HDU Wolf and Rabbit

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原文链接：http://www.cnblogs.com/newpanderking/archive/2011/07/29/2120985.html

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#java

本文介绍了一道经典的狼追兔子问题。狼从编号为0的洞开始按一定规律搜寻藏身于n个洞之一的兔子。通过算法判断是否存在安全洞使兔子能够逃脱。输入包含多组测试数据，每组由两个正整数组成，分别代表狼搜索的步长m和洞的数量n。

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 335 Accepted Submission(s): 214

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

2
1 2
2 2

Sample Output

NO
YES

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int t,m,n;
    cin>>t;
    while(t--)
    {
        cin>>m>>n;
        while(n>m?(n=n%m):(m=m%n));
        if(n+m==1)printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}