs of integers [*l*; *r*] (*l*<=≤<=*r*). He has a set that consists of *n* integer segments: [*l*1; *r*1],<=[*l*2; *r*2],<=...,<=[*l**n*; *r**n*]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [*l*; *r*] to either segment [*l*<=-<=1; *r*], or to segment [*l*; *r*<=+<=1]. The value of a set of segments that consists of *n* segments [*l*1; *r*1],<=[*l*2; *r*2],<=...,<=[*l**n*; *r**n*] is the number of integers *x*, such that there is integer *j*, for which the following inequality holds, *l**j*<=≤<=*x*<=≤<=*r**j*. Find the minimum number of moves needed to make the value of the set of Polo's segments divisible by *k*.
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). Each of the following *n* lines contain a segment as a pair of integers *l**i* and *r**i* (<=-<=105<=≤<=*l**i*<=≤<=*r**i*<=≤<=105), separated by a space. It is guaranteed that no two segments intersect. In other words, for any two integers *i*,<=*j* (1<=≤<=*i*<=&lt;<=*j*<=≤<=*n*) the following inequality holds, *min*(*r**i*,<=*r**j*)<=&lt;<=*max*(*l**i*,<=*l**j*).
In a single line print a single integer — the answer to the problem.
[ "2 3\n1 2\n3 4\n", "3 7\n1 2\n3 3\n4 7\n" ]
[ "2\n", "0\n" ]
none
500
[ { "input": "2 3\n1 2\n3 4", "output": "2" }, { "input": "3 7\n1 2\n3 3\n4 7", "output": "0" }, { "input": "3 7\n1 10\n11 47\n74 128", "output": "3" }, { "input": "5 4\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "3" }, { "input": "7 4\n2 2\n-1 -1\n0 1\n7 8\n-3 -2\n9 9\n4 6", "output": "0" }, { "input": "10 2\n92 92\n55 59\n70 73\n78 81\n62 65\n95 99\n74 75\n85 87\n51 51\n60 60", "output": "0" }, { "input": "10 474\n56 60\n82 82\n73 73\n105 109\n77 80\n51 51\n85 88\n97 100\n91 92\n64 68", "output": "442" }, { "input": "47 21\n3 5\n-422 -417\n60 60\n-348 -348\n-3 -3\n-364 -361\n-49 -41\n-436 -430\n-250 -244\n-33 -26\n-162 -158\n-90 -88\n-357 -352\n-339 -337\n-25 -19\n-69 -67\n-261 -260\n-292 -283\n12 18\n44 44\n-277 -275\n-301 -293\n-108 -98\n-180 -172\n-327 -318\n-314 -309\n-12 -7\n-134 -130\n33 35\n-190 -184\n-65 -55\n-242 -240\n-448 -444\n-408 -405\n53 57\n-145 -144\n-207 -200\n-110 -110\n-221 -216\n-122 -112\n26 27\n-271 -269\n-82 -79\n-235 -229\n-382 -373\n-397 -391\n-155 -153", "output": "18" }, { "input": "3 4587\n-49 368\n-734 -390\n-380 -117", "output": "3560" }, { "input": "1 100000\n-100000 100000", "output": "99999" }, { "input": "2 100000\n-100000 99999\n100000 100000", "output": "99999" }, { "input": "1 7\n0 0", "output": "6" }, { "input": "2 5848\n-100000 0\n1 100000", "output": "4679" }, { "input": "3 99999\n-100000 -100000\n-99999 99998\n99999 100000", "output": "99996" } ]
1,690,762,853
2,147,483,647
Python 3
OK
TESTS
28
342
0
#اللهم صل على نبينا محمد i,n=input().split() i,n=int(i),int(n) count=0 count2=0 for i in range(i): j,k=input().split() j,k=int(j),int(k) count=count+(k-j+1) istrue= True if count%n==0: print(0) exit(0) while istrue: if count%n!=0: count+=1 count2+=1 else: break print(count2)
Title: Polo the Penguin and Segments Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little penguin Polo adores integer segments, that is, pairs of integers [*l*; *r*] (*l*<=≤<=*r*). He has a set that consists of *n* integer segments: [*l*1; *r*1],<=[*l*2; *r*2],<=...,<=[*l**n*; *r**n*]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [*l*; *r*] to either segment [*l*<=-<=1; *r*], or to segment [*l*; *r*<=+<=1]. The value of a set of segments that consists of *n* segments [*l*1; *r*1],<=[*l*2; *r*2],<=...,<=[*l**n*; *r**n*] is the number of integers *x*, such that there is integer *j*, for which the following inequality holds, *l**j*<=≤<=*x*<=≤<=*r**j*. Find the minimum number of moves needed to make the value of the set of Polo's segments divisible by *k*. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). Each of the following *n* lines contain a segment as a pair of integers *l**i* and *r**i* (<=-<=105<=≤<=*l**i*<=≤<=*r**i*<=≤<=105), separated by a space. It is guaranteed that no two segments intersect. In other words, for any two integers *i*,<=*j* (1<=≤<=*i*<=&lt;<=*j*<=≤<=*n*) the following inequality holds, *min*(*r**i*,<=*r**j*)<=&lt;<=*max*(*l**i*,<=*l**j*). Output Specification: In a single line print a single integer — the answer to the problem. Demo Input: ['2 3\n1 2\n3 4\n', '3 7\n1 2\n3 3\n4 7\n'] Demo Output: ['2\n', '0\n'] Note: none
```python #اللهم صل على نبينا محمد i,n=input().split() i,n=int(i),int(n) count=0 count2=0 for i in range(i): j,k=input().split() j,k=int(j),int(k) count=count+(k-j+1) istrue= True if count%n==0: print(0) exit(0) while istrue: if count%n!=0: count+=1 count2+=1 else: break print(count2) ```
3
69
A
Young Physicist
PROGRAMMING
1,000
[ "implementation", "math" ]
A. Young Physicist
2
256
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]
[ "NO", "YES" ]
none
500
[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]
1,624,946,740
2,147,483,647
Python 3
OK
TESTS
81
154
0
n=int(input()) s=0 d=0 for i in range(n): a,b,c=map(int,input().split()) s+=(a+b+c) d+=c if s==0 and d==0: print("YES") else: print("NO")
Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none
```python n=int(input()) s=0 d=0 for i in range(n): a,b,c=map(int,input().split()) s+=(a+b+c) d+=c if s==0 and d==0: print("YES") else: print("NO") ```
3.9615
136
A
Presents
PROGRAMMING
800
[ "implementation" ]
null
null
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]
[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]
none
500
[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71", "output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90" }, { "input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9", "output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9" }, { "input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44", "output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20" }, { "input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 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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" } ]
1,693,231,426
2,147,483,647
PyPy 3-64
OK
TESTS
101
124
0
n = int(input()) arr = list(map(int,input().split(" "))) for i in range(1, n + 1): print(arr.index(i) + 1, end=" ")
Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none
```python n = int(input()) arr = list(map(int,input().split(" "))) for i in range(1, n + 1): print(arr.index(i) + 1, end=" ") ```
3
379
A
New Year Candles
PROGRAMMING
1,000
[ "implementation" ]
null
null
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Print a single integer — the number of hours Vasily can light up the room for.
[ "4 2\n", "6 3\n" ]
[ "7\n", "8\n" ]
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
500
[ { "input": "4 2", "output": "7" }, { "input": "6 3", "output": "8" }, { "input": "1000 1000", "output": "1001" }, { "input": "123 5", "output": "153" }, { "input": "1000 2", "output": "1999" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "1 4", "output": "1" }, { "input": "2 2", "output": "3" }, { "input": "3 2", "output": "5" }, { "input": "3 3", "output": "4" }, { "input": "999 2", "output": "1997" }, { "input": "1000 3", "output": "1499" }, { "input": "1000 4", "output": "1333" }, { "input": "1 1000", "output": "1" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "10 4", "output": "13" }, { "input": "4 3", "output": "5" }, { "input": "91 5", "output": "113" }, { "input": "777 17", "output": "825" }, { "input": "100 4", "output": "133" }, { "input": "5 3", "output": "7" }, { "input": "6 4", "output": "7" }, { "input": "26 8", "output": "29" }, { "input": "9 4", "output": "11" }, { "input": "20 3", "output": "29" }, { "input": "17 3", "output": "25" } ]
1,674,814,581
2,147,483,647
Python 3
OK
TESTS
32
46
0
n , m = map(int,input().split()) ans = n + n//m x = n while(x!=ans): x = ans ans = n + ans//m print(ans)
Title: New Year Candles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. Input Specification: The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). Output Specification: Print a single integer — the number of hours Vasily can light up the room for. Demo Input: ['4 2\n', '6 3\n'] Demo Output: ['7\n', '8\n'] Note: Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
```python n , m = map(int,input().split()) ans = n + n//m x = n while(x!=ans): x = ans ans = n + ans//m print(ans) ```
3
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" } ]
1,681,111,577
2,147,483,647
Python 3
OK
TESTS
18
92
0
n, m = map(int, input().split()) puzzles = list(map(int, input().split())) # sort the puzzles in non-decreasing order puzzles.sort() # initialize the minimum difference with a large value min_diff = float('inf') # iterate through all possible pairs of puzzles for i in range(m - n + 1): diff = puzzles[i+n-1] - puzzles[i] if diff < min_diff: min_diff = diff print(min_diff)
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 n, m = map(int, input().split()) puzzles = list(map(int, input().split())) # sort the puzzles in non-decreasing order puzzles.sort() # initialize the minimum difference with a large value min_diff = float('inf') # iterate through all possible pairs of puzzles for i in range(m - n + 1): diff = puzzles[i+n-1] - puzzles[i] if diff < min_diff: min_diff = diff print(min_diff) ```
3
454
A
Little Pony and Crystal Mine
PROGRAMMING
800
[ "implementation" ]
null
null
Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=&gt;<=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it. You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).
Output a crystal of size *n*.
[ "3\n", "5\n", "7\n" ]
[ "*D*\nDDD\n*D*\n", "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n", "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n" ]
none
500
[ { "input": "3", "output": "*D*\nDDD\n*D*" }, { "input": "5", "output": "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**" }, { "input": "7", "output": "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***" }, { "input": "11", "output": "*****D*****\n****DDD****\n***DDDDD***\n**DDDDDDD**\n*DDDDDDDDD*\nDDDDDDDDDDD\n*DDDDDDDDD*\n**DDDDDDD**\n***DDDDD***\n****DDD****\n*****D*****" }, { "input": "15", "output": "*******D*******\n******DDD******\n*****DDDDD*****\n****DDDDDDD****\n***DDDDDDDDD***\n**DDDDDDDDDDD**\n*DDDDDDDDDDDDD*\nDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDD*\n**DDDDDDDDDDD**\n***DDDDDDDDD***\n****DDDDDDD****\n*****DDDDD*****\n******DDD******\n*******D*******" }, { "input": "21", "output": "**********D**********\n*********DDD*********\n********DDDDD********\n*******DDDDDDD*******\n******DDDDDDDDD******\n*****DDDDDDDDDDD*****\n****DDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDD**\n*DDDDDDDDDDDDDDDDDDD*\nDDDDDDDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDDDDDDDD*\n**DDDDDDDDDDDDDDDDD**\n***DDDDDDDDDDDDDDD***\n****DDDDDDDDDDDDD****\n*****DDDDDDDDDDD*****\n******DDDDDDDDD******\n*******DDDDDDD*******\n********DDDDD********\n*********DDD*********\n**********D**********" }, { "input": "33", "output": "****************D****************\n***************DDD***************\n**************DDDDD**************\n*************DDDDDDD*************\n************DDDDDDDDD************\n***********DDDDDDDDDDD***********\n**********DDDDDDDDDDDDD**********\n*********DDDDDDDDDDDDDDD*********\n********DDDDDDDDDDDDDDDDD********\n*******DDDDDDDDDDDDDDDDDDD*******\n******DDDDDDDDDDDDDDDDDDDDD******\n*****DDDDDDDDDDDDDDDDDDDDDDD*****\n****DDDDDDDDDDDDDDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDDDD..." }, { "input": "57", "output": "****************************D****************************\n***************************DDD***************************\n**************************DDDDD**************************\n*************************DDDDDDD*************************\n************************DDDDDDDDD************************\n***********************DDDDDDDDDDD***********************\n**********************DDDDDDDDDDDDD**********************\n*********************DDDDDDDDDDDDDDD*********************\n********************DDDDDDDDDDDDDDDDD**..." }, { "input": "69", "output": "**********************************D**********************************\n*********************************DDD*********************************\n********************************DDDDD********************************\n*******************************DDDDDDD*******************************\n******************************DDDDDDDDD******************************\n*****************************DDDDDDDDDDD*****************************\n****************************DDDDDDDDDDDDD****************************\n**************..." }, { "input": "81", "output": "****************************************D****************************************\n***************************************DDD***************************************\n**************************************DDDDD**************************************\n*************************************DDDDDDD*************************************\n************************************DDDDDDDDD************************************\n***********************************DDDDDDDDDDD***********************************\n*************..." }, { "input": "91", "output": "*********************************************D*********************************************\n********************************************DDD********************************************\n*******************************************DDDDD*******************************************\n******************************************DDDDDDD******************************************\n*****************************************DDDDDDDDD*****************************************\n****************************************DDDDDD..." }, { "input": "101", "output": "**************************************************D**************************************************\n*************************************************DDD*************************************************\n************************************************DDDDD************************************************\n***********************************************DDDDDDD***********************************************\n**********************************************DDDDDDDDD********************************************..." } ]
1,613,799,293
2,147,483,647
Python 3
OK
TESTS
12
62
0
n = int(input()) x = 'D' z='' for i in range(0,n+1): if i%2: print((x * i).center(n, '*')) for i in range(n-1,0,-1): if i%2: print((x * i).center(n, '*'))
Title: Little Pony and Crystal Mine Time Limit: None seconds Memory Limit: None megabytes Problem Description: Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=&gt;<=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it. You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input Specification: The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd). Output Specification: Output a crystal of size *n*. Demo Input: ['3\n', '5\n', '7\n'] Demo Output: ['*D*\nDDD\n*D*\n', '**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n', '***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n'] Note: none
```python n = int(input()) x = 'D' z='' for i in range(0,n+1): if i%2: print((x * i).center(n, '*')) for i in range(n-1,0,-1): if i%2: print((x * i).center(n, '*')) ```
3
580
A
Kefa and First Steps
PROGRAMMING
900
[ "brute force", "dp", "implementation" ]
null
null
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order. Help Kefa cope with this task!
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
[ "6\n2 2 1 3 4 1\n", "3\n2 2 9\n" ]
[ "3", "3" ]
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one. In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
750
[ { "input": "6\n2 2 1 3 4 1", "output": "3" }, { "input": "3\n2 2 9", "output": "3" }, { "input": "5\n10 100 111 1 2", "output": "3" }, { "input": "10\n1 2 3 4 1 2 3 4 5 6", "output": "6" }, { "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": "50" }, { "input": "100\n1 838 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 605 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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 27 533 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 835 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 992", "output": "42" }, { "input": "1\n1", "output": "1" }, { "input": "2\n1 1", "output": "2" }, { "input": "6\n5 4 3 2 1 2", "output": "2" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "9" }, { "input": "7\n99 100 1 2 3 4 5", "output": "5" }, { "input": "5\n3 3 1 2 3", "output": "3" }, { "input": "1\n100", "output": "1" }, { "input": "1\n5", "output": "1" } ]
1,691,063,411
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
8
62
3,379,200
n, my_list = int(input()), [] s = input().split() for i in range (len(s)): s[i] = int(s[i]) my_list.append(s[i]) count = 1 max_count = 0 first = my_list[0] for j in range(1, len(my_list)): if my_list[j] >= first: count += 1 first = my_list[j] if count >= max_count: max_count = count else: first = my_list[j] count = 1 print(max_count)
Title: Kefa and First Steps Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order. Help Kefa cope with this task! Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*. Demo Input: ['6\n2 2 1 3 4 1\n', '3\n2 2 9\n'] Demo Output: ['3', '3'] Note: In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one. In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
```python n, my_list = int(input()), [] s = input().split() for i in range (len(s)): s[i] = int(s[i]) my_list.append(s[i]) count = 1 max_count = 0 first = my_list[0] for j in range(1, len(my_list)): if my_list[j] >= first: count += 1 first = my_list[j] if count >= max_count: max_count = count else: first = my_list[j] count = 1 print(max_count) ```
0
61
A
Ultra-Fast Mathematician
PROGRAMMING
800
[ "implementation" ]
A. Ultra-Fast Mathematician
2
256
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Write one line — the corresponding answer. Do not omit the leading 0s.
[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]
[ "1110001\n", "111\n", "0100\n", "00010\n" ]
none
500
[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110", "output": "001100101000011111111101111011101010111001" }, { "input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001", "output": "0111010010100110110101100010000100010100000" }, { "input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100", "output": "11111110000000100101000100110111001100011001" }, { "input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011", "output": "101011011100100010100011011001101010100100010" }, { "input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001", "output": "1101001100111011010111110110101111001011110111" }, { "input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001", "output": "10010101000101000000011010011110011110011110001" }, { "input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100", "output": "011011011100000000010101110010000000101000111101" }, { "input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100", "output": "0101010111101001011011110110011101010101010100011" }, { "input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011", "output": "11001011010010111000010110011101100100001110111111" }, { "input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011", "output": "111011101010011100001111101001101011110010010110001" }, { "input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001", "output": "0100111110110011111110010010010000110111100101101101" }, { "input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100", "output": "01011001110111010111001100010011010100010000111011000" }, { "input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111", "output": "100011101001001000011011011001111000100000010100100100" }, { "input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110", "output": "1100110010000101101010111111101001001001110101110010110" }, { "input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110", "output": "01000111100111001011110010100011111111110010101100001101" }, { "input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010", "output": "110001010001000011000101110101000100001011111001011001001" }, { "input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111", "output": "1110100010111000101001001011101110011111100111000011011011" }, { "input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110", "output": "01110110101110100100110011010000001000101100101111000111011" }, { "input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011", "output": "111100101000000011101011011001110010101111000110010010000000" }, { "input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111", "output": "0100100010111110010011101010000011111110001110010110010111001" }, { "input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111", "output": "00110100000011001101101100100010110010001100000001100110011101" }, { "input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011", "output": "000000011000111011110011101000010000010100101000000011010110010" }, { "input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010", "output": "0010100110110100111100100100101101010100100111011010001001010101" }, { "input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111", "output": "11010110111100101111101001100001110100010110010110110111100110100" }, { "input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111", "output": "111111010011011100101110100110111111111001111110011010111111110000" }, { "input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110", "output": "1010101010100010001001001001100000111000010010010100010011000100000" }, { "input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000", "output": "00011111011111001000011100010011100011010100101011011000001001111110" }, { "input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111", "output": "001111000011001110100111010101111111011100110011001010010010000111011" }, { "input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101", "output": "0110001100110100010000110111000010011010011000011001010011010100010100" }, { "input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010", "output": "00010000000110110101000011001000000100100110111010011111101010001010000" }, { "input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001", "output": "000100100000000110011100100001010110101001100101110010010011111001110111" }, { "input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000", "output": "1000111100010011010110011101000000101010101100011111100001101111001010010" }, { "input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011", "output": "01000010011110111001011011110000001011000111101101101010010110001010100100" }, { "input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101", "output": "101101110110110010011100001011111100100001110000101100110000100011011100110" }, { "input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001", "output": "1110111111110010111000001100101010101011010100101010100101100011001001111111" }, { "input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111", "output": "10111011000111000101110100101000100111011011100011110110000101010001111010111" }, { "input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110", "output": "110100100110000101010010011010011001100110000111010000010100001011110111111101" }, { "input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111", "output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111" }, { "input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001", "output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001" }, { "input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110", "output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011" }, { "input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]
1,628,360,138
2,147,483,647
Python 3
OK
TESTS
102
77
6,758,400
# -*- coding: utf-8 -*- """ Created on Sun Aug 8 00:06:02 2021 @author: nijhum """ n = list(input()) m=list(input()) for i in range(len(n)) : if n[i] == m[i]: print('0',end='') else: print('1',end='')
Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none
```python # -*- coding: utf-8 -*- """ Created on Sun Aug 8 00:06:02 2021 @author: nijhum """ n = list(input()) m=list(input()) for i in range(len(n)) : if n[i] == m[i]: print('0',end='') else: print('1',end='') ```
3.968161
707
A
Brain's Photos
PROGRAMMING
800
[ "implementation" ]
null
null
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead. As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such). Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour! As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white. Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors: - 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black) The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively. Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Print the "#Black&amp;White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
[ "2 2\nC M\nY Y\n", "3 2\nW W\nW W\nB B\n", "1 1\nW\n" ]
[ "#Color", "#Black&amp;White", "#Black&amp;White" ]
none
500
[ { "input": "2 2\nC M\nY Y", "output": "#Color" }, { "input": "3 2\nW W\nW W\nB B", "output": "#Black&White" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "2 3\nW W W\nB G Y", "output": "#Color" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "5 5\nW G B Y M\nG B Y M C\nB Y M C W\nY M C W G\nM C W G B", "output": "#Color" }, { "input": "1 6\nC M Y W G B", "output": "#Color" }, { "input": "1 3\nW G B", "output": "#Black&White" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "5 5\nW G B W G\nG B W G B\nB W G B W\nW G B W G\nG B W G B", "output": "#Black&White" }, { "input": "2 3\nW W W\nB G C", "output": "#Color" }, { "input": "2 3\nW W W\nB G M", "output": "#Color" }, { "input": "3 3\nC B W\nB Y M\nB B W", "output": "#Color" }, { "input": "1 3\nW C W", "output": "#Color" }, { "input": "3 3\nB W B\nB C W\nB W W", "output": "#Color" }, { "input": "1 2\nW Y", "output": "#Color" }, { "input": "1 1\nG", "output": "#Black&White" }, { "input": "1 3\nB G W", "output": "#Black&White" }, { "input": "2 2\nW W\nB C", "output": "#Color" }, { "input": "1 1\nM", "output": "#Color" }, { "input": "1 2\nW C", "output": "#Color" }, { "input": "2 3\nW W M\nW W M", "output": "#Color" }, { "input": "3 2\nW W\nW W\nB C", "output": "#Color" }, { "input": "2 3\nW W C\nW W W", "output": "#Color" }, { "input": "1 3\nG G G", "output": "#Black&White" }, { "input": "1 1\nC", "output": "#Color" }, { "input": "1 2\nC W", "output": "#Color" }, { "input": "1 3\nW W C", "output": "#Color" }, { "input": "2 2\nW B\nB G", "output": "#Black&White" }, { "input": "2 2\nB B\nY Y", "output": "#Color" }, { "input": "2 2\nG G\nC C", "output": "#Color" }, { "input": "1 2\nB C", "output": "#Color" }, { "input": "1 4\nG G G C", "output": "#Color" }, { "input": "1 1\nB", "output": "#Black&White" }, { "input": "2 1\nY\nB", "output": "#Color" } ]
1,693,718,301
2,147,483,647
Python 3
OK
TESTS
50
46
102,400
a,b= map(int,input().split()) l=[] while(a!=0): x=list(map(str,input().split())) l.extend(x) a=a-1 #print(l) if ('C' in l) or('M' in l) or('Y' in l): print("#Color") else: print("#Black&White")
Title: Brain's Photos Time Limit: None seconds Memory Limit: None megabytes Problem Description: Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead. As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such). Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour! As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white. Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors: - 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black) The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively. Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. Output Specification: Print the "#Black&amp;White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. Demo Input: ['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n'] Demo Output: ['#Color', '#Black&amp;White', '#Black&amp;White'] Note: none
```python a,b= map(int,input().split()) l=[] while(a!=0): x=list(map(str,input().split())) l.extend(x) a=a-1 #print(l) if ('C' in l) or('M' in l) or('Y' in l): print("#Color") else: print("#Black&White") ```
3
0
none
none
none
0
[ "none" ]
null
null
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research. First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad. In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
[ "5\n1 3 2 1 2\n", "6\n2 1 2 2 4 1\n" ]
[ "3\n", "2\n" ]
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer. In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
0
[ { "input": "5\n1 3 2 1 2", "output": "3" }, { "input": "6\n2 1 2 2 4 1", "output": "2" }, { "input": "1\n0", "output": "0" }, { "input": "1\n200000", "output": "200000" }, { "input": "2\n2018 2017", "output": "2018" }, { "input": "5\n100 1000 1000 1000 1000", "output": "100" }, { "input": "8\n200000 1 200000 1 200000 1 200000 2", "output": "1" }, { "input": "5\n20000 1 2 3 4", "output": "20000" }, { "input": "2\n200000 1", "output": "200000" }, { "input": "3\n2 100 2", "output": "100" }, { "input": "2\n10 1", "output": "10" }, { "input": "11\n1 1 1 1 1 1 1 1 1 1 1", "output": "1" }, { "input": "3\n5 5 5", "output": "5" }, { "input": "2\n5 1", "output": "5" } ]
1,510,510,375
7,675
Python 3
OK
TESTS
33
202
18,329,600
#B問題 N = int(input()) a = list(map(int,input().split(" "))) index = {} for i in range(N): index[a[i]] = i print(min(index.items(), key=lambda x:x[1])[0])
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research. First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe. Input Specification: In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad. In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted. Output Specification: Print one integer — index of the cafe that Vlad hasn't visited for as long as possible. Demo Input: ['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n'] Demo Output: ['3\n', '2\n'] Note: In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer. In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
```python #B問題 N = int(input()) a = list(map(int,input().split(" "))) index = {} for i in range(N): index[a[i]] = i print(min(index.items(), key=lambda x:x[1])[0]) ```
3
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" } ]
1,698,238,799
2,147,483,647
Python 3
OK
TESTS
56
46
0
n=input() flag=True arr=[] for num in n: arr.append(num) for i in range(1,len(arr)): if arr[i].islower(): flag=False if flag==True: if arr[0].isupper(): for i in range(0,len(arr)): arr[i]=arr[i].lower() else: arr[0]=arr[0].upper() for i in range (1,len(arr)): arr[i]=arr[i].lower() brr="" for b in arr : brr=brr+b print(brr)
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 n=input() flag=True arr=[] for num in n: arr.append(num) for i in range(1,len(arr)): if arr[i].islower(): flag=False if flag==True: if arr[0].isupper(): for i in range(0,len(arr)): arr[i]=arr[i].lower() else: arr[0]=arr[0].upper() for i in range (1,len(arr)): arr[i]=arr[i].lower() brr="" for b in arr : brr=brr+b print(brr) ```
3
651
B
Beautiful Paintings
PROGRAMMING
1,200
[ "greedy", "sortings" ]
null
null
There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one. We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=&gt;<=*a**i*.
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting.
Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=&gt;<=*a**i*, after the optimal rearrangement.
[ "5\n20 30 10 50 40\n", "4\n200 100 100 200\n" ]
[ "4\n", "2\n" ]
In the first sample, the optimal order is: 10, 20, 30, 40, 50. In the second sample, the optimal order is: 100, 200, 100, 200.
1,000
[ { "input": "5\n20 30 10 50 40", "output": "4" }, { "input": "4\n200 100 100 200", "output": "2" }, { "input": "10\n2 2 2 2 2 2 2 2 2 2", "output": "0" }, { "input": "1\n1000", "output": "0" }, { "input": "2\n444 333", "output": "1" }, { "input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14", "output": "95" }, { "input": "1\n995", "output": "0" }, { "input": "10\n103 101 103 103 101 102 100 100 101 104", "output": "7" }, { "input": "20\n102 100 102 104 102 101 104 103 100 103 105 105 100 105 100 100 101 105 105 102", "output": "15" }, { "input": "20\n990 994 996 999 997 994 990 992 990 993 992 990 999 999 992 994 997 990 993 998", "output": "15" }, { "input": "100\n1 8 3 8 10 8 5 3 10 3 5 8 4 5 5 5 10 3 6 6 6 6 6 7 2 7 2 4 7 8 3 8 7 2 5 6 1 5 5 7 9 7 6 9 1 8 1 3 6 5 1 3 6 9 5 6 8 4 8 6 10 9 2 9 3 8 7 5 2 10 2 10 3 6 5 5 3 5 10 2 3 7 10 8 8 4 3 4 9 6 10 7 6 6 6 4 9 9 8 9", "output": "84" } ]
1,585,931,040
2,147,483,647
Python 3
OK
TESTS
31
109
307,200
picture_quantity = int(input()) picture_beauty_list = list(map(int, input().split())) c = [] # try before b.sort() while len(picture_beauty_list) > 0: a = set(picture_beauty_list) b = list(a) b.sort() c += b for del_el in b: picture_beauty_list.remove(del_el) b = [] satisfied = 0 for i in range(len(c)): if i != len(c) - 1: if c[i] < c[i + 1]: satisfied += 1 print(satisfied)
Title: Beautiful Paintings Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one. We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=&gt;<=*a**i*. Input Specification: The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting. Output Specification: Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=&gt;<=*a**i*, after the optimal rearrangement. Demo Input: ['5\n20 30 10 50 40\n', '4\n200 100 100 200\n'] Demo Output: ['4\n', '2\n'] Note: In the first sample, the optimal order is: 10, 20, 30, 40, 50. In the second sample, the optimal order is: 100, 200, 100, 200.
```python picture_quantity = int(input()) picture_beauty_list = list(map(int, input().split())) c = [] # try before b.sort() while len(picture_beauty_list) > 0: a = set(picture_beauty_list) b = list(a) b.sort() c += b for del_el in b: picture_beauty_list.remove(del_el) b = [] satisfied = 0 for i in range(len(c)): if i != len(c) - 1: if c[i] < c[i + 1]: satisfied += 1 print(satisfied) ```
3
926
H
Endless Roses Most Beautiful
PROGRAMMING
2,200
[]
null
null
Arkady decided to buy roses for his girlfriend. A flower shop has white, orange and red roses, and the total amount of them is *n*. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white roses. Also, Arkady won't buy a bouquet where all roses have the same color. Arkady wants to buy exactly *k* roses. For each rose in the shop he knows its beauty and color: the beauty of the *i*-th rose is *b**i*, and its color is *c**i* ('W' for a white rose, 'O' for an orange rose and 'R' for a red rose). Compute the maximum possible total beauty of a bouquet of *k* roses satisfying the constraints above or determine that it is not possible to make such a bouquet.
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000) — the number of roses in the show and the number of roses Arkady wants to buy. The second line contains a sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=10<=000), where *b**i* equals the beauty of the *i*-th rose. The third line contains a string *c* of length *n*, consisting of uppercase English letters 'W', 'O' and 'R', where *c**i* denotes the color of the *i*-th rose: 'W' denotes white, 'O'  — orange, 'R' — red.
Print the maximum possible total beauty of a bouquet of *k* roses that satisfies the constraints above. If it is not possible to make a single such bouquet, print -1.
[ "5 3\n4 3 4 1 6\nRROWW\n", "5 2\n10 20 14 20 11\nRRRRR\n", "11 5\n5 6 3 2 3 4 7 5 4 5 6\nRWOORWORROW\n" ]
[ "11\n", "-1\n", "28\n" ]
In the first example Arkady wants to buy 3 roses. He can, for example, buy both red roses (their indices are 1 and 2, and their total beauty is 7) and the only orange rose (its index is 3, its beauty is 4). This way the total beauty of the bouquet is 11. In the second example Arkady can not buy a bouquet because all roses have the same color.
0
[ { "input": "5 3\n4 3 4 1 6\nRROWW", "output": "11" }, { "input": "5 2\n10 20 14 20 11\nRRRRR", "output": "-1" }, { "input": "11 5\n5 6 3 2 3 4 7 5 4 5 6\nRWOORWORROW", "output": "28" }, { "input": "15 10\n8560 6244 9607 5137 7187 3217 5527 9919 282 8748 3529 6110 5767 521 3393\nOWRWOORWRORWWRO", "output": "64282" }, { "input": "10 4\n1208 5835 2637 5827 3722 6837 3499 6438 43 5333\nWRRWRWRWRW", "output": "-1" }, { "input": "13 3\n9675 8988 5499 6356 5083 6067 5580 4580 6735 3617 9536 8218 3265\nRRWRRROWRWWWW", "output": "24243" }, { "input": "13 7\n8543 3460 1282 3956 8203 762 6059 9361 4427 8868 5849 3439 8891\nWWOOOOWOWWRWO", "output": "54352" }, { "input": "30 15\n7926 577 5009 7237 4395 3239 8994 4429 8126 2925 139 320 4442 3397 1292 2800 9505 6043 5946 8058 4031 6871 4689 1977 73 440 5320 5290 4707 387\nOOWOWWORRWOWORWRRRRWORROOWWROW", "output": "91633" }, { "input": "1 1\n100\nO", "output": "-1" }, { "input": "1 1\n1059\nO", "output": "-1" }, { "input": "2 2\n9907 4483\nOO", "output": "-1" }, { "input": "1 1\n6750\nW", "output": "-1" }, { "input": "2 2\n144 174\nOW", "output": "318" }, { "input": "3 2\n776 4797 9449\nOWO", "output": "14246" }, { "input": "2 2\n3486 8968\nWW", "output": "-1" }, { "input": "3 2\n2330 2140 3440\nWOW", "output": "5580" }, { "input": "4 2\n1175 8186 4321 1810\nWWOO", "output": "12507" }, { "input": "1 1\n6479\nR", "output": "-1" }, { "input": "2 2\n8512 9903\nOR", "output": "18415" }, { "input": "3 2\n7035 5046 7357\nOOR", "output": "14392" }, { "input": "2 2\n6442 4558\nWR", "output": "-1" }, { "input": "3 2\n9700 698 2122\nOWR", "output": "11822" }, { "input": "4 3\n254 4510 2194 9543\nWOOR", "output": "16247" }, { "input": "3 2\n517 6744 2364\nRWW", "output": "-1" }, { "input": "4 2\n2884 2918 8629 9905\nRWOW", "output": "18534" }, { "input": "5 2\n7882 871 789 4432 7664\nOWORW", "output": "15546" }, { "input": "2 2\n2926 8428\nRR", "output": "-1" }, { "input": "3 2\n7926 1770 3255\nORR", "output": "11181" }, { "input": "4 2\n2578 7910 108 3809\nOROR", "output": "10488" }, { "input": "3 2\n5920 9303 7542\nWRR", "output": "-1" }, { "input": "4 2\n5909 4286 5444 6473\nOWRR", "output": "12382" }, { "input": "5 2\n96 6627 8780 3764 970\nRROWO", "output": "15407" }, { "input": "4 2\n6657 1489 9138 4273\nRRWW", "output": "-1" }, { "input": "5 2\n1598 6700 334 6455 9292\nWORWR", "output": "15992" }, { "input": "6 2\n6231 9178 9845 5932 5477 6659\nORRWOW", "output": "16076" }, { "input": "1 1\n780\nO", "output": "-1" }, { "input": "1 1\n3214\nW", "output": "-1" }, { "input": "2 2\n8455 5432\nOW", "output": "13887" }, { "input": "1 1\n6555\nR", "output": "-1" }, { "input": "2 2\n1120 5670\nOR", "output": "6790" }, { "input": "3 2\n8884 4514 1673\nORW", "output": "13398" }, { "input": "1 1\n6908\nO", "output": "-1" }, { "input": "1 1\n3934\nW", "output": "-1" }, { "input": "2 2\n8856 7299\nWO", "output": "16155" }, { "input": "1 1\n2683\nR", "output": "-1" }, { "input": "2 2\n9094 5772\nRO", "output": "14866" }, { "input": "3 2\n518 9438 7938\nWOR", "output": "17376" }, { "input": "10 4\n9513 754 5917 1337 2337 1387 3499 9873 9138 7647\nWROWWOWWRO", "output": "32950" }, { "input": "20 5\n3747 219 7826 7713 6886 466 1136 7069 1234 7556 3657 6017 9965 2847 6880 362 3179 4676 5934 4913\nWWWOWOWOWOROWOWWWOWW", "output": "40129" } ]
1,521,304,502
3,602
Python 3
WRONG_ANSWER
TESTS
16
826
21,299,200
n,k=map(int,input().split()) cost=list(map(int,input().split())) s=input() #exit(0) a=[] b=[] c=[] #print(n,k,cost) kek1=0 kek2=0 kek3=0 for i in range(n): if (s[i]==chr(ord("W"))): kek1+=1 a.append(cost[i]) if (s[i]==chr(ord("O"))): kek2+=1 b.append(cost[i]) if (s[i]==chr(ord("R"))): kek3+=1 c.append(cost[i]) a=sorted(a) a=list(reversed(a)) b=sorted(b) b=list(reversed(b)) c=sorted(c) c=list(reversed(c)) pref_a=[] pref_b=[] pref_c=[] if (kek1!=0): pref_a.append(a[0]) for i in range(1,len(a)): pref_a.append(pref_a[len(pref_a)-1]+a[i]) if (kek2!=0): pref_b.append(b[0]) for i in range(1,len(b)): pref_b.append(pref_b[len(pref_b)-1]+b[i]) if (kek3!=0): pref_c.append(c[0]) for i in range(1,len(c)): pref_c.append(pref_c[len(pref_c)-1]+c[i]) inf=10**20 ans=-inf for i in range(0,min(k,kek2)): cur=pref_b[i] left_k=k-(i+1) res_a=-inf res_c=-inf if (kek1): if (left_k>0): res_a=pref_a[min(kek1,left_k)-1] if (kek3): if (left_k>0): res_c=pref_c[min(kek3,left_k)-1] cur+=max(res_a,res_c) ans=max(ans,cur) if (ans<-(10**18)): ans=-1 print(ans)
Title: Endless Roses Most Beautiful Time Limit: None seconds Memory Limit: None megabytes Problem Description: Arkady decided to buy roses for his girlfriend. A flower shop has white, orange and red roses, and the total amount of them is *n*. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white roses. Also, Arkady won't buy a bouquet where all roses have the same color. Arkady wants to buy exactly *k* roses. For each rose in the shop he knows its beauty and color: the beauty of the *i*-th rose is *b**i*, and its color is *c**i* ('W' for a white rose, 'O' for an orange rose and 'R' for a red rose). Compute the maximum possible total beauty of a bouquet of *k* roses satisfying the constraints above or determine that it is not possible to make such a bouquet. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000) — the number of roses in the show and the number of roses Arkady wants to buy. The second line contains a sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=10<=000), where *b**i* equals the beauty of the *i*-th rose. The third line contains a string *c* of length *n*, consisting of uppercase English letters 'W', 'O' and 'R', where *c**i* denotes the color of the *i*-th rose: 'W' denotes white, 'O'  — orange, 'R' — red. Output Specification: Print the maximum possible total beauty of a bouquet of *k* roses that satisfies the constraints above. If it is not possible to make a single such bouquet, print -1. Demo Input: ['5 3\n4 3 4 1 6\nRROWW\n', '5 2\n10 20 14 20 11\nRRRRR\n', '11 5\n5 6 3 2 3 4 7 5 4 5 6\nRWOORWORROW\n'] Demo Output: ['11\n', '-1\n', '28\n'] Note: In the first example Arkady wants to buy 3 roses. He can, for example, buy both red roses (their indices are 1 and 2, and their total beauty is 7) and the only orange rose (its index is 3, its beauty is 4). This way the total beauty of the bouquet is 11. In the second example Arkady can not buy a bouquet because all roses have the same color.
```python n,k=map(int,input().split()) cost=list(map(int,input().split())) s=input() #exit(0) a=[] b=[] c=[] #print(n,k,cost) kek1=0 kek2=0 kek3=0 for i in range(n): if (s[i]==chr(ord("W"))): kek1+=1 a.append(cost[i]) if (s[i]==chr(ord("O"))): kek2+=1 b.append(cost[i]) if (s[i]==chr(ord("R"))): kek3+=1 c.append(cost[i]) a=sorted(a) a=list(reversed(a)) b=sorted(b) b=list(reversed(b)) c=sorted(c) c=list(reversed(c)) pref_a=[] pref_b=[] pref_c=[] if (kek1!=0): pref_a.append(a[0]) for i in range(1,len(a)): pref_a.append(pref_a[len(pref_a)-1]+a[i]) if (kek2!=0): pref_b.append(b[0]) for i in range(1,len(b)): pref_b.append(pref_b[len(pref_b)-1]+b[i]) if (kek3!=0): pref_c.append(c[0]) for i in range(1,len(c)): pref_c.append(pref_c[len(pref_c)-1]+c[i]) inf=10**20 ans=-inf for i in range(0,min(k,kek2)): cur=pref_b[i] left_k=k-(i+1) res_a=-inf res_c=-inf if (kek1): if (left_k>0): res_a=pref_a[min(kek1,left_k)-1] if (kek3): if (left_k>0): res_c=pref_c[min(kek3,left_k)-1] cur+=max(res_a,res_c) ans=max(ans,cur) if (ans<-(10**18)): ans=-1 print(ans) ```
0
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" } ]
1,594,962,793
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
7
202
22,323,200
n,m = [int(i) for i in input().split()] s = [] s1 = [] s2 = [] for j in range(m): s = s + [str(i) for i in input().split()[:m]] for i in range(len(s)): if(i%2==0): s1.append(s[i]) else: s2.append(s[i]) f = [str(i) for i in input().split()[:n]] # print(s1) # print(s2) v = [] for i in range(m): if(len(s1[i])<len(s2[i])): v.append(s1[i]) elif(len(s1[i])==len(s2[i])): v.append(s1[i]) else: v.append(s2[i]) for i in range(n): if(f[i] in s1): c = s1.index(f[i]) elif(f[i] in s2): c = s2.index(f[i]) #print(f[i],len(f[i]),c,v[c],len(v[c])) if(len(f[i])>len(v[c])): f[i] = v[c] #print(f[i]) #print(f) g = ' ' print(g.join(f))
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 = [int(i) for i in input().split()] s = [] s1 = [] s2 = [] for j in range(m): s = s + [str(i) for i in input().split()[:m]] for i in range(len(s)): if(i%2==0): s1.append(s[i]) else: s2.append(s[i]) f = [str(i) for i in input().split()[:n]] # print(s1) # print(s2) v = [] for i in range(m): if(len(s1[i])<len(s2[i])): v.append(s1[i]) elif(len(s1[i])==len(s2[i])): v.append(s1[i]) else: v.append(s2[i]) for i in range(n): if(f[i] in s1): c = s1.index(f[i]) elif(f[i] in s2): c = s2.index(f[i]) #print(f[i],len(f[i]),c,v[c],len(v[c])) if(len(f[i])>len(v[c])): f[i] = v[c] #print(f[i]) #print(f) g = ' ' print(g.join(f)) ```
-1
617
A
Elephant
PROGRAMMING
800
[ "math" ]
null
null
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
[ "5\n", "12\n" ]
[ "1\n", "3\n" ]
In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
500
[ { "input": "5", "output": "1" }, { "input": "12", "output": "3" }, { "input": "999999", "output": "200000" }, { "input": "41", "output": "9" }, { "input": "1000000", "output": "200000" }, { "input": "1", "output": "1" }, { "input": "2", "output": "1" }, { "input": "3", "output": "1" }, { "input": "4", "output": "1" }, { "input": "534204", "output": "106841" }, { "input": "469569", "output": "93914" }, { "input": "502877", "output": "100576" }, { "input": "942212", "output": "188443" }, { "input": "97", "output": "20" }, { "input": "53", "output": "11" }, { "input": "89", "output": "18" }, { "input": "574", "output": "115" }, { "input": "716", "output": "144" }, { "input": "729", "output": "146" }, { "input": "8901", "output": "1781" }, { "input": "3645", "output": "729" }, { "input": "4426", "output": "886" }, { "input": "46573", "output": "9315" }, { "input": "86380", "output": "17276" }, { "input": "94190", "output": "18838" }, { "input": "999990", "output": "199998" }, { "input": "999991", "output": "199999" }, { "input": "999992", "output": "199999" }, { "input": "999993", "output": "199999" }, { "input": "999994", "output": "199999" }, { "input": "999995", "output": "199999" }, { "input": "999996", "output": "200000" }, { "input": "999997", "output": "200000" }, { "input": "999998", "output": "200000" } ]
1,698,165,923
2,147,483,647
Python 3
OK
TESTS
34
62
0
x=int(input()) s=0 while(x>=5): x=x-5 s+=1 if(x>=1 and x<=4): s+=1 print(s)
Title: Elephant Time Limit: None seconds Memory Limit: None megabytes Problem Description: An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=&gt;<=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. Input Specification: The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. Output Specification: Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. Demo Input: ['5\n', '12\n'] Demo Output: ['1\n', '3\n'] Note: In the first sample the elephant needs to make one step of length 5 to reach the point *x*. In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
```python x=int(input()) s=0 while(x>=5): x=x-5 s+=1 if(x>=1 and x<=4): s+=1 print(s) ```
3
463
B
Caisa and Pylons
PROGRAMMING
1,100
[ "brute force", "implementation", "math" ]
null
null
Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=&gt;<=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Print a single number representing the minimum number of dollars paid by Caisa.
