HDU 1222 Wolf and Rabbit（gcd）

HDU 1222   Wolf and Rabbit   （最大公约数）解题报告

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88159#problem/G

题目：

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：

山周围有从0~n-1的n个山洞，狼按逆时针的顺序抓兔子，判断兔子是否能活下来。

 

分析：

这是一个求最大公约数的问题。可以用书上的gcd算法--辗转相除法。gcd(a,b)=gcd(b,a%b),还要考虑它和边界条件gcd(a,0)=a。

 

代码：

#include<cstdio>
#include<iostream>
using namespace std;

int gcd(int a,int b)//辗转相除法
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    int p;
    int m,n;
    scanf("%d",&p);
    while(p--)
    {
        scanf("%d%d",&m,&n);
        gcd(m,n);
        if(gcd(m,n)==1)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}