暑期集训之最大公约数问题

最新推荐文章于 2021-01-12 23:48:19 发布

转载最新推荐文章于 2021-01-12 23:48:19 发布 · 310 阅读

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本文探讨了一个数学问题：一只兔子如何在狼寻找它的过程中找到安全的洞穴。通过输入两个正整数m和n，我们能够判断是否存在兔子可以安全藏身的洞穴，并使用最大公约数算法来解决这个问题。

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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
本题的实质就是求2个数是否互质，若是，则则兔子逃不的
知道原理后就好办了，判断互质就得用到最大公约数的算法，利用最大公约数是否是一来判断互质情况，代码如下
#include<stdio.h>
int gcd(int a,int b)
{
	int tep;
	while(b!=0)
	{
		int r=b;
		b=a%b;
		a=r;
	}
	return a;
}
main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int m,n;
		scanf("%d%d",&m,&n);
		if(gcd(n,m)==1)
		printf("NO\n");
		else
		printf("YES\n");
	}
}