[ "5\n3 4 3 2 4\n", "3\n4 4 4\n" ]
[ "4\n", "4\n" ]
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
1,000
[ { "input": "5\n3 4 3 2 4", "output": "4" }, { "input": "3\n4 4 4", "output": "4" }, { "input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730", "output": "4000" }, { "input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621", "output": "3993" }, { "input": "30\n30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "30" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n69", "output": "69" } ]
1,584,523,568
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
77
0
n = input() a = [int(i) for i in input().split()] c = 0 for i in range(len(a)-1): c+=a[i+1]-a[i] print(a[0]+abs(c))
Title: Caisa and Pylons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=&gt;<=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons. Output Specification: Print a single number representing the minimum number of dollars paid by Caisa. Demo Input: ['5\n3 4 3 2 4\n', '3\n4 4 4\n'] Demo Output: ['4\n', '4\n'] Note: In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
```python n = input() a = [int(i) for i in input().split()] c = 0 for i in range(len(a)-1): c+=a[i+1]-a[i] print(a[0]+abs(c)) ```
0
922
C
Cave Painting
PROGRAMMING
1,600
[ "brute force", "number theory" ]
null
null
Imp is watching a documentary about cave painting. Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number *n* by all integers *i* from 1 to *k*. Unfortunately, there are too many integers to analyze for Imp. Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all , 1<=≤<=*i*<=≤<=*k*, are distinct, i. e. there is no such pair (*i*,<=*j*) that: - 1<=≤<=*i*<=&lt;<=*j*<=≤<=*k*, - , where is the remainder of division *x* by *y*.
The only line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=1018).
Print "Yes", if all the remainders are distinct, and "No" otherwise. You can print each letter in arbitrary case (lower or upper).
[ "4 4\n", "5 3\n" ]
[ "No\n", "Yes\n" ]
In the first sample remainders modulo 1 and 4 coincide.
1,250
[ { "input": "4 4", "output": "No" }, { "input": "5 3", "output": "Yes" }, { "input": "1 1", "output": "Yes" }, { "input": "744 18", "output": "No" }, { "input": "47879 10", "output": "Yes" }, { "input": "1000000000000000000 1000000000000000000", "output": "No" }, { "input": "657180569218773599 42", "output": "Yes" }, { "input": "442762254977842799 30", "output": "Yes" }, { "input": "474158606260730555 1", "output": "Yes" }, { "input": "807873101233533988 39", "output": "No" }, { "input": "423 7", "output": "No" }, { "input": "264306177888923090 5", "output": "No" }, { "input": "998857801526481788 87", "output": "No" }, { "input": "999684044704565212 28", "output": "No" }, { "input": "319575605003866172 71", "output": "No" }, { "input": "755804560577415016 17", "output": "No" }, { "input": "72712630136142067 356370939", "output": "No" }, { "input": "807264258068668062 33080422", "output": "No" }, { "input": "808090496951784190 311661970", "output": "No" }, { "input": "808916740129867614 180178111", "output": "No" }, { "input": "1 2", "output": "Yes" }, { "input": "2 1", "output": "Yes" }, { "input": "57334064998850639 19", "output": "Yes" }, { "input": "144353716412182199 11", "output": "Yes" }, { "input": "411002215096001759 11", "output": "Yes" }, { "input": "347116374613371527 3", "output": "Yes" }, { "input": "518264351335130399 37", "output": "Yes" }, { "input": "192435891235905239 11", "output": "Yes" }, { "input": "491802505049361659 7", "output": "Yes" }, { "input": "310113769227703889 3", "output": "Yes" }, { "input": "876240758958364799 41", "output": "Yes" }, { "input": "173284263472319999 33", "output": "Yes" }, { "input": "334366426725130799 29", "output": "Yes" }, { "input": "415543470272330399 26", "output": "Yes" }, { "input": "631689521541558479 22", "output": "Yes" }, { "input": "581859366558790319 14", "output": "Yes" }, { "input": "224113913709159599 10", "output": "Yes" }, { "input": "740368848764104559 21", "output": "Yes" }, { "input": "895803074828822159 17", "output": "Yes" }, { "input": "400349974997012039 13", "output": "Yes" }, { "input": "205439024252247599 5", "output": "Yes" }, { "input": "197688463911338399 39", "output": "Yes" }, { "input": "283175367224349599 39", "output": "Yes" }, { "input": "893208176423362799 31", "output": "Yes" }, { "input": "440681012669897999 27", "output": "Yes" }, { "input": "947403664618451039 19", "output": "Yes" }, { "input": "232435556779345919 19", "output": "Yes" }, { "input": "504428493840551279 23", "output": "Yes" }, { "input": "30019549241681999 20", "output": "Yes" }, { "input": "648000813924303839 16", "output": "Yes" }, { "input": "763169499725761451 488954176053755860", "output": "No" }, { "input": "199398459594277592 452260924647536414", "output": "No" }, { "input": "635627415167826436 192195636386541160", "output": "No" }, { "input": "71856370741375281 155502380685354417", "output": "No" }, { "input": "731457367464667229 118809129279134971", "output": "No" }, { "input": "167686318743248777 858743836723172421", "output": "No" }, { "input": "603915274316797622 822050585316952974", "output": "No" }, { "input": "647896534275160623 65689274138731296", "output": "No" }, { "input": "648722777453244047 501918229712280140", "output": "No" }, { "input": "649549020631327471 41923378183538525", "output": "No" }, { "input": "650375259514443599 597748177714153637", "output": "No" }, { "input": "651201506987494319 33977137582669778", "output": "No" }, { "input": "652027745870610447 470206093156218622", "output": "No" }, { "input": "652853989048693871 906435048729767466", "output": "No" }, { "input": "653680227931809999 342664004303316311", "output": "No" }, { "input": "654506475404860719 375019787446735639", "output": "No" }, { "input": "655332714287976847 438493956600157103", "output": "No" }, { "input": "166512305365727033 900267947832156186", "output": "No" }, { "input": "167338548543810457 336496907700672326", "output": "No" }, { "input": "168164787426926585 772725863274221171", "output": "No" }, { "input": "523 3", "output": "No" }, { "input": "39211 6", "output": "No" }, { "input": "22151 9", "output": "No" }, { "input": "1 3", "output": "No" }, { "input": "47 5", "output": "No" }, { "input": "999999998999999999 1000000000", "output": "No" }, { "input": "11 6", "output": "No" }, { "input": "7 4", "output": "No" }, { "input": "1 10", "output": "No" }, { "input": "9 5", "output": "No" }, { "input": "2519 20", "output": "No" }, { "input": "700001 3", "output": "Yes" }, { "input": "13 7", "output": "No" }, { "input": "999999 10000", "output": "No" }, { "input": "1 4", "output": "No" }, { "input": "232792559 30", "output": "No" }, { "input": "1 5", "output": "No" }, { "input": "5 4", "output": "No" }, { "input": "5 8", "output": "No" }, { "input": "55 4", "output": "No" } ]
1,603,468,382
2,147,483,647
PyPy 3
OK
TESTS
90
155
0
from sys import stdin,stdout n,k=map(int,input().split()) print(["Yes","No"][any((n+1)%i for i in range(2,k+1))])
Title: Cave Painting Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imp is watching a documentary about cave painting. Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number *n* by all integers *i* from 1 to *k*. Unfortunately, there are too many integers to analyze for Imp. Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all , 1<=≤<=*i*<=≤<=*k*, are distinct, i. e. there is no such pair (*i*,<=*j*) that: - 1<=≤<=*i*<=&lt;<=*j*<=≤<=*k*, - , where is the remainder of division *x* by *y*. Input Specification: The only line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=1018). Output Specification: Print "Yes", if all the remainders are distinct, and "No" otherwise. You can print each letter in arbitrary case (lower or upper). Demo Input: ['4 4\n', '5 3\n'] Demo Output: ['No\n', 'Yes\n'] Note: In the first sample remainders modulo 1 and 4 coincide.
```python from sys import stdin,stdout n,k=map(int,input().split()) print(["Yes","No"][any((n+1)%i for i in range(2,k+1))]) ```
3
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" } ]
1,519,032,705
2,147,483,647
Python 3
OK
TESTS
37
124
5,734,400
import math n=int(input()) a=list(map(int,input().split())) b=[0]*1005 for i in a: b[i]+=1 if max(b)<=(math.ceil(n/2)): print("YES") else: print("NO")
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 import math n=int(input()) a=list(map(int,input().split())) b=[0]*1005 for i in a: b[i]+=1 if max(b)<=(math.ceil(n/2)): print("YES") else: print("NO") ```
3
586
A
Alena's Schedule
PROGRAMMING
900
[ "implementation" ]
null
null
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying. One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes). The University works in such a way that every day it holds exactly *n* lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks). The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the *n* pairs she knows if there will be a class at that time or not. Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university. Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home. Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of lessons at the university. The second line contains *n* numbers *a**i* (0<=≤<=*a**i*<=≤<=1). Number *a**i* equals 0, if Alena doesn't have the *i*-th pairs, otherwise it is equal to 1. Numbers *a*1,<=*a*2,<=...,<=*a**n* are separated by spaces.
Print a single number — the number of pairs during which Alena stays at the university.
[ "5\n0 1 0 1 1\n", "7\n1 0 1 0 0 1 0\n", "1\n0\n" ]
[ "4\n", "4\n", "0\n" ]
In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair. In the last sample Alena doesn't have a single pair, so she spends all the time at home.
500
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1 0 0 1 0 1 1", "output": "51" }, { "input": "88\n0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 1 0", "output": "44" }, { "input": "88\n0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 1 0 1 1 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 1", "output": "59" }, { "input": "88\n1 0 0 0 1 1 1 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 0 1 0 1 1 1 0 1 0 1 1 1 1 0 1 0 1 1 1 0 0 0", "output": "53" }, { "input": "88\n1 1 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 0 1 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 1 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 1 1 0 1 1 0 1", "output": "63" }, { "input": "99\n0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 1 1 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 1 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 1 0 1 0 1 0 0 0 1 1 0 0 0 0", "output": "56" }, { "input": "99\n0 0 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 1 1 1", "output": "58" }, { "input": "99\n1 1 0 0 1 1 1 0 0 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 1 1 1 0 0 1 0", "output": "65" }, { "input": "99\n1 1 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1", "output": "77" }, { "input": "90\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", "output": "0" }, { "input": "90\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "90" }, { "input": "95\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", "output": "0" }, { "input": "95\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", "output": "95" }, { "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\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" } ]
1,444,644,762
3,762
Python 3
OK
TESTS
60
61
0
n = int(input()) P = list(map(int, input().split())) kol = 0 if P[0] == 1: kol += 1 for i in range(1, n-1): if P[i] != 0: kol += 1 elif P[i-1] != 0 and P[i] == 0 and P[i+1] != 0: kol += 1 if P[n-1] != 0 and n != 1: kol += 1 print(kol)
Title: Alena's Schedule Time Limit: None seconds Memory Limit: None megabytes Problem Description: Alena has successfully passed the entrance exams to the university and is now looking forward to start studying. One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes). The University works in such a way that every day it holds exactly *n* lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks). The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the *n* pairs she knows if there will be a class at that time or not. Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university. Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home. Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair. Input Specification: The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of lessons at the university. The second line contains *n* numbers *a**i* (0<=≤<=*a**i*<=≤<=1). Number *a**i* equals 0, if Alena doesn't have the *i*-th pairs, otherwise it is equal to 1. Numbers *a*1,<=*a*2,<=...,<=*a**n* are separated by spaces. Output Specification: Print a single number — the number of pairs during which Alena stays at the university. Demo Input: ['5\n0 1 0 1 1\n', '7\n1 0 1 0 0 1 0\n', '1\n0\n'] Demo Output: ['4\n', '4\n', '0\n'] Note: In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair. In the last sample Alena doesn't have a single pair, so she spends all the time at home.
```python n = int(input()) P = list(map(int, input().split())) kol = 0 if P[0] == 1: kol += 1 for i in range(1, n-1): if P[i] != 0: kol += 1 elif P[i-1] != 0 and P[i] == 0 and P[i+1] != 0: kol += 1 if P[n-1] != 0 and n != 1: kol += 1 print(kol) ```
3
32
B
Borze
PROGRAMMING
800
[ "expression parsing", "implementation" ]
B. Borze
2
256
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output the decoded ternary number. It can have leading zeroes.
[ ".-.--\n", "--.\n", "-..-.--\n" ]
[ "012", "20", "1012" ]
none
1,000
[ { "input": ".-.--", "output": "012" }, { "input": "--.", "output": "20" }, { "input": "-..-.--", "output": "1012" }, { "input": "---..", "output": "210" }, { "input": "..--.---..", "output": "0020210" }, { "input": "-.....----.", "output": "10000220" }, { "input": ".", "output": "0" }, { "input": "-.", "output": "1" }, { "input": "--", "output": "2" }, { "input": "..", "output": "00" }, { "input": "--.", "output": "20" }, { "input": ".--.", "output": "020" }, { "input": ".-.-..", "output": "0110" }, { "input": "----.-.", "output": "2201" }, { "input": "-..--.-.", "output": "10201" }, { "input": "..--..--.", "output": "0020020" }, { "input": "-.-.---.--..-..-.-.-..-..-.--.", "output": "112120010111010120" }, { "input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.", "output": "21112220010101011012011011221011" }, { "input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------", "output": "11020111110111222212021020002022111100201121222020012022110010222" }, { "input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..", "output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110" }, { "input": ".-..-.-.---.-----.--.---...-.--.-.-....-..", "output": "01011212212021001201100010" }, { "input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..", "output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210" }, { "input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.", "output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220" }, { "input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----", "output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122" }, { "input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..", "output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110" }, { "input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.", "output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020" }, { "input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..", "output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200" }, { "input": "-.---...----...--.--...-.--.----", "output": "121002200020200012022" }, { "input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.", "output": "202020212012111010201021100001021221200011222112012201012220" } ]
1,617,476,586
2,147,483,647
Python 3
OK
TESTS
30
124
0
x = input() ans = "" move = False for i in range(len(x)) : if move : move = False continue if x[i] == "." : ans += "0" else : move = True if x[i] == "-" and x[i+1] == "." : ans += "1" elif x[i] == "-" and x[i+1] == "-" : ans += "2" print(ans)
Title: Borze Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. Input Specification: The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). Output Specification: Output the decoded ternary number. It can have leading zeroes. Demo Input: ['.-.--\n', '--.\n', '-..-.--\n'] Demo Output: ['012', '20', '1012'] Note: none
```python x = input() ans = "" move = False for i in range(len(x)) : if move : move = False continue if x[i] == "." : ans += "0" else : move = True if x[i] == "-" and x[i+1] == "." : ans += "1" elif x[i] == "-" and x[i+1] == "-" : ans += "2" print(ans) ```
3.969
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" } ]
1,586,286,702
2,147,483,647
Python 3
OK
TESTS
40
109
0
s=input() compare="hello" i, j = 0, 0 while i < len(s) and j < 5: if s[i] != compare[j]: i+=1 else: i, j = i+1, j+1 check=j==5 if check: 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() compare="hello" i, j = 0, 0 while i < len(s) and j < 5: if s[i] != compare[j]: i+=1 else: i, j = i+1, j+1 check=j==5 if check: print("YES") else: print("NO") ```
3.9455
186
A
Comparing Strings
PROGRAMMING
1,100
[ "implementation", "strings" ]
null
null
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters. Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
[ "ab\nba\n", "aa\nab\n" ]
[ "YES\n", "NO\n" ]
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
500
[ { "input": "ab\nba", "output": "YES" }, { "input": "aa\nab", "output": "NO" }, { "input": "a\nza", "output": "NO" }, { "input": "vvea\nvvae", "output": "YES" }, { "input": "rtfabanpc\natfabrnpc", "output": "YES" }, { "input": "mt\ntm", "output": "YES" }, { "input": "qxolmbkkt\naovlajmlf", "output": "NO" }, { "input": "b\ng", "output": "NO" }, { "input": "ab\naba", "output": "NO" }, { "input": "ba\na", "output": "NO" }, { "input": "a\nab", "output": "NO" }, { "input": "a\naa", "output": "NO" }, { "input": "a\nz", "output": "NO" }, { "input": "aabb\nbbaa", "output": "NO" }, { "input": "ab\nbd", "output": "NO" }, { "input": "bac\ndae", "output": "NO" }, { "input": "abc\nakl", "output": "NO" }, { "input": "cb\naa", "output": "NO" }, { "input": "abaab\naabba", "output": "NO" }, { "input": "aab\naaa", "output": "NO" }, { "input": "abcde\nedcba", "output": "NO" }, { "input": "abab\nbaba", "output": "NO" }, { "input": "ab\nbac", "output": "NO" }, { "input": "abcd\naqcb", "output": "NO" }, { "input": "abc\nbad", "output": "NO" }, { "input": "ab\nca", "output": "NO" }, { "input": "abc\nab", "output": "NO" }, { "input": "ab\nbae", "output": "NO" }, { "input": "aaaabcccca\naaaadccccb", "output": "NO" }, { "input": "acaa\nabca", "output": "NO" }, { "input": "aba\naab", "output": "YES" }, { "input": "ah\nla", "output": "NO" }, { "input": "aab\naa", "output": "NO" }, { "input": "abab\nabcd", "output": "NO" }, { "input": "ed\nab", "output": "NO" } ]
1,589,484,326
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
4
248
20,172,800
a=str(input()) b=str(input()) d='' flag=0 for i in range(0,len(a)-1): d=a[0:i]+a[i+1]+a[i]+a[i+2:] if(d==b): flag=1 break if(flag>0): print("YES") else: print("NO")
Title: Comparing Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters. Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not. Input Specification: The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters. The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length. Output Specification: Print "YES", if the dwarves belong to the same race. Otherwise, print "NO". Demo Input: ['ab\nba\n', 'aa\nab\n'] Demo Output: ['YES\n', 'NO\n'] Note: - First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
```python a=str(input()) b=str(input()) d='' flag=0 for i in range(0,len(a)-1): d=a[0:i]+a[i+1]+a[i]+a[i+2:] if(d==b): flag=1 break if(flag>0): print("YES") else: print("NO") ```
0
937
A
Olympiad
PROGRAMMING
800
[ "implementation", "sortings" ]
null
null
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score.
Print a single integer — the desired number of ways.
[ "4\n1 3 3 2\n", "3\n1 1 1\n", "4\n42 0 0 42\n" ]
[ "3\n", "1\n", "1\n" ]
There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.
500
[ { "input": "4\n1 3 3 2", "output": "3" }, { "input": "3\n1 1 1", "output": "1" }, { "input": "4\n42 0 0 42", "output": "1" }, { "input": "10\n1 0 1 0 1 0 0 0 0 1", "output": "1" }, { "input": "10\n572 471 540 163 50 30 561 510 43 200", "output": "10" }, { "input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600", "output": "94" }, { "input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600", "output": "1" }, { "input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12", "output": "13" }, { "input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391", "output": "33" }, { "input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3", "output": "3" }, { "input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55", "output": "93" }, { "input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13", "output": "13" }, { "input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1", "output": "1" }, { "input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37", "output": "61" }, { "input": "2\n0 1", "output": "1" }, { "input": "2\n0 600", "output": "1" }, { "input": "4\n1 1 1 2", "output": "2" }, { "input": "4\n0 0 1 2", "output": "2" }, { "input": "1\n5", "output": "1" }, { "input": "2\n0 5", "output": "1" }, { "input": "5\n1 0 0 1 2", "output": "2" } ]
1,694,144,422
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
0
61
2,969,600
n=int(input()) li=list(map(int, input().split())) s=set() for i in range(n): if li[i]>0: li.add(li[i]) print(len(li))
Title: Olympiad Time Limit: None seconds Memory Limit: None megabytes Problem Description: The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score. Output Specification: Print a single integer — the desired number of ways. Demo Input: ['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n'] Demo Output: ['3\n', '1\n', '1\n'] Note: There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.
```python n=int(input()) li=list(map(int, input().split())) s=set() for i in range(n): if li[i]>0: li.add(li[i]) print(len(li)) ```
-1
670
D1
Magic Powder - 1
PROGRAMMING
1,400
[ "binary search", "brute force", "implementation" ]
null
null
This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only. Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs *n* ingredients, and for each ingredient she knows the value *a**i* — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all *n* ingredients. Apollinaria has *b**i* gram of the *i*-th ingredient. Also she has *k* grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the *n* ingredients and can be used for baking cookies. Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
The first line of the input contains two positive integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1000) — the number of ingredients and the number of grams of the magic powder. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, needed to bake one cookie. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=1000), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, which Apollinaria has.
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
[ "3 1\n2 1 4\n11 3 16\n", "4 3\n4 3 5 6\n11 12 14 20\n" ]
[ "4\n", "3\n" ]
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies. In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
1,000
[ { "input": "3 1\n2 1 4\n11 3 16", "output": "4" }, { "input": "4 3\n4 3 5 6\n11 12 14 20", "output": "3" }, { "input": "10 926\n5 6 8 1 2 5 1 8 4 4\n351 739 998 725 953 970 906 691 707 1000", "output": "137" }, { "input": "20 925\n7 3 1 2 1 3 1 3 1 2 3 1 5 8 1 3 7 3 4 2\n837 898 965 807 786 670 626 873 968 745 878 359 760 781 829 882 777 740 907 779", "output": "150" }, { "input": "30 300\n1 4 2 1 2 5 6 4 1 3 2 1 1 1 1 1 2 3 1 3 4 2 2 3 2 2 2 1 1 1\n997 817 767 860 835 809 817 565 630 804 586 953 977 356 905 890 958 916 740 583 902 945 313 956 871 729 976 707 516 788", "output": "164" }, { "input": "40 538\n1 3 3 1 4 1 1 1 1 5 3 3 4 1 4 2 7 1 4 1 1 2 2 1 1 1 1 4 1 4 2 3 3 3 1 3 4 1 3 5\n975 635 795 835 982 965 639 787 688 796 988 779 839 942 491 696 396 995 718 810 796 879 957 783 844 765 968 783 647 214 995 868 318 453 989 889 504 962 945 925", "output": "104" }, { "input": "50 530\n2 3 3 1 1 1 3 4 4 2 4 2 5 1 3 1 2 6 1 1 2 5 3 2 1 5 1 3 3 2 1 1 1 1 2 1 1 2 2 1 4 2 1 3 1 2 1 1 4 2\n959 972 201 990 675 679 972 268 976 886 488 924 795 959 647 994 969 862 898 646 763 797 978 763 995 641 923 856 829 921 934 883 904 986 728 980 1000 775 716 745 833 832 999 651 571 626 827 456 636 795", "output": "133" }, { "input": "60 735\n3 1 4 7 1 7 3 1 1 5 4 7 3 3 3 2 5 3 1 2 3 6 1 1 1 1 1 2 5 3 2 1 3 5 2 1 2 2 2 2 1 3 3 3 6 4 3 5 1 3 2 2 1 3 1 1 1 7 1 2\n596 968 975 493 665 571 598 834 948 941 737 649 923 848 950 907 929 865 227 836 956 796 861 801 746 667 539 807 405 355 501 879 994 890 573 848 597 873 130 985 924 426 999 550 586 924 601 807 994 878 410 817 922 898 982 525 611 685 806 847", "output": "103" }, { "input": "1 1\n1\n1", "output": "2" }, { "input": "70 130\n2 1 2 2 3 3 2 5 2 2 3 3 3 1 1 4 3 5 3 2 1 3 7 1 2 7 5 2 1 6 3 4 1 2 1 1 1 1 3 6 4 2 2 8 2 2 4 1 4 2 1 4 4 3 5 1 1 1 1 1 2 3 1 5 1 3 3 4 2 2\n473 311 758 768 797 572 656 898 991 534 989 702 934 767 777 799 1000 655 806 727 718 948 834 965 832 778 706 861 799 874 745 970 772 967 984 886 835 795 832 837 950 952 475 891 947 952 903 929 689 478 725 945 585 943 771 631 729 887 557 738 824 758 999 786 669 992 918 762 964 941", "output": "119" }, { "input": "80 979\n2 1 1 1 2 1 1 1 3 1 4 4 2 1 1 3 1 1 2 1 4 1 1 2 5 4 8 1 3 6 5 7 2 3 4 1 2 2 6 1 2 2 4 1 1 2 3 2 8 1 1 3 3 4 1 1 2 1 4 4 1 4 3 2 6 5 2 1 4 1 2 3 2 1 3 3 1 2 1 3\n498 976 513 869 917 914 664 656 957 893 981 947 985 693 576 958 987 822 981 718 884 729 295 683 485 998 730 894 731 975 739 854 906 740 987 976 606 689 990 775 522 994 920 893 529 651 989 799 643 215 946 987 297 868 425 810 694 908 736 903 970 751 625 904 955 945 839 777 977 974 905 900 666 680 799 873 565 919 536 686", "output": "128" }, { "input": "1 1000\n1000\n1000", "output": "2" }, { "input": "1 1000\n1\n1000", "output": "2000" }, { "input": "1 1\n4\n6", "output": "1" }, { "input": "1 1\n10\n2", "output": "0" }, { "input": "2 1\n2 2\n1 1", "output": "0" }, { "input": "2 6\n1 3\n6 2", "output": "2" } ]
1,611,585,018
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
46
307,200
n,k=map(int,input().split()) need=list(map(int,input().split())) have=list(map(int,input().split())) get=[have[i]//need[i] for i in range(n)] total=min(get) left=[have[i]-total*need[i] for i in range(n)] # print(left) br=False while(not br): for i in range(n): if(left[i]>=need[i]): left[i]-=need[i] else: want=need[i]-left[i] k-=want # print(want,k) if(k<0): br=True if(not br): total+=1 print(total)
Title: Magic Powder - 1 Time Limit: None seconds Memory Limit: None megabytes Problem Description: This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only. Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs *n* ingredients, and for each ingredient she knows the value *a**i* — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all *n* ingredients. Apollinaria has *b**i* gram of the *i*-th ingredient. Also she has *k* grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the *n* ingredients and can be used for baking cookies. Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder. Input Specification: The first line of the input contains two positive integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1000) — the number of ingredients and the number of grams of the magic powder. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, needed to bake one cookie. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=1000), where the *i*-th number is equal to the number of grams of the *i*-th ingredient, which Apollinaria has. Output Specification: Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder. Demo Input: ['3 1\n2 1 4\n11 3 16\n', '4 3\n4 3 5 6\n11 12 14 20\n'] Demo Output: ['4\n', '3\n'] Note: In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies. In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
```python n,k=map(int,input().split()) need=list(map(int,input().split())) have=list(map(int,input().split())) get=[have[i]//need[i] for i in range(n)] total=min(get) left=[have[i]-total*need[i] for i in range(n)] # print(left) br=False while(not br): for i in range(n): if(left[i]>=need[i]): left[i]-=need[i] else: want=need[i]-left[i] k-=want # print(want,k) if(k<0): br=True if(not br): total+=1 print(total) ```
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 ..." } ]
1,652,364,358
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
5
92
0
s = input() c = 0 a = dict() for i in range(len(s)): if s == "": break s1 = s[0] + s[1] # print(s1) if s1[1] == "/": s1 = s1 + s[2] + s[3] s = s[4:] print(" "*(a.pop("<" + s1[2] + ">")) + s1) if c > 1: c -= 2 else: pass else: s1 = s1 + s[2] s = s[3:] print(" "*(c) + s1) a.update({s1:c}) c += 2 # print(s1) #print(a) # Thu May 12 2022 14:05:56 GMT+0000 (Coordinated Universal Time)
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 s = input() c = 0 a = dict() for i in range(len(s)): if s == "": break s1 = s[0] + s[1] # print(s1) if s1[1] == "/": s1 = s1 + s[2] + s[3] s = s[4:] print(" "*(a.pop("<" + s1[2] + ">")) + s1) if c > 1: c -= 2 else: pass else: s1 = s1 + s[2] s = s[3:] print(" "*(c) + s1) a.update({s1:c}) c += 2 # print(s1) #print(a) # Thu May 12 2022 14:05:56 GMT+0000 (Coordinated Universal Time) ```
-1
1
B
Spreadsheet
PROGRAMMING
1,600
[ "implementation", "math" ]
B. Spreadsheets
10
64
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc. The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23. Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example. Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
Write *n* lines, each line should contain a cell coordinates in the other numeration system.
[ "2\nR23C55\nBC23\n" ]
[ "BC23\nR23C55\n" ]
none
0
[ { "input": "2\nR23C55\nBC23", "output": "BC23\nR23C55" }, { "input": "1\nA1", "output": "R1C1" }, { "input": "5\nR8C3\nD1\nR7C2\nR8C9\nR8C9", "output": "C8\nR1C4\nB7\nI8\nI8" }, { "input": "4\nR4C25\nR90C35\nAP55\nX83", "output": "Y4\nAI90\nR55C42\nR83C24" }, { "input": "10\nR50C12\nR23C47\nY96\nR44C13\nR19C21\nR95C73\nBK12\nR51C74\nAY34\nR63C25", "output": "L50\nAU23\nR96C25\nM44\nU19\nBU95\nR12C63\nBV51\nR34C51\nY63" } ]
1,687,414,547
2,147,483,647
PyPy 3-64
OK
TESTS
10
1,622
19,251,200
import math import re def iteracion(num, base): if num % base == 0: residuo = base else: residuo = num % base cociente = (num-residuo)/base return int(cociente), int(residuo) def tranformar_letter(num, base): lista = [] while (True): num, residuo = iteracion(num, base) lista = [residuo]+lista if num < base: lista = [num]+lista break letters = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ" letter = "" for i in lista: if i != 0: letter = letter+letters[i] return letter def tranformar_num(text, base): letters = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ" n = len(text) num = 0 for i in text: a = letters.index(str(i)) num = num+a*(base**(n-1)) n = n-1 return str(num) n = int(input()) data = [] for i in range(n): data.append(input()) for i in data: resultado = re.findall(r'R(\d+)C(\d+)', i) if resultado: print(tranformar_letter(int(resultado[0][1]), 26)+resultado[0][0]) else: resultado = re.findall(r'([A-Za-z]+)(\d+)', i) num = tranformar_num(resultado[0][0], 26) print("R"+resultado[0][1]+"C"+num)
Title: Spreadsheet Time Limit: 10 seconds Memory Limit: 64 megabytes Problem Description: In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc. The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23. Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example. Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system. Input Specification: The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 . Output Specification: Write *n* lines, each line should contain a cell coordinates in the other numeration system. Demo Input: ['2\nR23C55\nBC23\n'] Demo Output: ['BC23\nR23C55\n'] Note: none
```python import math import re def iteracion(num, base): if num % base == 0: residuo = base else: residuo = num % base cociente = (num-residuo)/base return int(cociente), int(residuo) def tranformar_letter(num, base): lista = [] while (True): num, residuo = iteracion(num, base) lista = [residuo]+lista if num < base: lista = [num]+lista break letters = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ" letter = "" for i in lista: if i != 0: letter = letter+letters[i] return letter def tranformar_num(text, base): letters = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ" n = len(text) num = 0 for i in text: a = letters.index(str(i)) num = num+a*(base**(n-1)) n = n-1 return str(num) n = int(input()) data = [] for i in range(n): data.append(input()) for i in data: resultado = re.findall(r'R(\d+)C(\d+)', i) if resultado: print(tranformar_letter(int(resultado[0][1]), 26)+resultado[0][0]) else: resultado = re.findall(r'([A-Za-z]+)(\d+)', i) num = tranformar_num(resultado[0][0], 26) print("R"+resultado[0][1]+"C"+num) ```
3.775467
982
A
Row
PROGRAMMING
1,200
[ "brute force", "constructive algorithms" ]
null
null
You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold: 1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule. The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal". Note that the first and last seats are not adjacent (if $n \ne 2$).
The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs. The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating.
Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No". You are allowed to print letters in whatever case you'd like (uppercase or lowercase).
[ "3\n101\n", "4\n1011\n", "5\n10001\n" ]
[ "Yes\n", "No\n", "No\n" ]
In sample case one the given seating is maximal. In sample case two the person at chair three has a neighbour to the right. In sample case three it is possible to seat yet another person into chair three.
500
[ { "input": "3\n101", "output": "Yes" }, { "input": "4\n1011", "output": "No" }, { "input": "5\n10001", "output": "No" }, { "input": "1\n0", "output": "No" }, { "input": "1\n1", "output": "Yes" }, { "input": "100\n0101001010101001010010010101001010100101001001001010010101010010101001001010101001001001010100101010", "output": "Yes" }, { "input": "4\n0100", "output": "No" }, { "input": "42\n011000100101001001101011011010100010011010", "output": "No" }, { "input": "3\n001", "output": "No" }, { "input": "64\n1001001010010010100101010010010100100101001001001001010100101001", "output": "Yes" }, { "input": "3\n111", "output": "No" }, { "input": "4\n0000", "output": "No" }, { "input": "4\n0001", "output": "No" }, { "input": "4\n0010", "output": "No" }, { "input": "4\n0011", "output": "No" }, { "input": "4\n0101", "output": "Yes" }, { "input": "4\n0110", "output": "No" }, { "input": "4\n0111", "output": "No" }, { "input": "4\n1000", "output": "No" }, { "input": "4\n1001", "output": "Yes" }, { "input": "4\n1010", "output": "Yes" }, { "input": "4\n1100", "output": "No" }, { "input": "4\n1101", "output": "No" }, { "input": "4\n1110", "output": "No" }, { "input": "4\n1111", "output": "No" }, { "input": "2\n00", "output": "No" }, { "input": "2\n01", "output": "Yes" }, { "input": "2\n10", "output": "Yes" }, { "input": "2\n11", "output": "No" }, { "input": "3\n000", "output": "No" }, { "input": "3\n010", "output": "Yes" }, { "input": "3\n011", "output": "No" }, { "input": "3\n100", "output": "No" }, { "input": "3\n110", "output": "No" }, { "input": "100\n0111001010101110001100000010011000100101110010001100111110101110001110101010111000010010011000000110", "output": "No" }, { "input": "357\n100101010010010010010100101001001010101010100100100100101001010101001010010100101001010100101001010010100100101001010101010101001001010100101010010100101001010100100100101010010010010100101010010010101001010010010101001001010010010101010100100101010010100100101001010100101001010100101001010010010010100101001010100100100100100100100101010101010010010100101", "output": "Yes" }, { "input": "459\n000111000101010000100001001010111110011011010001100101111010111011101110111101111101100101100011011001100110001001111001101000111001011100110100011111011111000010000110010011100110011011111110011100001101001111000100111011001000001011111100110100001001001100101011100001110110100101011011110100100111101011000101110000100110100100010000000100001001111111000011101010010011001111010111001100000100111001010111011010000011000011100101101011101000011011000110011", "output": "No" }, { "input": "3\n001", "output": "No" }, { "input": "4\n0010", "output": "No" }, { "input": "3\n111", "output": "No" }, { "input": "4\n0110", "output": "No" }, { "input": "2\n11", "output": "No" }, { "input": "2\n10", "output": "Yes" }, { "input": "2\n01", "output": "Yes" }, { "input": "4\n1101", "output": "No" }, { "input": "2\n00", "output": "No" }, { "input": "4\n1111", "output": "No" }, { "input": "15\n000010101010000", "output": "No" }, { "input": "8\n10010100", "output": "No" }, { "input": "5\n00100", "output": "No" }, { "input": "6\n100100", "output": "No" }, { "input": "7\n1000001", "output": "No" }, { "input": "98\n10101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "Yes" } ]
1,530,195,738
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
3
93
0
n=int(input()) s=input() def find(l): for i in range(1,len(l)-1): if l[i-1]=='1' and l[i]=='0' and l[i+1]=='1': return True return False if n==1: print('Yes') elif n==2: if s=='11': print('No') else: print('Yes') else: flag1,flag2=True,True for i in range(n-1): if s[i]=='1' and s[i+1]=='1': flag1=False break if not flag1: print('No') else: l=list(s) for i in range(n): if l[i]=='0': l[i]='1' f=find(l) if f: print('No') exit(0) else: l[i]='0' print('Yes')
Title: Row Time Limit: None seconds Memory Limit: None megabytes Problem Description: You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold: 1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule. The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal". Note that the first and last seats are not adjacent (if $n \ne 2$). Input Specification: The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs. The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating. Output Specification: Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No". You are allowed to print letters in whatever case you'd like (uppercase or lowercase). Demo Input: ['3\n101\n', '4\n1011\n', '5\n10001\n'] Demo Output: ['Yes\n', 'No\n', 'No\n'] Note: In sample case one the given seating is maximal. In sample case two the person at chair three has a neighbour to the right. In sample case three it is possible to seat yet another person into chair three.
```python n=int(input()) s=input() def find(l): for i in range(1,len(l)-1): if l[i-1]=='1' and l[i]=='0' and l[i+1]=='1': return True return False if n==1: print('Yes') elif n==2: if s=='11': print('No') else: print('Yes') else: flag1,flag2=True,True for i in range(n-1): if s[i]=='1' and s[i+1]=='1': flag1=False break if not flag1: print('No') else: l=list(s) for i in range(n): if l[i]=='0': l[i]='1' f=find(l) if f: print('No') exit(0) else: l[i]='0' print('Yes') ```
0
867
A
Between the Offices
PROGRAMMING
800
[ "implementation" ]
null
null
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane. You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days. The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise. You can print each letter in any case (upper or lower).
[ "4\nFSSF\n", "2\nSF\n", "10\nFFFFFFFFFF\n", "10\nSSFFSFFSFF\n" ]
[ "NO\n", "YES\n", "NO\n", "YES\n" ]
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO". In the second example you just flew from Seattle to San Francisco, so the answer is "YES". In the third example you stayed the whole period in San Francisco, so the answer is "NO". In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
500
[ { "input": "4\nFSSF", "output": "NO" }, { "input": "2\nSF", "output": "YES" }, { "input": "10\nFFFFFFFFFF", "output": "NO" }, { "input": "10\nSSFFSFFSFF", "output": "YES" }, { "input": "20\nSFSFFFFSSFFFFSSSSFSS", "output": "NO" }, { "input": "20\nSSFFFFFSFFFFFFFFFFFF", "output": "YES" }, { "input": "20\nSSFSFSFSFSFSFSFSSFSF", "output": "YES" }, { "input": "20\nSSSSFSFSSFSFSSSSSSFS", "output": "NO" }, { "input": "100\nFFFSFSFSFSSFSFFSSFFFFFSSSSFSSFFFFSFFFFFSFFFSSFSSSFFFFSSFFSSFSFFSSFSSSFSFFSFSFFSFSFFSSFFSFSSSSFSFSFSS", "output": "NO" }, { "input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF", "output": "NO" }, { "input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFSS", "output": "NO" }, { "input": "100\nFFFFFFFFFFFFFSFFFFFFFFFSFSSFFFFFFFFFFFFFFFFFFFFFFSFFSFFFFFSFFFFFFFFSFFFFFFFFFFFFFSFFFFFFFFSFFFFFFFSF", "output": "NO" }, { "input": "100\nSFFSSFFFFFFSSFFFSSFSFFFFFSSFFFSFFFFFFSFSSSFSFSFFFFSFSSFFFFFFFFSFFFFFSFFFFFSSFFFSFFSFSFFFFSFFSFFFFFFF", "output": "YES" }, { "input": "100\nFFFFSSSSSFFSSSFFFSFFFFFSFSSFSFFSFFSSFFSSFSFFFFFSFSFSFSFFFFFFFFFSFSFFSFFFFSFSFFFFFFFFFFFFSFSSFFSSSSFF", "output": "NO" }, { "input": "100\nFFFFFFFFFFFFSSFFFFSFSFFFSFSSSFSSSSSFSSSSFFSSFFFSFSFSSFFFSSSFFSFSFSSFSFSSFSFFFSFFFFFSSFSFFFSSSFSSSFFS", "output": "NO" }, { "input": "100\nFFFSSSFSFSSSSFSSFSFFSSSFFSSFSSFFSSFFSFSSSSFFFSFFFSFSFSSSFSSFSFSFSFFSSSSSFSSSFSFSFFSSFSFSSFFSSFSFFSFS", "output": "NO" }, { "input": "100\nFFSSSSFSSSFSSSSFSSSFFSFSSFFSSFSSSFSSSFFSFFSSSSSSSSSSSSFSSFSSSSFSFFFSSFFFFFFSFSFSSSSSSFSSSFSFSSFSSFSS", "output": "NO" }, { "input": "100\nSSSFFFSSSSFFSSSSSFSSSSFSSSFSSSSSFSSSSSSSSFSFFSSSFFSSFSSSSFFSSSSSSFFSSSSFSSSSSSFSSSFSSSSSSSFSSSSFSSSS", "output": "NO" }, { "input": "100\nFSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSSSSSSFSSSSSSSSSSSSSFSSFSSSSSFSSFSSSSSSSSSFFSSSSSFSFSSSFFSSSSSSSSSSSSS", "output": "NO" }, { "input": "100\nSSSSSSSSSSSSSFSSSSSSSSSSSSFSSSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSFS", "output": "NO" }, { "input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS", "output": "NO" }, { "input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF", "output": "YES" }, { "input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFSFFFFFFFFFFFSFSFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFF", "output": "YES" }, { "input": "100\nSFFFFFFFFFFFFSSFFFFSFFFFFFFFFFFFFFFFFFFSFFFSSFFFFSFSFFFSFFFFFFFFFFFFFFFSSFFFFFFFFSSFFFFFFFFFFFFFFSFF", "output": "YES" }, { "input": "100\nSFFSSSFFSFSFSFFFFSSFFFFSFFFFFFFFSFSFFFSFFFSFFFSFFFFSFSFFFFFFFSFFFFFFFFFFSFFSSSFFSSFFFFSFFFFSFFFFSFFF", "output": "YES" }, { "input": "100\nSFFFSFFFFSFFFSSFFFSFSFFFSFFFSSFSFFFFFSFFFFFFFFSFSFSFFSFFFSFSSFSFFFSFSFFSSFSFSSSFFFFFFSSFSFFSFFFFFFFF", "output": "YES" }, { "input": "100\nSSSSFFFFSFFFFFFFSFFFFSFSFFFFSSFFFFFFFFFSFFSSFFFFFFSFSFSSFSSSFFFFFFFSFSFFFSSSFFFFFFFSFFFSSFFFFSSFFFSF", "output": "YES" }, { "input": "100\nSSSFSSFFFSFSSSSFSSFSSSSFSSFFFFFSFFSSSSFFSSSFSSSFSSSSFSSSSFSSSSSSSFSFSSFFFSSFFSFFSSSSFSSFFSFSSFSFFFSF", "output": "YES" }, { "input": "100\nSFFSFSSSSSSSFFSSSFSSSSFSFSSFFFSSSSSSFSSSSFSSFSSSFSSSSSSSFSSFSFFFSSFSSFSFSFSSSSSSSSSSSSFFFFSSSSSFSFFF", "output": "YES" }, { "input": "100\nSSSFSFFSFSFFSSSSSFSSSFSSSFFFSSSSSSSSSFSFSFSSSSFSFSSFFFFFSSSSSSSSSSSSSSSSSSSFFSSSSSFSFSSSSFFSSSSFSSSF", "output": "YES" }, { "input": "100\nSSSFSSSSSSSSSSFSSSSFSSSSSSFSSSSSSFSSSSSSSSSSSSSSFSSSFSSSFSSSSSSSSSSSFSSSSSSFSFSSSSFSSSSSSFSSSSSSSSFF", "output": "YES" }, { "input": "100\nSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFFSSSSSSSSSFSSSSSSSSSSSSSSSSSF", "output": "YES" }, { "input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSF", "output": "YES" }, { "input": "2\nSS", "output": "NO" } ]
1,579,508,359
2,147,483,647
Python 3
OK
TESTS
34
124
307,200
cnt = int(input()) locs = input() pre, sf, fs = locs[0], 0, 0 for i in range(1, len(locs)): if pre != locs[i]: if locs[i] == 'S': fs += 1 elif locs[i] == 'F': sf += 1 pre = locs[i] if sf > fs: print("YES") else: print("NO")
Title: Between the Offices Time Limit: None seconds Memory Limit: None megabytes Problem Description: As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane. You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. Input Specification: The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days. The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. Output Specification: Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise. You can print each letter in any case (upper or lower). Demo Input: ['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n', 'YES\n'] Note: In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO". In the second example you just flew from Seattle to San Francisco, so the answer is "YES". In the third example you stayed the whole period in San Francisco, so the answer is "NO". In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
```python cnt = int(input()) locs = input() pre, sf, fs = locs[0], 0, 0 for i in range(1, len(locs)): if pre != locs[i]: if locs[i] == 'S': fs += 1 elif locs[i] == 'F': sf += 1 pre = locs[i] if sf > fs: print("YES") else: print("NO") ```
3
856
A
Set Theory
PROGRAMMING
1,600
[ "brute force", "constructive algorithms" ]
null
null
Masha and Grisha like studying sets of positive integers. One day Grisha has written a set *A* containing *n* different integers *a**i* on a blackboard. Now he asks Masha to create a set *B* containing *n* different integers *b**j* such that all *n*2 integers that can be obtained by summing up *a**i* and *b**j* for all possible pairs of *i* and *j* are different. Both Masha and Grisha don't like big numbers, so all numbers in *A* are from 1 to 106, and all numbers in *B* must also be in the same range. Help Masha to create the set *B* that satisfies Grisha's requirement.
