hdu 1222 狼和兔子

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

 

 

 

题意：有n个洞，这些洞围成一个圆形，狼去找兔子，按照一个规律去找，问最后还有没有剩下狼没找过洞.....

若果有就输出YES，没有就输出NO...

 

 

 

解题思路：第一眼看上去我就觉得只要狼不是按照一个一个的找的就一定会有剩下的....然后我发现我煞笔了....

  　　　　这题其实是找n和m有没有公约数，如果有公约数那么浪找完一圈之后他就会一直找这么几个洞，如果没有公约数，那么狼就总会找到的只是时间问题

 

 

 

代码如下：

 

 

#include <stdio.h>
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int m,n;
        scanf("%d%d",&m,&n);
        int j=gcd(n,m);
        //printf("%d
",j);
        if(j==1)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}