Wolf and Rabbit

最新推荐文章于 2019-04-06 20:09:51 发布

原创最新推荐文章于 2019-04-06 20:09:51 发布 · 264 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#GCD #数学小技巧 #水题

数学公式+小窍门专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个有趣的数学问题：一只兔子如何在多个洞中找到安全的藏身之处以躲避狼的搜索。通过分析狼搜索洞穴的规律，文章提供了一种简单有效的方法来判断是否存在安全洞。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

YES

思路：   思考一下，每次到的位置依次是  m%n,2m%n,3m%n,4m%n...km%n  尝试一下可知 当两个数互质的时候，就可以遍历所有的洞
代码
#include<stdio.h>
int gcd(int a,int b) //递归
{
	return b==0?a:gcd(b,a%b);// 
}
unsigned int gcd(unsigned int a,unsigned int b) //  不用递归 即 辗转相除法
{
    int r;
    while(b>0)
    {
         r=a%b;
         a=b;
         b=r;
    }
    return a;
}
int main()
{
	int p;
	scanf("%d",&p);
	while(p--)
	{
		int n,m;
		scanf("%d%d",&n,&m);
		if(gcd(n,m)==1) printf("NO\n");
		else printf("YES\n");
	}
	return 0;
}