Input data contains multiple test cases. The first line contains an integer *t* — the number of test cases (1<=≤<=*t*<=≤<=100). Each test case is described in the following way: the first line of the description contains one integer *n* — the number of elements in *A* (1<=≤<=*n*<=≤<=100). The second line contains *n* integers *a**i* — the elements of *A* (1<=≤<=*a**i*<=≤<=106).
For each test first print the answer: - NO, if Masha's task is impossible to solve, there is no way to create the required set *B*. - YES, if there is the way to create the required set. In this case the second line must contain *n* different positive integers *b**j* — elements of *B* (1<=≤<=*b**j*<=≤<=106). If there are several possible sets, output any of them.
[ "3\n3\n1 10 100\n1\n1\n2\n2 4\n" ]
[ "YES\n1 2 3 \nYES\n1 \nYES\n1 2 \n" ]
none
0
[ { "input": "3\n3\n1 10 100\n1\n1\n2\n2 4", "output": "YES\n1 2 3 \nYES\n1 \nYES\n1 2 " }, { "input": "1\n100\n74 14 24 45 22 9 49 78 79 20 60 1 31 91 32 39 90 5 42 57 30 58 64 68 12 11 86 8 3 38 76 17 98 26 85 92 56 65 89 66 36 87 23 67 13 48 15 47 81 73 63 50 34 93 82 44 77 69 96 100 41 19 35 16 88 27 99 40 62 95 70 18 46 21 53 59 37 6 61 71 2 4 52 28 97 25 29 51 7 33 80 83 72 10 75 94 43 84 54 55", "output": "YES\n1 101 201 301 401 501 601 701 801 901 1001 1101 1201 1301 1401 1501 1601 1701 1801 1901 2001 2101 2201 2301 2401 2501 2601 2701 2801 2901 3001 3101 3201 3301 3401 3501 3601 3701 3801 3901 4001 4101 4201 4301 4401 4501 4601 4701 4801 4901 5001 5101 5201 5301 5401 5501 5601 5701 5801 5901 6001 6101 6201 6301 6401 6501 6601 6701 6801 6901 7001 7101 7201 7301 7401 7501 7601 7701 7801 7901 8001 8101 8201 8301 8401 8501 8601 8701 8801 8901 9001 9101 9201 9301 9401 9501 9601 9701 9801 9901 " } ]
1,679,573,684
2,147,483,647
PyPy 3-64
TIME_LIMIT_EXCEEDED
TESTS
1
1,000
42,291,200
from sys import stdin input = lambda: stdin.readline()[:-1] t = int(input()) for i in range(t): n = int(input()) lst = list(map(int, input().split())) d = [True for j in range(1000000)] xd = [] last = 1 for i in range(n): for k in range(last,1000000): if d[k]: xd.append(k) for ik in lst: if ik+k < 1000000: d[ik + k] = False break last = xd[-1] + 1 print("YES") print(*xd)
Title: Set Theory Time Limit: None seconds Memory Limit: None megabytes Problem Description: Masha and Grisha like studying sets of positive integers. One day Grisha has written a set *A* containing *n* different integers *a**i* on a blackboard. Now he asks Masha to create a set *B* containing *n* different integers *b**j* such that all *n*2 integers that can be obtained by summing up *a**i* and *b**j* for all possible pairs of *i* and *j* are different. Both Masha and Grisha don't like big numbers, so all numbers in *A* are from 1 to 106, and all numbers in *B* must also be in the same range. Help Masha to create the set *B* that satisfies Grisha's requirement. Input Specification: Input data contains multiple test cases. The first line contains an integer *t* — the number of test cases (1<=≤<=*t*<=≤<=100). Each test case is described in the following way: the first line of the description contains one integer *n* — the number of elements in *A* (1<=≤<=*n*<=≤<=100). The second line contains *n* integers *a**i* — the elements of *A* (1<=≤<=*a**i*<=≤<=106). Output Specification: For each test first print the answer: - NO, if Masha's task is impossible to solve, there is no way to create the required set *B*. - YES, if there is the way to create the required set. In this case the second line must contain *n* different positive integers *b**j* — elements of *B* (1<=≤<=*b**j*<=≤<=106). If there are several possible sets, output any of them. Demo Input: ['3\n3\n1 10 100\n1\n1\n2\n2 4\n'] Demo Output: ['YES\n1 2 3 \nYES\n1 \nYES\n1 2 \n'] Note: none
```python from sys import stdin input = lambda: stdin.readline()[:-1] t = int(input()) for i in range(t): n = int(input()) lst = list(map(int, input().split())) d = [True for j in range(1000000)] xd = [] last = 1 for i in range(n): for k in range(last,1000000): if d[k]: xd.append(k) for ik in lst: if ik+k < 1000000: d[ik + k] = False break last = xd[-1] + 1 print("YES") print(*xd) ```
0
887
B
Cubes for Masha
PROGRAMMING
1,300
[ "brute force", "implementation" ]
null
null
Absent-minded Masha got set of *n* cubes for her birthday. At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*. To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number. The number can't contain leading zeros. It's not required to use all cubes to build a number. Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday. Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube.
Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1.
[ "3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n", "3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n" ]
[ "87", "98" ]
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.
1,000
[ { "input": "3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7", "output": "87" }, { "input": "3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9", "output": "98" }, { "input": "3\n0 1 2 3 4 5\n0 1 2 3 4 5\n0 1 2 3 4 5", "output": "5" }, { "input": "3\n1 2 3 7 8 9\n9 8 7 1 2 3\n7 9 2 3 1 8", "output": "3" }, { "input": "1\n5 2 2 5 6 7", "output": "0" }, { "input": "1\n7 6 5 8 9 0", "output": "0" }, { "input": "1\n2 5 9 6 7 9", "output": "0" }, { "input": "1\n6 3 1 9 4 9", "output": "1" }, { "input": "1\n1 9 8 3 7 8", "output": "1" }, { "input": "2\n1 7 2 0 4 3\n5 2 3 6 1 0", "output": "7" }, { "input": "2\n6 0 1 7 2 9\n1 3 4 6 7 0", "output": "4" }, { "input": "2\n8 6 4 1 2 0\n7 8 5 3 2 1", "output": "8" }, { "input": "2\n0 8 6 2 1 3\n5 2 7 1 0 9", "output": "3" }, { "input": "2\n0 9 5 7 6 2\n8 6 2 7 1 4", "output": "2" }, { "input": "3\n5 0 7 6 2 1\n2 7 4 6 1 9\n0 2 6 1 7 5", "output": "2" }, { "input": "3\n0 6 2 9 5 4\n3 8 0 1 6 9\n6 9 0 1 5 2", "output": "6" }, { "input": "3\n5 6 2 9 3 5\n5 4 1 5 9 8\n4 4 2 0 3 5", "output": "6" }, { "input": "3\n0 1 9 1 0 8\n9 9 3 5 6 2\n9 3 9 9 7 3", "output": "3" }, { "input": "3\n2 5 7 4 2 7\n1 5 5 9 0 3\n8 2 0 1 5 1", "output": "5" }, { "input": "1\n4 6 9 8 2 7", "output": "0" }, { "input": "1\n5 3 8 0 2 6", "output": "0" }, { "input": "1\n7 9 5 0 4 6", "output": "0" }, { "input": "1\n4 0 9 6 3 1", "output": "1" }, { "input": "1\n7 9 2 5 0 4", "output": "0" }, { "input": "1\n0 7 6 3 2 4", "output": "0" }, { "input": "1\n9 8 1 6 5 7", "output": "1" }, { "input": "1\n7 3 6 9 8 1", "output": "1" }, { "input": "1\n3 9 1 7 4 5", "output": "1" }, { "input": "1\n8 6 0 9 4 2", "output": "0" }, { "input": "1\n8 2 7 4 1 0", "output": "2" }, { "input": "1\n8 3 5 4 2 9", "output": "0" }, { "input": "1\n0 8 7 1 3 2", "output": "3" }, { "input": "1\n6 2 8 5 1 3", "output": "3" }, { "input": "1\n6 0 7 5 4 8", "output": "0" }, { "input": "1\n6 2 8 4 5 1", "output": "2" }, { "input": "1\n4 3 8 9 2 3", "output": "0" }, { "input": "1\n8 1 9 2 9 7", "output": "2" }, { "input": "1\n3 7 7 6 4 2", "output": "0" }, { "input": "1\n1 4 5 7 0 5", "output": "1" }, { "input": "2\n6 6 4 7 9 0\n2 1 2 8 6 4", "output": "2" }, { "input": "2\n5 3 2 9 8 2\n0 7 4 8 1 8", "output": "5" }, { "input": "2\n5 7 4 2 1 9\n2 2 7 1 1 8", "output": "2" }, { "input": "2\n9 3 3 6 7 2\n6 2 9 1 5 9", "output": "3" }, { "input": "2\n2 0 5 7 0 8\n4 5 1 5 4 9", "output": "2" }, { "input": "2\n2 6 8 1 3 1\n2 1 3 8 6 7", "output": "3" }, { "input": "2\n4 3 8 6 0 1\n4 7 1 8 9 0", "output": "1" }, { "input": "2\n0 2 9 1 8 5\n0 7 4 3 2 5", "output": "5" }, { "input": "2\n1 7 6 9 2 5\n1 6 7 0 9 2", "output": "2" }, { "input": "2\n0 2 9 8 1 7\n6 7 4 3 2 5", "output": "9" }, { "input": "2\n3 6 8 9 5 0\n6 7 0 8 2 3", "output": "0" }, { "input": "2\n5 1 2 3 0 8\n3 6 7 4 9 2", "output": "9" }, { "input": "2\n7 8 6 1 4 5\n8 6 4 3 2 5", "output": "8" }, { "input": "2\n2 3 5 1 9 6\n1 6 8 7 3 9", "output": "3" }, { "input": "2\n1 7 8 6 0 9\n3 2 1 7 4 9", "output": "4" }, { "input": "2\n2 4 0 3 7 6\n3 2 8 7 1 5", "output": "8" }, { "input": "2\n6 5 2 7 1 3\n3 7 8 1 0 9", "output": "3" }, { "input": "2\n5 8 4 7 1 2\n0 8 6 2 4 9", "output": "2" }, { "input": "2\n8 0 6 5 1 4\n7 1 0 8 3 4", "output": "1" }, { "input": "2\n2 3 9 1 6 7\n2 5 4 3 0 6", "output": "7" }, { "input": "3\n9 4 3 0 2 6\n7 0 5 3 3 9\n1 0 7 4 6 7", "output": "7" }, { "input": "3\n3 8 5 1 5 5\n1 5 7 2 6 9\n4 3 4 8 8 9", "output": "9" }, { "input": "3\n7 7 2 5 3 2\n3 0 0 6 4 4\n1 2 1 1 9 1", "output": "7" }, { "input": "3\n8 1 6 8 6 8\n7 0 2 5 8 4\n5 2 0 3 1 9", "output": "32" }, { "input": "3\n2 7 4 0 7 1\n5 5 4 9 1 4\n2 1 7 5 1 7", "output": "2" }, { "input": "3\n4 4 5 0 6 6\n7 1 6 9 5 4\n5 0 4 0 3 9", "output": "1" }, { "input": "3\n9 4 3 3 9 3\n1 0 3 4 5 3\n2 9 6 2 4 1", "output": "6" }, { "input": "3\n3 8 3 5 5 5\n3 0 1 6 6 3\n0 4 3 7 2 4", "output": "8" }, { "input": "3\n4 1 0 8 0 2\n1 5 3 5 0 7\n7 7 2 7 2 2", "output": "5" }, { "input": "3\n8 1 8 2 7 1\n9 1 9 9 4 7\n0 0 9 0 4 0", "output": "2" }, { "input": "3\n4 6 0 3 9 2\n8 6 9 0 7 2\n6 9 3 2 5 7", "output": "0" }, { "input": "3\n5 1 2 9 6 4\n9 0 6 4 2 8\n4 6 2 8 3 7", "output": "10" }, { "input": "3\n9 3 1 8 4 6\n6 9 1 2 0 7\n8 9 1 5 0 3", "output": "21" }, { "input": "3\n7 1 3 0 2 4\n2 4 3 0 9 5\n1 9 8 0 6 5", "output": "65" }, { "input": "3\n9 4 6 2 7 0\n3 7 1 9 6 4\n6 1 0 8 7 2", "output": "4" }, { "input": "3\n2 7 3 6 4 5\n0 2 1 9 4 8\n8 6 9 5 4 0", "output": "10" }, { "input": "3\n2 6 3 7 1 0\n9 1 2 4 7 6\n1 4 8 7 6 2", "output": "4" }, { "input": "3\n5 4 8 1 6 7\n0 9 3 5 8 6\n2 4 7 8 1 3", "output": "21" }, { "input": "3\n7 2 1 3 6 9\n0 3 8 4 7 6\n1 4 5 8 7 0", "output": "21" }, { "input": "3\n8 6 0 5 4 9\n1 8 5 3 9 7\n7 4 5 1 6 8", "output": "1" }, { "input": "1\n0 1 2 3 4 5", "output": "5" }, { "input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 8", "output": "9" }, { "input": "2\n0 1 2 3 4 5\n6 7 8 9 1 2", "output": "29" }, { "input": "3\n0 1 2 3 4 5\n6 7 8 9 1 2\n3 4 5 6 7 8", "output": "98" }, { "input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 1", "output": "19" }, { "input": "2\n0 1 2 3 4 5\n6 7 8 9 6 6", "output": "9" }, { "input": "2\n0 1 2 3 4 5\n4 5 6 7 8 9", "output": "9" }, { "input": "2\n1 8 9 1 1 0\n2 3 4 5 6 7", "output": "9" }, { "input": "2\n0 1 2 3 4 5\n9 8 7 6 5 4", "output": "9" }, { "input": "3\n2 3 4 5 6 7\n3 4 5 6 7 8\n9 1 2 3 4 5", "output": "9" }, { "input": "3\n1 1 2 3 4 5\n6 7 8 9 0 2\n3 4 5 6 7 8", "output": "10" }, { "input": "3\n1 1 1 0 2 3\n4 5 6 7 8 9\n0 0 0 0 0 0", "output": "10" }, { "input": "3\n0 1 2 2 4 5\n6 7 8 9 0 1\n3 3 4 5 6 7", "output": "21" }, { "input": "3\n2 0 1 3 4 5\n6 7 8 9 1 1\n3 4 5 6 6 7", "output": "19" }, { "input": "3\n1 1 1 1 1 1\n0 2 3 4 5 6\n7 8 9 2 3 4", "output": "10" }, { "input": "2\n0 1 2 3 4 5\n6 6 6 7 8 9", "output": "9" }, { "input": "3\n3 4 5 6 8 9\n1 1 1 1 1 1\n1 2 4 5 7 0", "output": "19" } ]
1,590,773,774
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
48
109
307,200
n = int(input()) l = [] for _ in range(n): l.extend(list(map(int, input().split()))) l.sort() ans = 0 i = 1 while True: st = sorted(list(str(i))) ind = -1 for j in st: if int(j) in l[ind + 1:]: ind = l.index(int(j)) else: print(ans) exit() ans = i i += 1 print(ans)
Title: Cubes for Masha Time Limit: None seconds Memory Limit: None megabytes Problem Description: Absent-minded Masha got set of *n* cubes for her birthday. At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*. To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number. The number can't contain leading zeros. It's not required to use all cubes to build a number. Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations. Input Specification: In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday. Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube. Output Specification: Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1. Demo Input: ['3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n', '3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n'] Demo Output: ['87', '98'] Note: In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.
```python n = int(input()) l = [] for _ in range(n): l.extend(list(map(int, input().split()))) l.sort() ans = 0 i = 1 while True: st = sorted(list(str(i))) ind = -1 for j in st: if int(j) in l[ind + 1:]: ind = l.index(int(j)) else: print(ans) exit() ans = i i += 1 print(ans) ```
0
660
A
Co-prime Array
PROGRAMMING
1,200
[ "greedy", "implementation", "math", "number theory" ]
null
null
You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible. In each move you can insert any positive integral number you want not greater than 109 in any place in the array. An array is co-prime if any two adjacent numbers of it are co-prime. In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1.
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime. The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it. If there are multiple answers you can print any one of them.
[ "3\n2 7 28\n" ]
[ "1\n2 7 9 28\n" ]
none
0
[ { "input": "3\n2 7 28", "output": "1\n2 7 1 28" }, { "input": "1\n1", "output": "0\n1" }, { "input": "1\n548", "output": "0\n548" }, { "input": "1\n963837006", "output": "0\n963837006" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "0\n1 1 1 1 1 1 1 1 1 1" }, { "input": "10\n26 723 970 13 422 968 875 329 234 983", "output": "2\n26 723 970 13 422 1 968 875 1 329 234 983" }, { "input": "10\n319645572 758298525 812547177 459359946 355467212 304450522 807957797 916787906 239781206 242840396", "output": "7\n319645572 1 758298525 1 812547177 1 459359946 1 355467212 1 304450522 807957797 916787906 1 239781206 1 242840396" }, { "input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1", "output": "19\n1 1 1 1 2 1 1 1 1 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1" }, { "input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917", "output": "38\n591 1 417 1 888 251 792 1 847 685 3 182 461 102 1 348 1 555 956 771 901 712 1 878 1 580 631 342 1 333 1 285 899 525 1 725 537 718 929 653 84 1 788 1 104 355 624 803 1 253 853 201 995 536 1 184 65 1 205 1 540 1 652 549 1 777 248 405 677 950 431 580 1 600 1 846 1 328 429 134 983 526 103 500 963 400 23 1 276 1 704 1 570 757 410 1 658 507 620 1 984 1 244 1 486 1 454 1 802 411 985 303 635 283 96 1 597 1 855 1 775 139 839 1 839 61 219 986 1 776 1 72 1 729 1 69 20 917" }, { "input": "5\n472882027 472882027 472882027 472882027 472882027", "output": "4\n472882027 1 472882027 1 472882027 1 472882027 1 472882027" }, { "input": "2\n1000000000 1000000000", "output": "1\n1000000000 1 1000000000" }, { "input": "2\n8 6", "output": "1\n8 1 6" }, { "input": "3\n100000000 1000000000 1000000000", "output": "2\n100000000 1 1000000000 1 1000000000" }, { "input": "5\n1 2 3 4 5", "output": "0\n1 2 3 4 5" }, { "input": "20\n2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000", "output": "19\n2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000" }, { "input": "2\n223092870 23", "output": "1\n223092870 1 23" }, { "input": "2\n100000003 100000003", "output": "1\n100000003 1 100000003" }, { "input": "2\n999999937 999999937", "output": "1\n999999937 1 999999937" }, { "input": "4\n999 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2 2 510510 510510 999999491 999999491 436077930 570018449", "output": "10\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491 436077930 1 570018449" }, { "input": "2\n3996017 3996017", "output": "1\n3996017 1 3996017" }, { "input": "2\n999983 999983", "output": "1\n999983 1 999983" }, { "input": "2\n618575685 773990454", "output": "1\n618575685 1 773990454" }, { "input": "3\n9699690 3 7", "output": "1\n9699690 1 3 7" }, { "input": "2\n999999999 999999996", "output": "1\n999999999 1 999999996" }, { "input": "2\n99999910 99999910", "output": "1\n99999910 1 99999910" }, { "input": "12\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491", "output": "9\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491" }, { "input": "3\n999999937 999999937 999999937", "output": "2\n999999937 1 999999937 1 999999937" }, { "input": "2\n99839 99839", "output": "1\n99839 1 99839" }, { "input": "3\n19999909 19999909 19999909", "output": "2\n19999909 1 19999909 1 19999909" }, { "input": "4\n1 1000000000 1 1000000000", "output": "0\n1 1000000000 1 1000000000" }, { "input": "2\n64006 64006", "output": "1\n64006 1 64006" }, { "input": "2\n1956955 1956955", "output": "1\n1956955 1 1956955" }, { "input": "3\n1 1000000000 1000000000", "output": "1\n1 1000000000 1 1000000000" }, { "input": "2\n982451707 982451707", "output": "1\n982451707 1 982451707" }, { "input": "2\n999999733 999999733", "output": "1\n999999733 1 999999733" }, { "input": "3\n999999733 999999733 999999733", "output": "2\n999999733 1 999999733 1 999999733" }, { "input": "2\n3257 3257", "output": "1\n3257 1 3257" }, { "input": "2\n223092870 181598", "output": "1\n223092870 1 181598" }, { "input": "3\n959919409 105935 105935", "output": "2\n959919409 1 105935 1 105935" }, { "input": "2\n510510 510510", "output": "1\n510510 1 510510" }, { "input": "3\n223092870 1000000000 1000000000", "output": "2\n223092870 1 1000000000 1 1000000000" }, { "input": "14\n1000000000 2 1000000000 3 1000000000 6 1000000000 1000000000 15 1000000000 1000000000 1000000000 100000000 1000", "output": "11\n1000000000 1 2 1 1000000000 3 1000000000 1 6 1 1000000000 1 1000000000 1 15 1 1000000000 1 1000000000 1 1000000000 1 100000000 1 1000" }, { "input": "7\n1 982451653 982451653 1 982451653 982451653 982451653", "output": "3\n1 982451653 1 982451653 1 982451653 1 982451653 1 982451653" }, { "input": "2\n100000007 100000007", "output": "1\n100000007 1 100000007" }, { "input": "3\n999999757 999999757 999999757", "output": "2\n999999757 1 999999757 1 999999757" }, { "input": "3\n99999989 99999989 99999989", "output": "2\n99999989 1 99999989 1 99999989" }, { "input": "5\n2 4 982451707 982451707 3", "output": "2\n2 1 4 982451707 1 982451707 3" }, { "input": "2\n20000014 20000014", "output": "1\n20000014 1 20000014" }, { "input": "2\n99999989 99999989", "output": "1\n99999989 1 99999989" }, { "input": "2\n111546435 111546435", "output": "1\n111546435 1 111546435" }, { "input": "2\n55288874 33538046", "output": "1\n55288874 1 33538046" }, { "input": "5\n179424673 179424673 179424673 179424673 179424673", "output": "4\n179424673 1 179424673 1 179424673 1 179424673 1 179424673" }, { "input": "2\n199999978 199999978", "output": "1\n199999978 1 199999978" }, { "input": "2\n1000000000 2", "output": "1\n1000000000 1 2" }, { "input": "3\n19999897 19999897 19999897", "output": "2\n19999897 1 19999897 1 19999897" }, { "input": "2\n19999982 19999982", "output": "1\n19999982 1 19999982" }, { "input": "2\n10000007 10000007", "output": "1\n10000007 1 10000007" }, { "input": "3\n999999937 999999937 2", "output": "1\n999999937 1 999999937 2" }, { "input": "5\n2017 2017 2017 2017 2017", "output": "4\n2017 1 2017 1 2017 1 2017 1 2017" }, { "input": "2\n19999909 39999818", "output": "1\n19999909 1 39999818" }, { "input": "2\n62615533 7919", "output": "1\n62615533 1 7919" }, { "input": "5\n39989 39989 33 31 29", "output": "1\n39989 1 39989 33 31 29" }, { "input": "2\n1000000000 100000", "output": "1\n1000000000 1 100000" }, { "input": "2\n1938 10010", "output": "1\n1938 1 10010" }, { "input": "2\n199999 199999", "output": "1\n199999 1 199999" }, { "input": "2\n107273 107273", "output": "1\n107273 1 107273" }, { "input": "3\n49999 49999 49999", "output": "2\n49999 1 49999 1 49999" }, { "input": "2\n1999966 1999958", "output": "1\n1999966 1 1999958" }, { "input": "2\n86020 300846", "output": "1\n86020 1 300846" }, { "input": "2\n999999997 213", "output": "1\n999999997 1 213" }, { "input": "2\n200000014 200000434", "output": "1\n200000014 1 200000434" } ]
1,640,303,452
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
46
0
def gcd(x, y): while y: x, y = y, x % y return x n = int(input()) entrada = list(map(int, input().split())) resultado = [str(entrada[0])] cont = 0 for i in range(1, n): if gcd(entrada[i], entrada[i - 1]) > 1: resultado.append("1") cont += 1 resultado.append(str(entrada[i])) cont += 1 print(cont - n) print(' '.join(resultado))
Title: Co-prime Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible. In each move you can insert any positive integral number you want not greater than 109 in any place in the array. An array is co-prime if any two adjacent numbers of it are co-prime. In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. Output Specification: Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime. The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it. If there are multiple answers you can print any one of them. Demo Input: ['3\n2 7 28\n'] Demo Output: ['1\n2 7 9 28\n'] Note: none
```python def gcd(x, y): while y: x, y = y, x % y return x n = int(input()) entrada = list(map(int, input().split())) resultado = [str(entrada[0])] cont = 0 for i in range(1, n): if gcd(entrada[i], entrada[i - 1]) > 1: resultado.append("1") cont += 1 resultado.append(str(entrada[i])) cont += 1 print(cont - n) print(' '.join(resultado)) ```
0
735
C
Tennis Championship
PROGRAMMING
1,600
[ "combinatorics", "constructive algorithms", "greedy", "math" ]
null
null
Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be *n* players participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately. Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament. Tournament hasn't started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this problem without your help.
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1018) — the number of players to participate in the tournament.
Print the maximum number of games in which the winner of the tournament can take part.
[ "2\n", "3\n", "4\n", "10\n" ]
[ "1\n", "2\n", "2\n", "4\n" ]
In all samples we consider that player number 1 is the winner. In the first sample, there would be only one game so the answer is 1. In the second sample, player 1 can consequently beat players 2 and 3. In the third sample, player 1 can't play with each other player as after he plays with players 2 and 3 he can't play against player 4, as he has 0 games played, while player 1 already played 2. Thus, the answer is 2 and to achieve we make pairs (1, 2) and (3, 4) and then clash the winners.
1,750
[ { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "4", "output": "2" }, { "input": "10", "output": "4" }, { "input": "1000", "output": "14" }, { "input": "2500", "output": "15" }, { "input": "690000", "output": "27" }, { "input": "3000000000", "output": "45" }, { "input": "123456789123456789", "output": "81" }, { "input": "5", "output": "3" }, { "input": "143", "output": "9" }, { "input": "144", "output": "10" }, { "input": "145", "output": "10" }, { "input": "232", "output": "10" }, { "input": "233", "output": "11" }, { "input": "234", "output": "11" }, { "input": "679891637638612257", "output": "84" }, { "input": "679891637638612258", "output": "85" }, { "input": "679891637638612259", "output": "85" }, { "input": "1000000000000000000", "output": "85" }, { "input": "10235439547", "output": "47" }, { "input": "1240723548", "output": "43" }, { "input": "92353046212453", "output": "66" }, { "input": "192403205846532", "output": "68" }, { "input": "13925230525389", "output": "62" }, { "input": "12048230592523", "output": "62" }, { "input": "19204385325853", "output": "63" }, { "input": "902353283921", "output": "56" }, { "input": "793056859214355", "output": "70" }, { "input": "982045466234565", "output": "71" }, { "input": "126743950353465", "output": "67" }, { "input": "12405430465", "output": "47" }, { "input": "10238439257768", "output": "61" }, { "input": "1728493055346", "output": "58" }, { "input": "927553829046", "output": "56" }, { "input": "62735129403", "output": "51" }, { "input": "71624823950223", "output": "65" }, { "input": "8902353464851212", "output": "75" }, { "input": "61824012598535", "output": "65" }, { "input": "1294902504603347", "output": "71" }, { "input": "6", "output": "3" }, { "input": "7", "output": "3" }, { "input": "8", "output": "4" }, { "input": "9", "output": "4" }, { "input": "11", "output": "4" }, { "input": "12", "output": "4" }, { "input": "13", "output": "5" }, { "input": "14", "output": "5" }, { "input": "15", "output": "5" }, { "input": "16", "output": "5" }, { "input": "17", "output": "5" }, { "input": "18", "output": "5" }, { "input": "19", "output": "5" }, { "input": "20", "output": "5" }, { "input": "21", "output": "6" }, { "input": "22", "output": "6" }, { "input": "23", "output": "6" }, { "input": "355687428096000", "output": "69" }, { "input": "576460752303423488", "output": "84" }, { "input": "32212254719", "output": "49" }, { "input": "26388279066623", "output": "63" }, { "input": "618473717761", "output": "56" }, { "input": "262406072477", "output": "54" } ]
1,528,647,696
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
77
0
# http://codeforces.com/problemset/problem/735/C players = int(input()) games = 0 game_list = [] def g_match(before): global games games += 1 if len(before) is 1: return games i = 0 if len(before) % 2 == 0: total = len(before) else: total = len(before) + 1 after = [] while i < total: try: after.append((before[i][0], before[i + 1][0])) except IndexError: after.append((before[i][0], before[i][1])) i += 2 return g_match(after) for i in range(players): if i % 2 == 0: game_list.append((i, i + 1)) print(g_match(game_list))
Title: Tennis Championship Time Limit: None seconds Memory Limit: None megabytes Problem Description: Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be *n* players participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately. Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament. Tournament hasn't started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this problem without your help. Input Specification: The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1018) — the number of players to participate in the tournament. Output Specification: Print the maximum number of games in which the winner of the tournament can take part. Demo Input: ['2\n', '3\n', '4\n', '10\n'] Demo Output: ['1\n', '2\n', '2\n', '4\n'] Note: In all samples we consider that player number 1 is the winner. In the first sample, there would be only one game so the answer is 1. In the second sample, player 1 can consequently beat players 2 and 3. In the third sample, player 1 can't play with each other player as after he plays with players 2 and 3 he can't play against player 4, as he has 0 games played, while player 1 already played 2. Thus, the answer is 2 and to achieve we make pairs (1, 2) and (3, 4) and then clash the winners.
```python # http://codeforces.com/problemset/problem/735/C players = int(input()) games = 0 game_list = [] def g_match(before): global games games += 1 if len(before) is 1: return games i = 0 if len(before) % 2 == 0: total = len(before) else: total = len(before) + 1 after = [] while i < total: try: after.append((before[i][0], before[i + 1][0])) except IndexError: after.append((before[i][0], before[i][1])) i += 2 return g_match(after) for i in range(players): if i % 2 == 0: game_list.append((i, i + 1)) print(g_match(game_list)) ```
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" } ]
1,691,237,674
2,147,483,647
Python 3
OK
TESTS
81
92
0
x=int(input()) a=0 b=0 c=0 for i in range(x): q,w,e=map(int,input().split()) a=a+q b=b+w c=c+e if(a==b==c==0): print("YES") else: print("NO")
Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none
```python x=int(input()) a=0 b=0 c=0 for i in range(x): q,w,e=map(int,input().split()) a=a+q b=b+w c=c+e if(a==b==c==0): print("YES") else: print("NO") ```
3.977
573
A
Bear and Poker
PROGRAMMING
1,300
[ "implementation", "math", "number theory" ]
null
null
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars. Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players. The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
[ "4\n75 150 75 50\n", "3\n100 150 250\n" ]
[ "Yes\n", "No\n" ]
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid. It can be shown that in the second sample test there is no way to make all bids equal.
500
[ { "input": "4\n75 150 75 50", "output": "Yes" }, { "input": "3\n100 150 250", "output": "No" }, { "input": "7\n34 34 68 34 34 68 34", "output": "Yes" }, { "input": "10\n72 96 12 18 81 20 6 2 54 1", "output": "No" }, { "input": "20\n958692492 954966768 77387000 724664764 101294996 614007760 202904092 555293973 707655552 108023967 73123445 612562357 552908390 914853758 915004122 466129205 122853497 814592742 373389439 818473058", "output": "No" }, { "input": "2\n1 1", "output": "Yes" }, { "input": "2\n72 72", "output": "Yes" }, { "input": "2\n49 42", "output": "No" }, { "input": "3\n1000000000 1000000000 1000000000", "output": "Yes" }, { "input": "6\n162000 96000 648000 1000 864000 432000", "output": "Yes" }, { "input": "8\n600000 100000 100000 100000 900000 600000 900000 600000", "output": "Yes" }, { "input": "12\n2048 1024 6144 1024 3072 3072 6144 1024 4096 2048 6144 3072", "output": "Yes" }, { "input": "20\n246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246", "output": "Yes" }, { "input": "50\n840868705 387420489 387420489 795385082 634350497 206851546 536870912 536870912 414927754 387420489 387420489 536870912 387420489 149011306 373106005 536870912 700746206 387420489 777952883 847215247 176645254 576664386 387420489 230876513 536870912 536870912 536870912 387420489 387420489 536870912 460495524 528643722 387420489 536870912 470369206 899619085 387420489 631148352 387420489 387420489 536870912 414666674 521349938 776784669 387420489 102428009 536870912 387420489 536870912 718311009", "output": "No" }, { "input": "2\n5 6", "output": "No" }, { "input": "3\n536870912 387420489 257407169", "output": "No" }, { "input": "4\n2 2 5 2", "output": "No" }, { "input": "2\n33554432 59049", "output": "Yes" }, { "input": "3\n536870912 387420489 387420489", "output": "Yes" }, { "input": "2\n1 5", "output": "No" }, { "input": "18\n2 3 5 7 11 13 17 19 23 29 31 37 43 47 53 59 67 71", "output": "No" }, { "input": "2\n1 30", "output": "No" }, { "input": "3\n335544320 71744535 71744535", "output": "Yes" }, { "input": "5\n1000000000 999999999 999999998 999999997 999999996", "output": "No" }, { "input": "2\n25 5", "output": "No" }, { "input": "4\n75 150 75 5", "output": "No" }, { "input": "3\n536870912 387420489 362797056", "output": "Yes" }, { "input": "3\n536870912 387420489 89", "output": "No" }, { "input": "4\n547 2606459 222763549 143466789", "output": "No" }, { "input": "3\n129140163 33554432 1", "output": "Yes" }, { "input": "10\n244140625 244140625 244140625 244140625 244140625 244140625 244140625 244140625 536870912 387420489", "output": "No" }, { "input": "3\n5 5 1", "output": "No" }, { "input": "5\n3 7 29 36760123 823996703", "output": "No" } ]
1,578,935,369
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
21
186
7,475,200
import math n=int(input()) a=list(map(int,input().split())) lcm=a[0] for i in a[1:]: lcm=(lcm*i)//math.gcd(lcm,i) fl=0 for i in a: k=lcm/i if k%6==0: continue z=math.log(k)/math.log(2) d=math.log(k)/math.log(3) if z==int(z): continue elif d==int(d): continue else: fl=1 break if fl: print("NO") else: print("YES")
Title: Bear and Poker Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars. Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? Input Specification: First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players. The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. Output Specification: Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. Demo Input: ['4\n75 150 75 50\n', '3\n100 150 250\n'] Demo Output: ['Yes\n', 'No\n'] Note: In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid. It can be shown that in the second sample test there is no way to make all bids equal.
```python import math n=int(input()) a=list(map(int,input().split())) lcm=a[0] for i in a[1:]: lcm=(lcm*i)//math.gcd(lcm,i) fl=0 for i in a: k=lcm/i if k%6==0: continue z=math.log(k)/math.log(2) d=math.log(k)/math.log(3) if z==int(z): continue elif d==int(d): continue else: fl=1 break if fl: print("NO") else: print("YES") ```
0
540
A
Combination Lock
PROGRAMMING
800
[ "implementation" ]
null
null
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
[ "5\n82195\n64723\n" ]
[ "13\n" ]
In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
500
[ { "input": "5\n82195\n64723", "output": "13" }, { "input": "12\n102021090898\n010212908089", "output": "16" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "10\n0728592530\n1362615763", "output": "27" }, { "input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746", "output": "245" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "3\n607\n684", "output": "5" }, { "input": "4\n0809\n0636", "output": "8" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762280548\n9519431339078678836940020", "output": "72" }, { "input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024", "output": "305" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762285484\n9519431339078678836940202", "output": "74" }, { "input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831", "output": "357" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8378525987476228048406972\n9194339078883694020217816", "output": "55" }, { "input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599", "output": "274" }, { "input": "1\n0\n0", "output": "0" }, { "input": "1\n7\n7", "output": "0" }, { "input": "1\n0\n5", "output": "5" }, { "input": "1\n2\n7", "output": "5" }, { "input": "1\n7\n9", "output": "2" }, { "input": "1\n9\n7", "output": "2" }, { "input": "1\n2\n9", "output": "3" }, { "input": "1\n9\n2", "output": "3" }, { "input": "25\n3164978461316464614169874\n9413979197249127496597357", "output": "66" }, { "input": "4\n9999\n9999", "output": "0" }, { "input": "2\n11\n11", "output": "0" } ]
1,591,796,988
2,147,483,647
Python 3
OK
TESTS
43
124
0
n = int(input()) a = input() b = input() c =0 for i in range(n): d = abs(int(a[i]) - int(b[i])) m = min(d,(10-d)) #print(m) c =c+m print(c)
Title: Combination Lock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock. Output Specification: Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock. Demo Input: ['5\n82195\n64723\n'] Demo Output: ['13\n'] Note: In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
```python n = int(input()) a = input() b = input() c =0 for i in range(n): d = abs(int(a[i]) - int(b[i])) m = min(d,(10-d)) #print(m) c =c+m print(c) ```
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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"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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1,699,556,090
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
62
0
a=input() b=input() a=a.lower() b=b.lower() if a==b: print(0) j=0 length=len(a) sum1 = 0 sum2 = 0 for j in range(length): sum1 = sum1 + ord(a[j]) sum2 = sum2 + ord(b[j]) if sum1<sum2: print(-1) elif sum1>sum2: print(1)
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 a=input() b=input() a=a.lower() b=b.lower() if a==b: print(0) j=0 length=len(a) sum1 = 0 sum2 = 0 for j in range(length): sum1 = sum1 + ord(a[j]) sum2 = sum2 + ord(b[j]) if sum1<sum2: print(-1) elif sum1>sum2: print(1) ```
0
580
C
Kefa and Park
PROGRAMMING
1,500
[ "dfs and similar", "graphs", "trees" ]
null
null
Kefa decided to celebrate his first big salary by going to the restaurant. He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them. The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats. Your task is to help Kefa count the number of restaurants where he can go.
The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat). Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge. It is guaranteed that the given set of edges specifies a tree.
A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats.
[ "4 1\n1 1 0 0\n1 2\n1 3\n1 4\n", "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n" ]
[ "2\n", "2\n" ]
Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children. Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2. Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7.
1,500
[ { "input": "4 1\n1 1 0 0\n1 2\n1 3\n1 4", "output": "2" }, { "input": "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7", "output": "2" }, { "input": "3 2\n1 1 1\n1 2\n2 3", "output": "0" }, { "input": "5 2\n1 1 0 1 1\n1 2\n2 3\n3 4\n4 5", "output": "1" }, { "input": "6 1\n1 0 1 1 0 0\n1 2\n1 3\n1 4\n1 5\n1 6", "output": "3" }, { "input": "7 3\n1 1 1 1 1 0 1\n1 2\n1 3\n2 4\n3 5\n5 6\n6 7", "output": "2" }, { "input": "15 2\n1 0 1 0 1 0 0 0 0 0 0 0 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n4 8\n4 9\n5 10\n5 11\n6 12\n6 13\n7 14\n7 15", "output": "8" }, { "input": "2 1\n1 1\n2 1", "output": "0" }, { "input": "12 3\n1 0 1 0 1 1 1 1 0 0 0 0\n6 7\n12 1\n9 7\n1 4\n10 7\n7 1\n11 8\n5 1\n3 7\n5 8\n4 2", "output": "7" } ]
1,685,571,968
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
2
46
0
import io, os, sys input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline print = lambda x: sys.stdout.write(str(x) + "\n") II = lambda: int(input()) MII = lambda: map(int, input().split()) LMII = lambda: list(MII()) #SLMII = lambda: sorted(LMII()) ans = 0 n, m = MII() a = LMII() adj = [[] for i in range(n)] #leafs = [] for _ in range(n-1): l, r = MII() adj[l-1].append(r-1) adj[r-1].append(l-1) def tree_dfs(v, adj_v, par_v, cat_v): global ans if len(adj_v) == 1 and cat_v <= m: #leafs.append(v) ans += 1 for son_v in adj_v: cat_son_v = cat_v + 1 if a[son_v] else 0 if son_v != par_v and cat_son_v <= m: tree_dfs(son_v, adj[son_v], v, cat_son_v) tree_dfs(0, adj[0], None, a[0]) print(ans)
Title: Kefa and Park Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kefa decided to celebrate his first big salary by going to the restaurant. He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them. The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats. Your task is to help Kefa count the number of restaurants where he can go. Input Specification: The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat). Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge. It is guaranteed that the given set of edges specifies a tree. Output Specification: A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats. Demo Input: ['4 1\n1 1 0 0\n1 2\n1 3\n1 4\n', '7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n'] Demo Output: ['2\n', '2\n'] Note: Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children. Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2. Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7.
```python import io, os, sys input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline print = lambda x: sys.stdout.write(str(x) + "\n") II = lambda: int(input()) MII = lambda: map(int, input().split()) LMII = lambda: list(MII()) #SLMII = lambda: sorted(LMII()) ans = 0 n, m = MII() a = LMII() adj = [[] for i in range(n)] #leafs = [] for _ in range(n-1): l, r = MII() adj[l-1].append(r-1) adj[r-1].append(l-1) def tree_dfs(v, adj_v, par_v, cat_v): global ans if len(adj_v) == 1 and cat_v <= m: #leafs.append(v) ans += 1 for son_v in adj_v: cat_son_v = cat_v + 1 if a[son_v] else 0 if son_v != par_v and cat_son_v <= m: tree_dfs(son_v, adj[son_v], v, cat_son_v) tree_dfs(0, adj[0], None, a[0]) print(ans) ```
0
832
A
Sasha and Sticks
PROGRAMMING
800
[ "games", "math" ]
null
null
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends. Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes). You can print each letter in arbitrary case (upper of lower).
