Wolf and Rabbit （GCD）【HDU】-1222

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                                              Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8811    Accepted Submission(s): 4484

Problem Description

There is a hill with n holes around. The holes are signed from 0 to n-1.



A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

 

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n(0<m,n<2147483648).

 

Output

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

 

Sample Input

21 22 2

 

Sample Output

NOYES

 

Author

weigang Lee

 

Source

杭州电子科技大学第三届程序设计大赛

 

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Ignatius.L

题意：GCD运用经典例题，一狼一兔，狼围绕一个均匀分布着n个兔子洞的山转圈，狼每经过m个洞口，就会进入洞中，那么这个洞就是不安全的。问n个洞中，是否存在安全的洞让兔子藏身才能不是可爱的兔子惨遭毒手呢。

解析：如果有公约数（1除外），就会有重复的，即不能遍历完，就输出YES，反之最大公约数为1即（gcd(m,n)==1）的时候，就输出

代码如下：

#include <stdio.h>int gcd(int a,int b){if(b==0)return a;return gcd (b,a%b);}int main(){int t;scanf("%d",&t);while(t--){int m,n;scanf("%d %d",&m,&n);if(gcd(m,n)==1)printf("NO\n");elseprintf("YES\n");}return 0;}