[ "1 1\n", "10 4\n" ]
[ "YES\n", "NO\n" ]
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins. In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
500
[ { "input": "1 1", "output": "YES" }, { "input": "10 4", "output": "NO" }, { "input": "251656215122324104 164397544865601257", "output": "YES" }, { "input": "963577813436662285 206326039287271924", "output": "NO" }, { "input": "1000000000000000000 1", "output": "NO" }, { "input": "253308697183523656 25332878317796706", "output": "YES" }, { "input": "669038685745448997 501718093668307460", "output": "YES" }, { "input": "116453141993601660 87060381463547965", "output": "YES" }, { "input": "766959657 370931668", "output": "NO" }, { "input": "255787422422806632 146884995820359999", "output": "YES" }, { "input": "502007866464507926 71266379084204128", "output": "YES" }, { "input": "257439908778973480 64157133126869976", "output": "NO" }, { "input": "232709385 91708542", "output": "NO" }, { "input": "252482458300407528 89907711721009125", "output": "NO" }, { "input": "6 2", "output": "YES" }, { "input": "6 3", "output": "NO" }, { "input": "6 4", "output": "YES" }, { "input": "6 5", "output": "YES" }, { "input": "6 6", "output": "YES" }, { "input": "258266151957056904 30153168463725364", "output": "NO" }, { "input": "83504367885565783 52285355047292458", "output": "YES" }, { "input": "545668929424440387 508692735816921376", "output": "YES" }, { "input": "547321411485639939 36665750286082900", "output": "NO" }, { "input": "548973893546839491 183137237979822911", "output": "NO" }, { "input": "544068082 193116851", "output": "NO" }, { "input": "871412474 749817171", "output": "YES" }, { "input": "999999999 1247", "output": "NO" }, { "input": "851941088 712987048", "output": "YES" }, { "input": "559922900 418944886", "output": "YES" }, { "input": "293908937 37520518", "output": "YES" }, { "input": "650075786 130049650", "output": "NO" }, { "input": "1000000000 1000000000", "output": "YES" }, { "input": "548147654663723363 107422751713800746", "output": "YES" }, { "input": "828159210 131819483", "output": "NO" }, { "input": "6242634 4110365", "output": "YES" }, { "input": "458601973 245084155", "output": "YES" }, { "input": "349593257 18089089", "output": "YES" }, { "input": "814768821 312514745", "output": "NO" }, { "input": "697884949 626323363", "output": "YES" }, { "input": "667011589 54866795", "output": "NO" }, { "input": "1000000000000000000 2", "output": "NO" }, { "input": "1000000000000000000 3", "output": "YES" }, { "input": "1000000000000000000 4", "output": "NO" }, { "input": "999999999999999 1", "output": "YES" }, { "input": "17 4", "output": "NO" }, { "input": "2 2", "output": "YES" }, { "input": "1000000000000000 2", "output": "NO" }, { "input": "12 4", "output": "YES" }, { "input": "6 1", "output": "NO" }, { "input": "2 1", "output": "NO" }, { "input": "10000000005 1", "output": "YES" }, { "input": "10000000000000009 2", "output": "NO" }, { "input": "12457895452123 1", "output": "YES" }, { "input": "999999999999999999 9", "output": "YES" }, { "input": "1000000000000 3", "output": "YES" }, { "input": "13099714659575475 6549849616514894", "output": "NO" }, { "input": "100000000000000001 1", "output": "YES" }, { "input": "825175814723458 324", "output": "YES" }, { "input": "20 4", "output": "YES" }, { "input": "100000176877 4", "output": "YES" }, { "input": "100000 3", "output": "YES" }, { "input": "946744073709551614 10", "output": "YES" } ]
1,630,381,564
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
2
2,000
6,758,400
n,k = map(int,input().split()) i = 0 temp = 0 while i <= n: if i == k: temp = i break else: i += 1 if temp % 2 != 0: print("YES") else: print("NO")
Title: Sasha and Sticks Time Limit: None seconds Memory Limit: None megabytes Problem Description: It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends. Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn. Output Specification: If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes). You can print each letter in arbitrary case (upper of lower). Demo Input: ['1 1\n', '10 4\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins. In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
```python n,k = map(int,input().split()) i = 0 temp = 0 while i <= n: if i == k: temp = i break else: i += 1 if temp % 2 != 0: print("YES") else: print("NO") ```
0
136
A
Presents
PROGRAMMING
800
[ "implementation" ]
null
null
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]
[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]
none
500
[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71", "output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90" }, { "input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9", "output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9" }, { "input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44", "output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20" }, { "input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17", "output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28" }, { "input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7", "output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5" }, { "input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16", "output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43" }, { "input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65", "output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43" }, { "input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16", "output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45" }, { "input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51", "output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64" }, { "input": "12\n12 3 1 5 11 6 7 10 2 8 9 4", "output": "3 9 2 12 4 6 7 10 11 8 5 1" }, { "input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17", "output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44" }, { "input": "8\n1 3 5 2 4 8 6 7", "output": "1 4 2 5 3 7 8 6" }, { "input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14", "output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12" }, { "input": "10\n2 9 4 6 10 1 7 5 3 8", "output": "6 1 9 3 8 4 7 10 2 5" }, { "input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3", "output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7" }, { "input": "10\n2 4 9 3 6 8 10 5 1 7", "output": "9 1 4 2 8 5 10 6 3 7" }, { "input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15", "output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23" }, { "input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1", "output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27" }, { "input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30", "output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23" }, { "input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52", "output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50" }, { "input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18", "output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98" }, { "input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16", "output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25" }, { "input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37", "output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54" }, { "input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71", "output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71" }, { "input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22", "output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8" }, { "input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12", "output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51" }, { "input": "7\n2 1 5 7 3 4 6", "output": "2 1 5 6 3 7 4" }, { "input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35", "output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31" }, { "input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16", "output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21" }, { "input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52", "output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11" }, { "input": "9\n8 5 2 6 1 9 4 7 3", "output": "5 3 9 7 2 4 8 1 6" }, { "input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27", "output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13" }, { "input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34", "output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6" }, { "input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29", "output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14" }, { "input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7", "output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16" }, { "input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34", "output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24" }, { "input": "9\n4 8 6 5 3 9 2 7 1", "output": "9 7 5 1 4 3 8 2 6" }, { "input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29", "output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32" }, { "input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47", "output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11" }, { "input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6", "output": "10 4 1 8 6 13 12 5 3 11 9 2 7" }, { "input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29", "output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37" }, { "input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3", "output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25" }, { "input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20", "output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24" }, { "input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42", "output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59" }, { "input": "6\n4 3 6 5 1 2", "output": "5 6 2 1 4 3" }, { "input": "9\n7 8 5 3 1 4 2 9 6", "output": "5 7 4 6 3 9 1 2 8" }, { "input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37", "output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13" }, { "input": "1\n1", "output": "1" }, { "input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11", "output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13" }, { "input": "2\n2 1", "output": "2 1" }, { "input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28", "output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18" }, { "input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13", "output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1" }, { "input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14", "output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36" }, { "input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29", "output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32" }, { "input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5", "output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40" }, { "input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54", "output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21" }, { "input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29", "output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4" }, { "input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23", "output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50" }, { "input": "10\n2 10 7 4 1 5 8 6 3 9", "output": "5 1 9 4 6 8 3 7 10 2" }, { "input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13", "output": "3 2 11 12 13 5 10 4 9 8 7 6 14 1" }, { "input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2", "output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11" }, { "input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53", "output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37" }, { "input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53", "output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53" }, { "input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64", "output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49" }, { "input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86", "output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38" }, { "input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75", "output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2" }, { "input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53", "output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5" }, { "input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39", "output": "4 82 16 9 13 28 42 93 3 92 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" } ]
1,691,240,045
2,147,483,647
Python 3
OK
TESTS
101
92
0
n = int(input()) p = list(map(int, input().split())) l = [0 for _ in range(n)] for i in range(n): l[p[i]-1] = str(i+1) print(" ".join(l))
Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none
```python n = int(input()) p = list(map(int, input().split())) l = [0 for _ in range(n)] for i in range(n): l[p[i]-1] = str(i+1) print(" ".join(l)) ```
3
25
A
IQ test
PROGRAMMING
1,300
[ "brute force" ]
A. IQ test
2
256
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
[ "5\n2 4 7 8 10\n", "4\n1 2 1 1\n" ]
[ "3\n", "2\n" ]
none
0
[ { "input": "5\n2 4 7 8 10", "output": "3" }, { "input": "4\n1 2 1 1", "output": "2" }, { "input": "3\n1 2 2", "output": "1" }, { "input": "3\n100 99 100", "output": "2" }, { "input": "3\n5 3 2", "output": "3" }, { "input": "4\n43 28 1 91", "output": "2" }, { "input": "4\n75 13 94 77", "output": "3" }, { "input": "4\n97 8 27 3", "output": "2" }, { "input": "10\n95 51 12 91 85 3 1 31 25 7", "output": "3" }, { "input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46", "output": "4" }, { "input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36", "output": "26" }, { "input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83", "output": "48" }, { "input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25", "output": "1" }, { "input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46", "output": "25" }, { "input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73", "output": "13" }, { "input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76", "output": "99" }, { "input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14", "output": "1" }, { "input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22", "output": "93" }, { "input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23", "output": "1" }, { "input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98", "output": "1" }, { "input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94", "output": "93" }, { "input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29", "output": "32" }, { "input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47", "output": "26" }, { "input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72", "output": "56" }, { "input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86", "output": "97" }, { "input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42", "output": "79" }, { "input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52", "output": "25" }, { "input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86", "output": "78" }, { "input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68", "output": "70" }, { "input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81", "output": "39" }, { "input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83", "output": "88" }, { "input": "3\n100 100 1", "output": "3" } ]
1,584,077,968
2,147,483,647
PyPy 3
OK
TESTS
32
312
0
n = int(input()) e=o=0 a = list(map(int,input().split())) for i in range(len(a)): if(a[i]%2 == 0): e+=1 if(e>1): break; else: o+=1 if(e>o): for i in range(len(a)): if(a[i]%2 == 1): print(i+1) else: for i in range(len(a)): if(a[i]%2 == 0): print(i+1)
Title: IQ test Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. Output Specification: Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. Demo Input: ['5\n2 4 7 8 10\n', '4\n1 2 1 1\n'] Demo Output: ['3\n', '2\n'] Note: none
```python n = int(input()) e=o=0 a = list(map(int,input().split())) for i in range(len(a)): if(a[i]%2 == 0): e+=1 if(e>1): break; else: o+=1 if(e>o): for i in range(len(a)): if(a[i]%2 == 1): print(i+1) else: for i in range(len(a)): if(a[i]%2 == 0): print(i+1) ```
3.922
570
A
Elections
PROGRAMMING
1,100
[ "implementation" ]
null
null
The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate. The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index. At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index. Determine who will win the elections.
The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively. Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*. It is guaranteed that the total number of people in all the cities does not exceed 109.
Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.
[ "3 3\n1 2 3\n2 3 1\n1 2 1\n", "3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n" ]
[ "2", "1" ]
Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes. Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.
500
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1,498,213,798
2,147,483,647
Python 3
OK
TESTS
62
62
5,529,600
n,m=map(int,input().split(" ")) a=list() b=list() c=[0]*(n+1) for i in range(0,m): a=list(map(int,input().split(" "))) k=max(a) if a.count(k)==1: b.append(a.index(k)+1) c[a.index(k)+1]=c[a.index(k)+1]+1 else: for j in range(0,n): if a[j]==k: b.append(j+1) c[j+1]=c[j+1]+1 break s=max(c) print(c.index(s))
Title: Elections Time Limit: None seconds Memory Limit: None megabytes Problem Description: The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate. The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index. At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index. Determine who will win the elections. Input Specification: The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively. Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*. It is guaranteed that the total number of people in all the cities does not exceed 109. Output Specification: Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one. Demo Input: ['3 3\n1 2 3\n2 3 1\n1 2 1\n', '3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n'] Demo Output: ['2', '1'] Note: Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes. Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.
```python n,m=map(int,input().split(" ")) a=list() b=list() c=[0]*(n+1) for i in range(0,m): a=list(map(int,input().split(" "))) k=max(a) if a.count(k)==1: b.append(a.index(k)+1) c[a.index(k)+1]=c[a.index(k)+1]+1 else: for j in range(0,n): if a[j]==k: b.append(j+1) c[j+1]=c[j+1]+1 break s=max(c) print(c.index(s)) ```
3
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,583,759,345
2,147,483,647
Python 3
OK
TESTS
40
216
0
def A41(string1,string2): reversed_string = string1[::-1] if reversed_string == string2: print('YES') else: print('NO') string1 = input() string2 = input() A41(string1,string2)
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python def A41(string1,string2): reversed_string = string1[::-1] if reversed_string == string2: print('YES') else: print('NO') string1 = input() string2 = input() A41(string1,string2) ```
3.946
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" } ]
1,642,945,857
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
26
1,000
512,000
n,m=map(int,input().split()) lt=[] while(m>0): a=[str(a) for a in input().split()] lt.append(a) m=m-1 b=[str(b) for b in input().split()] for i in b: for j in lt: if(i in j): k=j if(len(k[0])<len(k[1])): print(k[0],end=' ') elif(len(k[0])==len(k[1])): print(i,end=' ') else: print(k[1],end=' ') break;
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()) lt=[] while(m>0): a=[str(a) for a in input().split()] lt.append(a) m=m-1 b=[str(b) for b in input().split()] for i in b: for j in lt: if(i in j): k=j if(len(k[0])<len(k[1])): print(k[0],end=' ') elif(len(k[0])==len(k[1])): print(i,end=' ') else: print(k[1],end=' ') break; ```
0
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" } ]
1,692,419,936
2,147,483,647
Python 3
OK
TESTS
81
93
13,209,600
lis=list(map(int,input().strip().split())) n=lis[0] x=lis[1] lis=list(map(int,input().strip().split())) c=0 for i in range(1,len(lis)): #print(lis[i]-lis[i-1],x) if lis[i]-lis[i-1]>x: c=0 else: c=c+1 print(c+1)
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 lis=list(map(int,input().strip().split())) n=lis[0] x=lis[1] lis=list(map(int,input().strip().split())) c=0 for i in range(1,len(lis)): #print(lis[i]-lis[i-1],x) if lis[i]-lis[i-1]>x: c=0 else: c=c+1 print(c+1) ```
3
916
A
Jamie and Alarm Snooze
PROGRAMMING
900
[ "brute force", "implementation", "math" ]
null
null
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button. A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky. Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*. Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'. Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60). The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
Print the minimum number of times he needs to press the button.
[ "3\n11 23\n", "5\n01 07\n" ]
[ "2\n", "0\n" ]
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20. In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
500
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"output": "2" }, { "input": "1\n02 30", "output": "3" }, { "input": "1\n18 54", "output": "7" }, { "input": "1\n02 19", "output": "2" }, { "input": "1\n05 25", "output": "8" }, { "input": "60\n23 55", "output": "6" }, { "input": "60\n08 19", "output": "1" }, { "input": "60\n00 00", "output": "7" }, { "input": "60\n08 24", "output": "1" }, { "input": "60\n16 13", "output": "9" }, { "input": "60\n08 21", "output": "1" }, { "input": "60\n16 45", "output": "9" }, { "input": "60\n08 26", "output": "1" }, { "input": "60\n08 50", "output": "1" }, { "input": "60\n05 21", "output": "12" }, { "input": "60\n13 29", "output": "6" }, { "input": "60\n05 18", "output": "12" }, { "input": "60\n13 42", "output": "6" }, { "input": "60\n05 07", "output": "0" }, { "input": "60\n05 47", "output": "0" }, { "input": "60\n21 55", "output": "4" }, { "input": "60\n05 36", "output": "12" }, { "input": "60\n21 08", "output": "4" }, { "input": "60\n21 32", "output": "4" }, { "input": "60\n16 31", "output": "9" }, { "input": "5\n00 00", "output": "73" }, { "input": "2\n06 58", "output": "390" }, { "input": "60\n00 00", "output": "7" }, { "input": "2\n00 00", "output": "181" }, { "input": "10\n00 00", "output": "37" }, { "input": "60\n01 00", "output": "8" }, { "input": "12\n00 06", "output": "31" }, { "input": "1\n00 01", "output": "4" }, { "input": "5\n00 05", "output": "74" }, { "input": "60\n01 01", "output": "8" }, { "input": "11\n18 11", "output": "2" }, { "input": "60\n01 15", "output": "8" }, { "input": "10\n00 16", "output": "38" }, { "input": "60\n00 59", "output": "7" }, { "input": "30\n00 00", "output": "13" }, { "input": "60\n01 05", "output": "8" }, { "input": "4\n00 03", "output": "4" }, { "input": "4\n00 00", "output": "91" }, { "input": "60\n00 01", "output": "7" }, { "input": "6\n00 03", "output": "1" }, { "input": "13\n00 00", "output": "1" }, { "input": "1\n18 01", "output": "2" }, { "input": "5\n06 00", "output": "145" }, { "input": "60\n04 08", "output": "11" }, { "input": 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"output": "9" }, { "input": "6\n08 06", "output": "2" }, { "input": "48\n06 24", "output": "16" }, { "input": "8\n06 58", "output": "98" }, { "input": "3\n12 00", "output": "1" }, { "input": "5\n01 06", "output": "86" }, { "input": "2\n00 08", "output": "185" }, { "input": "3\n18 03", "output": "2" }, { "input": "1\n17 00", "output": "0" }, { "input": "59\n00 48", "output": "7" }, { "input": "5\n12 01", "output": "49" }, { "input": "55\n01 25", "output": "9" }, { "input": "2\n07 23", "output": "0" }, { "input": "10\n01 10", "output": "44" }, { "input": "2\n00 01", "output": "2" }, { "input": "59\n00 01", "output": "6" }, { "input": "5\n00 02", "output": "1" }, { "input": "4\n01 02", "output": "106" }, { "input": "5\n00 06", "output": "74" }, { "input": "42\n00 08", "output": "9" }, { "input": "60\n01 20", "output": "8" }, { "input": "3\n06 00", "output": "1" }, { "input": "4\n00 01", "output": "1" }, { "input": "2\n00 06", "output": "184" }, { "input": "1\n00 57", "output": "0" }, { "input": "6\n00 00", "output": "61" }, { "input": "5\n08 40", "output": "9" }, { "input": "58\n00 55", "output": "1" }, { "input": "2\n00 02", "output": "182" }, { "input": "1\n08 01", "output": "2" }, { "input": "10\n10 10", "output": "14" }, { "input": "60\n01 11", "output": "8" }, { "input": "2\n07 00", "output": "0" }, { "input": "15\n00 03", "output": "25" }, { "input": "6\n04 34", "output": "106" }, { "input": "16\n00 16", "output": "24" }, { "input": "2\n00 59", "output": "1" }, { "input": "59\n00 08", "output": "7" }, { "input": "10\n03 10", "output": "56" }, { "input": "3\n08 03", "output": "2" }, { "input": "20\n06 11", "output": "37" }, { "input": "4\n01 00", "output": "106" }, { "input": "38\n01 08", "output": "12" }, { "input": "60\n00 06", "output": "7" }, { "input": "5\n12 00", "output": "49" }, { "input": "6\n01 42", "output": "78" }, { "input": "4\n00 04", "output": "92" }, { "input": "60\n04 05", "output": "11" }, { "input": "1\n00 53", "output": "6" }, { "input": "5\n08 05", "output": "2" }, { "input": "60\n18 45", "output": "1" }, { "input": "60\n06 23", "output": "13" }, { "input": "6\n00 15", "output": "3" }, { "input": "58\n00 06", "output": "7" }, { "input": "2\n06 44", "output": "383" }, { "input": "1\n08 00", "output": "1" }, { "input": "10\n06 58", "output": "78" }, { "input": "59\n00 58", "output": "8" }, { "input": "1\n18 00", "output": "1" }, { "input": "50\n00 42", "output": "9" }, { "input": "30\n18 30", "output": "2" }, { "input": "60\n21 59", "output": "4" }, { "input": "2\n10 52", "output": "87" }, { "input": "56\n00 00", "output": "7" }, { "input": "16\n18 16", "output": "2" }, { "input": "5\n01 05", "output": "86" }, { "input": "5\n05 00", "output": "133" }, { "input": "5\n23 59", "output": "72" }, { "input": "7\n17 13", "output": "0" }, { "input": "58\n00 00", "output": "7" }, { "input": "15\n00 07", "output": "0" }, { "input": "59\n08 00", "output": "1" }, { "input": "46\n00 00", "output": "8" }, { "input": "59\n01 05", "output": "2" }, { "input": "2\n01 00", "output": "211" }, { "input": "60\n00 24", "output": "7" }, { "input": "10\n00 08", "output": "37" }, { "input": "10\n00 06", "output": "37" }, { "input": "60\n01 24", "output": "8" }, { "input": "50\n00 10", "output": "8" }, { "input": "2\n03 00", "output": "271" }, { "input": "4\n19 04", "output": "17" }, { "input": "25\n00 23", "output": "16" }, { "input": "10\n01 01", "output": "43" } ]
1,551,793,796
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
8
93
0
def f(h, m): return True if '7' in str(h) + str(m) else False x = int(input()) h, m = map(int, input().split()) c = 0 while not f(h, m): c += 1 m -= x if m < 0: m += 60 h -= 1 if h < 0: f += 24 print(c)
Title: Jamie and Alarm Snooze Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button. A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky. Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*. Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'. Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00. Input Specification: The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60). The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59). Output Specification: Print the minimum number of times he needs to press the button. Demo Input: ['3\n11 23\n', '5\n01 07\n'] Demo Output: ['2\n', '0\n'] Note: In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20. In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
```python def f(h, m): return True if '7' in str(h) + str(m) else False x = int(input()) h, m = map(int, input().split()) c = 0 while not f(h, m): c += 1 m -= x if m < 0: m += 60 h -= 1 if h < 0: f += 24 print(c) ```
-1
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" } ]
1,611,932,504
2,147,483,647
Python 3
OK
TESTS
28
436
0
n = int(input()) s = input() l = len(s) for i in range(1, n): si = input() j = 0 while s[j] == si[j] and j < l: j += 1 l = j print(l)
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()) s = input() l = len(s) for i in range(1, n): si = input() j = 0 while s[j] == si[j] and j < l: j += 1 l = j print(l) ```
3
10
A
Power Consumption Calculation
PROGRAMMING
900
[ "implementation" ]
A. Power Consumption Calculation
1
256
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=&lt;<=*r**i*<=≤<=1440, *r**i*<=&lt;<=*l**i*<=+<=1 for *i*<=&lt;<=*n*), which stand for the start and the end of the *i*-th period of work.
Output the answer to the problem.
[ "1 3 2 1 5 10\n0 10\n", "2 8 4 2 5 10\n20 30\n50 100\n" ]
[ "30", "570" ]
none
0
[ { "input": "1 3 2 1 5 10\n0 10", "output": "30" }, { "input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570" }, { "input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436", "output": "8445" }, { "input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945", "output": "52870" }, { "input": "5 41 20 33 43 4\n46 465\n598 875\n967 980\n1135 1151\n1194 1245", "output": "46995" }, { "input": "6 88 28 100 53 36\n440 445\n525 614\n644 844\n1238 1261\n1305 1307\n1425 1434", "output": "85540" }, { "input": "7 46 61 55 28 59\n24 26\n31 61\n66 133\n161 612\n741 746\n771 849\n1345 1357", "output": "67147" }, { "input": "8 83 18 30 28 5\n196 249\n313 544\n585 630\n718 843\n1040 1194\n1207 1246\n1268 1370\n1414 1422", "output": "85876" }, { "input": "9 31 65 27 53 54\n164 176\n194 210\n485 538\n617 690\n875 886\n888 902\n955 957\n1020 1200\n1205 1282", "output": "38570" }, { "input": "30 3 1 58 44 7\n11 13\n14 32\n37 50\n70 74\n101 106\n113 129\n184 195\n197 205\n213 228\n370 394\n443 446\n457 460\n461 492\n499 585\n602 627\n709 776\n812 818\n859 864\n910 913\n918 964\n1000 1010\n1051 1056\n1063 1075\n1106 1145\n1152 1189\n1211 1212\n1251 1259\n1272 1375\n1412 1417\n1430 1431", "output": "11134" }, { "input": "30 42 3 76 28 26\n38 44\n55 66\n80 81\n84 283\n298 314\n331 345\n491 531\n569 579\n597 606\n612 617\n623 701\n723 740\n747 752\n766 791\n801 827\n842 846\n853 891\n915 934\n945 949\n955 964\n991 1026\n1051 1059\n1067 1179\n1181 1191\n1214 1226\n1228 1233\n1294 1306\n1321 1340\n1371 1374\n1375 1424", "output": "59043" }, { "input": "30 46 5 93 20 46\n12 34\n40 41\n54 58\n100 121\n162 182\n220 349\n358 383\n390 398\n401 403\n408 409\n431 444\n466 470\n471 535\n556 568\n641 671\n699 709\n767 777\n786 859\n862 885\n912 978\n985 997\n1013 1017\n1032 1038\n1047 1048\n1062 1080\n1094 1097\n1102 1113\n1122 1181\n1239 1280\n1320 1369", "output": "53608" }, { "input": "30 50 74 77 4 57\n17 23\n24 61\n67 68\n79 87\n93 101\n104 123\n150 192\n375 377\n398 414\n461 566\n600 633\n642 646\n657 701\n771 808\n812 819\n823 826\n827 833\n862 875\n880 891\n919 920\n928 959\n970 1038\n1057 1072\n1074 1130\n1165 1169\n1171 1230\n1265 1276\n1279 1302\n1313 1353\n1354 1438", "output": "84067" }, { "input": "30 54 76 95 48 16\n9 11\n23 97\n112 116\n126 185\n214 223\n224 271\n278 282\n283 348\n359 368\n373 376\n452 463\n488 512\n532 552\n646 665\n681 685\n699 718\n735 736\n750 777\n791 810\n828 838\n841 858\n874 1079\n1136 1171\n1197 1203\n1210 1219\n1230 1248\n1280 1292\n1324 1374\n1397 1435\n1438 1439", "output": "79844" }, { "input": "30 58 78 12 41 28\n20 26\n27 31\n35 36\n38 99\n103 104\n106 112\n133 143\n181 246\n248 251\n265 323\n350 357\n378 426\n430 443\n466 476\n510 515\n517 540\n542 554\n562 603\n664 810\n819 823\n826 845\n869 895\n921 973\n1002 1023\n1102 1136\n1143 1148\n1155 1288\n1316 1388\n1394 1403\n1434 1437", "output": "82686" }, { "input": "30 62 80 97 25 47\n19 20\n43 75\n185 188\n199 242\n252 258\n277 310\n316 322\n336 357\n398 399\n404 436\n443 520\n549 617\n637 649\n679 694\n705 715\n725 730\n731 756\n768 793\n806 833\n834 967\n1003 1079\n1088 1097\n1100 1104\n1108 1121\n1127 1164\n1240 1263\n1274 1307\n1367 1407\n1419 1425\n1433 1437", "output": "92356" }, { "input": "30 100 48 14 9 7\n26 55\n75 107\n145 146\n158 167\n197 199\n201 213\n222 238\n257 265\n271 338\n357 439\n454 462\n463 562\n633 660\n670 691\n696 829\n844 864\n880 888\n891 894\n895 900\n950 973\n1004 1007\n1018 1021\n1045 1049\n1062 1073\n1138 1184\n1240 1272\n1278 1315\n1403 1410\n1412 1418\n1426 1434", "output": "107490" }, { "input": "30 3 50 32 1 18\n46 60\n66 94\n99 141\n162 239\n309 334\n400 550\n594 630\n659 663\n664 701\n735 736\n741 755\n756 801\n823 854\n865 915\n937 964\n984 986\n990 1007\n1008 1031\n1052 1064\n1073 1080\n1190 1191\n1192 1254\n1265 1266\n1272 1273\n1274 1277\n1284 1302\n1308 1312\n1358 1383\n1388 1389\n1390 1401", "output": "25258" }, { "input": "30 7 52 49 46 37\n2 73\n129 130\n131 140\n143 150\n187 190\n217 238\n248 271\n375 426\n465 470\n475 503\n506 541\n569 571\n578 585\n680 683\n690 738\n742 761\n868 877\n890 892\n904 915\n935 961\n1076 1100\n1103 1114\n1232 1244\n1246 1247\n1257 1261\n1271 1276\n1280 1298\n1346 1360\n1373 1382\n1407 1419", "output": "23992" } ]
1,602,869,939
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
186
307,200
number_of_lines,p1,p2,p3,t1,t2 = map(int,input().split()) di = {} consume = 0 previous_last = -1 for _ in range(number_of_lines): l,r = map(int,input().split()) consume = consume+p1*(r-l) if previous_last == -1: previous_last = r else: difference = l-previous_last if difference<=t1: consume = consume+difference*p1 difference = 0 else: consume = consume+t1*p1 difference = difference - t1 if difference <= t2: consume = consume+difference*p2 difference = 0 else: consume = consume+t2*p2 difference = difference - t2 consume = consume+ difference*p3 print(consume)
Title: Power Consumption Calculation Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*]. Input Specification: The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=&lt;<=*r**i*<=≤<=1440, *r**i*<=&lt;<=*l**i*<=+<=1 for *i*<=&lt;<=*n*), which stand for the start and the end of the *i*-th period of work. Output Specification: Output the answer to the problem. Demo Input: ['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n'] Demo Output: ['30', '570'] Note: none
```python number_of_lines,p1,p2,p3,t1,t2 = map(int,input().split()) di = {} consume = 0 previous_last = -1 for _ in range(number_of_lines): l,r = map(int,input().split()) consume = consume+p1*(r-l) if previous_last == -1: previous_last = r else: difference = l-previous_last if difference<=t1: consume = consume+difference*p1 difference = 0 else: consume = consume+t1*p1 difference = difference - t1 if difference <= t2: consume = consume+difference*p2 difference = 0 else: consume = consume+t2*p2 difference = difference - t2 consume = consume+ difference*p3 print(consume) ```
0
801
A
Vicious Keyboard
PROGRAMMING
1,100
[ "brute force" ]
null
null
Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
[ "VK\n", "VV\n", "V\n", "VKKKKKKKKKVVVVVVVVVK\n", "KVKV\n" ]
[ "1\n", "1\n", "0\n", "3\n", "1\n" ]
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear. For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring. For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
500
[ { "input": "VK", "output": "1" }, { "input": "VV", "output": "1" }, { "input": "V", "output": "0" }, { "input": "VKKKKKKKKKVVVVVVVVVK", "output": "3" }, { "input": "KVKV", "output": "1" }, { "input": "VKKVVVKVKVK", "output": "5" }, { "input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV", "output": "14" }, { "input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK", "output": "32" }, { "input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK", "output": "32" }, { "input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK", "output": "21" }, { "input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV", "output": "25" }, { "input": "KKVVKVVKVVKKVVKKVKVVKKV", "output": "7" }, { "input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV", "output": "24" }, { "input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV", "output": "35" }, { "input": "VKVVVKKKVKVVKVKVKVKVKVV", "output": "9" }, { "input": "KKKKVKKVKVKVKKKVVVVKK", "output": "6" }, { "input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK", "output": "9" }, { "input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK", "output": "22" }, { "input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK", "output": "25" }, { "input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK", "output": "29" }, { "input": "KKVKVVVKKVV", "output": "3" }, { "input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK", "output": "16" }, { "input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV", "output": "13" }, { "input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV", "output": "26" }, { "input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV", "output": "26" }, { "input": "K", "output": "0" }, { "input": "VKVK", "output": "2" }, { "input": "VKVV", "output": "2" }, { "input": "KV", "output": "0" }, { "input": "KK", "output": "1" }, { "input": "KKVK", "output": "2" }, { "input": "KKKK", "output": "1" }, { "input": "KKV", "output": "1" }, { "input": "KKVKVK", "output": "3" }, { "input": "VKKVK", "output": "2" }, { "input": "VKKK", "output": "2" }, { "input": "KKK", "output": "1" }, { "input": "KVV", "output": "1" }, { "input": "KKVKV", "output": "2" }, { "input": "VVK", "output": "1" }, { "input": "VVVKVKVKVKVKVKVK", "output": "8" }, { "input": "KVVVK", "output": "2" }, { "input": "VVVKK", "output": "2" }, { "input": "KKVV", "output": "1" }, { "input": "KKKKKKK", "output": "1" }, { "input": "VKKKVK", "output": "3" }, { "input": "KKVVV", "output": "1" }, { "input": "VVVVVV", "output": "1" }, { "input": "KKKV", "output": "1" }, { "input": "VVKVV", "output": "2" }, { "input": "VKVKKK", "output": "3" }, { "input": "VKKV", "output": "1" }, { "input": "VKKVV", "output": "2" }, { "input": "VVKKVV", "output": "2" }, { "input": "KKVVKKV", "output": "2" }, { "input": "KKKKK", "output": "1" }, { "input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK", "output": "10" }, { "input": "VKVKVV", "output": "3" }, { "input": "VKVVKVV", "output": "3" }, { "input": "VVV", "output": "1" }, { "input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV", "output": "1" }, { "input": "VVKKKKKKVKK", "output": "3" }, { "input": "KVKVKVV", "output": "3" } ]
1,499,077,571
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
6
62
5,529,600
s = input().strip() c = 0 if len(s) > 2: c += s.count('VK') l = s.split('VK') for i in l: if 'VV' in i or 'KK' in i: c += 1 elif len(s) == 2: c = 1 print(c)
Title: Vicious Keyboard Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string. Input Specification: The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100. Output Specification: Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character. Demo Input: ['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n'] Demo Output: ['1\n', '1\n', '0\n', '3\n', '1\n'] Note: For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear. For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring. For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
```python s = input().strip() c = 0 if len(s) > 2: c += s.count('VK') l = s.split('VK') for i in l: if 'VV' in i or 'KK' in i: c += 1 elif len(s) == 2: c = 1 print(c) ```
0
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,633,256,832
2,147,483,647
Python 3
OK
TESTS
40
124
6,758,400
x=input() y=list(input()) y.reverse() z="".join(y) if z==x: print("YES") else:print("NO")
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python x=input() y=list(input()) y.reverse() z="".join(y) if z==x: print("YES") else:print("NO") ```
3.956411
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" } ]
1,678,455,354
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
1
62
2,867,200
n=int(input()) a=list(map(int,input().split())) a=sorted(list(set(a))) for i in range(1,n-1): if a[i+1]-a[i-1]<=2: print('YES') break else: print('NO')
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=int(input()) a=list(map(int,input().split())) a=sorted(list(set(a))) for i in range(1,n-1): if a[i+1]-a[i-1]<=2: print('YES') break else: print('NO') ```
-1
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" } ]
1,598,889,904
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
93
307,200
list1 = [] list2 = [] num = int(input()) for i in range(num): n,m = map(int,input().split()) list1.append(n) list2.append(m) if(len(set(list1))!=len(list1)): print("maybe") elif(list1==list2): print("unrated") else: print("rated")
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 list1 = [] list2 = [] num = int(input()) for i in range(num): n,m = map(int,input().split()) list1.append(n) list2.append(m) if(len(set(list1))!=len(list1)): print("maybe") elif(list1==list2): print("unrated") else: print("rated") ```
0
160
A
Twins
PROGRAMMING
900
[ "greedy", "sortings" ]
null
null
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like. Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally. As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
In the single line print the single number — the minimum needed number of coins.
[ "2\n3 3\n", "3\n2 1 2\n" ]
[ "2\n", "2\n" ]
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum. In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
500
[ { "input": "2\n3 3", "output": "2" }, { "input": "3\n2 1 2", "output": "2" }, { "input": "1\n5", "output": "1" }, { "input": "5\n4 2 2 2 2", "output": "3" }, { "input": "7\n1 10 1 2 1 1 1", "output": "1" }, { "input": "5\n3 2 3 3 1", "output": "3" }, { "input": "2\n2 1", "output": "1" }, { "input": "3\n2 1 3", "output": "2" }, { "input": "6\n1 1 1 1 1 1", "output": "4" }, { "input": "7\n10 10 5 5 5 5 1", "output": "3" }, { "input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1", "output": "8" }, { "input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4", "output": "8" }, { "input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40", "output": "8" }, { "input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90", "output": "6" }, { "input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28", "output": "14" }, { "input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42", "output": "22" }, { "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": "51" }, { "input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2", "output": "37" }, { "input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2", "output": "36" }, { "input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3", "output": "33" }, { "input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2", "output": "30" }, { "input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2", "output": "29" }, { "input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1", "output": "30" }, { "input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52", "output": "29" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100", "output": "30" }, { "input": "1\n1", "output": "1" }, { "input": "4\n4 1 2 3", "output": "2" }, { "input": "5\n2 2 2 4 4", "output": "2" }, { "input": "4\n100 99 99 100", "output": "2" } ]
1,699,385,458
2,147,483,647
Python 3
OK
TESTS
29
62
0
coin_num = int(input()) coin_arr = [int(i) for i in input().split()] coin_arr.sort(reverse = True) total = sum(coin_arr) nijer = 0 counter = 0 while total>=nijer: nijer+= coin_arr[counter] total-=coin_arr[counter] counter+=1 print(counter)
Title: Twins Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like. Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally. As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces. Output Specification: In the single line print the single number — the minimum needed number of coins. Demo Input: ['2\n3 3\n', '3\n2 1 2\n'] Demo Output: ['2\n', '2\n'] Note: In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum. In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
```python coin_num = int(input()) coin_arr = [int(i) for i in input().split()] coin_arr.sort(reverse = True) total = sum(coin_arr) nijer = 0 counter = 0 while total>=nijer: nijer+= coin_arr[counter] total-=coin_arr[counter] counter+=1 print(counter) ```
3
25
A
IQ test
PROGRAMMING
1,300
[ "brute force" ]
A. IQ test
2
256
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
[ "5\n2 4 7 8 10\n", "4\n1 2 1 1\n" ]
[ "3\n", "2\n" ]
none
0
[ { "input": "5\n2 4 7 8 10", "output": "3" }, { "input": "4\n1 2 1 1", "output": "2" }, { "input": "3\n1 2 2", "output": "1" }, { "input": "3\n100 99 100", "output": "2" }, { "input": "3\n5 3 2", "output": "3" }, { "input": "4\n43 28 1 91", "output": "2" }, { "input": "4\n75 13 94 77", "output": "3" }, { "input": "4\n97 8 27 3", "output": "2" }, { "input": "10\n95 51 12 91 85 3 1 31 25 7", "output": "3" }, { "input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46", "output": "4" }, { "input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36", "output": "26" }, { "input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83", "output": "48" }, { "input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25", "output": "1" }, { "input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46", "output": "25" }, { "input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73", "output": "13" }, { "input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76", "output": "99" }, { "input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14", "output": "1" }, { "input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22", "output": "93" }, { "input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23", "output": "1" }, { "input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98", "output": "1" }, { "input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94", "output": "93" }, { "input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29", "output": "32" }, { "input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47", "output": "26" }, { "input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72", "output": "56" }, { "input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86", "output": "97" }, { "input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42", "output": "79" }, { "input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52", "output": "25" }, { "input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86", "output": "78" }, { "input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68", "output": "70" }, { "input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81", "output": "39" }, { "input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83", "output": "88" }, { "input": "3\n100 100 1", "output": "3" } ]
1,665,156,178
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
30
0
t = int(input()) evens = [] odds = [] for x in range(t): lst = input().split() numE = 0 numO = 0 for i in lst: if int(i) % 2 == 0: evens.append(int(i)) numE = lst.index(i) else: odds.append(int(i)) numO = lst.index(i) if len(evens) == 1: print(numE+1) if len(odds) == 1: print(numO+1)
Title: IQ test Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. Output Specification: Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. Demo Input: ['5\n2 4 7 8 10\n', '4\n1 2 1 1\n'] Demo Output: ['3\n', '2\n'] Note: none
```python t = int(input()) evens = [] odds = [] for x in range(t): lst = input().split() numE = 0 numO = 0 for i in lst: if int(i) % 2 == 0: evens.append(int(i)) numE = lst.index(i) else: odds.append(int(i)) numO = lst.index(i) if len(evens) == 1: print(numE+1) if len(odds) == 1: print(numO+1) ```
-1
686
A
Free Ice Cream
PROGRAMMING
800
[ "constructive algorithms", "implementation" ]
null
null
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
[ "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n", "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n" ]
[ "22 1\n", "3 2\n" ]
Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
500
[ { "input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20", "output": "22 1" }, { "input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98", "output": "3 2" }, { "input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000", "output": "7000000000 0" }, { "input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46", "output": "46 0" }, { "input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100", "output": "700 0" }, { "input": "1 0\n- 526403222", "output": "0 1" }, { "input": "1 897986543\n- 371188251", "output": "526798292 0" }, { "input": "1 0\n+ 1", "output": "1 0" }, { "input": "1 0\n- 1", "output": "0 1" }, { "input": "1 10\n+ 10", "output": "20 0" }, { "input": "1 3\n- 5", "output": "3 1" }, { "input": "1 0\n- 5", "output": "0 1" }, { "input": "1 0\n+ 5", "output": "5 0" } ]
1,600,286,608
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
7
155
1,433,600
n , x = list(map(int,input().split(' '))) m = x c = 0 for i in range(n): s = input() if(s[0] == '+'): m += int(s[2:]) else: d = int(s[2:]) if(m > d): m -= d else: c += 1 print(str(m) + " " + str(c))
Title: Free Ice Cream Time Limit: None seconds Memory Limit: None megabytes Problem Description: After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. Input Specification: The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. Output Specification: Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. Demo Input: ['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n'] Demo Output: ['22 1\n', '3 2\n'] Note: Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
```python n , x = list(map(int,input().split(' '))) m = x c = 0 for i in range(n): s = input() if(s[0] == '+'): m += int(s[2:]) else: d = int(s[2:]) if(m > d): m -= d else: c += 1 print(str(m) + " " + str(c)) ```
0
743
B
Chloe and the sequence
PROGRAMMING
1,200
[ "binary search", "bitmasks", "constructive algorithms", "implementation" ]
null
null
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem!
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Print single integer — the integer at the *k*-th position in the obtained sequence.
[ "3 2\n", "4 8\n" ]
[ "2", "4" ]
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
1,000
[ { "input": "3 2", "output": "2" }, { "input": "4 8", "output": "4" }, { "input": "5 27", "output": "1" }, { "input": "7 44", "output": "3" }, { "input": "15 18432", "output": "12" }, { "input": "20 259676", "output": "3" }, { "input": "30 671088640", "output": "28" }, { "input": "38 137438953472", "output": "38" }, { "input": "1 1", "output": "1" }, { "input": "3 3", "output": "1" }, { "input": "4 12", "output": "3" }, { "input": "8 224", "output": "6" }, { "input": "13 1368", "output": "4" }, { "input": "16 49152", "output": "15" }, { "input": "19 1024", "output": "11" }, { "input": "24 15204352", "output": "20" }, { "input": "27 6586544", "output": "5" }, { "input": "31 536870912", "output": "30" }, { "input": "38 94489280512", "output": "34" }, { "input": "42 215268308020", "output": "3" }, { "input": "42 3452074721280", "output": "21" }, { "input": "46 34394312982528", "output": "15" }, { "input": "48 133865540681728", "output": "39" }, { "input": "49 76541041985542", "output": "2" }, { "input": "49 104822971826176", "output": "34" }, { "input": "49 351843720888320", "output": "47" }, { "input": "50 743042492421629", "output": "1" }, { "input": "50 666266740436818", "output": "2" }, { "input": "50 704239287953456", "output": "5" }, { "input": "50 116938486513664", "output": "17" }, { "input": "50 806423059496960", "output": "37" }, { "input": "50 985162418487296", "output": "48" }, { "input": "50 844424930131968", "output": "49" }, { "input": "50 562949953421312", "output": "50" }, { "input": "50 1", "output": "1" }, { "input": "50 1125899906842623", "output": "1" }, { "input": "50 1125899906842620", "output": "3" }, { "input": "39 549755813887", "output": "1" }, { "input": "50 100000000000", "output": "12" } ]
1,647,545,722
2,147,483,647
Python 3
OK
TESTS
39
31
0
n,k = [int(i) for i in input().split(' ')] #ajustamos el index k = k-1 j = 1 while True: #Miramos si es par if k%2 == 0: print(j) break #realizamos unda division entera k = k // 2 j += 1
Title: Chloe and the sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem! Input Specification: The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). Output Specification: Print single integer — the integer at the *k*-th position in the obtained sequence. Demo Input: ['3 2\n', '4 8\n'] Demo Output: ['2', '4'] Note: In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
```python n,k = [int(i) for i in input().split(' ')] #ajustamos el index k = k-1 j = 1 while True: #Miramos si es par if k%2 == 0: print(j) break #realizamos unda division entera k = k // 2 j += 1 ```
3
305
B
Continued Fractions
PROGRAMMING
1,700
[ "brute force", "implementation", "math" ]
null
null
A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal.
The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction. The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Print "YES" if these fractions are equal and "NO" otherwise.
[ "9 4\n2\n2 4\n", "9 4\n3\n2 3 1\n", "9 4\n3\n1 2 4\n" ]
[ "YES\n", "YES\n", "NO\n" ]
In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
1,000
[ { "input": "9 4\n2\n2 4", "output": "YES" }, { "input": "9 4\n3\n2 3 1", "output": "YES" }, { "input": "9 4\n3\n1 2 4", "output": "NO" }, { "input": "39088169 24157817\n36\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2", "output": "YES" }, { "input": "39088169 24157817\n36\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4", "output": "NO" }, { "input": "61305790721611591 37889062373143906\n80\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4", "output": "NO" }, { "input": "61305790721611591 37889062373143906\n80\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2", "output": "YES" }, { "input": "565049485241691020 228217260073568804\n40\n2 2 9 1 7 1 2 1 2 1 1 1 9 1 2 1 9 1 3 2 3 10 13 2 1 2 7 1 1 2 2 2 1 1 2 1 6 5 3 2", "output": "YES" }, { "input": "2 1\n4\n2 1 1 1", "output": "NO" }, { "input": "4 1\n2\n3 1", "output": "YES" }, { "input": "72723460248141 1597\n1\n45537545554", "output": "NO" }, { "input": "14930352 13\n6\n1148488 1 1 1 1 2", "output": "YES" }, { "input": "86267571272 102334155\n6\n842 1 841 1 842 145", "output": "NO" }, { "input": "72723460248141 121393\n7\n599074578 122 1 122 2 1 2", "output": "YES" }, { "input": "168455988218483660 53310571951833359\n32\n3 6 3 1 14 1 48 1 3 2 1 1 39 2 1 3 13 23 4 1 11 1 1 23 1 3 3 2 1 1 1 3", "output": "NO" }, { "input": "382460255113156464 275525972692563593\n37\n1 2 1 1 2 1 3 4 5 5 1 4 2 1 1 1 4 2 2 1 2 1 1 2 3 3 1 2 2 50 4 1 4 2 5 109 8", "output": "YES" }, { "input": "1000000000000000000 1\n1\n1000000000000000000", "output": "YES" }, { "input": "362912509915545727 266073193475139553\n30\n1 2 1 2 1 25 75 1 14 6 6 9 1 1 1 1 210 2 2 2 5 2 1 3 1 1 13 3 14 3", "output": "NO" }, { "input": "933329105990871495 607249523603826772\n33\n1 1 1 6 3 1 5 24 3 55 1 15 2 2 1 12 2 2 3 109 1 1 4 1 4 1 7 2 4 1 3 3 2", "output": "YES" }, { "input": "790637895857383456 679586240913926415\n40\n1 6 8 2 1 2 1 7 2 4 1 1 1 10 1 10 1 4 1 4 41 1 1 7 1 1 2 1 2 4 1 2 1 63 1 2 1 1 4 3", "output": "NO" }, { "input": "525403371166594848 423455864168639615\n38\n1 4 6 1 1 32 3 1 14 1 3 1 2 4 5 4 1 2 1 5 8 1 3 1 2 1 46 1 1 1 3 1 4 1 11 1 2 4", "output": "YES" }, { "input": "1 1\n1\n1", "output": "YES" }, { "input": "2 1\n2\n1 2", "output": "NO" }, { "input": "531983955813463755 371380136962341468\n38\n1 2 3 4 1 37 1 12 1 3 2 1 6 3 1 7 3 2 8 1 2 1 1 7 1 1 1 7 1 47 2 1 3 1 1 5 1 2", "output": "YES" }, { "input": "32951280099 987\n7\n33385288 1 5 1 5 1 6", "output": "YES" }, { "input": "6557470319842 86267571272\n6\n76 76 76 76 76 76", "output": "YES" }, { "input": "934648630114363087 6565775686518446\n31\n142 2 1 5 2 2 1 1 3 1 2 8 1 3 12 2 1 23 5 1 10 1 863 1 1 1 2 1 14 2 3", "output": "YES" }, { "input": "61305790721611591 37889062373143906\n81\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "4 1\n1\n4", "output": "YES" }, { "input": "500000000000000001 5\n2\n100000000000000000 5", "output": "YES" }, { "input": "1000000000000000000 3\n3\n3 4 5", "output": "NO" }, { "input": "822981258385599125 28316248989464296\n39\n29 15 1 1 1 4 4 4 1 3 1 5 12 1 1 1 1 1 6 5 2 1 11 1 1 26 1 2 2 2 14 1 1 1 3 2 4 1 1", "output": "NO" }, { "input": "823443107025550834 331822464812968648\n42\n2 2 13 14 4 4 1 1 1 1 2 1 1 1 1 113 1 1 8 1 1 1 1 2 2 1 15 1 5 1 1 2 1 1 1 14 4 3 1 5 1 1", "output": "NO" }, { "input": "226137305050296073 27076290603746056\n30\n8 2 1 5 3 67 2 1 6 1 2 1 5 1 11 8 43 2 1 7 1 95 2 3 1 11 5 2 1 1", "output": "NO" }, { "input": "524928871965838747 313083111434773473\n35\n1 1 2 10 1 4 12 3 28 1 23 1 1 1 4 1 4 3 1 3 2 3 1 4 3 1 3 2 3 11 21 1 35 1 1", "output": "NO" }, { "input": "633468529243155234 4\n90\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "742143496299253703 2\n90\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "550736960584023286 3\n90\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "2 1\n10\n99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999 99999999999999999", "output": "NO" }, { "input": "262882295792523313 105000000000078855\n1\n105000000000078855", "output": "NO" }, { "input": "990130967049151695 166430169817556175\n1\n564668656008429569", "output": "NO" }, { "input": "9 4\n2\n2 3", "output": "NO" }, { "input": "529824479480396864 4705882352941177\n2\n80000000000000007 80000000000000009", "output": "NO" }, { "input": "985625905209512860 565433601688714177\n10\n6423 24947 27507 13031 16414 29169 901 32592 18763 1656", "output": "NO" }, { "input": "913255926290448385 4400000000\n2\n4400000000 4400000000", "output": "NO" }, { "input": "7 2\n2\n2 1", "output": "NO" }, { "input": "10 3\n1\n3", "output": "NO" }, { "input": "4 2\n1\n2", "output": "YES" }, { "input": "1337 42\n1\n31", "output": "NO" } ]
1,368,969,792
1,392
Python 3
OK
TESTS
49
78
102,400
p,q=tuple(map(int,input().split())) n1=input().split() n=int(n1[0]) a=list(map(int,input().split())) x=1 y=a[n-1] for i in range (n-2,-1,-1): x1=a[i]*y+x x=y y=x1 if p*x==q*y: print('YES') else: print('NO')
Title: Continued Fractions Time Limit: None seconds Memory Limit: None megabytes Problem Description: A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal. Input Specification: The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction. The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print "YES" if these fractions are equal and "NO" otherwise. Demo Input: ['9 4\n2\n2 4\n', '9 4\n3\n2 3 1\n', '9 4\n3\n1 2 4\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
```python p,q=tuple(map(int,input().split())) n1=input().split() n=int(n1[0]) a=list(map(int,input().split())) x=1 y=a[n-1] for i in range (n-2,-1,-1): x1=a[i]*y+x x=y y=x1 if p*x==q*y: print('YES') else: print('NO') ```
3
270
A
Fancy Fence
PROGRAMMING
1,100
[ "geometry", "implementation", "math" ]
null
null
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
The first line of input contains an integer *t* (0<=&lt;<=*t*<=&lt;<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=&lt;<=*a*<=&lt;<=180) — the angle the robot can make corners at measured in degrees.
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
[ "3\n30\n60\n90\n" ]
[ "NO\nYES\nYES\n" ]
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.
500
[ { "input": "3\n30\n60\n90", "output": "NO\nYES\nYES" }, { "input": "6\n1\n2\n3\n170\n179\n25", "output": "NO\nNO\nNO\nYES\nYES\nNO" } ]
1,632,821,223
2,147,483,647
Python 3
OK
TESTS
3
92
6,758,400
a=int(input()) list=[] while a!=0: b=int(input()) if 360%(180-b)!=0: list.append("NO") else: list.append("YES") a=a-1 for i in list: print(i)
Title: Fancy Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? Input Specification: The first line of input contains an integer *t* (0<=&lt;<=*t*<=&lt;<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=&lt;<=*a*<=&lt;<=180) — the angle the robot can make corners at measured in degrees. Output Specification: For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. Demo Input: ['3\n30\n60\n90\n'] Demo Output: ['NO\nYES\nYES\n'] Note: In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.
```python a=int(input()) list=[] while a!=0: b=int(input()) if 360%(180-b)!=0: list.append("NO") else: list.append("YES") a=a-1 for i in list: print(i) ```
3
869
B
The Eternal Immortality
PROGRAMMING
1,100
[ "math" ]
null
null
Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
[ "2 4\n", "0 10\n", "107 109\n" ]
[ "2\n", "0\n", "2\n" ]
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
1,000
[ { "input": "2 4", "output": "2" }, { "input": "0 10", "output": "0" }, { "input": "107 109", "output": "2" }, { "input": "10 13", "output": "6" }, { "input": "998244355 998244359", "output": "4" }, { "input": "999999999000000000 1000000000000000000", "output": "0" }, { "input": "2 3", "output": "3" }, { "input": "3 15", "output": "0" }, { "input": "24 26", "output": "0" }, { "input": "14 60", "output": "0" }, { "input": "11 79", "output": "0" }, { "input": "1230 1232", "output": "2" }, { "input": "2633 2634", "output": "4" }, { "input": "535 536", "output": "6" }, { "input": "344319135 396746843", "output": "0" }, { "input": "696667767 696667767", "output": "1" }, { "input": "419530302 610096911", "output": "0" }, { "input": "238965115 821731161", "output": "0" }, { "input": "414626436 728903812", "output": "0" }, { "input": "274410639 293308324", "output": "0" }, { "input": "650636673091305697 650636673091305702", "output": "0" }, { "input": "651240548333620923 651240548333620924", "output": "4" }, { "input": "500000000000000000 1000000000000000000", "output": "0" }, { "input": "999999999999999999 1000000000000000000", "output": "0" }, { "input": "1000000000000000000 1000000000000000000", "output": "1" }, { "input": "0 4", "output": "4" }, { "input": "50000000062000007 50000000062000011", "output": "0" }, { "input": "0 0", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "0 2", "output": "2" }, { "input": "10000000000012 10000000000015", "output": "0" }, { "input": "5 5", "output": "1" }, { "input": "12 23", "output": "0" }, { "input": "0 11", "output": "0" }, { "input": "11111234567890 11111234567898", "output": "0" }, { "input": "0 3", "output": "6" }, { "input": "1 2", "output": "2" }, { "input": "999999999999999997 999999999999999999", "output": "2" }, { "input": "4 5", "output": "5" }, { "input": "0 1", "output": "1" }, { "input": "101 1002", "output": "0" }, { "input": "0 100000000000000001", "output": "0" }, { "input": "99999999999999997 99999999999999999", "output": "2" }, { "input": "14 15", "output": "5" }, { "input": "8 19", "output": "0" }, { "input": "12 22", "output": "0" }, { "input": "999999999999996 999999999999999", "output": "4" }, { "input": "1 3", "output": "6" }, { "input": "124 125", "output": "5" }, { "input": "11 32", "output": "0" }, { "input": "0 5", "output": "0" }, { "input": "0 999999", "output": "0" }, { "input": "151151151515 151151151526", "output": "0" }, { "input": "6 107", "output": "0" }, { "input": "5 16", "output": "0" }, { "input": "7 16", "output": "0" }, { "input": "6 19", "output": "0" }, { "input": "11113111111111 13111111111111", "output": "0" }, { "input": "1 1000", "output": "0" }, { "input": "24 25", "output": "5" }, { "input": "0 100000000000", "output": "0" }, { "input": "1 22", "output": "0" }, { "input": "999999999999999996 999999999999999999", "output": "4" } ]
1,671,809,789
2,147,483,647
Python 3
OK
TESTS
63
46
0
c = 1 x , y = [int(i) for i in input().split()] # input -> x , input -> y if(y - x >= 5): # Here ?? print(0) # like 0 10 else: for _ in range(x + 1, y + 1): # 1 2 3 4 5 6 7 8 9 10 11 == num > num[-1] c *= _ print(c % 10) # num[-1] =>= c % 10 =>= str(c)[-1]
Title: The Eternal Immortality Time Limit: None seconds Memory Limit: None megabytes Problem Description: Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge. Input Specification: The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018). Output Specification: Output one line containing a single decimal digit — the last digit of the value that interests Koyomi. Demo Input: ['2 4\n', '0 10\n', '107 109\n'] Demo Output: ['2\n', '0\n', '2\n'] Note: In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
```python c = 1 x , y = [int(i) for i in input().split()] # input -> x , input -> y if(y - x >= 5): # Here ?? print(0) # like 0 10 else: for _ in range(x + 1, y + 1): # 1 2 3 4 5 6 7 8 9 10 11 == num > num[-1] c *= _ print(c % 10) # num[-1] =>= c % 10 =>= str(c)[-1] ```
3
595
A
Vitaly and Night
PROGRAMMING
800
[ "constructive algorithms", "implementation" ]
null
null
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment. Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on. Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively. Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
[ "2 2\n0 0 0 1\n1 0 1 1\n", "1 3\n1 1 0 1 0 0\n" ]
[ "3\n", "2\n" ]
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off. In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
500
[ { "input": "2 2\n0 0 0 1\n1 0 1 1", "output": "3" }, { "input": "1 3\n1 1 0 1 0 0", "output": "2" }, { "input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1", "output": "8" }, { "input": "1 5\n1 0 1 1 1 0 1 1 1 1", "output": "5" }, { "input": "1 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99" }, { "input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "6" }, { "input": "1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "100" }, { "input": "100 1\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0", "output": "8" }, { "input": "100 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "1 1\n0 0", "output": "0" }, { "input": "1 1\n0 1", "output": "1" }, { "input": "1 1\n1 0", "output": "1" }, { "input": "1 1\n1 1", "output": "1" } ]
1,553,338,530
2,147,483,647
Python 3
OK
TESTS
36
124
102,400
# import sys # sys.stdin=open("input.in",'r') # sys.stdout=open("outp.out",'w') n,m=map(int,input().split()) c=0 for i in range(n): s=list(map(int,input().split())) x=0 while x<2*m: if s[x]==1 or s[x+1]==1: c+=1 x+=2 print(c)
Title: Vitaly and Night Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment. Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on. Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping. Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively. Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'. Output Specification: Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping. Demo Input: ['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n'] Demo Output: ['3\n', '2\n'] Note: In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off. In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
```python # import sys # sys.stdin=open("input.in",'r') # sys.stdout=open("outp.out",'w') n,m=map(int,input().split()) c=0 for i in range(n): s=list(map(int,input().split())) x=0 while x<2*m: if s[x]==1 or s[x+1]==1: c+=1 x+=2 print(c) ```
3
0
none
none
none
0
[ "none" ]
null
null
Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order. After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in lexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical! She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order. Lexicographical order is defined in following way. When we compare *s* and *t*, first we find the leftmost position with differing characters: *s**i*<=≠<=*t**i*. If there is no such position (i. e. *s* is a prefix of *t* or vice versa) the shortest string is less. Otherwise, we compare characters *s**i* and *t**i* according to their order in alphabet.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100): number of names. Each of the following *n* lines contain one string *name**i* (1<=≤<=|*name**i*|<=≤<=100), the *i*-th name. Each name contains only lowercase Latin letters. All names are different.
If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on). Otherwise output a single word "Impossible" (without quotes).
[ "3\nrivest\nshamir\nadleman\n", "10\ntourist\npetr\nwjmzbmr\nyeputons\nvepifanov\nscottwu\noooooooooooooooo\nsubscriber\nrowdark\ntankengineer\n", "10\npetr\negor\nendagorion\nfeferivan\nilovetanyaromanova\nkostka\ndmitriyh\nmaratsnowbear\nbredorjaguarturnik\ncgyforever\n", "7\ncar\ncare\ncareful\ncarefully\nbecarefuldontforgetsomething\notherwiseyouwillbehacked\ngoodluck\n" ]
[ "bcdefghijklmnopqrsatuvwxyz\n", "Impossible\n", "aghjlnopefikdmbcqrstuvwxyz\n", "acbdefhijklmnogpqrstuvwxyz\n" ]
none
0
[ { "input": "3\nrivest\nshamir\nadleman", "output": "bcdefghijklmnopqrsatuvwxyz" }, { "input": "10\ntourist\npetr\nwjmzbmr\nyeputons\nvepifanov\nscottwu\noooooooooooooooo\nsubscriber\nrowdark\ntankengineer", "output": "Impossible" }, { "input": "10\npetr\negor\nendagorion\nfeferivan\nilovetanyaromanova\nkostka\ndmitriyh\nmaratsnowbear\nbredorjaguarturnik\ncgyforever", "output": "aghjlnopefikdmbcqrstuvwxyz" }, { "input": "7\ncar\ncare\ncareful\ncarefully\nbecarefuldontforgetsomething\notherwiseyouwillbehacked\ngoodluck", "output": "acbdefhijklmnogpqrstuvwxyz" }, { "input": "2\na\naa", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "6\nax\nay\nby\nbz\ncz\ncx", "output": "Impossible" }, { "input": "4\nax\nay\nby\nbx", "output": "Impossible" }, { "input": "4\nax\nay\nby\nbz", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "1\na", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "1\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "2\naa\na", "output": "Impossible" }, { "input": "5\naaaaa\naaaa\naaa\naa\na", "output": "Impossible" }, { "input": "2\nanud\nanu", "output": "Impossible" }, { "input": "8\nwa\nwb\nxc\nxd\nyb\nyc\nzd\nza", "output": "Impossible" } ]
1,598,086,755
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
0
109
20,172,800
#for i in range(int(input())): n=int(input()) g=[[] for i in range(26)] in_degree=[0]*26 s=input() f=0 for i in range(n-1): t=input() if len(s)>len(t) and s[:len(t)]==t: f=1 break for j in range(min(len(t),len(s))): if t[j]!=s[j]: g[ord(s[j])-97].append(t[j]) break s=t for i in range(26): for j in g[i]: in_degree[ord(j)-97]+=1 q=[] for i in range(26): if in_degree[i]==0: q.append(i) cnt=0 top_order=[] while q: u=q.pop() #print("u",u) top_order.append(chr(97+u)) for i in g[u]: in_degree[ord(i)-97]-=1 if in_degree[ord(i)-97]==0: q.append(ord(i)-97) cnt+=1 if cnt!=26: print("Impossible") else: top_order=top_order[::-1] print("".join(top_order))
Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order. After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in lexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical! She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order. Lexicographical order is defined in following way. When we compare *s* and *t*, first we find the leftmost position with differing characters: *s**i*<=≠<=*t**i*. If there is no such position (i. e. *s* is a prefix of *t* or vice versa) the shortest string is less. Otherwise, we compare characters *s**i* and *t**i* according to their order in alphabet. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100): number of names. Each of the following *n* lines contain one string *name**i* (1<=≤<=|*name**i*|<=≤<=100), the *i*-th name. Each name contains only lowercase Latin letters. All names are different. Output Specification: If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on). Otherwise output a single word "Impossible" (without quotes). Demo Input: ['3\nrivest\nshamir\nadleman\n', '10\ntourist\npetr\nwjmzbmr\nyeputons\nvepifanov\nscottwu\noooooooooooooooo\nsubscriber\nrowdark\ntankengineer\n', '10\npetr\negor\nendagorion\nfeferivan\nilovetanyaromanova\nkostka\ndmitriyh\nmaratsnowbear\nbredorjaguarturnik\ncgyforever\n', '7\ncar\ncare\ncareful\ncarefully\nbecarefuldontforgetsomething\notherwiseyouwillbehacked\ngoodluck\n'] Demo Output: ['bcdefghijklmnopqrsatuvwxyz\n', 'Impossible\n', 'aghjlnopefikdmbcqrstuvwxyz\n', 'acbdefhijklmnogpqrstuvwxyz\n'] Note: none
```python #for i in range(int(input())): n=int(input()) g=[[] for i in range(26)] in_degree=[0]*26 s=input() f=0 for i in range(n-1): t=input() if len(s)>len(t) and s[:len(t)]==t: f=1 break for j in range(min(len(t),len(s))): if t[j]!=s[j]: g[ord(s[j])-97].append(t[j]) break s=t for i in range(26): for j in g[i]: in_degree[ord(j)-97]+=1 q=[] for i in range(26): if in_degree[i]==0: q.append(i) cnt=0 top_order=[] while q: u=q.pop() #print("u",u) top_order.append(chr(97+u)) for i in g[u]: in_degree[ord(i)-97]-=1 if in_degree[ord(i)-97]==0: q.append(ord(i)-97) cnt+=1 if cnt!=26: print("Impossible") else: top_order=top_order[::-1] print("".join(top_order)) ```
0
44
A
Indian Summer
PROGRAMMING
900
[ "implementation" ]
A. Indian Summer
2
256
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
Output the single number — the number of Alyona's leaves.
[ "5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n", "3\noak yellow\noak yellow\noak yellow\n" ]
[ "4\n", "1\n" ]
none
0
[ { "input": "5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green", "output": "4" }, { "input": "3\noak yellow\noak yellow\noak yellow", "output": "1" }, { "input": "5\nxbnbkzn hp\nkaqkl vrgzbvqstu\nj aqidx\nhos gyul\nwefxmh tygpluae", "output": "5" }, { "input": "1\nqvwli hz", "output": "1" }, { "input": "4\nsrhk x\nsrhk x\nqfoe vnrjuab\nqfoe vnrjuab", "output": "2" }, { "input": "4\nsddqllmmpk syded\nfprsq fnenjnaz\nn hdej\nsddqllmmpk syded", "output": "3" }, { "input": "17\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw", "output": "1" }, { "input": "18\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nb d\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp", "output": "2" }, { "input": "13\nsvpzwtwn rykzfdce\nqweiyeck jkreouy\nhk nnli\ntwxrnbbdt vtuv\nxokqjtylly sz\nesdt dbfidjslq\ng ybqgomvw\nxcpfjmf zcqvz\nifajadhj invzueip\nvdemdnxifb hckwebmi\nsdpnhipam wvowzavh\nuqdlfskhgo vunbpghae\ne dtigwnb", "output": "13" }, { "input": "20\nm vkfh\nvptikamead mvx\nitu mf\nklruxckw aqega\nekqkd enjllwol\ncc uybfdh\nimrfdngvo u\ne uh\ntwt jsslcfuogk\nbljwqsag tuqbdn\nqcv q\nasx gzhvwwmajj\nqcv q\nekqkd enjllwol\nasx gzhvwwmajj\nks vv\nkzyfi cn\ncc uybfdh\nitu mf\ncjbjhtbyvk vatwfmux", "output": "15" }, { "input": "2\nab ab\na bab", "output": "2" } ]
1,647,105,706
2,147,483,647
Python 3
OK
TESTS
29
92
0
ll=set() for _ in range(int(input())): ll.add(input()) print(len(ll))
Title: Indian Summer Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space. Output Specification: Output the single number — the number of Alyona's leaves. Demo Input: ['5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n', '3\noak yellow\noak yellow\noak yellow\n'] Demo Output: ['4\n', '1\n'] Note: none
```python ll=set() for _ in range(int(input())): ll.add(input()) print(len(ll)) ```
3.977
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,600,049,392
2,147,483,647
Python 3
OK
TESTS
40
218
0
s = input() t = input() tl = len(t)-1 ans = 1 for i in range(len(s)): if(s[i] != t[tl]): ans = 0 break tl -= 1 if(ans == 1): print("YES") else: print("NO")
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python s = input() t = input() tl = len(t)-1 ans = 1 for i in range(len(s)): if(s[i] != t[tl]): ans = 0 break tl -= 1 if(ans == 1): print("YES") else: print("NO") ```
3.9455
460
A
Vasya and Socks
PROGRAMMING
900
[ "brute force", "implementation", "math" ]
null
null
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Print a single integer — the answer to the problem.
[ "2 2\n", "9 3\n" ]
[ "3\n", "13\n" ]
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two. In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
500
[ { "input": "2 2", "output": "3" }, { "input": "9 3", "output": "13" }, { "input": "1 2", "output": "1" }, { "input": "2 3", "output": "2" }, { "input": "1 99", "output": "1" }, { "input": "4 4", "output": "5" }, { "input": "10 2", "output": "19" }, { "input": "10 9", "output": "11" }, { "input": "100 100", "output": "101" }, { "input": "2 27", "output": "2" }, { "input": "99 100", "output": "99" }, { "input": "99 2", "output": "197" }, { "input": "100 3", "output": "149" }, { "input": "98 3", "output": "146" }, { "input": "4 4", "output": "5" }, { "input": "100 2", "output": "199" }, { "input": "62 4", "output": "82" }, { "input": "99 10", "output": "109" }, { "input": "100 5", "output": "124" }, { "input": "80 80", "output": "81" }, { "input": "95 16", "output": "101" }, { "input": "75 16", "output": "79" }, { "input": "99 74", "output": "100" }, { "input": "20 21", "output": "20" }, { "input": "52 96", "output": "52" }, { "input": "24 5", "output": "29" } ]
1,676,578,704
2,147,483,647
Python 3
OK
TESTS
26
46
0
n , m = map(int,input().split()) days = 0 while n!=0: days+=1 if days%m==0: pass else: n-=1 print(days)
Title: Vasya and Socks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks? Input Specification: The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space. Output Specification: Print a single integer — the answer to the problem. Demo Input: ['2 2\n', '9 3\n'] Demo Output: ['3\n', '13\n'] Note: In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two. In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
```python n , m = map(int,input().split()) days = 0 while n!=0: days+=1 if days%m==0: pass else: n-=1 print(days) ```
3
747
A
Display Size
PROGRAMMING
800
[ "brute force", "math" ]
null
null
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels. Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that: - there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
Print two integers — the number of rows and columns on the display.
[ "8\n", "64\n", "5\n", "999999\n" ]
[ "2 4\n", "8 8\n", "1 5\n", "999 1001\n" ]
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels. In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels. In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
500
[ { "input": "8", "output": "2 4" }, { "input": "64", "output": "8 8" }, { "input": "5", "output": "1 5" }, { "input": "999999", "output": "999 1001" }, { "input": "716539", "output": "97 7387" }, { "input": "1", "output": "1 1" }, { "input": "2", "output": "1 2" }, { "input": "3", "output": "1 3" }, { "input": "4", "output": "2 2" }, { "input": "6", "output": "2 3" }, { "input": "7", "output": "1 7" }, { "input": "9", "output": "3 3" }, { "input": "10", "output": "2 5" }, { "input": "11", "output": "1 11" }, { "input": "12", "output": "3 4" }, { "input": "15", "output": "3 5" }, { "input": "100", "output": "10 10" }, { "input": "101", "output": "1 101" }, { "input": "169", "output": "13 13" }, { "input": "179", "output": "1 179" }, { "input": "190", "output": "10 19" }, { "input": "1000", "output": "25 40" }, { "input": "10000", "output": "100 100" }, { "input": "10001", "output": "73 137" }, { "input": "100000", "output": "250 400" }, { "input": "100001", "output": "11 9091" }, { "input": "1000000", "output": "1000 1000" }, { "input": "999983", "output": "1 999983" }, { "input": "524288", "output": "512 1024" }, { "input": "954493", "output": "971 983" }, { "input": "966289", "output": "983 983" }, { "input": "944663", "output": "961 983" }, { "input": "912673", "output": "97 9409" }, { "input": "732641", "output": "679 1079" }, { "input": "232897", "output": "343 679" }, { "input": "16807", "output": "49 343" }, { "input": "999958", "output": "2 499979" }, { "input": "990151", "output": "1 990151" }, { "input": "997002", "output": "998 999" }, { "input": "20", "output": "4 5" }, { "input": "20261", "output": "1 20261" }, { "input": "999123", "output": "3 333041" }, { "input": "901841", "output": "1 901841" } ]
1,660,334,531
2,147,483,647
PyPy 3
OK
TESTS
43
77
0
from cmath import sqrt def solves(): n=int(input()) m=1 temp=int(n**0.5) for i in range(1,temp+1): if n%i==0: m=max(m,i) print(m,n//m) for _ in range(1): solves()
Title: Display Size Time Limit: None seconds Memory Limit: None megabytes Problem Description: A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels. Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that: - there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible. Input Specification: The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have. Output Specification: Print two integers — the number of rows and columns on the display. Demo Input: ['8\n', '64\n', '5\n', '999999\n'] Demo Output: ['2 4\n', '8 8\n', '1 5\n', '999 1001\n'] Note: In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels. In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels. In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.
```python from cmath import sqrt def solves(): n=int(input()) m=1 temp=int(n**0.5) for i in range(1,temp+1): if n%i==0: m=max(m,i) print(m,n//m) for _ in range(1): solves() ```
3
755
B
PolandBall and Game
PROGRAMMING
1,100
[ "binary search", "data structures", "games", "greedy", "sortings", "strings" ]
null
null
PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses. You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally?
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively. Then *n* strings follow, one per line — words familiar to PolandBall. Then *m* strings follow, one per line — words familiar to EnemyBall. Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players. Each word is non-empty and consists of no more than 500 lowercase English alphabet letters.
In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally.
[ "5 1\npolandball\nis\na\ncool\ncharacter\nnope\n", "2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n", "1 2\na\na\nb\n" ]
[ "YES", "YES", "NO" ]
In the first example PolandBall knows much more words and wins effortlessly. In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins.
1,000
[ { "input": "5 1\npolandball\nis\na\ncool\ncharacter\nnope", "output": "YES" }, { "input": "2 2\nkremowka\nwadowicka\nkremowka\nwiedenska", "output": "YES" }, { "input": "1 2\na\na\nb", "output": "NO" }, { "input": "2 2\na\nb\nb\nc", "output": "YES" }, { "input": "2 1\nc\na\na", "output": "YES" }, { "input": "3 3\nab\nbc\ncd\ncd\ndf\nfg", "output": "YES" }, { "input": "3 3\nc\na\nb\na\nd\ng", "output": "YES" }, { "input": "1 1\naa\naa", "output": "YES" }, { "input": "2 1\na\nb\na", "output": "YES" }, { "input": "6 5\na\nb\nc\nd\ne\nf\nf\ne\nd\nz\ny", "output": "YES" }, { "input": "3 2\na\nb\nc\nd\ne", "output": "YES" } ]
1,673,113,537
2,147,483,647
Python 3
OK
TESTS
33
46
512,000
import sys input = lambda: sys.stdin.readline().strip() n, m = map(int, input().split()) pol, com = set(), 0 for _ in range(n): pol.add(input()) for _ in range(m): com += input() in pol n += com & 1 if n > m: print("YES") else: print("NO")
Title: PolandBall and Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses. You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally? Input Specification: The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively. Then *n* strings follow, one per line — words familiar to PolandBall. Then *m* strings follow, one per line — words familiar to EnemyBall. Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players. Each word is non-empty and consists of no more than 500 lowercase English alphabet letters. Output Specification: In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally. Demo Input: ['5 1\npolandball\nis\na\ncool\ncharacter\nnope\n', '2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n', '1 2\na\na\nb\n'] Demo Output: ['YES', 'YES', 'NO'] Note: In the first example PolandBall knows much more words and wins effortlessly. In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins.
```python import sys input = lambda: sys.stdin.readline().strip() n, m = map(int, input().split()) pol, com = set(), 0 for _ in range(n): pol.add(input()) for _ in range(m): com += input() in pol n += com & 1 if n > m: print("YES") else: print("NO") ```
3
1,010
A
Fly
PROGRAMMING
1,500
[ "binary search", "math" ]
null
null
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$. Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet. The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet. For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons. Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well. Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets. The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload. The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel. The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel. It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$. It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel. The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$.
[ "2\n12\n11 8\n7 5\n", "3\n1\n1 4 1\n2 5 3\n", "6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n" ]
[ "10.0000000000\n", "-1\n", "85.4800000000\n" ]
Let's consider the first example. Initially, the mass of a rocket with fuel is $22$ tons. - At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel. In the second case, the rocket will not be able even to take off from Earth.
500
[ { "input": "2\n12\n11 8\n7 5", "output": "10.0000000000" }, { "input": "3\n1\n1 4 1\n2 5 3", "output": "-1" }, { "input": "6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3", "output": "85.4800000000" }, { "input": "3\n3\n1 2 1\n2 2 2", "output": "-1" }, { "input": "4\n4\n2 3 2 2\n2 3 4 3", "output": "284.0000000000" }, { "input": "5\n2\n1 2 2 1 2\n4 5 1 4 1", "output": "-1" }, { "input": "7\n7\n3 2 6 2 2 2 5\n4 7 5 6 2 2 2", "output": "4697.0000000000" }, { "input": "2\n1000\n12 34\n56 78", "output": "159.2650775220" }, { "input": "8\n4\n1 1 4 1 3 1 8 1\n1 1 1 1 1 3 1 2", "output": "-1" }, { "input": "9\n2\n8 7 1 1 3 7 1 2 4\n4 1 1 8 7 7 1 1 5", "output": "-1" }, { "input": "10\n10\n9 8 8 7 2 10 2 9 2 4\n3 10 6 2 6 6 5 9 4 5", "output": "3075.7142857143" }, { "input": "20\n12\n3 9 12 13 16 18 9 9 19 7 2 5 17 14 7 7 15 16 5 7\n16 9 13 5 14 10 4 3 16 16 12 20 17 11 4 5 5 14 6 15", "output": "4670.8944493007" }, { "input": "30\n5\n25 1 28 1 27 25 24 1 28 1 12 1 29 16 1 1 1 1 27 1 24 1 1 1 1 1 1 1 30 3\n1 22 1 1 24 2 13 1 16 21 1 27 14 16 1 1 7 1 1 18 1 23 10 1 15 16 16 15 10 1", "output": "-1" }, { "input": "40\n13\n1 1 1 23 21 1 1 1 1 1 40 32 1 21 1 8 1 1 36 15 33 1 30 1 1 37 22 1 4 39 7 1 9 37 1 1 1 28 1 1\n1 34 17 1 38 20 8 14 1 18 29 3 21 21 18 14 1 11 1 1 23 1 25 1 14 1 7 31 9 20 25 1 1 1 1 8 26 12 1 1", "output": "-1" }, { "input": "50\n19\n17 7 13 42 19 25 10 25 2 36 17 40 30 48 34 43 34 20 5 15 8 7 43 35 21 40 40 19 30 11 49 7 24 23 43 30 38 49 10 8 30 11 28 50 48 25 25 20 48 24\n49 35 10 22 24 50 50 7 6 13 16 35 12 43 50 44 35 33 38 49 26 18 23 37 7 38 23 20 28 48 41 16 6 32 32 34 11 39 38 9 38 23 16 31 37 47 33 20 46 30", "output": "7832.1821424977" }, { "input": "60\n21\n11 35 1 28 39 13 19 56 13 13 21 25 1 1 23 1 52 26 53 1 1 1 30 39 1 7 1 1 3 1 1 10 1 1 37 1 1 25 1 1 1 53 1 3 48 1 6 5 4 15 1 14 25 53 25 38 27 1 1 1\n1 1 1 35 40 58 10 22 1 56 1 59 1 6 33 1 1 1 1 18 14 1 1 40 25 47 1 34 1 1 53 1 1 25 1 45 1 1 25 34 3 1 1 1 53 27 11 58 1 1 1 10 12 1 1 1 31 52 1 1", "output": "-1" }, { "input": "70\n69\n70 66 57 58 24 60 39 2 48 61 65 22 10 26 68 62 48 25 12 14 45 57 6 30 48 15 46 33 42 28 69 42 64 25 24 8 62 12 68 53 55 20 32 70 3 5 41 49 16 26 2 34 34 20 39 65 18 47 62 31 39 28 61 67 7 14 31 31 53 54\n40 33 24 20 68 20 22 39 53 56 48 38 59 45 47 46 7 69 11 58 61 40 35 38 62 66 18 36 44 48 67 24 14 27 67 63 68 30 50 6 58 7 6 35 20 58 6 12 12 23 14 2 63 27 29 22 49 16 55 40 70 27 27 70 42 38 66 55 69 47", "output": "217989.4794743629" }, { "input": "80\n21\n65 4 26 25 1 1 1 1 1 1 60 1 29 43 48 6 48 13 29 1 1 62 1 1 1 1 1 1 1 26 9 1 22 1 35 13 66 36 1 1 1 38 55 21 70 1 58 70 1 1 38 1 1 20 1 1 51 1 1 28 1 23 11 1 39 47 1 52 41 1 63 1 1 52 1 45 11 10 80 1\n1 1 25 30 1 1 55 54 1 48 10 37 22 1 74 1 78 13 1 65 32 1 1 1 1 69 5 59 1 1 65 1 40 1 31 1 1 75 54 1 60 1 1 1 1 1 1 1 11 29 36 1 72 71 52 1 1 1 37 1 1 75 43 9 53 1 62 1 29 1 40 27 59 74 41 53 19 30 1 73", "output": "-1" }, { "input": "90\n35\n1 68 16 30 24 1 1 1 35 1 1 67 1 1 1 1 33 16 37 77 83 1 77 26 1 1 68 67 70 62 1 47 1 1 1 84 1 65 1 32 83 1 1 1 28 1 71 76 84 1 1 5 1 74 10 1 1 1 38 87 13 1 7 66 81 49 1 9 1 11 1 25 1 1 1 1 7 1 1 36 61 47 51 1 1 69 40 1 37 1\n40 1 21 1 19 51 37 52 64 1 86 1 5 24 1 1 1 19 36 1 1 77 24 4 1 18 89 1 1 1 1 1 29 22 1 80 32 36 6 1 63 1 30 1 1 1 86 79 73 52 9 1 1 11 7 1 25 20 1 20 1 49 1 37 1 41 1 1 1 1 54 55 1 10 1 1 1 1 1 1 66 1 68 1 1 1 1 53 1 1", "output": "-1" }, { "input": "2\n1\n1 1\n1 1", "output": "-1" }, { "input": "2\n1\n1 1\n2 2", "output": "-1" }, { "input": "2\n1\n2 2\n1 1", "output": "-1" }, { "input": "2\n1\n2 2\n2 2", "output": "15.0000000000" }, { "input": "2\n2\n1 1\n1 1", "output": "-1" }, { "input": "2\n2\n1 1\n2 2", "output": "-1" }, { "input": "2\n2\n2 2\n1 1", "output": "-1" }, { "input": "2\n2\n2 2\n2 2", "output": "30.0000000000" }, { "input": "40\n55\n1 382 1 1 1 629 111 689 396 614 1 1 995 148 7 820 913 1 1 169 157 1 702 1 159 1 1 226 1 253 1 319 1 130 1 1 1 466 1 756\n1 23 555 1 412 1 1 373 316 234 888 1 112 818 33 443 313 1 235 1 1 610 110 535 1 445 1 386 1 1 758 1 292 1 862 1 244 428 530 1", "output": "-1" }, { "input": "49\n1\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\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "695580114.6380882263" }, { "input": "2\n12\n11 8\n1 1", "output": "-1" }, { "input": "3\n3\n7 11 17\n19 31 33", "output": "1.6012429470" } ]
1,569,572,145
2,147,483,647
Python 3
OK
TESTS
76
109
307,200
n = int(input()) p = int(input()) R = lambda :map(int,input().split()) a = list(R()) b = list(R()) t = 1 for i in range(n) : t *= (1- (1/a[i]))*(1- (1/b[i])) if t >= 1 or t == 0 : print(-1) else : print(p*((1-t)/t))
Title: Fly Time Limit: None seconds Memory Limit: None megabytes Problem Description: Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$. Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet. The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet. For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons. Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well. Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly. Input Specification: The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets. The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload. The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel. The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel. It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel. Output Specification: If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$. It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel. The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$. Demo Input: ['2\n12\n11 8\n7 5\n', '3\n1\n1 4 1\n2 5 3\n', '6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n'] Demo Output: ['10.0000000000\n', '-1\n', '85.4800000000\n'] Note: Let's consider the first example. Initially, the mass of a rocket with fuel is $22$ tons. - At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel. In the second case, the rocket will not be able even to take off from Earth.
```python n = int(input()) p = int(input()) R = lambda :map(int,input().split()) a = list(R()) b = list(R()) t = 1 for i in range(n) : t *= (1- (1/a[i]))*(1- (1/b[i])) if t >= 1 or t == 0 : print(-1) else : print(p*((1-t)/t)) ```
3
607
A
Chain Reaction
PROGRAMMING
1,600
[ "binary search", "dp" ]
null
null
There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated. Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons. The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.
Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.
[ "4\n1 9\n3 1\n6 1\n7 4\n", "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n" ]
[ "1\n", "3\n" ]
For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2. For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
500
[ { "input": "4\n1 9\n3 1\n6 1\n7 4", "output": "1" }, { "input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1", "output": "3" }, { "input": "1\n0 1", "output": "0" }, { "input": "1\n0 1000000", "output": "0" }, { "input": "1\n1000000 1000000", "output": "0" }, { "input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 6\n7 7", "output": "4" }, { "input": "5\n1 1\n3 1\n5 1\n7 10\n8 10", "output": "2" }, { "input": "11\n110 90\n100 70\n90 10\n80 10\n70 1\n60 1\n50 10\n40 1\n30 1\n10 1\n20 1", "output": "4" } ]
1,569,098,204
2,147,483,647
Python 3
OK
TESTS
41
608
6,246,400
# your code goes here n = int(input()) bb = [0] * 1000001 for i in range(n): a, b = map(int, input().split()) bb[a] = b a = 0 m = 0 for index, value in enumerate(bb): if value > 0: if (index - value) > 0: a = (1 + bb[index - value -1]) else: a = 1 bb[index] = a print(n - max(bb))
Title: Chain Reaction Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated. Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed. Input Specification: The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons. The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*. Output Specification: Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added. Demo Input: ['4\n1 9\n3 1\n6 1\n7 4\n', '7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n'] Demo Output: ['1\n', '3\n'] Note: For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2. For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.
```python # your code goes here n = int(input()) bb = [0] * 1000001 for i in range(n): a, b = map(int, input().split()) bb[a] = b a = 0 m = 0 for index, value in enumerate(bb): if value > 0: if (index - value) > 0: a = (1 + bb[index - value -1]) else: a = 1 bb[index] = a print(n - max(bb)) ```
3
239
A
Two Bags of Potatoes
PROGRAMMING
1,200
[ "greedy", "implementation", "math" ]
null
null
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*. Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once. If there are no such values of *x* print a single integer -1.
[ "10 1 10\n", "10 6 40\n" ]
[ "-1\n", "2 8 14 20 26 \n" ]
none
500
[ { "input": "10 1 10", "output": "-1" }, { "input": "10 6 40", "output": "2 8 14 20 26 " }, { "input": "10 1 20", "output": "1 2 3 4 5 6 7 8 9 10 " }, { "input": "1 10000 1000000000", "output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 129999 139999 149999 159999 169999 179999 189999 199999 209999 219999 229999 239999 249999 259999 269999 279999 289999 299999 309999 319999 329999 339999 349999 359999 369999 379999 389999 399999 409999 419999 429999 439999 449999 459999 469999 479999 489999 499999 509999 519999 529999 539999 549999 559999 569999 579999 589999 599999 609999 619999 629999 639999 649999 659999 669999 679999 689999 699999 709999 719999 729999 739999 7499..." }, { "input": "84817 1 33457", "output": "-1" }, { "input": "21 37 99", "output": "16 53 " }, { "input": "78 7 15", "output": "-1" }, { "input": "74 17 27", "output": "-1" }, { "input": "79 23 43", "output": "-1" }, { "input": "32 33 3", "output": "-1" }, { "input": "55 49 44", "output": "-1" }, { "input": "64 59 404", "output": "54 113 172 231 290 " }, { "input": "61 69 820", "output": "8 77 146 215 284 353 422 491 560 629 698 " }, { "input": "17 28 532", "output": "11 39 67 95 123 151 179 207 235 263 291 319 347 375 403 431 459 487 515 " }, { "input": "46592 52 232", "output": "-1" }, { "input": "1541 58 648", "output": "-1" }, { "input": "15946 76 360", "output": "-1" }, { "input": "30351 86 424", "output": "-1" }, { "input": "1 2 37493", "output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..." }, { "input": "1 3 27764", "output": "2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 95 98 101 104 107 110 113 116 119 122 125 128 131 134 137 140 143 146 149 152 155 158 161 164 167 170 173 176 179 182 185 188 191 194 197 200 203 206 209 212 215 218 221 224 227 230 233 236 239 242 245 248 251 254 257 260 263 266 269 272 275 278 281 284 287 290 293 296 299 302 305 308 311 314 317 320 323 326 329 332 335 338 341 344 347 350 353 356 359 362 365 368 371 374 377 380 383 386 389 392 395 398 401 404 407 410..." }, { "input": "10 4 9174", "output": "2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 78 82 86 90 94 98 102 106 110 114 118 122 126 130 134 138 142 146 150 154 158 162 166 170 174 178 182 186 190 194 198 202 206 210 214 218 222 226 230 234 238 242 246 250 254 258 262 266 270 274 278 282 286 290 294 298 302 306 310 314 318 322 326 330 334 338 342 346 350 354 358 362 366 370 374 378 382 386 390 394 398 402 406 410 414 418 422 426 430 434 438 442 446 450 454 458 462 466 470 474 478 482 486 490 494 498 502 506 510 514 518 522 526 530 534 53..." }, { "input": "33 7 4971", "output": "2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128 135 142 149 156 163 170 177 184 191 198 205 212 219 226 233 240 247 254 261 268 275 282 289 296 303 310 317 324 331 338 345 352 359 366 373 380 387 394 401 408 415 422 429 436 443 450 457 464 471 478 485 492 499 506 513 520 527 534 541 548 555 562 569 576 583 590 597 604 611 618 625 632 639 646 653 660 667 674 681 688 695 702 709 716 723 730 737 744 751 758 765 772 779 786 793 800 807 814 821 828 835 842 849 856 863 870 877 884 891 898 905 912 919..." }, { "input": "981 1 3387", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "386 1 2747", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "123 2 50000", "output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..." }, { "input": "3123 100 10000000", "output": "77 177 277 377 477 577 677 777 877 977 1077 1177 1277 1377 1477 1577 1677 1777 1877 1977 2077 2177 2277 2377 2477 2577 2677 2777 2877 2977 3077 3177 3277 3377 3477 3577 3677 3777 3877 3977 4077 4177 4277 4377 4477 4577 4677 4777 4877 4977 5077 5177 5277 5377 5477 5577 5677 5777 5877 5977 6077 6177 6277 6377 6477 6577 6677 6777 6877 6977 7077 7177 7277 7377 7477 7577 7677 7777 7877 7977 8077 8177 8277 8377 8477 8577 8677 8777 8877 8977 9077 9177 9277 9377 9477 9577 9677 9777 9877 9977 10077 10177 10277 1037..." }, { "input": "2 10000 1000000000", "output": "9998 19998 29998 39998 49998 59998 69998 79998 89998 99998 109998 119998 129998 139998 149998 159998 169998 179998 189998 199998 209998 219998 229998 239998 249998 259998 269998 279998 289998 299998 309998 319998 329998 339998 349998 359998 369998 379998 389998 399998 409998 419998 429998 439998 449998 459998 469998 479998 489998 499998 509998 519998 529998 539998 549998 559998 569998 579998 589998 599998 609998 619998 629998 639998 649998 659998 669998 679998 689998 699998 709998 719998 729998 739998 7499..." }, { "input": "3 10000 1000000000", "output": "9997 19997 29997 39997 49997 59997 69997 79997 89997 99997 109997 119997 129997 139997 149997 159997 169997 179997 189997 199997 209997 219997 229997 239997 249997 259997 269997 279997 289997 299997 309997 319997 329997 339997 349997 359997 369997 379997 389997 399997 409997 419997 429997 439997 449997 459997 469997 479997 489997 499997 509997 519997 529997 539997 549997 559997 569997 579997 589997 599997 609997 619997 629997 639997 649997 659997 669997 679997 689997 699997 709997 719997 729997 739997 7499..." }, { "input": "12312223 10000 1000000000", "output": "7777 17777 27777 37777 47777 57777 67777 77777 87777 97777 107777 117777 127777 137777 147777 157777 167777 177777 187777 197777 207777 217777 227777 237777 247777 257777 267777 277777 287777 297777 307777 317777 327777 337777 347777 357777 367777 377777 387777 397777 407777 417777 427777 437777 447777 457777 467777 477777 487777 497777 507777 517777 527777 537777 547777 557777 567777 577777 587777 597777 607777 617777 627777 637777 647777 657777 667777 677777 687777 697777 707777 717777 727777 737777 7477..." }, { "input": "500000000 1000000000 1000000000", "output": "500000000 " }, { "input": "1 1000000000 1000000000", "output": "999999999 " }, { "input": "10 6 11", "output": "-1" }, { "input": "2 100 10", "output": "-1" }, { "input": "1 100000007 1000000000", "output": "100000006 200000013 300000020 400000027 500000034 600000041 700000048 800000055 900000062 " }, { "input": "1 999999999 1000000000", "output": "999999998 " }, { "input": "100000000 1000000000 1000000000", "output": "900000000 " }, { "input": "11 2 12", "output": "1 " }, { "input": "31 10 39", "output": "-1" }, { "input": "48 6 50", "output": "-1" }, { "input": "500000000 500000000 1000000000", "output": "500000000 " }, { "input": "1 1000000000 999999999", "output": "-1" }, { "input": "4 2 10", "output": "2 4 6 " }, { "input": "1000000000 1 1", "output": "-1" }, { "input": "1000000000 1 100000", "output": "-1" }, { "input": "1000000000 1 10", "output": "-1" }, { "input": "10 5 14", "output": "-1" }, { "input": "500000000 499999999 1000000000", "output": "499999998 " }, { "input": "1 999999997 1000000000", "output": "999999996 " } ]
1,655,704,342
2,147,483,647
PyPy 3
TIME_LIMIT_EXCEEDED
TESTS
3
1,000
4,710,400
y,k,n = input().split() w=0 for i in range(1,int(n)+1-int(y)): if (i+int(y))%int(k)==0: w=1 print(i,end=' ') if w==0: print(-1)
Title: Two Bags of Potatoes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*. Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. Input Specification: The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). Output Specification: Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once. If there are no such values of *x* print a single integer -1. Demo Input: ['10 1 10\n', '10 6 40\n'] Demo Output: ['-1\n', '2 8 14 20 26 \n'] Note: none
```python y,k,n = input().split() w=0 for i in range(1,int(n)+1-int(y)): if (i+int(y))%int(k)==0: w=1 print(i,end=' ') if w==0: print(-1) ```
0
190
A
Vasya and the Bus
PROGRAMMING
1,100
[ "greedy", "math" ]
null
null
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..." The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride. The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=&gt;<=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble. We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups. Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly. Otherwise, print "Impossible" (without the quotes).
[ "1 2\n", "0 5\n", "2 2\n" ]
[ "2 2", "Impossible", "2 3" ]
In the first sample a grown-up rides with two children and pays two rubles. In the second sample there are only children in the bus, so the situation is impossible. In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
500
[ { "input": "1 2", "output": "2 2" }, { "input": "0 5", "output": "Impossible" }, { "input": "2 2", "output": "2 3" }, { "input": "2 7", "output": "7 8" }, { "input": "4 10", "output": "10 13" }, { "input": "6 0", "output": "6 6" }, { "input": "7 1", "output": "7 7" }, { "input": "0 0", "output": "0 0" }, { "input": "71 24", "output": "71 94" }, { "input": "16 70", "output": "70 85" }, { "input": "0 1", "output": "Impossible" }, { "input": "1 0", "output": "1 1" }, { "input": "1 1", "output": "1 1" }, { "input": "63 82", "output": "82 144" }, { "input": "8 26", "output": "26 33" }, { "input": "21 27", "output": "27 47" }, { "input": "0 38", "output": "Impossible" }, { "input": "46 84", "output": "84 129" }, { "input": "59 96", "output": "96 154" }, { "input": "63028 0", "output": "63028 63028" }, { "input": "9458 0", "output": "9458 9458" }, { "input": "80236 0", "output": "80236 80236" }, { "input": "26666 0", "output": "26666 26666" }, { "input": "59617 0", "output": "59617 59617" }, { "input": "0 6048", "output": "Impossible" }, { "input": "63028 28217", "output": "63028 91244" }, { "input": "9458 39163", "output": "39163 48620" }, { "input": "80236 14868", "output": "80236 95103" }, { "input": "26666 52747", "output": "52747 79412" }, { "input": "59617 28452", "output": "59617 88068" }, { "input": "6048 4158", "output": "6048 10205" }, { "input": "76826 4210", "output": "76826 81035" }, { "input": "23256 15156", "output": "23256 38411" }, { "input": "56207 53035", "output": "56207 109241" }, { "input": "2637 28740", "output": "28740 31376" }, { "input": "73415 4445", "output": "73415 77859" }, { "input": "82019 4498", "output": "82019 86516" }, { "input": "28449 80204", "output": "80204 108652" }, { "input": "99227 53323", "output": "99227 152549" }, { "input": "45657 29028", "output": "45657 74684" }, { "input": "78608 4733", "output": "78608 83340" }, { "input": "25038 4786", "output": "25038 29823" }, { "input": "95816 80492", "output": "95816 176307" }, { "input": "42246 94024", "output": "94024 136269" }, { "input": "0 100000", "output": "Impossible" }, { "input": "100000 0", "output": "100000 100000" }, { "input": "1 100000", "output": "100000 100000" }, { "input": "100000 1", "output": "100000 100000" }, { "input": "63028 63028", "output": "63028 126055" }, { "input": "9458 9458", "output": "9458 18915" }, { "input": "80236 80236", "output": "80236 160471" }, { "input": "26666 26666", "output": "26666 53331" }, { "input": "59617 59617", "output": "59617 119233" }, { "input": "100000 100000", "output": "100000 199999" } ]
1,656,149,640
2,147,483,647
Python 3
OK
TESTS
54
92
0
a,b=map(int,input().split()) if a<1 and b:print('Impossible');exit() print(max(a,b),a+max(b-1,0))
Title: Vasya and the Bus Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..." The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride. The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=&gt;<=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble. We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups. Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total. Input Specification: The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly. Output Specification: If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly. Otherwise, print "Impossible" (without the quotes). Demo Input: ['1 2\n', '0 5\n', '2 2\n'] Demo Output: ['2 2', 'Impossible', '2 3'] Note: In the first sample a grown-up rides with two children and pays two rubles. In the second sample there are only children in the bus, so the situation is impossible. In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
```python a,b=map(int,input().split()) if a<1 and b:print('Impossible');exit() print(max(a,b),a+max(b-1,0)) ```
3
651
B
Beautiful Paintings
PROGRAMMING
1,200
[ "greedy", "sortings" ]
null
null
There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one. We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=&gt;<=*a**i*.
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting.
Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=&gt;<=*a**i*, after the optimal rearrangement.
[ "5\n20 30 10 50 40\n", "4\n200 100 100 200\n" ]
[ "4\n", "2\n" ]
In the first sample, the optimal order is: 10, 20, 30, 40, 50. In the second sample, the optimal order is: 100, 200, 100, 200.
1,000
[ { "input": "5\n20 30 10 50 40", "output": "4" }, { "input": "4\n200 100 100 200", "output": "2" }, { "input": "10\n2 2 2 2 2 2 2 2 2 2", "output": "0" }, { "input": "1\n1000", "output": "0" }, { "input": "2\n444 333", "output": "1" }, { "input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14", "output": "95" }, { "input": "1\n995", "output": "0" }, { "input": "10\n103 101 103 103 101 102 100 100 101 104", "output": "7" }, { "input": "20\n102 100 102 104 102 101 104 103 100 103 105 105 100 105 100 100 101 105 105 102", "output": "15" }, { "input": "20\n990 994 996 999 997 994 990 992 990 993 992 990 999 999 992 994 997 990 993 998", "output": "15" }, { "input": "100\n1 8 3 8 10 8 5 3 10 3 5 8 4 5 5 5 10 3 6 6 6 6 6 7 2 7 2 4 7 8 3 8 7 2 5 6 1 5 5 7 9 7 6 9 1 8 1 3 6 5 1 3 6 9 5 6 8 4 8 6 10 9 2 9 3 8 7 5 2 10 2 10 3 6 5 5 3 5 10 2 3 7 10 8 8 4 3 4 9 6 10 7 6 6 6 4 9 9 8 9", "output": "84" } ]
1,658,296,059
2,147,483,647
Python 3
OK
TESTS
31
93
0
n=int(input()) l=list(map(int,input().split())) f,b=0,l.copy() l.sort() for i in range(len(l)-1): if l[i+1]>l[i]:f+=1 s=list(set(l)) while 1: for i in s: b.remove(i) if len(b)==0: break b.sort() for i in range(len(b)-1): if b[i+1]>b[i]: f+=1 s=list(set(b)) print(f)
Title: Beautiful Paintings Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one. We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=&gt;<=*a**i*. Input Specification: The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting. Output Specification: Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=&gt;<=*a**i*, after the optimal rearrangement. Demo Input: ['5\n20 30 10 50 40\n', '4\n200 100 100 200\n'] Demo Output: ['4\n', '2\n'] Note: In the first sample, the optimal order is: 10, 20, 30, 40, 50. In the second sample, the optimal order is: 100, 200, 100, 200.
```python n=int(input()) l=list(map(int,input().split())) f,b=0,l.copy() l.sort() for i in range(len(l)-1): if l[i+1]>l[i]:f+=1 s=list(set(l)) while 1: for i in s: b.remove(i) if len(b)==0: break b.sort() for i in range(len(b)-1): if b[i+1]>b[i]: f+=1 s=list(set(b)) print(f) ```
3
680
B
Bear and Finding Criminals
PROGRAMMING
1,000
[ "constructive algorithms", "implementation" ]
null
null
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|. Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city. Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal. You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Print the number of criminals Limak will catch.
[ "6 3\n1 1 1 0 1 0\n", "5 2\n0 0 0 1 0\n" ]
[ "3\n", "1\n" ]
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red. Using the BCD gives Limak the following information: - There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance. So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total. In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
1,000
[ { "input": "6 3\n1 1 1 0 1 0", "output": "3" }, { "input": "5 2\n0 0 0 1 0", "output": "1" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 1\n0", "output": "0" }, { "input": "9 3\n1 1 1 1 1 1 1 1 0", "output": "8" }, { "input": "9 5\n1 0 1 0 1 0 1 0 1", "output": "5" }, { "input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0", "output": "10" }, { "input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0", "output": "27" }, { "input": "8 1\n1 0 1 1 0 0 1 0", "output": "4" }, { "input": "11 11\n0 1 0 0 1 1 1 0 0 0 0", "output": "4" }, { "input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1", "output": "4" }, { "input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "1" }, { "input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "3" }, { "input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24" }, { "input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24" }, { "input": "98 70\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 0 0 0 0 0 0 0 0 0 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": "41" }, { "input": "99 70\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 1 1 1 1 1 1 1 1 1 1 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": "9" }, { "input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1", "output": "34" }, { "input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1", "output": "39" }, { "input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" }, { "input": "100 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "2 1\n0 1", "output": "1" } ]
1,587,754,354
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
109
307,200
def findcriminal(arr,c): thief=0 if arr[c]==1: thief+=1 if(c==0): i=0 else: i=c-1 if(c==len(arr)-1): j=len(arr)-1 else: j=c+1 while(i>=0 and j<=len(arr)-1): if arr[i]==1 and arr[j]==1: thief+=2 i-=1 j+=1 if (i>-1): for k in range(i+1): if arr[k]==1: thief+=1 if j<len(arr): for k in range(j,len(arr)): if arr[k]==1: thief+=1 return thief inp1=[int(i) for i in input().split()] arr=[int(i) for i in input().split()] print(findcriminal(arr,inp1[1]-1))
Title: Bear and Finding Criminals Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|. Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city. Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal. You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD. Input Specification: The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city. Output Specification: Print the number of criminals Limak will catch. Demo Input: ['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n'] Demo Output: ['3\n', '1\n'] Note: In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red. Using the BCD gives Limak the following information: - There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance. So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total. In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
```python def findcriminal(arr,c): thief=0 if arr[c]==1: thief+=1 if(c==0): i=0 else: i=c-1 if(c==len(arr)-1): j=len(arr)-1 else: j=c+1 while(i>=0 and j<=len(arr)-1): if arr[i]==1 and arr[j]==1: thief+=2 i-=1 j+=1 if (i>-1): for k in range(i+1): if arr[k]==1: thief+=1 if j<len(arr): for k in range(j,len(arr)): if arr[k]==1: thief+=1 return thief inp1=[int(i) for i in input().split()] arr=[int(i) for i in input().split()] print(findcriminal(arr,inp1[1]-1)) ```
0
281
A
Word Capitalization
PROGRAMMING
800
[ "implementation", "strings" ]
null
null
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged.
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output the given word after capitalization.
[ "ApPLe\n", "konjac\n" ]
[ "ApPLe\n", "Konjac\n" ]
none
500
[ { "input": "ApPLe", "output": "ApPLe" }, { "input": "konjac", "output": "Konjac" }, { "input": "a", "output": "A" }, { "input": "A", "output": "A" }, { "input": "z", "output": "Z" }, { "input": "ABACABA", "output": "ABACABA" }, { "input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX", "output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX" }, { "input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO", "output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO" }, { "input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD", "output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD" }, { "input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW", "output": 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1,696,542,714
2,147,483,647
Python 3
OK
TESTS
25
62
0
a = input() l = list(a) l[0] = l[0].upper() b = ''.join(l) print(b)
Title: Word Capitalization Time Limit: None seconds Memory Limit: None megabytes Problem Description: Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged. Input Specification: A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. Output Specification: Output the given word after capitalization. Demo Input: ['ApPLe\n', 'konjac\n'] Demo Output: ['ApPLe\n', 'Konjac\n'] Note: none
```python a = input() l = list(a) l[0] = l[0].upper() b = ''.join(l) print(b) ```
3
389
A
Fox and Number Game
PROGRAMMING
1,000
[ "greedy", "math" ]
null
null
Fox Ciel is playing a game with numbers now. Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* &gt; *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible. Please help Ciel to find this minimal sum.
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output a single integer — the required minimal sum.
[ "2\n1 2\n", "3\n2 4 6\n", "2\n12 18\n", "5\n45 12 27 30 18\n" ]
[ "2\n", "6\n", "12\n", "15\n" ]
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
500
[ { "input": "2\n1 2", "output": "2" }, { "input": "3\n2 4 6", "output": "6" }, { "input": "2\n12 18", "output": "12" }, { "input": "5\n45 12 27 30 18", "output": "15" }, { "input": "2\n1 1", "output": "2" }, { "input": "2\n100 100", "output": "200" }, { "input": "2\n87 58", "output": "58" }, { "input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52", "output": "2028" }, { "input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96", "output": "5664" }, { "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": "10000" }, { "input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98", "output": "700" }, { "input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42", "output": "819" }, { "input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70", "output": "4130" }, { "input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22", "output": "1914" }, { "input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63", "output": "945" }, { "input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91", "output": "273" }, { "input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36", "output": "324" }, { "input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71", "output": "3266" }, { "input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33", "output": "770" }, { "input": "10\n60 12 96 48 60 24 60 36 60 60", "output": "120" }, { "input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51", "output": "1020" }, { "input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58", "output": "2900" }, { "input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90", "output": "980" }, { "input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82", "output": "8200" }, { "input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70", "output": "100" }, { "input": "2\n100 1", "output": "2" }, { "input": "2\n18 30", "output": "12" }, { "input": "2\n3 5", "output": "2" }, { "input": "2\n1 10", "output": "2" }, { "input": "2\n8 5", "output": "2" }, { "input": "5\n2 3 5 8 18", "output": "5" }, { "input": "5\n2 4 1 6 8", "output": "5" }, { "input": "3\n12 10 5", "output": "3" }, { "input": "3\n6 10 15", "output": "3" } ]
1,461,748,619
2,147,483,647
Python 3
OK
TESTS
34
77
5,222,400
def f(x,y): while True: t=x%y x=y y=t if t==0: return x n=int(input()) x=list(map(int,input().split())) j=x[0] for i in x: j=f(j,i) #print(j) print(j*n)
Title: Fox and Number Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Fox Ciel is playing a game with numbers now. Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* &gt; *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible. Please help Ciel to find this minimal sum. Input Specification: The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100). Output Specification: Output a single integer — the required minimal sum. Demo Input: ['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n'] Demo Output: ['2\n', '6\n', '12\n', '15\n'] Note: In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
```python def f(x,y): while True: t=x%y x=y y=t if t==0: return x n=int(input()) x=list(map(int,input().split())) j=x[0] for i in x: j=f(j,i) #print(j) print(j*n) ```
3
770
B
Maximize Sum of Digits
PROGRAMMING
1,300
[ "*special", "implementation", "math" ]
null
null
Anton has the integer *x*. He is interested what positive integer, which doesn't exceed *x*, has the maximum sum of digits. Your task is to help Anton and to find the integer that interests him. If there are several such integers, determine the biggest of them.
The first line contains the positive integer *x* (1<=≤<=*x*<=≤<=1018) — the integer which Anton has.
Print the positive integer which doesn't exceed *x* and has the maximum sum of digits. If there are several such integers, print the biggest of them. Printed integer must not contain leading zeros.
[ "100\n", "48\n", "521\n" ]
[ "99\n", "48\n", "499\n" ]
none
1,000
[ { "input": "100", "output": "99" }, { "input": "48", "output": "48" }, { "input": "521", "output": "499" }, { "input": "1", "output": "1" }, { "input": "2", "output": "2" }, { "input": "3", "output": "3" }, { "input": "39188", "output": "38999" }, { "input": "5", "output": "5" }, { "input": "6", "output": "6" }, { "input": "7", "output": "7" }, { "input": "8", "output": "8" }, { "input": "9", "output": "9" }, { "input": "10", "output": "9" }, { "input": "59999154", "output": "59998999" }, { "input": "1000", "output": "999" }, { "input": "10000", "output": "9999" }, { "input": "100000", "output": "99999" }, { "input": "1000000", "output": "999999" }, { "input": "10000000", "output": "9999999" }, { "input": "100000000", "output": "99999999" }, { "input": "1000000000", "output": "999999999" }, { "input": "10000000000", "output": "9999999999" }, { "input": "100000000000", "output": "99999999999" }, { "input": "1000000000000", "output": "999999999999" }, { "input": "10000000000000", "output": "9999999999999" }, { "input": "100000000000000", "output": "99999999999999" }, { "input": "1000000000000000", "output": "999999999999999" }, { "input": "10000000000000000", "output": "9999999999999999" }, { "input": "100000000000000000", "output": "99999999999999999" }, { "input": "1000000000000000000", "output": "999999999999999999" }, { "input": "999999990", "output": "999999989" }, { "input": "666666899789879", "output": "599999999999999" }, { "input": "65499992294999000", "output": "59999999999999999" }, { "input": "9879100000000099", "output": "8999999999999999" }, { "input": "9991919190909919", "output": "9989999999999999" }, { "input": "978916546899999999", "output": "899999999999999999" }, { "input": "5684945999999999", "output": "4999999999999999" }, { "input": "999999999999999999", "output": "999999999999999999" }, { "input": "999999999999990999", "output": "999999999999989999" }, { "input": "999999999999999990", "output": "999999999999999989" }, { "input": "909999999999999999", "output": "899999999999999999" }, { "input": "199999999999999999", "output": "199999999999999999" }, { "input": "299999999999999999", "output": "299999999999999999" }, { "input": "999999990009999999", "output": "999999989999999999" }, { "input": "999000000001999999", "output": "998999999999999999" }, { "input": "999999999991", "output": "999999999989" }, { "input": "999999999992", "output": "999999999989" }, { "input": "79320", "output": "78999" }, { "input": "99004", "output": "98999" }, { "input": "99088", "output": "98999" }, { "input": "99737", "output": "98999" }, { "input": "29652", "output": "28999" }, { "input": "59195", "output": "58999" }, { "input": "19930", "output": "19899" }, { "input": "49533", "output": "48999" }, { "input": "69291", "output": "68999" }, { "input": "59452", "output": "58999" }, { "input": "11", "output": "9" }, { "input": "110", "output": "99" }, { "input": "111", "output": "99" }, { "input": "119", "output": "99" }, { "input": "118", "output": "99" }, { "input": "1100", "output": "999" }, { "input": "1199", "output": "999" }, { "input": "1109", "output": "999" }, { "input": "1190", "output": "999" }, { "input": "12", "output": "9" }, { "input": "120", "output": "99" }, { "input": "121", "output": "99" }, { "input": "129", "output": "99" }, { "input": "128", "output": "99" }, { "input": "1200", "output": "999" }, { "input": "1299", "output": "999" }, { "input": "1209", "output": "999" }, { "input": "1290", "output": "999" }, { "input": "13", "output": "9" }, { "input": "130", "output": "99" }, { "input": "131", "output": "99" }, { "input": "139", "output": "99" }, { "input": "138", "output": "99" }, { "input": "1300", "output": "999" }, { "input": "1399", "output": "999" }, { "input": "1309", "output": "999" }, { "input": "1390", "output": "999" }, { "input": "14", "output": "9" }, { "input": "140", "output": "99" }, { "input": "141", "output": "99" }, { "input": "149", "output": "99" }, { "input": "148", "output": "99" }, { "input": "1400", "output": "999" }, { "input": "1499", "output": "999" }, { "input": "1409", "output": "999" }, { "input": "1490", "output": "999" }, { "input": "15", "output": "9" }, { "input": "150", "output": "99" }, { "input": "151", "output": "99" }, { "input": "159", "output": "99" }, { "input": "158", "output": "99" }, { "input": "1500", "output": "999" }, { "input": "1599", "output": "999" }, { "input": "1509", "output": "999" }, { "input": "1590", "output": "999" }, { "input": "16", "output": "9" }, { "input": "160", "output": "99" }, { "input": "161", "output": "99" }, { "input": "169", "output": "99" }, { "input": "168", "output": "99" }, { "input": "1600", "output": "999" }, { "input": "1699", "output": "999" }, { "input": "1609", "output": "999" }, { "input": "1690", "output": "999" }, { "input": "17", "output": "9" }, { "input": "170", "output": "99" }, { "input": "171", "output": "99" }, { "input": "179", "output": "99" }, { "input": "178", "output": "99" }, { "input": "1700", "output": "999" }, { "input": "1799", "output": "999" }, { "input": "1709", "output": "999" }, { "input": "1790", "output": "999" }, { "input": "18", "output": "18" }, { "input": "180", "output": "99" }, { "input": "181", "output": "99" }, { "input": "189", "output": "189" }, { "input": "188", "output": "99" }, { "input": "1800", "output": "999" }, { "input": "1899", "output": "1899" }, { "input": "1809", "output": "999" }, { "input": "1890", "output": "999" }, { "input": "19", "output": "19" }, { "input": "190", "output": "189" }, { "input": "191", "output": "189" }, { "input": "199", "output": "199" }, { "input": "198", "output": "198" }, { "input": "1900", "output": "1899" }, { "input": "1999", "output": "1999" }, { "input": "1909", "output": "1899" }, { "input": "1990", "output": "1989" }, { "input": "20", "output": "19" }, { "input": "200", "output": "199" }, { "input": "201", "output": "199" }, { "input": "209", "output": "199" }, { "input": "208", "output": "199" }, { "input": "2000", "output": "1999" }, { "input": "2099", "output": "1999" }, { "input": "2009", "output": "1999" }, { "input": "2090", "output": "1999" }, { "input": "21", "output": "19" }, { "input": "210", "output": "199" }, { "input": "211", "output": "199" }, { "input": "219", "output": "199" }, { "input": "218", "output": "199" }, { "input": "2100", "output": "1999" }, { "input": "2199", "output": "1999" }, { "input": "2109", "output": "1999" }, { "input": "2190", "output": "1999" }, { "input": "22", "output": "19" }, { "input": "220", "output": "199" }, { "input": "221", "output": "199" }, { "input": "229", "output": "199" }, { "input": "228", "output": "199" }, { "input": "2200", "output": "1999" }, { "input": "2299", "output": "1999" }, { "input": "2209", "output": "1999" }, { "input": "2290", "output": "1999" }, { "input": "23", "output": "19" }, { "input": "230", "output": "199" }, { "input": "231", "output": "199" }, { "input": "239", "output": "199" }, { "input": "238", "output": "199" }, { "input": "2300", "output": "1999" }, { "input": "2399", "output": "1999" }, { "input": "2309", "output": "1999" }, { "input": "2390", "output": "1999" }, { "input": "24", "output": "19" }, { "input": "240", "output": "199" }, { "input": "241", "output": "199" }, { "input": "249", "output": "199" }, { "input": "248", "output": "199" }, { "input": "2400", "output": "1999" }, { "input": "2499", "output": "1999" }, { "input": "2409", "output": "1999" }, { "input": "2490", "output": "1999" }, { "input": "25", "output": "19" }, { "input": "250", "output": "199" }, { "input": "251", "output": "199" }, { "input": "259", "output": "199" }, { "input": "258", "output": "199" }, { "input": "2500", "output": "1999" }, { "input": "2599", "output": "1999" }, { "input": "2509", "output": "1999" }, { "input": "2590", "output": "1999" }, { "input": "26", "output": "19" }, { "input": "260", "output": "199" }, { "input": "261", "output": "199" }, { "input": "269", "output": "199" }, { "input": "268", "output": "199" }, { "input": "2600", "output": "1999" }, { "input": "2699", "output": "1999" }, { "input": "2609", "output": "1999" }, { "input": "2690", "output": "1999" }, { "input": "27", "output": "19" }, { "input": "270", "output": "199" }, { "input": "271", "output": "199" }, { "input": "279", "output": "199" }, { "input": "278", "output": "199" }, { "input": "2700", "output": "1999" }, { "input": "2799", "output": "1999" }, { "input": "2709", "output": "1999" }, { "input": "2790", "output": "1999" }, { "input": "28", "output": "28" }, { "input": "280", "output": "199" }, { "input": "281", "output": "199" }, { "input": "289", "output": "289" }, { "input": "288", "output": "199" }, { "input": "2800", "output": "1999" }, { "input": "2899", "output": "2899" }, { "input": "2809", "output": "1999" }, { "input": "2890", "output": "1999" }, { "input": "29", "output": "29" }, { "input": "290", "output": "289" }, { "input": "291", "output": "289" }, { "input": "299", "output": "299" }, { "input": "298", "output": "298" }, { "input": "2900", "output": "2899" }, { "input": "2999", "output": "2999" }, { "input": "2909", "output": "2899" }, { "input": "2990", "output": "2989" }, { "input": "999", "output": "999" }, { "input": "999", "output": "999" }, { "input": "890", "output": "889" }, { "input": "995", "output": "989" }, { "input": "999", "output": "999" }, { "input": "989", "output": "989" }, { "input": "999", "output": "999" }, { "input": "999", "output": "999" }, { "input": "991", "output": "989" }, { "input": "999", "output": "999" }, { "input": "9929", "output": "9899" }, { "input": "4999", "output": "4999" }, { "input": "9690", "output": "8999" }, { "input": "8990", "output": "8989" }, { "input": "9982", "output": "9899" }, { "input": "9999", "output": "9999" }, { "input": "1993", "output": "1989" }, { "input": "9367", "output": "8999" }, { "input": "8939", "output": "8899" }, { "input": "9899", "output": "9899" }, { "input": "99999", "output": "99999" }, { "input": "93929", "output": "89999" }, { "input": "99999", "output": "99999" }, { "input": "38579", "output": "29999" }, { "input": "79096", "output": "78999" }, { "input": "72694", "output": "69999" }, { "input": "99999", "output": "99999" }, { "input": "99999", "output": "99999" }, { "input": "99992", "output": "99989" }, { "input": "27998", "output": "19999" }, { "input": "460999", "output": "399999" }, { "input": "999999", "output": "999999" }, { "input": "999999", "output": "999999" }, { "input": "998999", "output": "998999" }, { "input": "999999", "output": "999999" }, { "input": "999929", "output": "999899" }, { "input": "999999", "output": "999999" }, { "input": "999999", "output": "999999" }, { "input": "979199", "output": "899999" }, { "input": "999999", "output": "999999" }, { "input": "9899999", "output": "9899999" }, { "input": "9699959", "output": "8999999" }, { "input": "9999999", "output": "9999999" }, { "input": "9997099", "output": "9989999" }, { "input": "8992091", "output": "8989999" }, { "input": "9599295", "output": "8999999" }, { "input": "2999902", "output": "2999899" }, { "input": "9999953", "output": "9999899" }, { "input": "9999999", "output": "9999999" }, { "input": "9590999", "output": "8999999" } ]
1,624,757,193
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
6
62
0
num = input() if num[1:] != "9"*(len(num)-1): if num[1:] == "8"+"9"*(len(num)-2): print(num) else: print(int(str(int(num[0])-1)+"9"*(len(num)-1))) else: print(num)
Title: Maximize Sum of Digits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton has the integer *x*. He is interested what positive integer, which doesn't exceed *x*, has the maximum sum of digits. Your task is to help Anton and to find the integer that interests him. If there are several such integers, determine the biggest of them. Input Specification: The first line contains the positive integer *x* (1<=≤<=*x*<=≤<=1018) — the integer which Anton has. Output Specification: Print the positive integer which doesn't exceed *x* and has the maximum sum of digits. If there are several such integers, print the biggest of them. Printed integer must not contain leading zeros. Demo Input: ['100\n', '48\n', '521\n'] Demo Output: ['99\n', '48\n', '499\n'] Note: none
```python num = input() if num[1:] != "9"*(len(num)-1): if num[1:] == "8"+"9"*(len(num)-2): print(num) else: print(int(str(int(num[0])-1)+"9"*(len(num)-1))) else: print(num) ```
0
322
B
Ciel and Flowers
PROGRAMMING
1,600
[ "combinatorics", "math" ]
null
null
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets: - To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower. Help Fox Ciel to find the maximal number of bouquets she can make.
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Print the maximal number of bouquets Fox Ciel can make.
[ "3 6 9\n", "4 4 4\n", "0 0 0\n" ]
[ "6\n", "4\n", "0\n" ]
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets. In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
1,000
[ { "input": "3 6 9", "output": "6" }, { "input": "4 4 4", "output": "4" }, { "input": "0 0 0", "output": "0" }, { "input": "0 3 6", "output": "3" }, { "input": "7 8 9", "output": "7" }, { "input": "8 8 9", "output": "8" }, { "input": "15 3 999", "output": "339" }, { "input": "32 62 92", "output": "62" }, { "input": "123456789 123456789 123456789", "output": "123456789" }, { "input": "3 5 5", "output": "4" }, { "input": "666806767 385540591 357848286", "output": "470065214" }, { "input": "80010646 727118126 817880463", "output": "541669744" }, { "input": "829651016 732259171 572879931", "output": "711596705" }, { "input": "242854896 442432924 180395753", "output": "288561190" }, { "input": "139978911 5123031 935395222", "output": "360165721" }, { "input": "553182792 10264076 395427398", "output": "319624755" }, { "input": "597790453 720437830 855459575", "output": "724562619" }, { "input": "494914467 356982656 757942689", "output": "536613270" }, { "input": "908118348 67156409 217974865", "output": "397749873" }, { "input": "952726009 629846517 972974334", "output": "851848953" }, { "input": "775140200 616574841 630329230", "output": "674014756" }, { "input": "524780569 326748594 90361407", "output": "313963523" }, { "input": "937984449 184405994 992844522", "output": "705078321" }, { "input": "835108464 525983528 452876698", "output": "604656229" }, { "input": "879716125 531124573 207876166", "output": "539572288" }, { "input": "292920005 241298326 667908343", "output": "400708891" }, { "input": "1000000000 1000000000 1000000000", "output": "1000000000" }, { "input": "1000000000 999999999 999999998", "output": "999999998" }, { "input": "999999998 999999998 999999999", "output": "999999998" }, { "input": "0 1 1", "output": "0" }, { "input": "0 1000000000 0", "output": "333333333" }, { "input": "0 1 0", "output": "0" }, { "input": "1 0 0", "output": "0" }, { "input": "0 2 2", "output": "0" }, { "input": "3 3 5", "output": "3" }, { "input": "2 2 0", "output": "0" }, { "input": "0 5 5", "output": "2" }, { "input": "2 0 11", "output": "3" }, { "input": "9 9 7", "output": "8" }, { "input": "65 30 74", "output": "56" } ]
1,490,562,302
2,147,483,647
Python 3
OK
TESTS
40
124
4,608,000
r,g,b=map(int,input().split()) a,d,c=0,0,0 a=r//3+b//3+g//3 if(r and g and b):r,g,b=r-1,g-1,b-1;a=max(a,r//3+b//3+g//3+1) if(r and g and b):r,g,b=r-1,g-1,b-1;a=max(a,r//3+b//3+g//3+2) print(a)
Title: Ciel and Flowers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets: - To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower. Help Fox Ciel to find the maximal number of bouquets she can make. Input Specification: The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers. Output Specification: Print the maximal number of bouquets Fox Ciel can make. Demo Input: ['3 6 9\n', '4 4 4\n', '0 0 0\n'] Demo Output: ['6\n', '4\n', '0\n'] Note: In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets. In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
```python r,g,b=map(int,input().split()) a,d,c=0,0,0 a=r//3+b//3+g//3 if(r and g and b):r,g,b=r-1,g-1,b-1;a=max(a,r//3+b//3+g//3+1) if(r and g and b):r,g,b=r-1,g-1,b-1;a=max(a,r//3+b//3+g//3+2) print(a) ```
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" } ]
1,679,998,889
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
15
0
word = input() num_word = len(word) if num_word >= 12: print(f"{word[0]}{num_word-2}{word[-1]}") elif num_word > 1 and num_word < 12: print(word) else: pass
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 word = input() num_word = len(word) if num_word >= 12: print(f"{word[0]}{num_word-2}{word[-1]}") elif num_word > 1 and num_word < 12: print(word) else: pass ```
0
801
B
Valued Keys
PROGRAMMING
900
[ "constructive algorithms", "greedy", "strings" ]
null
null
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length. The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2. For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel". You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
The first line of input contains the string *x*. The second line of input contains the string *y*. Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1. Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
[ "ab\naa\n", "nzwzl\nniwel\n", "ab\nba\n" ]
[ "ba\n", "xiyez\n", "-1\n" ]
The first case is from the statement. Another solution for the second case is "zizez" There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) =  "ba".
1,000
[ { "input": "ab\naa", "output": "ba" }, { "input": "nzwzl\nniwel", "output": "xiyez" }, { "input": "ab\nba", "output": "-1" }, { "input": "r\nl", "output": "l" }, { "input": "d\ny", "output": "-1" }, { "input": "yvowz\ncajav", "output": "cajav" }, { "input": "lwzjp\ninjit", "output": "-1" }, { "input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm", "output": "eodnlemiicdedmkcgavqbnqmm" }, { "input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs", "output": "-1" }, { "input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf", "output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf" }, { "input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen", "output": "-1" }, { "input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb", "output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb" }, { "input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym", "output": "-1" }, { "input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib", "output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib" }, { "input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew", "output": "-1" }, { "input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv", "output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv" }, { "input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns", "output": "-1" }, { "input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf", "output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf" }, { "input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp", "output": "-1" }, { "input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp", "output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp" }, { "input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej", "output": "-1" }, { "input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe", "output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe" }, { "input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang", "output": "-1" }, { "input": "ftfr\nftfr", "output": "ftfr" }, { "input": "ftr\nftr", "output": "ftr" }, { "input": "shftr\nshftr", "output": "shftr" }, { "input": "vkvkkv\nvkvkkv", "output": "vkvkkv" }, { "input": "ftrd\nftrd", "output": "ftrd" }, { "input": "fztr\nfztr", "output": "fztr" }, { "input": "frtr\nfrtr", "output": "frtr" }, { "input": "shdftr\nshdftr", "output": "shdftr" }, { "input": "zzz\nzzz", "output": "zzz" }, { "input": "shtr\nshtr", "output": "shtr" }, { "input": "aaaaa\nzzzzz", "output": "-1" }, { "input": "efr\nefr", "output": "efr" } ]
1,502,874,959
2,147,483,647
Python 3
OK
TESTS
35
77
4,608,000
s=input() r=input() x=0 for i in range(len(s)): if(s[i]<r[i]): x=1 break if(x): print(-1) else: print(r)
Title: Valued Keys Time Limit: None seconds Memory Limit: None megabytes Problem Description: You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length. The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2. For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel". You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists. Input Specification: The first line of input contains the string *x*. The second line of input contains the string *y*. Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100. Output Specification: If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1. Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters. Demo Input: ['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n'] Demo Output: ['ba\n', 'xiyez\n', '-1\n'] Note: The first case is from the statement. Another solution for the second case is "zizez" There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) =  "ba".
```python s=input() r=input() x=0 for i in range(len(s)): if(s[i]<r[i]): x=1 break if(x): print(-1) else: print(r) ```
3
75
A
Life Without Zeros
PROGRAMMING
1,000
[ "implementation" ]
A. Life Without Zeros
2
256
Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
[ "101\n102\n", "105\n106\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "101\n102", "output": "YES" }, { "input": "105\n106", "output": "NO" }, { "input": "544\n397", "output": "YES" }, { "input": "822\n280", "output": "NO" }, { "input": "101\n413", "output": "NO" }, { "input": "309\n139", "output": "NO" }, { "input": "693\n970", "output": "NO" }, { "input": "981\n1", "output": "YES" }, { "input": "352\n276", "output": "YES" }, { "input": "164\n691", "output": "YES" }, { "input": "110036\n43", "output": "YES" }, { "input": "100\n1000", "output": "NO" }, { "input": "1000000000\n1000000000", "output": "YES" }, { "input": "999999999\n999999999", "output": "YES" }, { "input": "6\n4", "output": "NO" }, { "input": "123456\n876543", "output": "YES" }, { "input": "1234567\n9876543", "output": "NO" }, { "input": "1111111\n1119111", "output": "NO" }, { "input": "10502405\n40403403", "output": "YES" }, { "input": "1\n1", "output": "YES" }, { "input": "5\n4", "output": "YES" }, { "input": "1\n6", "output": "YES" }, { "input": "40\n17", "output": "NO" }, { "input": "35\n81", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "2081\n7450", "output": "NO" }, { "input": "84355694\n373629503", "output": "NO" }, { "input": "885992048\n510468672", "output": "NO" }, { "input": "540144752\n647307841", "output": "NO" }, { "input": "341781108\n784147010", "output": "NO" }, { "input": "960440938\n572334653", "output": "NO" }, { "input": "657414510\n768771903", "output": "NO" }, { "input": "501871728\n725074574", "output": "NO" }, { "input": "968\n851", "output": "YES" }, { "input": "943\n939", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "611\n41", "output": "YES" }, { "input": "737\n289", "output": "NO" }, { "input": "168\n571", "output": "YES" }, { "input": "868\n463", "output": "YES" }, { "input": "815732007\n913882008", "output": "NO" }, { "input": "459039901\n453033850", "output": "NO" }, { "input": "975305404\n484601256", "output": "NO" } ]
1,696,390,170
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
92
0
# https://codeforces.com/problemset/problem/75/A n= int(input()) m = int(input()) pre_total = n + m new_n = "" for i in str(n): if i != str(0): new_n += i new_m = "" for i in str(m): if i != str(0): new_m += i if int(new_n) + int(new_m) == pre_total: print("YES") else: print("NO")
Title: Life Without Zeros Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation. Input Specification: The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*. Output Specification: The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise. Demo Input: ['101\n102\n', '105\n106\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python # https://codeforces.com/problemset/problem/75/A n= int(input()) m = int(input()) pre_total = n + m new_n = "" for i in str(n): if i != str(0): new_n += i new_m = "" for i in str(m): if i != str(0): new_m += i if int(new_n) + int(new_m) == pre_total: print("YES") else: print("NO") ```
0
344
A
Magnets
PROGRAMMING
800
[ "implementation" ]
null
null
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
On the single line of the output print the number of groups of magnets.
[ "6\n10\n10\n10\n01\n10\n10\n", "4\n01\n01\n10\n10\n" ]
[ "3\n", "2\n" ]
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.
500
[ { "input": "6\n10\n10\n10\n01\n10\n10", "output": "3" }, { "input": "4\n01\n01\n10\n10", "output": "2" }, { "input": "1\n10", "output": "1" }, { "input": "2\n01\n10", "output": "2" }, { "input": "2\n10\n10", "output": "1" }, { "input": "3\n10\n01\n10", "output": "3" }, { "input": "1\n01", "output": "1" }, { "input": "2\n01\n01", "output": "1" }, { "input": "2\n10\n01", "output": "2" }, { "input": "3\n01\n01\n01", "output": "1" }, { "input": "3\n10\n10\n01", "output": "2" }, { "input": "3\n01\n10\n10", "output": "2" }, { "input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10", "output": "55" } ]
1,697,009,917
2,147,483,647
Python 3
OK
TESTS
31
218
0
LastMagnet = None Groups = 0 for i in range(int(input())): Magnet = input() if Magnet != LastMagnet: Groups += 1 LastMagnet = Magnet print(Groups)
Title: Magnets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. Output Specification: On the single line of the output print the number of groups of magnets. Demo Input: ['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n'] Demo Output: ['3\n', '2\n'] Note: The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.
```python LastMagnet = None Groups = 0 for i in range(int(input())): Magnet = input() if Magnet != LastMagnet: Groups += 1 LastMagnet = Magnet print(Groups) ```
3
459
B
Pashmak and Flowers
PROGRAMMING
1,300
[ "combinatorics", "implementation", "sortings" ]
null
null
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible! Your task is to write a program which calculates two things: 1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
[ "2\n1 2\n", "3\n1 4 5\n", "5\n3 1 2 3 1\n" ]
[ "1 1", "4 1", "2 4" ]
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this: 1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
500
[ { "input": "2\n1 2", "output": "1 1" }, { "input": "3\n1 4 5", "output": "4 1" }, { "input": "5\n3 1 2 3 1", "output": "2 4" }, { "input": "2\n1 1", "output": "0 1" }, { "input": "3\n1 1 1", "output": "0 3" }, { "input": "4\n1 1 1 1", "output": "0 6" }, { "input": "5\n1 1 1 1 1", "output": "0 10" }, { "input": "5\n2 2 2 2 2", "output": "0 10" }, { "input": "10\n2 2 2 2 2 2 2 2 2 2", "output": "0 45" }, { "input": "3\n2 2 2", "output": "0 3" }, { "input": "3\n3 3 3", "output": "0 3" }, { "input": "2\n10000000 100000000", "output": "90000000 1" }, { "input": "5\n5 5 5 5 5", "output": "0 10" }, { "input": "5\n3 3 3 3 3", "output": "0 10" }, { "input": "6\n1 1 1 1 1 1", "output": "0 15" }, { "input": "2\n5 6", "output": "1 1" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "0 45" }, { "input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0 45" }, { "input": "4\n4 4 4 4", "output": "0 6" }, { "input": "7\n1 1 1 1 1 1 1", "output": "0 21" }, { "input": "11\n1 1 1 1 1 1 1 1 1 1 1", "output": "0 55" }, { "input": "8\n8 8 8 8 8 8 8 8", "output": "0 28" }, { "input": "2\n3 2", "output": "1 1" } ]
1,688,031,747
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
5
109
26,931,200
length = int(input()) nums = list(map(int, input().split())) diff = max(nums) - min(nums) if diff == 0: print("0 1") exit(0) mins = nums.count(min(nums)) maxs = nums.count(max(nums)) print(diff, mins * maxs)
Title: Pashmak and Flowers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible! Your task is to write a program which calculates two things: 1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. Input Specification: The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). Output Specification: The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. Demo Input: ['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n'] Demo Output: ['1 1', '4 1', '2 4'] Note: In the third sample the maximum beauty difference is 2 and there are 4 ways to do this: 1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
```python length = int(input()) nums = list(map(int, input().split())) diff = max(nums) - min(nums) if diff == 0: print("0 1") exit(0) mins = nums.count(min(nums)) maxs = nums.count(max(nums)) print(diff, mins * maxs) ```
0
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" } ]
1,641,617,444
2,147,483,647
Python 3
OK
TESTS
28
216
1,638,400
def solve(): n = int(input()) numbers = [] for i in range(n): numbers.append(input()) numbers_length = len(numbers) number_length = len(numbers[0]) code = 0 for i in range(number_length): first_number = numbers[0][i] for j in range(numbers_length): if numbers[j][i] != first_number: print(code) return code = code + 1 print(code) solve()
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 def solve(): n = int(input()) numbers = [] for i in range(n): numbers.append(input()) numbers_length = len(numbers) number_length = len(numbers[0]) code = 0 for i in range(number_length): first_number = numbers[0][i] for j in range(numbers_length): if numbers[j][i] != first_number: print(code) return code = code + 1 print(code) solve() ```
3
922
C
Cave Painting
PROGRAMMING
1,600
[ "brute force", "number theory" ]
null
null
Imp is watching a documentary about cave painting. Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number *n* by all integers *i* from 1 to *k*. Unfortunately, there are too many integers to analyze for Imp. Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all , 1<=≤<=*i*<=≤<=*k*, are distinct, i. e. there is no such pair (*i*,<=*j*) that: - 1<=≤<=*i*<=&lt;<=*j*<=≤<=*k*, - , where is the remainder of division *x* by *y*.
The only line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=1018).
Print "Yes", if all the remainders are distinct, and "No" otherwise. You can print each letter in arbitrary case (lower or upper).
[ "4 4\n", "5 3\n" ]
[ "No\n", "Yes\n" ]
In the first sample remainders modulo 1 and 4 coincide.
1,250
[ { "input": "4 4", "output": "No" }, { "input": "5 3", "output": "Yes" }, { "input": "1 1", "output": "Yes" }, { "input": "744 18", "output": "No" }, { "input": "47879 10", "output": "Yes" }, { "input": "1000000000000000000 1000000000000000000", "output": "No" }, { "input": "657180569218773599 42", "output": "Yes" }, { "input": "442762254977842799 30", "output": "Yes" }, { "input": "474158606260730555 1", "output": "Yes" }, { "input": "807873101233533988 39", "output": "No" }, { "input": "423 7", "output": "No" }, { "input": "264306177888923090 5", "output": "No" }, { "input": "998857801526481788 87", "output": "No" }, { "input": "999684044704565212 28", "output": "No" }, { "input": "319575605003866172 71", "output": "No" }, { "input": "755804560577415016 17", "output": "No" }, { "input": "72712630136142067 356370939", "output": "No" }, { "input": "807264258068668062 33080422", "output": "No" }, { "input": "808090496951784190 311661970", "output": "No" }, { "input": "808916740129867614 180178111", "output": "No" }, { "input": "1 2", "output": "Yes" }, { "input": "2 1", "output": "Yes" }, { "input": "57334064998850639 19", "output": "Yes" }, { "input": "144353716412182199 11", "output": "Yes" }, { "input": "411002215096001759 11", "output": "Yes" }, { "input": "347116374613371527 3", "output": "Yes" }, { "input": "518264351335130399 37", "output": "Yes" }, { "input": "192435891235905239 11", "output": "Yes" }, { "input": "491802505049361659 7", "output": "Yes" }, { "input": "310113769227703889 3", "output": "Yes" }, { "input": "876240758958364799 41", "output": "Yes" }, { "input": "173284263472319999 33", "output": "Yes" }, { "input": "334366426725130799 29", "output": "Yes" }, { "input": "415543470272330399 26", "output": "Yes" }, { "input": "631689521541558479 22", "output": "Yes" }, { "input": "581859366558790319 14", "output": "Yes" }, { "input": "224113913709159599 10", "output": "Yes" }, { "input": "740368848764104559 21", "output": "Yes" }, { "input": "895803074828822159 17", "output": "Yes" }, { "input": "400349974997012039 13", "output": "Yes" }, { "input": "205439024252247599 5", "output": "Yes" }, { "input": "197688463911338399 39", "output": "Yes" }, { "input": "283175367224349599 39", "output": "Yes" }, { "input": "893208176423362799 31", "output": "Yes" }, { "input": "440681012669897999 27", "output": "Yes" }, { "input": "947403664618451039 19", "output": "Yes" }, { "input": "232435556779345919 19", "output": "Yes" }, { "input": "504428493840551279 23", "output": "Yes" }, { "input": "30019549241681999 20", "output": "Yes" }, { "input": "648000813924303839 16", "output": "Yes" }, { "input": "763169499725761451 488954176053755860", "output": "No" }, { "input": "199398459594277592 452260924647536414", "output": "No" }, { "input": "635627415167826436 192195636386541160", "output": "No" }, { "input": "71856370741375281 155502380685354417", "output": "No" }, { "input": "731457367464667229 118809129279134971", "output": "No" }, { "input": "167686318743248777 858743836723172421", "output": "No" }, { "input": "603915274316797622 822050585316952974", "output": "No" }, { "input": "647896534275160623 65689274138731296", "output": "No" }, { "input": "648722777453244047 501918229712280140", "output": "No" }, { "input": "649549020631327471 41923378183538525", "output": "No" }, { "input": "650375259514443599 597748177714153637", "output": "No" }, { "input": "651201506987494319 33977137582669778", "output": "No" }, { "input": "652027745870610447 470206093156218622", "output": "No" }, { "input": "652853989048693871 906435048729767466", "output": "No" }, { "input": "653680227931809999 342664004303316311", "output": "No" }, { "input": "654506475404860719 375019787446735639", "output": "No" }, { "input": "655332714287976847 438493956600157103", "output": "No" }, { "input": "166512305365727033 900267947832156186", "output": "No" }, { "input": "167338548543810457 336496907700672326", "output": "No" }, { "input": "168164787426926585 772725863274221171", "output": "No" }, { "input": "523 3", "output": "No" }, { "input": "39211 6", "output": "No" }, { "input": "22151 9", "output": "No" }, { "input": "1 3", "output": "No" }, { "input": "47 5", "output": "No" }, { "input": "999999998999999999 1000000000", "output": "No" }, { "input": "11 6", "output": "No" }, { "input": "7 4", "output": "No" }, { "input": "1 10", "output": "No" }, { "input": "9 5", "output": "No" }, { "input": "2519 20", "output": "No" }, { "input": "700001 3", "output": "Yes" }, { "input": "13 7", "output": "No" }, { "input": "999999 10000", "output": "No" }, { "input": "1 4", "output": "No" }, { "input": "232792559 30", "output": "No" }, { "input": "1 5", "output": "No" }, { "input": "5 4", "output": "No" }, { "input": "5 8", "output": "No" }, { "input": "55 4", "output": "No" } ]
1,614,531,826
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
2
77
0
n,k=map(int,input().split(' ')) print('YES' if k<n else 'NO')
Title: Cave Painting Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imp is watching a documentary about cave painting. Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number *n* by all integers *i* from 1 to *k*. Unfortunately, there are too many integers to analyze for Imp. Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all , 1<=≤<=*i*<=≤<=*k*, are distinct, i. e. there is no such pair (*i*,<=*j*) that: - 1<=≤<=*i*<=&lt;<=*j*<=≤<=*k*, - , where is the remainder of division *x* by *y*. Input Specification: The only line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=1018). Output Specification: Print "Yes", if all the remainders are distinct, and "No" otherwise. You can print each letter in arbitrary case (lower or upper). Demo Input: ['4 4\n', '5 3\n'] Demo Output: ['No\n', 'Yes\n'] Note: In the first sample remainders modulo 1 and 4 coincide.
```python n,k=map(int,input().split(' ')) print('YES' if k<n else 'NO') ```
0
342
A
Xenia and Divisors
PROGRAMMING
1,200
[ "greedy", "implementation" ]
null
null
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held: - *a*<=&lt;<=*b*<=&lt;<=*c*; - *a* divides *b*, *b* divides *c*. Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three. Help Xenia, find the required partition or else say that it doesn't exist.
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7. It is guaranteed that *n* is divisible by 3.
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them. If there is no solution, print -1.
[ "6\n1 1 1 2 2 2\n", "6\n2 2 1 1 4 6\n" ]
[ "-1\n", "1 2 4\n1 2 6\n" ]
none
500
[ { "input": "6\n1 1 1 2 2 2", "output": "-1" }, { "input": "6\n2 2 1 1 4 6", "output": "1 2 4\n1 2 6" }, { "input": "3\n1 2 3", "output": "-1" }, { "input": "3\n7 5 7", "output": "-1" }, { "input": "3\n1 3 4", "output": "-1" }, { "input": "3\n1 1 1", "output": "-1" }, { "input": "9\n1 3 6 6 3 1 3 1 6", "output": "1 3 6\n1 3 6\n1 3 6" }, { "input": "6\n1 2 4 1 3 5", "output": "-1" }, { "input": "3\n1 3 7", "output": "-1" }, { "input": "3\n1 1 1", "output": "-1" }, { "input": "9\n1 2 4 1 2 4 1 3 6", "output": "1 2 4\n1 2 4\n1 3 6" }, { "input": "12\n3 6 1 1 3 6 1 1 2 6 2 6", "output": "1 3 6\n1 3 6\n1 2 6\n1 2 6" }, { "input": "9\n1 1 1 4 4 4 6 2 2", "output": "-1" }, { "input": "9\n1 2 4 6 3 1 3 1 5", "output": "-1" }, { "input": "15\n2 1 2 1 3 6 1 2 1 6 1 3 4 6 4", "output": "1 2 4\n1 2 4\n1 3 6\n1 3 6\n1 2 6" }, { "input": "3\n2 3 6", "output": "-1" }, { "input": "3\n2 4 6", "output": "-1" }, { "input": "3\n2 5 6", "output": "-1" }, { "input": "3\n2 4 7", "output": "-1" }, { "input": "6\n1 2 3 4 5 6", "output": "-1" }, { "input": "3\n7 7 7", "output": "-1" }, { "input": "6\n1 2 4 7 7 7", "output": "-1" }, { "input": "6\n1 1 2 6 6 6", "output": "-1" }, { "input": "9\n1 1 1 3 3 2 4 4 6", "output": "-1" }, { "input": "6\n1 2 4 5 5 5", "output": "-1" }, { "input": "15\n1 1 1 1 1 2 2 2 2 4 4 6 6 6 6", "output": "-1" }, { "input": "6\n1 1 5 5 7 7", "output": "-1" }, { "input": "9\n1 1 1 2 3 4 5 6 7", "output": "-1" }, { "input": "6\n1 1 4 4 7 7", "output": "-1" }, { "input": "24\n1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 4 4 4 6 6 6", "output": "-1" }, { "input": "3\n1 7 6", "output": "-1" }, { "input": "6\n1 1 2 4 7 7", "output": "-1" }, { "input": "9\n1 1 1 7 7 7 7 7 7", "output": "-1" }, { "input": "9\n1 1 1 2 3 4 6 5 5", "output": "-1" } ]
1,667,422,005
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
6
46
2,867,200
import heapq n = int(input()) arr = [int(k) for k in input().split()] def helper(): obj = {} for el in arr: if el == 5 or el == 7: return -1 obj[el] = obj.get(el, 0) + 1 num_groups = n // 3 big = [] if obj.get(1, 0) != num_groups: return -1 for _ in range(num_groups): big.append([1, -1, -1]) if obj.get(6, 0) + obj.get(4, 0) != num_groups: return -1 if obj.get(2, 0) + obj.get(3, 0) != num_groups: return -1 i = 0 while obj.get(6, 0) > 0: big[i][2] = 6 obj[6] -= 1 if obj.get(3, 0) >= 1: big[i][1] = 3 obj[3] -= 1 elif obj.get(2, 0) >= 1: big[i][1] = 2 obj[2] -= 1 else: return -1 i += 1 while obj[4] > 0: big[i][2] = 4 obj[4] -= 1 if obj.get(2, 0) >= 1: big[i][1] = 2 obj[2] -= 1 else: return -1 i += 1 for el in big: if any(k == -1 for k in el): return -1 return big res = helper() if res == -1: print(-1) else: for el in res: print(' '.join([str(j) for j in el]))
Title: Xenia and Divisors Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held: - *a*<=&lt;<=*b*<=&lt;<=*c*; - *a* divides *b*, *b* divides *c*. Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three. Help Xenia, find the required partition or else say that it doesn't exist. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7. It is guaranteed that *n* is divisible by 3. Output Specification: If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them. If there is no solution, print -1. Demo Input: ['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n'] Demo Output: ['-1\n', '1 2 4\n1 2 6\n'] Note: none
```python import heapq n = int(input()) arr = [int(k) for k in input().split()] def helper(): obj = {} for el in arr: if el == 5 or el == 7: return -1 obj[el] = obj.get(el, 0) + 1 num_groups = n // 3 big = [] if obj.get(1, 0) != num_groups: return -1 for _ in range(num_groups): big.append([1, -1, -1]) if obj.get(6, 0) + obj.get(4, 0) != num_groups: return -1 if obj.get(2, 0) + obj.get(3, 0) != num_groups: return -1 i = 0 while obj.get(6, 0) > 0: big[i][2] = 6 obj[6] -= 1 if obj.get(3, 0) >= 1: big[i][1] = 3 obj[3] -= 1 elif obj.get(2, 0) >= 1: big[i][1] = 2 obj[2] -= 1 else: return -1 i += 1 while obj[4] > 0: big[i][2] = 4 obj[4] -= 1 if obj.get(2, 0) >= 1: big[i][1] = 2 obj[2] -= 1 else: return -1 i += 1 for el in big: if any(k == -1 for k in el): return -1 return big res = helper() if res == -1: print(-1) else: for el in res: print(' '.join([str(j) for j in el])) ```
-1
34
D
Road Map
PROGRAMMING
1,600
[ "dfs and similar", "graphs" ]
D. Road Map
2
256
There are *n* cities in Berland. Each city has its index — an integer number from 1 to *n*. The capital has index *r*1. All the roads in Berland are two-way. The road system is such that there is exactly one path from the capital to each city, i.e. the road map looks like a tree. In Berland's chronicles the road map is kept in the following way: for each city *i*, different from the capital, there is kept number *p**i* — index of the last city on the way from the capital to *i*. Once the king of Berland Berl XXXIV decided to move the capital from city *r*1 to city *r*2. Naturally, after this the old representation of the road map in Berland's chronicles became incorrect. Please, help the king find out a new representation of the road map in the way described above.
The first line contains three space-separated integers *n*, *r*1, *r*2 (2<=≤<=*n*<=≤<=5·104,<=1<=≤<=*r*1<=≠<=*r*2<=≤<=*n*) — amount of cities in Berland, index of the old capital and index of the new one, correspondingly. The following line contains *n*<=-<=1 space-separated integers — the old representation of the road map. For each city, apart from *r*1, there is given integer *p**i* — index of the last city on the way from the capital to city *i*. All the cities are described in order of increasing indexes.
Output *n*<=-<=1 numbers — new representation of the road map in the same format.
[ "3 2 3\n2 2\n", "6 2 4\n6 1 2 4 2\n" ]
[ "2 3 ", "6 4 1 4 2 " ]
none
2,000
[ { "input": "3 2 3\n2 2", "output": "2 3 " }, { "input": "6 2 4\n6 1 2 4 2", "output": "6 4 1 4 2 " }, { "input": "7 7 6\n7 7 5 5 7 7", "output": "7 7 5 5 7 6 " }, { "input": "4 2 3\n2 1 3", "output": "3 1 3 " }, { "input": "5 5 4\n5 4 1 5", "output": "5 4 1 4 " }, { "input": "6 2 1\n4 1 2 2 2", "output": "4 1 1 2 2 " }, { "input": "2 1 2\n1", "output": "2 " }, { "input": "2 2 1\n2", "output": "1 " }, { "input": "3 1 3\n1 2", "output": "2 3 " }, { "input": "10 4 3\n5 6 9 10 1 9 6 8 4", "output": "6 6 10 1 8 9 9 3 5 " }, { "input": "100 62 42\n45 66 13 18 93 47 83 19 41 66 5 70 16 29 54 28 65 91 92 24 90 54 7 40 77 53 68 58 70 65 68 30 94 80 16 18 69 21 94 1 99 61 97 2 69 15 79 38 61 46 88 32 57 84 35 18 17 77 38 91 84 2 53 54 68 19 13 78 62 16 63 17 76 49 28 45 94 13 35 69 37 56 12 41 95 70 57 24 45 84 87 65 14 76 95 94 31 67 5", "output": "45 66 13 18 93 47 83 19 41 66 5 84 16 29 54 28 65 91 92 24 90 54 7 40 77 53 68 58 70 65 68 30 94 80 16 18 69 21 94 1 99 97 2 69 15 79 38 61 46 88 32 57 84 35 18 17 77 38 91 42 70 2 53 54 68 19 13 78 12 16 63 17 76 49 28 45 94 13 35 69 37 56 61 41 95 70 57 24 45 84 87 65 14 76 95 94 31 67 5 " } ]
1,629,918,146
2,147,483,647
PyPy 3
COMPILATION_ERROR
TESTS
0
0
0
/* * Rishabh Rao (https://github.com/rishabhrao) */ #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include<ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; #define MOD 1000000007 #define MOD1 998244353 #define PI 3.141592653589793238462 #define INF 1e18 #define nline "\n" typedef long long ll; typedef unsigned long long ull; typedef long double lld; typedef long long int lli; #define pb push_back #define pf push_front #define ppb pop_back #define mp make_pair #define cntOnes(x) __builtin_popcount(x) #define llCntOnes(x) __builtin_popcountll(x) #define hasOddNumOnes(x) _builtin_parity(x) #define llHasOddNumOnes(x) _builtin_parityll(x) #define cntStartZeros(x) __builtin_clz(x) #define llCntStartZeros(x) __builtin_clzll(x) #define cntEndZeros(x) __builtin_ctz(x) #define llCntEndZeros(x) __builtin_ctzll(x) #define sz(x) ((int)(x).size()) #define all(x) begin(x), end(x) #define uniq(x) x.erase(unique(all(x)),x.end()); #define rep(x,start,end) for(auto x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--)) // typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key #define fastio() \ ios_base::sync_with_stdio(false); \ cin.tie(NULL); \ cout.tie(NULL) // #pragma GCC optimization("unroll-loops") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") // #pragma GCC target("fpmath=387") // #pragma GCC optimize("-O2") // #pragma GCC optimize("Ofast") /*---------------------------------------------------------------------------------------------------------------------------*/ template<typename T> ostream& operator<<(ostream& os, unordered_map<T, T> m) { for (auto& p : m) os << p.first << ": " << p.second << "\n"; return os; } template<typename T> ostream& operator<<(ostream& os, map<T, T> m) { for (auto& p : m) os << p.first << ": " << p.second << "\n"; return os; } template<typename T> istream& operator>>(istream& is, deque<T>& d) { for (auto& i : d) is >> i; return is; } template<typename T> ostream& operator<<(ostream& os, deque<T> d) { for (auto& i : d) os << i << ' '; return os; } template<typename T> istream& operator>>(istream& is, vector<T>& v) { for (auto& i : v) is >> i; return is; } template<typename T> ostream& operator<<(ostream& os, vector<T> v) { for (auto& i : v) os << i << ' '; return os; } const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count(); struct chash { // http://xorshift.di.unimi.it/splitmix64.c static uint64_t splitmix64(uint64_t x) { x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb; return x ^ (x >> 31); } size_t operator()(uint64_t x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x + FIXED_RANDOM); } size_t operator()(pair<uint64_t, uint64_t> x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x.first + FIXED_RANDOM) ^ (splitmix64(x.second + FIXED_RANDOM) >> 1); } }; int minvi(vector<int>& a) { return *min_element(a.begin(), a.end()); } ll minvl(vector<long long>& a) { return *min_element(a.begin(), a.end()); } int maxvi(vector<int>& a) { return *max_element(a.begin(), a.end()); } ll maxvl(vector<long long>& a) { return *max_element(a.begin(), a.end()); } string cins() { string inp; getline(cin, inp); return inp; } // vector<string> splitStr(string str, char delim = ' ') { vector<string> splits; for (int i = 0, start = 0, end = 0; i <= str.size(); i++) { if (str[i] == delim or i == str.size()) { splits.pb(str.substr(start, end)); start = end = i + 1; } else end++; }; return splits; } string strMult(string& str, int mul) { string out = ""; while (mul--) out += str; return out; }; unordered_map<int, int> unordNumberCounter(vector<int>& nums) { unordered_map<int, int> ctr; for (auto num : nums) ctr[num]++; return ctr; } map<int, int> ordNumberCounter(vector<int>& nums) { map<int, int> ctr; for (auto num : nums) ctr[num]++; return ctr; } string bin(uint_fast8_t i) { return !i ? "0" : i == 1 ? "1" : bin(i / 2) + (i % 2 ? '1' : '0'); } ll gcd(ll a, ll b) { if (b > a) { return gcd(b, a); } if (b == 0) { return a; } return gcd(b, a % b); } ll expo(ll a, ll b, ll mod) { ll res = 1; while (b > 0) { if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1; } return res; } void extendgcd(ll a, ll b, ll* v) { if (b == 0) { v[0] = 1; v[1] = 0; v[2] = a; return; } extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return; } //pass an arry of size1 3 ll mminv(ll a, ll b) { ll arr[3]; extendgcd(a, b, arr); return arr[0]; } //for non prime b ll mminvprime(ll a, ll b) { return expo(a, b - 2, b); } bool revsort(ll a, ll b) { return a > b; } void swap(int& x, int& y) { int temp = x; x = y; y = temp; } ll combination(ll n, ll r, ll m, ll* fact, ll* ifact) { ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m; } void google(int t) { cout << "Case #" << t << ": "; } vector<ll> sieve(int n) { int* arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) { vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1; } return vect; } ll mod_add(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a + b) % m) + m) % m; } ll mod_mul(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a * b) % m) + m) % m; } ll mod_sub(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a - b) % m) + m) % m; } ll mod_div(ll a, ll b, ll m) { a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m; } // Only for Prime m ll phin(ll n) { ll number = n; if (n % 2 == 0) { number /= 2; while (n % 2 == 0) n /= 2; } for (ll i = 3; i <= sqrt(n); i += 2) { if (n % i == 0) { while (n % i == 0)n /= i; number = (number / i * (i - 1)); } } if (n > 1)number = (number / n * (n - 1)); return number; } // O(sqrt(N)) int pow2(int n, int m = 1) { int ans = 1; while (n--) ans = (ans << 1) % m; return ans; } int pow2(ll n, int m = 1) { int ans = 1; while (n--) ans = (ans << 1) % m; return ans; } /*--------------------------------------------------------------------------------------------------------------------------*/ int main() { fastio(); int n, r1, r2; cin >> n >> r1 >> r2; vector<vector<int>> graph(n + 1, vector<int>()); for (int i = 1; i <= n; i++) { if (i == r1) continue; int par; cin >> par; graph[par].pb(i); graph[i].pb(par); } vector<int> parents(n + 1, -1); stack<int> st; st.push(r2); while (not st.empty()) { int curr = st.top(); st.pop(); for (int& child : graph[curr]) if (child != r2 and parents[child] == -1) { parents[child] = curr; st.push(child); } } for (int i = 1; i <= n; i++) if (parents[i] != -1) cout << parents[i] << " "; cout << nline; return 0; }
Title: Road Map Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There are *n* cities in Berland. Each city has its index — an integer number from 1 to *n*. The capital has index *r*1. All the roads in Berland are two-way. The road system is such that there is exactly one path from the capital to each city, i.e. the road map looks like a tree. In Berland's chronicles the road map is kept in the following way: for each city *i*, different from the capital, there is kept number *p**i* — index of the last city on the way from the capital to *i*. Once the king of Berland Berl XXXIV decided to move the capital from city *r*1 to city *r*2. Naturally, after this the old representation of the road map in Berland's chronicles became incorrect. Please, help the king find out a new representation of the road map in the way described above. Input Specification: The first line contains three space-separated integers *n*, *r*1, *r*2 (2<=≤<=*n*<=≤<=5·104,<=1<=≤<=*r*1<=≠<=*r*2<=≤<=*n*) — amount of cities in Berland, index of the old capital and index of the new one, correspondingly. The following line contains *n*<=-<=1 space-separated integers — the old representation of the road map. For each city, apart from *r*1, there is given integer *p**i* — index of the last city on the way from the capital to city *i*. All the cities are described in order of increasing indexes. Output Specification: Output *n*<=-<=1 numbers — new representation of the road map in the same format. Demo Input: ['3 2 3\n2 2\n', '6 2 4\n6 1 2 4 2\n'] Demo Output: ['2 3 ', '6 4 1 4 2 '] Note: none
```python /* * Rishabh Rao (https://github.com/rishabhrao) */ #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include<ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; #define MOD 1000000007 #define MOD1 998244353 #define PI 3.141592653589793238462 #define INF 1e18 #define nline "\n" typedef long long ll; typedef unsigned long long ull; typedef long double lld; typedef long long int lli; #define pb push_back #define pf push_front #define ppb pop_back #define mp make_pair #define cntOnes(x) __builtin_popcount(x) #define llCntOnes(x) __builtin_popcountll(x) #define hasOddNumOnes(x) _builtin_parity(x) #define llHasOddNumOnes(x) _builtin_parityll(x) #define cntStartZeros(x) __builtin_clz(x) #define llCntStartZeros(x) __builtin_clzll(x) #define cntEndZeros(x) __builtin_ctz(x) #define llCntEndZeros(x) __builtin_ctzll(x) #define sz(x) ((int)(x).size()) #define all(x) begin(x), end(x) #define uniq(x) x.erase(unique(all(x)),x.end()); #define rep(x,start,end) for(auto x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--)) // typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key #define fastio() \ ios_base::sync_with_stdio(false); \ cin.tie(NULL); \ cout.tie(NULL) // #pragma GCC optimization("unroll-loops") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") // #pragma GCC target("fpmath=387") // #pragma GCC optimize("-O2") // #pragma GCC optimize("Ofast") /*---------------------------------------------------------------------------------------------------------------------------*/ template<typename T> ostream& operator<<(ostream& os, unordered_map<T, T> m) { for (auto& p : m) os << p.first << ": " << p.second << "\n"; return os; } template<typename T> ostream& operator<<(ostream& os, map<T, T> m) { for (auto& p : m) os << p.first << ": " << p.second << "\n"; return os; } template<typename T> istream& operator>>(istream& is, deque<T>& d) { for (auto& i : d) is >> i; return is; } template<typename T> ostream& operator<<(ostream& os, deque<T> d) { for (auto& i : d) os << i << ' '; return os; } template<typename T> istream& operator>>(istream& is, vector<T>& v) { for (auto& i : v) is >> i; return is; } template<typename T> ostream& operator<<(ostream& os, vector<T> v) { for (auto& i : v) os << i << ' '; return os; } const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count(); struct chash { // http://xorshift.di.unimi.it/splitmix64.c static uint64_t splitmix64(uint64_t x) { x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb; return x ^ (x >> 31); } size_t operator()(uint64_t x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x + FIXED_RANDOM); } size_t operator()(pair<uint64_t, uint64_t> x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x.first + FIXED_RANDOM) ^ (splitmix64(x.second + FIXED_RANDOM) >> 1); } }; int minvi(vector<int>& a) { return *min_element(a.begin(), a.end()); } ll minvl(vector<long long>& a) { return *min_element(a.begin(), a.end()); } int maxvi(vector<int>& a) { return *max_element(a.begin(), a.end()); } ll maxvl(vector<long long>& a) { return *max_element(a.begin(), a.end()); } string cins() { string inp; getline(cin, inp); return inp; } // vector<string> splitStr(string str, char delim = ' ') { vector<string> splits; for (int i = 0, start = 0, end = 0; i <= str.size(); i++) { if (str[i] == delim or i == str.size()) { splits.pb(str.substr(start, end)); start = end = i + 1; } else end++; }; return splits; } string strMult(string& str, int mul) { string out = ""; while (mul--) out += str; return out; }; unordered_map<int, int> unordNumberCounter(vector<int>& nums) { unordered_map<int, int> ctr; for (auto num : nums) ctr[num]++; return ctr; } map<int, int> ordNumberCounter(vector<int>& nums) { map<int, int> ctr; for (auto num : nums) ctr[num]++; return ctr; } string bin(uint_fast8_t i) { return !i ? "0" : i == 1 ? "1" : bin(i / 2) + (i % 2 ? '1' : '0'); } ll gcd(ll a, ll b) { if (b > a) { return gcd(b, a); } if (b == 0) { return a; } return gcd(b, a % b); } ll expo(ll a, ll b, ll mod) { ll res = 1; while (b > 0) { if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1; } return res; } void extendgcd(ll a, ll b, ll* v) { if (b == 0) { v[0] = 1; v[1] = 0; v[2] = a; return; } extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return; } //pass an arry of size1 3 ll mminv(ll a, ll b) { ll arr[3]; extendgcd(a, b, arr); return arr[0]; } //for non prime b ll mminvprime(ll a, ll b) { return expo(a, b - 2, b); } bool revsort(ll a, ll b) { return a > b; } void swap(int& x, int& y) { int temp = x; x = y; y = temp; } ll combination(ll n, ll r, ll m, ll* fact, ll* ifact) { ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m; } void google(int t) { cout << "Case #" << t << ": "; } vector<ll> sieve(int n) { int* arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) { vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1; } return vect; } ll mod_add(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a + b) % m) + m) % m; } ll mod_mul(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a * b) % m) + m) % m; } ll mod_sub(ll a, ll b, ll m) { a = a % m; b = b % m; return (((a - b) % m) + m) % m; } ll mod_div(ll a, ll b, ll m) { a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m; } // Only for Prime m ll phin(ll n) { ll number = n; if (n % 2 == 0) { number /= 2; while (n % 2 == 0) n /= 2; } for (ll i = 3; i <= sqrt(n); i += 2) { if (n % i == 0) { while (n % i == 0)n /= i; number = (number / i * (i - 1)); } } if (n > 1)number = (number / n * (n - 1)); return number; } // O(sqrt(N)) int pow2(int n, int m = 1) { int ans = 1; while (n--) ans = (ans << 1) % m; return ans; } int pow2(ll n, int m = 1) { int ans = 1; while (n--) ans = (ans << 1) % m; return ans; } /*--------------------------------------------------------------------------------------------------------------------------*/ int main() { fastio(); int n, r1, r2; cin >> n >> r1 >> r2; vector<vector<int>> graph(n + 1, vector<int>()); for (int i = 1; i <= n; i++) { if (i == r1) continue; int par; cin >> par; graph[par].pb(i); graph[i].pb(par); } vector<int> parents(n + 1, -1); stack<int> st; st.push(r2); while (not st.empty()) { int curr = st.top(); st.pop(); for (int& child : graph[curr]) if (child != r2 and parents[child] == -1) { parents[child] = curr; st.push(child); } } for (int i = 1; i <= n; i++) if (parents[i] != -1) cout << parents[i] << " "; cout << nline; return 0; } ```
-1
281
A
Word Capitalization
PROGRAMMING
800
[ "implementation", "strings" ]
null
null
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged.
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output the given word after capitalization.
[ "ApPLe\n", "konjac\n" ]
[ "ApPLe\n", "Konjac\n" ]
none
500
[ { "input": "ApPLe", "output": "ApPLe" }, { "input": "konjac", "output": "Konjac" }, { "input": "a", "output": "A" }, { "input": "A", "output": "A" }, { "input": "z", "output": "Z" }, { "input": "ABACABA", "output": "ABACABA" }, { "input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX", "output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX" }, { "input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO", "output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO" }, { "input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD", "output": 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"UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX" }, { "input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ", "output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ" }, { "input": "P", "output": "P" }, { "input": "Xyzzy", "output": "Xyzzy" }, { "input": "Zzz", "output": "Zzz" }, { "input": "Zp", "output": "Zp" } ]
1,699,377,965
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
import java.util.*; public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); String s=sc.next(); String str; str=s.substring(0,1).toUpperCase()+s.substring(1); System.out.println(str); #output } }
Title: Word Capitalization Time Limit: None seconds Memory Limit: None megabytes Problem Description: Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged. Input Specification: A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. Output Specification: Output the given word after capitalization. Demo Input: ['ApPLe\n', 'konjac\n'] Demo Output: ['ApPLe\n', 'Konjac\n'] Note: none
```python import java.util.*; public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); String s=sc.next(); String str; str=s.substring(0,1).toUpperCase()+s.substring(1); System.out.println(str); #output } } ```
-1
862
A
Mahmoud and Ehab and the MEX
PROGRAMMING
1,000
[ "greedy", "implementation" ]
null
null
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go. Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 . Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX. The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
[ "5 3\n0 4 5 6 7\n", "1 0\n0\n", "5 0\n1 2 3 4 5\n" ]
[ "2\n", "1\n", "0\n" ]
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations. For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0. In the third test case the set is already evil.
500
[ { "input": "5 3\n0 4 5 6 7", "output": "2" }, { "input": "1 0\n0", "output": "1" }, { "input": "5 0\n1 2 3 4 5", "output": "0" }, { "input": "10 5\n57 1 47 9 93 37 76 70 78 15", "output": "4" }, { "input": "10 5\n99 98 93 97 95 100 92 94 91 96", "output": "5" }, { "input": "10 5\n1 2 3 4 59 45 0 58 51 91", "output": "0" }, { "input": "100 100\n79 13 21 11 3 87 28 40 29 4 96 34 8 78 61 46 33 45 99 30 92 67 22 97 39 86 73 31 74 44 62 55 57 2 54 63 80 69 25 48 77 98 17 93 15 16 89 12 43 23 37 95 14 38 83 90 49 56 72 10 20 0 50 71 70 88 19 1 76 81 52 41 82 68 85 47 6 7 35 60 18 64 75 84 27 9 65 91 94 42 53 24 66 26 59 36 51 32 5 58", "output": "0" }, { "input": "100 50\n95 78 46 92 80 18 79 58 30 72 19 89 39 29 44 65 15 100 59 8 96 9 62 67 41 42 82 14 57 32 71 77 40 5 7 51 28 53 85 23 16 35 3 91 6 11 75 61 17 66 13 47 36 56 10 22 83 60 48 24 26 97 4 33 76 86 70 0 34 64 52 43 21 49 55 74 1 73 81 25 54 63 94 84 20 68 87 12 31 88 38 93 37 90 98 69 99 45 27 2", "output": "0" }, { "input": "100 33\n28 11 79 92 88 62 77 72 7 41 96 97 67 84 44 8 81 35 38 1 64 68 46 17 98 83 31 12 74 21 2 22 47 6 36 75 65 61 37 26 25 45 59 48 100 51 93 76 78 49 3 57 16 4 87 29 55 82 70 39 53 0 60 15 24 71 58 20 66 89 95 42 13 43 63 90 85 52 50 30 54 40 56 23 27 34 32 18 10 19 69 9 99 73 91 14 5 80 94 86", "output": "0" }, { "input": "99 33\n25 76 41 95 55 20 47 59 58 84 87 92 16 27 35 65 72 63 93 54 36 96 15 86 5 69 24 46 67 73 48 60 40 6 61 74 97 10 100 8 52 26 77 18 7 62 37 2 14 66 11 56 68 91 0 64 75 99 30 21 53 1 89 81 3 98 12 88 39 38 29 83 22 90 9 28 45 43 78 44 32 57 4 50 70 17 13 51 80 85 71 94 82 19 34 42 23 79 49", "output": "1" }, { "input": "100 100\n65 56 84 46 44 33 99 74 62 72 93 67 43 92 75 88 38 34 66 12 55 76 58 90 78 8 14 45 97 59 48 32 64 18 39 89 31 51 54 81 29 36 70 77 40 22 49 27 3 1 73 13 98 42 87 37 2 57 4 6 50 25 23 79 28 86 68 61 80 17 19 10 15 63 52 11 35 60 21 16 24 85 30 91 7 5 69 20 71 82 53 94 41 95 96 9 26 83 0 47", "output": "0" }, { "input": "100 100\n58 88 12 71 22 1 40 19 73 20 67 48 57 17 69 36 100 35 33 37 72 55 52 8 89 85 47 42 78 70 81 86 11 9 68 99 6 16 21 61 53 98 23 62 32 59 51 0 87 24 50 30 65 10 80 95 7 92 25 74 60 79 91 5 13 31 75 38 90 94 46 66 93 34 14 41 28 2 76 84 43 96 3 56 49 82 27 77 64 63 4 45 18 29 54 39 15 26 83 44", "output": "2" }, { "input": "89 100\n58 96 17 41 86 34 28 84 18 40 8 77 87 89 68 79 33 35 53 49 0 6 22 12 72 90 48 55 21 50 56 62 75 2 37 95 69 74 14 20 44 46 27 32 31 59 63 60 10 85 71 70 38 52 94 30 61 51 80 26 36 23 39 47 76 45 100 57 15 78 97 66 54 13 99 16 93 73 24 4 83 5 98 81 92 25 29 88 65", "output": "13" }, { "input": "100 50\n7 95 24 76 81 78 60 69 83 84 100 1 65 31 48 92 73 39 18 89 38 97 10 42 8 55 98 51 21 90 62 77 16 91 0 94 4 37 19 17 67 35 45 41 56 20 15 85 75 28 59 27 12 54 61 68 36 5 79 93 66 11 70 49 50 34 30 25 96 46 64 14 32 22 47 40 58 23 43 9 87 82 26 53 80 52 3 86 13 99 33 71 6 88 57 74 2 44 72 63", "output": "2" }, { "input": "77 0\n27 8 20 92 21 41 53 98 17 65 67 35 81 11 55 49 61 44 2 66 51 89 40 28 52 62 86 91 64 24 18 5 94 82 96 99 71 6 39 83 26 29 16 30 45 97 80 90 69 12 13 33 76 73 46 19 78 56 88 38 42 34 57 77 47 4 59 58 7 100 95 72 9 74 15 43 54", "output": "0" }, { "input": "100 50\n55 36 0 32 81 6 17 43 24 13 30 19 8 59 71 45 15 74 3 41 99 42 86 47 2 94 35 1 66 95 38 49 4 27 96 89 34 44 92 25 51 39 54 28 80 77 20 14 48 40 68 56 31 63 33 78 69 37 18 26 83 70 23 82 91 65 67 52 61 53 7 22 60 21 12 73 72 87 75 100 90 29 64 79 98 85 5 62 93 84 50 46 97 58 57 16 9 10 76 11", "output": "1" }, { "input": "77 0\n12 8 19 87 9 54 55 86 97 7 27 85 25 48 94 73 26 1 13 57 72 69 76 39 38 91 75 40 42 28 93 21 70 84 65 11 60 90 20 95 66 89 59 47 34 99 6 61 52 100 50 3 77 81 82 53 15 24 0 45 44 14 68 96 58 5 18 35 10 98 29 74 92 49 83 71 17", "output": "1" }, { "input": "100 70\n25 94 66 65 10 99 89 6 70 31 7 40 20 92 64 27 21 72 77 98 17 43 47 44 48 81 38 56 100 39 90 22 88 76 3 83 86 29 33 55 82 79 49 11 2 16 12 78 85 69 32 97 26 15 53 24 23 91 51 67 34 35 52 5 62 50 95 18 71 13 75 8 30 42 93 36 45 60 63 46 57 41 87 0 84 54 74 37 4 58 28 19 96 61 80 9 1 14 73 68", "output": "2" }, { "input": "89 19\n14 77 85 81 79 38 91 45 55 51 50 11 62 67 73 76 2 27 16 23 3 29 65 98 78 17 4 58 22 20 34 66 64 31 72 5 32 44 12 75 80 47 18 25 99 0 61 56 71 84 48 88 10 7 86 8 49 24 43 21 37 28 33 54 46 57 40 89 36 97 6 96 39 95 26 74 1 69 9 100 52 30 83 87 68 60 92 90 35", "output": "2" }, { "input": "89 100\n69 61 56 45 11 41 42 32 28 29 0 76 7 65 13 35 36 82 10 39 26 34 38 40 92 12 17 54 24 46 88 70 66 27 100 52 85 62 22 48 86 68 21 49 53 94 67 20 1 90 77 84 31 87 58 47 95 33 4 72 93 83 8 51 91 80 99 43 71 19 44 59 98 97 64 9 81 16 79 63 25 37 3 75 2 55 50 6 18", "output": "13" }, { "input": "77 0\n38 76 24 74 42 88 29 75 96 46 90 32 59 97 98 60 41 57 80 37 100 49 25 63 95 31 61 68 53 78 27 66 84 48 94 83 30 26 36 99 71 62 45 47 70 28 35 54 34 85 79 43 91 72 86 33 67 92 77 65 69 52 82 55 87 64 56 40 50 44 51 73 89 81 58 93 39", "output": "0" }, { "input": "89 100\n38 90 80 64 35 44 56 11 15 89 23 12 49 70 72 60 63 85 92 10 45 83 8 88 41 33 16 6 61 76 62 71 87 13 25 77 74 0 1 37 96 93 7 94 21 82 34 78 4 73 65 20 81 95 50 32 48 17 69 55 68 5 51 27 53 43 91 67 59 46 86 84 99 24 22 3 97 98 40 36 26 58 57 9 42 30 52 2 47", "output": "11" }, { "input": "77 0\n55 71 78 86 68 35 53 10 59 32 81 19 74 97 62 61 93 87 96 44 25 18 43 82 84 16 34 48 92 39 64 36 49 91 45 76 95 31 57 29 75 79 13 2 14 24 52 23 33 20 47 99 63 15 5 80 58 67 12 3 85 6 1 27 73 90 4 42 37 70 8 11 89 77 9 22 94", "output": "0" }, { "input": "77 0\n12 75 31 71 44 8 3 82 21 77 50 29 57 74 40 10 15 42 84 2 100 9 28 72 92 0 49 11 90 55 17 36 19 54 68 52 4 69 97 91 5 39 59 45 89 62 53 83 16 94 76 60 95 47 30 51 7 48 20 70 67 32 58 78 63 34 56 93 99 88 24 1 66 22 25 14 13", "output": "1" }, { "input": "100 70\n91 82 8 85 26 25 95 97 40 87 81 93 7 73 38 94 64 96 74 18 90 19 65 68 72 61 23 43 36 41 60 88 30 33 71 24 52 39 15 3 16 89 86 79 55 4 9 58 67 44 46 29 6 48 84 69 27 21 78 54 51 57 80 53 76 50 47 77 45 12 34 10 100 0 17 31 56 99 98 11 92 5 2 42 32 59 66 62 37 63 28 75 35 1 22 13 83 49 20 14", "output": "0" }, { "input": "77 0\n51 5 81 62 30 22 11 0 83 16 79 85 52 70 69 10 8 47 58 3 24 34 44 14 82 66 99 17 28 31 64 67 23 49 94 45 4 12 27 15 21 6 43 72 87 2 63 92 35 39 59 9 90 78 93 20 65 36 60 89 50 41 61 84 77 86 76 100 38 68 53 97 96 95 7 19 88", "output": "1" }, { "input": "1 100\n0", "output": "99" }, { "input": "1 0\n100", "output": "0" }, { "input": "1 100\n100", "output": "101" }, { "input": "2 100\n0 100", "output": "100" }, { "input": "5 5\n1 2 3 4 5", "output": "2" }, { "input": "5 3\n0 3 4 5 6", "output": "3" }, { "input": "7 10\n0 1 2 3 4 5 10", "output": "5" }, { "input": "2 2\n0 2", "output": "2" }, { "input": "2 1\n1 2", "output": "2" }, { "input": "5 1\n1 2 3 4 5", "output": "2" }, { "input": "5 2\n1 2 3 4 5", "output": "2" }, { "input": "5 6\n0 1 2 3 4", "output": "1" }, { "input": "3 2\n3 4 5", "output": "2" } ]
1,507,788,289
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
11
61
0
n, x = map(int, input().split()) set1 = set(map(int, input().split())) if x == 0 and x in set1: print(n) else: set2 = set() for i in range(0, x): set2.add(i) set3 = set2 - set1 if x in set3: print(len(set3) - 1) else: print(len(set3))
Title: Mahmoud and Ehab and the MEX Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go. Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 . Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil? Input Specification: The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX. The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set. Output Specification: The only line should contain one integer — the minimal number of operations Dr. Evil should perform. Demo Input: ['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n'] Demo Output: ['2\n', '1\n', '0\n'] Note: For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations. For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0. In the third test case the set is already evil.
```python n, x = map(int, input().split()) set1 = set(map(int, input().split())) if x == 0 and x in set1: print(n) else: set2 = set() for i in range(0, x): set2.add(i) set3 = set2 - set1 if x in set3: print(len(set3) - 1) else: print(len(set3)) ```
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" } ]
1,686,124,208
2,147,483,647
Python 3
OK
TESTS
40
46
0
s=input() j=[] o=0 while o<len (s): if s[o] == 'h' and len(j)==0: j.append('h') elif s[o] == 'e' and len(j)==1: j.append('e') elif s[o] == 'l' and (len(j)==2 or len(j)==3): j.append('l') elif s[o] == 'o' and len(j)==4: j.append('l') o+=1 if len(j)==5: 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() j=[] o=0 while o<len (s): if s[o] == 'h' and len(j)==0: j.append('h') elif s[o] == 'e' and len(j)==1: j.append('e') elif s[o] == 'l' and (len(j)==2 or len(j)==3): j.append('l') elif s[o] == 'o' and len(j)==4: j.append('l') o+=1 if len(j)==5: print('YES') else: print('NO') ```
3.977
55
B
Smallest number
PROGRAMMING
1,600
[ "brute force" ]
B. Smallest number
2
256
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)
Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
[ "1 1 1 1\n+ + *\n", "2 2 2 2\n* * +\n", "1 2 3 4\n* + +\n" ]
[ "3\n", "8\n", "9\n" ]
none
1,000
[ { "input": "1 1 1 1\n+ + *", "output": "3" }, { "input": "2 2 2 2\n* * +", "output": "8" }, { "input": "1 2 3 4\n* + +", "output": "9" }, { "input": "15 1 3 1\n* * +", "output": "18" }, { "input": "8 1 7 14\n+ + +", "output": "30" }, { "input": "7 17 3 25\n+ * +", "output": "63" }, { "input": "13 87 4 17\n* * *", "output": "76908" }, { "input": "7 0 8 15\n+ + *", "output": "0" }, { "input": "52 0 43 239\n+ + +", "output": "334" }, { "input": "1000 1000 999 1000\n* * *", "output": "999000000000" }, { "input": "720 903 589 804\n* * *", "output": "307887168960" }, { "input": "631 149 496 892\n* * +", "output": "445884" }, { "input": "220 127 597 394\n* + +", "output": "28931" }, { "input": "214 862 466 795\n+ + +", "output": "2337" }, { "input": "346 290 587 525\n* * *", "output": "30922279500" }, { "input": "323 771 559 347\n+ * *", "output": "149067730" }, { "input": "633 941 836 254\n* + +", "output": "162559" }, { "input": "735 111 769 553\n+ * *", "output": "92320032" }, { "input": "622 919 896 120\n* * +", "output": "667592" }, { "input": "652 651 142 661\n+ + +", "output": "2106" }, { "input": "450 457 975 35\n* * *", "output": "7017806250" }, { "input": "883 954 804 352\n* * +", "output": "1045740" }, { "input": "847 206 949 358\n* + *", "output": "62660050" }, { "input": "663 163 339 76\n+ + +", "output": "1241" }, { "input": "990 330 253 553\n+ * +", "output": "85033" }, { "input": "179 346 525 784\n* * *", "output": "25492034400" }, { "input": "780 418 829 778\n+ + *", "output": "997766" }, { "input": "573 598 791 124\n* * *", "output": "33608874936" }, { "input": "112 823 202 223\n* * +", "output": "137222" }, { "input": "901 166 994 315\n* + *", "output": "47278294" }, { "input": "393 342 840 486\n+ * *", "output": "178222356" }, { "input": "609 275 153 598\n+ + *", "output": "226746" }, { "input": "56 828 386 57\n+ * *", "output": "3875088" }, { "input": "944 398 288 986\n+ + *", "output": "670464" }, { "input": "544 177 162 21\n+ + *", "output": "18543" }, { "input": "105 238 316 265\n+ + +", "output": "924" }, { "input": "31 353 300 911\n* * *", "output": "2990721900" }, { "input": "46 378 310 194\n* * +", "output": "77528" }, { "input": "702 534 357 657\n+ * *", "output": "259077042" }, { "input": "492 596 219 470\n+ + *", "output": "341202" }, { "input": "482 842 982 902\n+ * +", "output": "407728" }, { "input": "827 578 394 351\n* * *", "output": "66105361764" }, { "input": "901 884 426 451\n* + *", "output": "170223210" }, { "input": "210 295 12 795\n* * +", "output": "71490" }, { "input": "40 734 948 202\n+ * *", "output": "13590560" }, { "input": "136 611 963 195\n+ + *", "output": "240584" }, { "input": "695 74 871 760\n+ * +", "output": "53061" }, { "input": "666 884 772 54\n* + +", "output": "37620" }, { "input": "975 785 753 224\n+ * +", "output": "170432" }, { "input": "35 187 126 596\n+ + +", "output": "944" }, { "input": "243 386 431 35\n* + *", "output": "3298015" }, { "input": "229 602 133 635\n* * +", "output": "222313" }, { "input": "916 207 238 891\n+ + *", "output": "423315" }, { "input": "922 145 883 357\n+ + *", "output": "313490" }, { "input": "69 355 762 111\n* + +", "output": "8776" }, { "input": "209 206 34 67\n* + *", "output": "476374" }, { "input": "693 824 375 361\n* * +", "output": "557339" }, { "input": "45 712 635 467\n* + +", "output": "22362" }, { "input": "426 283 179 211\n+ + +", "output": "1099" }, { "input": "802 387 686 12\n+ + +", "output": "1887" } ]
1,678,947,099
2,147,483,647
PyPy 3-64
OK
TESTS
60
124
0
nums = list(map(str,input().split())) operators = list(map(str,input().split())) ans = float("inf") def smallestnum(nums,m): global ans if len(nums)==1: ans = min(ans,int(nums[0])) return for i in range(len(nums)): for j in range(i+1,len(nums)): new_list = [v for k,v in enumerate(nums) if k!=i and k!=j] op = eval(nums[i]+operators[m]+nums[j]) new_list.append(str(op)) smallestnum(new_list,m+1) new_list.remove(str(op)) smallestnum(nums,0) print(ans)
Title: Smallest number Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations. Input Specification: First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication) Output Specification: Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d). Demo Input: ['1 1 1 1\n+ + *\n', '2 2 2 2\n* * +\n', '1 2 3 4\n* + +\n'] Demo Output: ['3\n', '8\n', '9\n'] Note: none
```python nums = list(map(str,input().split())) operators = list(map(str,input().split())) ans = float("inf") def smallestnum(nums,m): global ans if len(nums)==1: ans = min(ans,int(nums[0])) return for i in range(len(nums)): for j in range(i+1,len(nums)): new_list = [v for k,v in enumerate(nums) if k!=i and k!=j] op = eval(nums[i]+operators[m]+nums[j]) new_list.append(str(op)) smallestnum(new_list,m+1) new_list.remove(str(op)) smallestnum(nums,0) print(ans) ```
3.969
869
B
The Eternal Immortality
PROGRAMMING
1,100
[ "math" ]
null
null
Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
[ "2 4\n", "0 10\n", "107 109\n" ]
[ "2\n", "0\n", "2\n" ]
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
1,000
[ { "input": "2 4", "output": "2" }, { "input": "0 10", "output": "0" }, { "input": "107 109", "output": "2" }, { "input": "10 13", "output": "6" }, { "input": "998244355 998244359", "output": "4" }, { "input": "999999999000000000 1000000000000000000", "output": "0" }, { "input": "2 3", "output": "3" }, { "input": "3 15", "output": "0" }, { "input": "24 26", "output": "0" }, { "input": "14 60", "output": "0" }, { "input": "11 79", "output": "0" }, { "input": "1230 1232", "output": "2" }, { "input": "2633 2634", "output": "4" }, { "input": "535 536", "output": "6" }, { "input": "344319135 396746843", "output": "0" }, { "input": "696667767 696667767", "output": "1" }, { "input": "419530302 610096911", "output": "0" }, { "input": "238965115 821731161", "output": "0" }, { "input": "414626436 728903812", "output": "0" }, { "input": "274410639 293308324", "output": "0" }, { "input": "650636673091305697 650636673091305702", "output": "0" }, { "input": "651240548333620923 651240548333620924", "output": "4" }, { "input": "500000000000000000 1000000000000000000", "output": "0" }, { "input": "999999999999999999 1000000000000000000", "output": "0" }, { "input": "1000000000000000000 1000000000000000000", "output": "1" }, { "input": "0 4", "output": "4" }, { "input": "50000000062000007 50000000062000011", "output": "0" }, { "input": "0 0", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "0 2", "output": "2" }, { "input": "10000000000012 10000000000015", "output": "0" }, { "input": "5 5", "output": "1" }, { "input": "12 23", "output": "0" }, { "input": "0 11", "output": "0" }, { "input": "11111234567890 11111234567898", "output": "0" }, { "input": "0 3", "output": "6" }, { "input": "1 2", "output": "2" }, { "input": "999999999999999997 999999999999999999", "output": "2" }, { "input": "4 5", "output": "5" }, { "input": "0 1", "output": "1" }, { "input": "101 1002", "output": "0" }, { "input": "0 100000000000000001", "output": "0" }, { "input": "99999999999999997 99999999999999999", "output": "2" }, { "input": "14 15", "output": "5" }, { "input": "8 19", "output": "0" }, { "input": "12 22", "output": "0" }, { "input": "999999999999996 999999999999999", "output": "4" }, { "input": "1 3", "output": "6" }, { "input": "124 125", "output": "5" }, { "input": "11 32", "output": "0" }, { "input": "0 5", "output": "0" }, { "input": "0 999999", "output": "0" }, { "input": "151151151515 151151151526", "output": "0" }, { "input": "6 107", "output": "0" }, { "input": "5 16", "output": "0" }, { "input": "7 16", "output": "0" }, { "input": "6 19", "output": "0" }, { "input": "11113111111111 13111111111111", "output": "0" }, { "input": "1 1000", "output": "0" }, { "input": "24 25", "output": "5" }, { "input": "0 100000000000", "output": "0" }, { "input": "1 22", "output": "0" }, { "input": "999999999999999996 999999999999999999", "output": "4" } ]
1,643,957,061
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
15
0
import sys nums = [int(x) for x in input().split()] first_number, second_number = nums def factorial(num): if num <= 1: return 1 else: return num * factorial(num-1) first_number_factorial = factorial(first_number) second_number_factorial = factorial(second_number) value = second_number_factorial / first_number_factorial print(str(value[-1]))
Title: The Eternal Immortality Time Limit: None seconds Memory Limit: None megabytes Problem Description: Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this. The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1. Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer. As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge. Input Specification: The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018). Output Specification: Output one line containing a single decimal digit — the last digit of the value that interests Koyomi. Demo Input: ['2 4\n', '0 10\n', '107 109\n'] Demo Output: ['2\n', '0\n', '2\n'] Note: In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2; In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0; In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
```python import sys nums = [int(x) for x in input().split()] first_number, second_number = nums def factorial(num): if num <= 1: return 1 else: return num * factorial(num-1) first_number_factorial = factorial(first_number) second_number_factorial = factorial(second_number) value = second_number_factorial / first_number_factorial print(str(value[-1])) ```
-1
991
B
Getting an A
PROGRAMMING
900
[ "greedy", "sortings" ]
null
null
Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system. The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically  — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$. This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo. Help Vasya — calculate the minimum amount of lab works Vasya has to redo.
The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$). The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works.
Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$.
[ "3\n4 4 4\n", "4\n5 4 5 5\n", "4\n5 3 3 5\n" ]
[ "2\n", "0\n", "1\n" ]
In the first sample, it is enough to redo two lab works to make two $4$s into $5$s. In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$. In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$.
1,000
[ { "input": "3\n4 4 4", "output": "2" }, { "input": "4\n5 4 5 5", "output": "0" }, { "input": "4\n5 3 3 5", "output": "1" }, { "input": "1\n5", "output": "0" }, { "input": "4\n3 2 5 4", "output": "2" }, { "input": "5\n5 4 3 2 5", "output": "2" }, { "input": "8\n5 4 2 5 5 2 5 5", "output": "1" }, { "input": "5\n5 5 2 5 5", "output": "1" }, { "input": "6\n5 5 5 5 5 2", "output": "0" }, { "input": "6\n2 2 2 2 2 2", "output": "5" }, { "input": "100\n3 2 4 3 3 3 4 2 3 5 5 2 5 2 3 2 4 4 4 5 5 4 2 5 4 3 2 5 3 4 3 4 2 4 5 4 2 4 3 4 5 2 5 3 3 4 2 2 4 4 4 5 4 3 3 3 2 5 2 2 2 3 5 4 3 2 4 5 5 5 2 2 4 2 3 3 3 5 3 2 2 4 5 5 4 5 5 4 2 3 2 2 2 2 5 3 5 2 3 4", "output": "40" }, { "input": "1\n2", "output": "1" }, { "input": "1\n3", "output": "1" }, { "input": "1\n4", "output": "1" }, { "input": "4\n3 2 5 5", "output": "1" }, { "input": "6\n4 3 3 3 3 4", "output": "4" }, { "input": "8\n3 3 5 3 3 3 5 5", "output": "3" }, { "input": "10\n2 4 5 5 5 5 2 3 3 2", "output": "3" }, { "input": "20\n5 2 5 2 2 2 2 2 5 2 2 5 2 5 5 2 2 5 2 2", "output": "10" }, { "input": "25\n4 4 4 4 3 4 3 3 3 3 3 4 4 3 4 4 4 4 4 3 3 3 4 3 4", "output": "13" }, { "input": "30\n4 2 4 2 4 2 2 4 4 4 4 2 4 4 4 2 2 2 2 4 2 4 4 4 2 4 2 4 2 2", "output": "15" }, { "input": "52\n5 3 4 4 4 3 5 3 4 5 3 4 4 3 5 5 4 3 3 3 4 5 4 4 5 3 5 3 5 4 5 5 4 3 4 5 3 4 3 3 4 4 4 3 5 3 4 5 3 5 4 5", "output": "14" }, { "input": "77\n5 3 2 3 2 3 2 3 5 2 2 3 3 3 3 5 3 3 2 2 2 5 5 5 5 3 2 2 5 2 3 2 2 5 2 5 3 3 2 2 5 5 2 3 3 2 3 3 3 2 5 5 2 2 3 3 5 5 2 2 5 5 3 3 5 5 2 2 5 2 2 5 5 5 2 5 2", "output": "33" }, { "input": "55\n3 4 2 3 3 2 4 4 3 3 4 2 4 4 3 3 2 3 2 2 3 3 2 3 2 3 2 4 4 3 2 3 2 3 3 2 2 4 2 4 4 3 4 3 2 4 3 2 4 2 2 3 2 3 4", "output": "34" }, { "input": "66\n5 4 5 5 4 4 4 4 4 2 5 5 2 4 2 2 2 5 4 4 4 4 5 2 2 5 5 2 2 4 4 2 4 2 2 5 2 5 4 5 4 5 4 4 2 5 2 4 4 4 2 2 5 5 5 5 4 4 4 4 4 2 4 5 5 5", "output": "16" }, { "input": "99\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", "output": "83" }, { "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": "84" }, { "input": "99\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "75" }, { "input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "75" }, { "input": "99\n2 2 3 3 3 3 3 2 2 3 2 3 2 3 2 2 3 2 3 2 3 3 3 3 2 2 2 2 3 2 3 3 3 3 3 2 3 3 3 3 2 3 2 3 3 3 2 3 2 3 3 3 3 2 2 3 2 3 2 3 2 3 2 2 2 3 3 2 3 2 2 2 2 2 2 2 2 3 3 3 3 2 3 2 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3", "output": "75" }, { "input": "100\n3 2 3 3 2 2 3 2 2 3 3 2 3 2 2 2 2 2 3 2 2 2 3 2 3 3 2 2 3 2 2 2 2 3 2 3 3 2 2 3 2 2 3 2 3 2 2 3 2 3 2 2 3 2 2 3 3 3 3 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 2 3 3 2 2 3 3 3 3 2 3 2 2 2 3 3 2 2 3 2 2 2 2 3 2 2", "output": "75" }, { "input": "99\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 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": "50" }, { "input": "100\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 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 4", "output": "50" }, { "input": "99\n2 2 2 2 4 2 2 2 2 4 4 4 4 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 4 2 2 2 4 4 2 2 2 2 4 4 4 2 2 2 4 4 2 4 2 4 2 2 4 2 4 4 4 4 4 2 2 4 4 4 2 2 2 2 4 2 4 2 2 2 2 2 2 4 4 2 4 2 2 4 2 2 2 2 2 4 2 4 2 2 4 4 4", "output": "54" }, { "input": "100\n4 2 4 4 2 4 2 2 4 4 4 4 4 4 4 4 4 2 4 4 2 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 2 2 4 4 2 4 2 4 4 4 2 2 2 2 2 2 2 4 2 2 2 4 4 4 2 2 2 2 4 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 4 4 4 4 2 4 2 2 4", "output": "50" }, { "input": "99\n4 3 4 4 4 4 4 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 3 4 3 3 3 3 4 3 4 4 3 4 4 3 3 4 4 4 3 3 3 4 4 3 3 4 3 4 3 4 3 4 3 3 3 3 4 3 4 4 4 4 4 4 3 4 4 3 3 3 3 3 3 3 3 4 3 3 3 4 4 4 4 4 4 3 3 3 3 4 4 4 3 3 4 3", "output": "51" }, { "input": "100\n3 3 4 4 4 4 4 3 4 4 3 3 3 3 4 4 4 4 4 4 3 3 3 4 3 4 3 4 3 3 4 3 3 3 3 3 3 3 3 4 3 4 3 3 4 3 3 3 4 4 3 4 4 3 3 4 4 4 4 4 4 3 4 4 3 4 3 3 3 4 4 3 3 4 4 3 4 4 4 3 3 4 3 3 4 3 4 3 4 3 3 4 4 4 3 3 4 3 3 4", "output": "51" }, { "input": "99\n3 3 4 4 4 2 4 4 3 2 3 4 4 4 2 2 2 3 2 4 4 2 4 3 2 2 2 4 2 3 4 3 4 2 3 3 4 2 3 3 2 3 4 4 3 2 4 3 4 3 3 3 3 3 4 4 3 3 4 4 2 4 3 4 3 2 3 3 3 4 4 2 4 4 2 3 4 2 3 3 3 4 2 2 3 2 4 3 2 3 3 2 3 4 2 3 3 2 3", "output": "58" }, { "input": "100\n2 2 4 2 2 3 2 3 4 4 3 3 4 4 4 2 3 2 2 3 4 2 3 2 4 3 4 2 3 3 3 2 4 3 3 2 2 3 2 4 4 2 4 3 4 4 3 3 3 2 4 2 2 2 2 2 2 3 2 3 2 3 4 4 4 2 2 3 4 4 3 4 3 3 2 3 3 3 4 3 2 3 3 2 4 2 3 3 4 4 3 3 4 3 4 3 3 4 3 3", "output": "61" }, { "input": "99\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "0" }, { "input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "0" }, { "input": "99\n2 2 2 2 2 5 2 2 5 2 5 2 5 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 2 5 5 2 5 2 2 5 2 5 2 2 5 5 2 2 2 2 5 5 2 2 2 5 2 2 5 2 2 2 2 2 5 5 5 5 2 2 5 2 5 2 2 2 2 2 5 2 2 5 5 2 2 2 2 2 5 5 2 2 5 5 2 2 2 2 5 5 5 2 5", "output": "48" }, { "input": "100\n5 5 2 2 2 2 2 2 5 5 2 5 2 2 2 2 5 2 5 2 5 5 2 5 5 2 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 5 5 5 2 5 5 5 5 5 5 2 2 5 2 2 5 5 5 5 5 2 5 2 5 2 2 2 5 2 5 2 5 5 2 5 5 2 2 5 2 5 5 2 5 2 2 5 2 2 2 5 2 2 2 2 5 5 2 5", "output": "38" }, { "input": "99\n5 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 5 5 3 5 5 3 3 5 5 5 3 5 3 3 3 3 5 3 3 5 5 3 5 5 5 3 5 3 5 3 5 5 5 5 3 3 3 5 3 5 3 3 3 5 5 5 5 5 3 5 5 3 3 5 5 3 5 5 3 5 5 3 3 5 5 5 3 3 3 5 3 3 3", "output": "32" }, { "input": "100\n3 3 3 5 3 3 3 3 3 3 5 5 5 5 3 3 3 3 5 3 3 3 3 3 5 3 5 3 3 5 5 5 5 5 5 3 3 5 3 3 5 3 5 5 5 3 5 3 3 3 3 3 3 3 3 3 3 3 5 5 3 5 3 5 5 3 5 3 3 5 3 5 5 5 5 3 5 3 3 3 5 5 5 3 3 3 5 3 5 5 5 3 3 3 5 3 5 5 3 5", "output": "32" }, { "input": "99\n5 3 5 5 3 3 3 2 2 5 2 5 3 2 5 2 5 2 3 5 3 2 3 2 5 5 2 2 3 3 5 5 3 5 5 2 3 3 5 2 2 5 3 2 5 2 3 5 5 2 5 2 2 5 3 3 5 3 3 5 3 2 3 5 3 2 3 2 3 2 2 2 2 5 2 2 3 2 5 5 5 3 3 2 5 3 5 5 5 2 3 2 5 5 2 5 2 5 3", "output": "39" }, { "input": "100\n3 5 3 3 5 5 3 3 2 5 5 3 3 3 2 2 3 2 5 3 2 2 3 3 3 3 2 5 3 2 3 3 5 2 2 2 3 2 3 5 5 3 2 5 2 2 5 5 3 5 5 5 2 2 5 5 3 3 2 2 2 5 3 3 2 2 3 5 3 2 3 5 5 3 2 3 5 5 3 3 2 3 5 2 5 5 5 5 5 5 3 5 3 2 3 3 2 5 2 2", "output": "42" }, { "input": "99\n4 4 4 5 4 4 5 5 4 4 5 5 5 4 5 4 5 5 5 4 4 5 5 5 5 4 5 5 5 4 4 5 5 4 5 4 4 4 5 5 5 5 4 4 5 4 4 5 4 4 4 4 5 5 5 4 5 4 5 5 5 5 5 4 5 4 5 4 4 4 4 5 5 5 4 5 5 4 4 5 5 5 4 5 4 4 5 5 4 5 5 5 5 4 5 5 4 4 4", "output": "0" }, { "input": "100\n4 4 5 5 5 5 5 5 4 4 5 5 4 4 5 5 4 5 4 4 4 4 4 4 4 4 5 5 5 5 5 4 4 4 4 4 5 4 4 5 4 4 4 5 5 5 4 5 5 5 5 5 5 4 4 4 4 4 4 5 5 4 5 4 4 5 4 4 4 4 5 5 4 5 5 4 4 4 5 5 5 5 4 5 5 5 4 4 5 5 5 4 5 4 5 4 4 5 5 4", "output": "1" }, { "input": "99\n2 2 2 5 2 2 2 2 2 4 4 5 5 2 2 4 2 5 2 2 2 5 2 2 5 5 5 4 5 5 4 4 2 2 5 2 2 2 2 5 5 2 2 4 4 4 2 2 2 5 2 4 4 2 4 2 4 2 5 4 2 2 5 2 4 4 4 2 5 2 2 5 4 2 2 5 5 5 2 4 5 4 5 5 4 4 4 5 4 5 4 5 4 2 5 2 2 2 4", "output": "37" }, { "input": "100\n4 4 5 2 2 5 4 5 2 2 2 4 2 5 4 4 2 2 4 5 2 4 2 5 5 4 2 4 4 2 2 5 4 2 5 4 5 2 5 2 4 2 5 4 5 2 2 2 5 2 5 2 5 2 2 4 4 5 5 5 5 5 5 5 4 2 2 2 4 2 2 4 5 5 4 5 4 2 2 2 2 4 2 2 5 5 4 2 2 5 4 5 5 5 4 5 5 5 2 2", "output": "31" }, { "input": "99\n5 3 4 4 5 4 4 4 3 5 4 3 3 4 3 5 5 5 5 4 3 3 5 3 4 5 3 5 4 4 3 5 5 4 4 4 4 3 5 3 3 5 5 5 5 5 4 3 4 4 3 5 5 3 3 4 4 4 5 4 4 5 4 4 4 4 5 5 4 3 3 4 3 5 3 3 3 3 4 4 4 4 3 4 5 4 4 5 5 5 3 4 5 3 4 5 4 3 3", "output": "24" }, { "input": "100\n5 4 4 4 5 5 5 4 5 4 4 3 3 4 4 4 5 4 5 5 3 5 5 4 5 5 5 4 4 5 3 5 3 5 3 3 5 4 4 5 5 4 5 5 3 4 5 4 4 3 4 4 3 3 5 4 5 4 5 3 4 5 3 4 5 4 3 5 4 5 4 4 4 3 4 5 3 4 3 5 3 4 4 4 3 4 4 5 3 3 4 4 5 5 4 3 4 4 3 5", "output": "19" }, { "input": "99\n2 2 5 2 5 3 4 2 3 5 4 3 4 2 5 3 2 2 4 2 4 4 5 4 4 5 2 5 5 3 2 3 2 2 3 4 5 3 5 2 5 4 4 5 4 2 2 3 2 3 3 3 4 4 3 2 2 4 4 2 5 3 5 3 5 4 4 4 5 4 5 2 2 5 4 4 4 3 3 2 5 2 5 2 3 2 5 2 2 5 5 3 4 5 3 4 4 4 4", "output": "37" }, { "input": "2\n5 2", "output": "1" }, { "input": "5\n2 2 2 2 2", "output": "5" }, { "input": "100\n2 3 2 2 2 3 2 3 3 3 3 3 2 3 3 2 2 3 3 2 3 2 3 2 3 4 4 4 3 3 3 3 3 4 4 3 3 4 3 2 3 4 3 3 3 3 2 3 4 3 4 3 3 2 4 4 2 4 4 3 3 3 3 4 3 2 3 4 3 4 4 4 4 4 3 2 2 3 4 2 4 4 4 2 2 4 2 2 3 2 2 4 4 3 4 2 3 3 2 2", "output": "61" }, { "input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 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 4 4", "output": "1" }, { "input": "100\n5 4 3 5 3 5 4 2 3 3 4 5 4 5 5 4 2 4 2 2 5 2 5 3 4 4 4 5 5 5 3 4 4 4 3 5 3 2 5 4 3 3 3 5 2 3 4 2 5 4 3 4 5 2 2 3 4 4 2 3 3 3 2 5 2 3 4 3 3 3 2 5 4 3 4 5 4 2 5 4 5 2 2 4 2 2 5 5 4 5 2 2 2 2 5 2 4 4 4 5", "output": "35" }, { "input": "2\n2 2", "output": "2" }, { "input": "20\n4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5", "output": "1" } ]
1,580,109,171
2,147,483,647
PyPy 3
OK
TESTS
61
155
0
import math n=int(input()) a = list(map(int,input().split())) g=math.ceil(4.5*n)-sum(a) a.sort() sum=0 i=0 while(sum<g): sum+=5-a[i] i+=1 print(i)
Title: Getting an A Time Limit: None seconds Memory Limit: None megabytes Problem Description: Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system. The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically  — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$. This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo. Help Vasya — calculate the minimum amount of lab works Vasya has to redo. Input Specification: The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$). The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works. Output Specification: Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$. Demo Input: ['3\n4 4 4\n', '4\n5 4 5 5\n', '4\n5 3 3 5\n'] Demo Output: ['2\n', '0\n', '1\n'] Note: In the first sample, it is enough to redo two lab works to make two $4$s into $5$s. In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$. In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$.
```python import math n=int(input()) a = list(map(int,input().split())) g=math.ceil(4.5*n)-sum(a) a.sort() sum=0 i=0 while(sum<g): sum+=5-a[i] i+=1 print(i) ```
3
579
A
Raising Bacteria
PROGRAMMING
1,000
[ "bitmasks" ]
null
null
You are a lover of bacteria. You want to raise some bacteria in a box. Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment. What is the minimum number of bacteria you need to put into the box across those days?
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
The only line containing one integer: the answer.
[ "5\n", "8\n" ]
[ "2\n", "1\n" ]
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2. For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
250
[ { "input": "5", "output": "2" }, { "input": "8", "output": "1" }, { "input": "536870911", "output": "29" }, { "input": "1", "output": "1" }, { "input": "343000816", "output": "14" }, { "input": "559980448", "output": "12" }, { "input": "697681824", "output": "14" }, { "input": "41313494", "output": "14" }, { "input": "673935585", "output": "13" }, { "input": "117422204", "output": "19" }, { "input": "954746654", "output": "15" }, { "input": "536838144", "output": "14" }, { "input": "1000000000", "output": "13" }, { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "4", "output": "1" }, { "input": "6", "output": "2" }, { "input": "7", "output": "3" }, { "input": "9", "output": "2" }, { "input": "10", "output": "2" }, { "input": "999999999", "output": "21" }, { "input": "999999998", "output": "20" }, { "input": "999999997", "output": "20" }, { "input": "999999996", "output": "19" }, { "input": "999999995", "output": "20" }, { "input": "999999994", "output": "19" }, { "input": "999999993", "output": "19" }, { "input": "999999992", "output": "18" }, { "input": "999999991", "output": "20" }, { "input": "999999990", "output": "19" }, { "input": "536870910", "output": "28" }, { "input": "536870912", "output": "1" }, { "input": "536870913", "output": "2" }, { "input": "536870914", "output": "2" }, { "input": "22", "output": "3" }, { "input": "14", "output": "3" } ]
1,678,679,703
2,147,483,647
Python 3
OK
TESTS
36
46
0
x = int(input()) ones = 0 while x != 0: #print(temp) if x % 2 == 1: ones = ones + 1 x = int(x/2) print(ones)
Title: Raising Bacteria Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are a lover of bacteria. You want to raise some bacteria in a box. Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment. What is the minimum number of bacteria you need to put into the box across those days? Input Specification: The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). Output Specification: The only line containing one integer: the answer. Demo Input: ['5\n', '8\n'] Demo Output: ['2\n', '1\n'] Note: For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2. For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
```python x = int(input()) ones = 0 while x != 0: #print(temp) if x % 2 == 1: ones = ones + 1 x = int(x/2) print(ones) ```
3
615
A
Bulbs
PROGRAMMING
800
[ "implementation" ]
null
null
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs? If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively. Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
[ "3 4\n2 1 4\n3 1 3 1\n1 2\n", "3 3\n1 1\n1 2\n1 1\n" ]
[ "YES\n", "NO\n" ]
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
500
[ { "input": "3 4\n2 1 4\n3 1 3 1\n1 2", "output": "YES" }, { "input": "3 3\n1 1\n1 2\n1 1", "output": "NO" }, { "input": "3 4\n1 1\n1 2\n1 3", "output": "NO" }, { "input": "1 5\n5 1 2 3 4 5", "output": "YES" }, { "input": "1 5\n5 4 4 1 2 3", "output": "NO" }, { "input": "1 5\n5 1 1 1 1 5", "output": "NO" }, { "input": "2 5\n4 3 1 4 2\n4 2 3 4 5", "output": "YES" }, { "input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1", "output": "YES" }, { "input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6", "output": "YES" }, { "input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "NO" }, { "input": "1 4\n3 1 2 3", "output": "NO" }, { "input": "1 4\n3 2 3 4", "output": "NO" }, { "input": "2 4\n3 2 3 4\n1 1", "output": "YES" }, { "input": "2 4\n3 1 2 3\n1 4", "output": "YES" }, { "input": "5 1\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "1 1\n0", "output": "NO" }, { "input": "1 10\n10 1 2 3 4 5 6 7 8 9 10", "output": "YES" }, { "input": "1 1\n1 1", "output": "YES" }, { "input": "1 100\n99 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", "output": "NO" }, { "input": "1 3\n3 1 2 1", "output": "NO" }, { "input": "1 100\n100 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", "output": "YES" } ]
1,670,842,810
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
46
0
n,m = [int(a) for a in input().split()] ans = [1]*m for i in range(n): li = list(map(int,input().split())) for i in range(len(li)): ans[li[i]-1]=0 if ans.count(1)==0: print('YES') else: print('NO')
Title: Bulbs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs? If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. Input Specification: The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively. Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. Output Specification: If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". Demo Input: ['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
```python n,m = [int(a) for a in input().split()] ans = [1]*m for i in range(n): li = list(map(int,input().split())) for i in range(len(li)): ans[li[i]-1]=0 if ans.count(1)==0: print('YES') else: print('NO') ```
0
0
none
none
none
0
[ "none" ]
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*.
0
[ { "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" } ]
1,442,673,542
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
31
0
m, n = map(int, input().split()) s, u = [], 0 for i in range(m): s.append(input()) for j in range(n): v = 0 for i in range(m-1): v = v | (s[i][j] > s[i+1][j]) u+= v print(u)
Title: none 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 m, n = map(int, input().split()) s, u = [], 0 for i in range(m): s.append(input()) for j in range(n): v = 0 for i in range(m-1): v = v | (s[i][j] > s[i+1][j]) u+= v print(u) ```
0
607
B
Zuma
PROGRAMMING
1,900
[ "dp" ]
null
null
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible. In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line? Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones. The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line.
Print a single integer — the minimum number of seconds needed to destroy the entire line.
[ "3\n1 2 1\n", "3\n1 2 3\n", "7\n1 4 4 2 3 2 1\n" ]
[ "1\n", "3\n", "2\n" ]
In the first sample, Genos can destroy the entire line in one second. In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds. In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
1,250
[ { "input": "3\n1 2 1", "output": "1" }, { "input": "3\n1 2 3", "output": "3" }, { "input": "7\n1 4 4 2 3 2 1", "output": "2" }, { "input": "1\n1", "output": "1" }, { "input": "2\n1 1", "output": "1" }, { "input": "2\n1 2", "output": "2" }, { "input": "8\n1 2 1 3 4 1 2 1", "output": "2" }, { "input": "50\n5 7 5 10 7 9 1 9 10 2 8 3 5 7 3 10 2 3 7 6 2 7 1 2 2 2 4 7 3 5 8 3 4 4 1 6 7 10 5 4 8 1 9 5 5 3 4 4 8 3", "output": "21" }, { "input": "50\n13 17 20 5 14 19 4 17 9 13 10 19 16 13 17 2 18 3 1 9 19 4 19 10 17 12 16 20 10 11 15 10 3 19 8 6 2 8 9 15 13 7 8 8 5 8 15 18 9 4", "output": "28" }, { "input": "50\n22 19 14 22 20 11 16 28 23 15 3 23 6 16 30 15 15 10 24 28 19 19 22 30 28 1 27 12 12 14 17 30 17 26 21 26 27 1 11 23 9 30 18 19 17 29 11 20 29 24", "output": "25" }, { "input": "50\n30 17 31 15 10 3 39 36 5 29 16 11 31 2 38 1 32 40 7 15 39 34 24 11 4 23 9 35 39 32 4 5 14 37 10 34 11 33 30 14 4 34 23 10 34 34 26 34 26 16", "output": "36" }, { "input": "50\n19 25 46 17 1 41 50 19 7 1 43 8 19 38 42 32 38 22 8 5 5 31 29 35 43 12 23 48 40 29 30 9 46 3 39 24 36 36 32 22 21 29 43 33 36 49 48 22 47 37", "output": "36" }, { "input": "6\n1 2 1 1 3 1", "output": "2" } ]
1,524,607,690
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
13
2,000
8,294,400
import sys num_gemstones = int(sys.stdin.readline()) gemstones = [int(x) for x in sys.stdin.readline().split()] dp = [ [ 1 for j in range(num_gemstones) ] for i in range(num_gemstones) ] for i in range(num_gemstones): for j in reversed(range(i)): if gemstones[j] == gemstones[i]: dp[j][i] = dp[j + 1][i - 1] else: dp[j][i] = min([ dp[j][j + x] + dp[j + x + 1][i] for x in range(i - j) ]) print(dp[0][-1])
Title: Zuma Time Limit: None seconds Memory Limit: None megabytes Problem Description: Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible. In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line? Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on. Input Specification: The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones. The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line. Output Specification: Print a single integer — the minimum number of seconds needed to destroy the entire line. Demo Input: ['3\n1 2 1\n', '3\n1 2 3\n', '7\n1 4 4 2 3 2 1\n'] Demo Output: ['1\n', '3\n', '2\n'] Note: In the first sample, Genos can destroy the entire line in one second. In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds. In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
```python import sys num_gemstones = int(sys.stdin.readline()) gemstones = [int(x) for x in sys.stdin.readline().split()] dp = [ [ 1 for j in range(num_gemstones) ] for i in range(num_gemstones) ] for i in range(num_gemstones): for j in reversed(range(i)): if gemstones[j] == gemstones[i]: dp[j][i] = dp[j + 1][i - 1] else: dp[j][i] = min([ dp[j][j + x] + dp[j + x + 1][i] for x in range(i - j) ]) print(dp[0][-1]) ```
0
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,457,020,221
2,147,483,647
Python 3
OK
TESTS
40
154
0
word1 = input() word2 = input() word3 = word2[::-1] if (word1 == word3): print("YES") else: print("NO")
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python word1 = input() word2 = input() word3 = word2[::-1] if (word1 == word3): print("YES") else: print("NO") ```
3.9615
863
B
Kayaking
PROGRAMMING
1,500
[ "brute force", "greedy", "sortings" ]
null
null
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers. Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash. Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks. Help the party to determine minimum possible total instability!
The first line contains one number *n* (2<=≤<=*n*<=≤<=50). The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Print minimum possible total instability.
[ "2\n1 2 3 4\n", "4\n1 3 4 6 3 4 100 200\n" ]
[ "1\n", "5\n" ]
none
0
[ { "input": "2\n1 2 3 4", "output": "1" }, { "input": "4\n1 3 4 6 3 4 100 200", "output": "5" }, { "input": "3\n305 139 205 406 530 206", "output": "102" }, { "input": "3\n610 750 778 6 361 407", "output": "74" }, { "input": "5\n97 166 126 164 154 98 221 7 51 47", "output": "35" }, { "input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2", "output": "0" }, { "input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4", "output": "1" }, { "input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476", "output": "368" }, { "input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597", "output": "393" }, { "input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21", "output": "387" }, { "input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637", "output": "376" }, { "input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61", "output": "351" }, { "input": "3\n1 2 10 11 100 100", "output": "1" }, { "input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858", "output": "318" }, { "input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636", "output": "355" }, { "input": "2\n55 5 25 51", "output": "4" }, { "input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973", "output": "348" }, { "input": "4\n35 48 71 44 78 79 57 48", "output": "10" }, { "input": "3\n58 89 73 15 5 47", "output": "21" }, { "input": "2\n1 20 99 100", "output": "1" } ]
1,678,338,811
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
30
0
n = input() ppl = [int(i) for i in input().split(" ")] ppl.sort() ppl.remove(ppl[-1]) ppl.remove(ppl[-1]) #print(ppl) total_instability = 0 for i in range(len(ppl)): #print("w") for j in range(len(ppl)-1, 0, -1): #print("w2") if i == j: continue #print(abs(ppl[j]- ppl[i]), "W") total_instability += abs(ppl[j]- ppl[i]) print(total_instability)
Title: Kayaking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers. Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash. Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks. Help the party to determine minimum possible total instability! Input Specification: The first line contains one number *n* (2<=≤<=*n*<=≤<=50). The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000). Output Specification: Print minimum possible total instability. Demo Input: ['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n'] Demo Output: ['1\n', '5\n'] Note: none
```python n = input() ppl = [int(i) for i in input().split(" ")] ppl.sort() ppl.remove(ppl[-1]) ppl.remove(ppl[-1]) #print(ppl) total_instability = 0 for i in range(len(ppl)): #print("w") for j in range(len(ppl)-1, 0, -1): #print("w2") if i == j: continue #print(abs(ppl[j]- ppl[i]), "W") total_instability += abs(ppl[j]- ppl[i]) print(total_instability) ```
0
318
A
Even Odds
PROGRAMMING
900
[ "math" ]
null
null
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Print the number that will stand at the position number *k* after Volodya's manipulations.
[ "10 3\n", "7 7\n" ]
[ "5", "6" ]
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
500
[ { "input": "10 3", "output": "5" }, { "input": "7 7", "output": "6" }, { "input": "7 1", "output": "1" }, { "input": "7 2", "output": "3" }, { "input": "8 5", "output": "2" }, { "input": "8 3", "output": "5" }, { "input": "8 4", "output": "7" }, { "input": "1000000000000 500000000001", "output": "2" }, { "input": "999999999997 499999999999", "output": "999999999997" }, { "input": "999999999999 999999999999", "output": "999999999998" }, { "input": "1000000000000 1", "output": "1" }, { "input": "999999999999 1", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "1000000000000 1000000000000", "output": "1000000000000" }, { "input": "1000000000000 500000000000", "output": "999999999999" }, { "input": "1000000000000 499999999999", "output": "999999999997" }, { "input": "999999999997 499999999998", "output": "999999999995" }, { "input": "619234238 556154835", "output": "493075432" }, { "input": "38151981 36650624", "output": "35149266" }, { "input": "680402465 442571217", "output": "204739968" }, { "input": "109135284 9408714", "output": "18817427" }, { "input": "603701841 56038951", "output": "112077901" }, { "input": "356764822 321510177", "output": "286255532" }, { "input": "284911189 142190783", "output": "284381565" }, { "input": "91028405 61435545", "output": "31842684" } ]
1,694,321,080
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
7
1,000
87,244,800
a=input().split() b=int(a[0]) c=[] d=1 while d<=b: c.append(d) d+=2 if len(c)>=int(a[1]): break d=2 while d<=b: c.append(d) d+=2 if len(c)>=int(a[1]): break print(c[int(a[1])-1])
Title: Even Odds Time Limit: None seconds Memory Limit: None megabytes Problem Description: Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. Input Specification: The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print the number that will stand at the position number *k* after Volodya's manipulations. Demo Input: ['10 3\n', '7 7\n'] Demo Output: ['5', '6'] Note: In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
```python a=input().split() b=int(a[0]) c=[] d=1 while d<=b: c.append(d) d+=2 if len(c)>=int(a[1]): break d=2 while d<=b: c.append(d) d+=2 if len(c)>=int(a[1]): break print(c[int(a[1])-1]) ```
0
842
A
Kirill And The Game
PROGRAMMING
1,200
[ "brute force", "two pointers" ]
null
null
Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number. For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions). Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this?
First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107).
Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise. You can output each of the letters in any register.
[ "1 10 1 10 1\n", "1 5 6 10 1\n" ]
[ "YES", "NO" ]
none
500
[ { "input": "1 10 1 10 1", "output": "YES" }, { "input": "1 5 6 10 1", "output": "NO" }, { "input": "1 1 1 1 1", "output": "YES" }, { "input": "1 1 1 1 2", "output": "NO" }, { "input": "1 100000 1 100000 100000", "output": "YES" }, { "input": "1 100000 1 100000 100001", "output": "NO" }, { "input": "25 10000 200 10000 5", "output": "YES" }, { "input": "1 100000 10 100000 50000", "output": "NO" }, { "input": "91939 94921 10197 89487 1", "output": "NO" }, { "input": "30518 58228 74071 77671 1", "output": "NO" }, { "input": "46646 79126 78816 91164 5", "output": "NO" }, { "input": "30070 83417 92074 99337 2", "output": "NO" }, { "input": "13494 17544 96820 99660 6", "output": "NO" }, { "input": "96918 97018 10077 86510 9", "output": "YES" }, { "input": "13046 45594 14823 52475 1", "output": "YES" }, { "input": "29174 40572 95377 97669 4", "output": "NO" }, { "input": "79894 92433 8634 86398 4", "output": "YES" }, { "input": "96022 98362 13380 94100 6", "output": "YES" }, { "input": "79446 95675 93934 96272 3", "output": "NO" }, { "input": "5440 46549 61481 99500 10", "output": "NO" }, { "input": "21569 53580 74739 87749 3", "output": "NO" }, { "input": "72289 78297 79484 98991 7", "output": "NO" }, { "input": "88417 96645 92742 98450 5", "output": "NO" }, { "input": "71841 96625 73295 77648 8", "output": "NO" }, { "input": "87969 99230 78041 94736 4", "output": "NO" }, { "input": "4 4 1 2 3", "output": "NO" }, { "input": "150 150 1 2 100", "output": "NO" }, { "input": "99 100 1 100 50", "output": "YES" }, { "input": "7 7 3 6 2", "output": "NO" }, { "input": "10 10 1 10 1", "output": "YES" }, { "input": "36 36 5 7 6", "output": "YES" }, { "input": "73 96 1 51 51", "output": "NO" }, { "input": "3 3 1 3 2", "output": "NO" }, { "input": "10000000 10000000 1 100000 10000000", "output": "YES" }, { "input": "9222174 9829060 9418763 9955619 9092468", "output": "NO" }, { "input": "70 70 1 2 50", "output": "NO" }, { "input": "100 200 1 20 5", "output": "YES" }, { "input": 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1,504,020,326
1,226
Python 3
CHALLENGED
CHALLENGES
13
62
0
import math l, r, x, y, k = map(int, input().split()) if not int(r/x) >= k >= math.ceil(l/y): print("NO") else: print("YES")
Title: Kirill And The Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number. For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions). Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this? Input Specification: First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107). Output Specification: Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise. You can output each of the letters in any register. Demo Input: ['1 10 1 10 1\n', '1 5 6 10 1\n'] Demo Output: ['YES', 'NO'] Note: none
```python import math l, r, x, y, k = map(int, input().split()) if not int(r/x) >= k >= math.ceil(l/y): print("NO") else: print("YES") ```
-1