杭电-1222Wolf and Rabbit

最新推荐文章于 2019-09-17 17:01:26 发布

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#杭电 #1222 #Wolf and Rabbit #规律

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本文介绍了一种经典的狼追兔问题，通过数学方法判断兔子在特定条件下是否能躲避狼的追捕。输入包括多组测试数据，每组数据包含两个整数m和n，分别代表狼前进的步长和洞的数量。输出则是根据算法判断是否存在安全洞穴。

Wolf and Rabbit

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

Total Submission(s): 6445 Accepted Submission(s): 3212

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

你会发现一个规律，每次转完一圈的数和上一圈每个位置的数相比，都偏移了一个m%n，只有当每次偏移1的时候，才能全部都访问，否则兔子存活！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int gcd(int x,int y)
{
	int t;
	while(y)
	{
		int t=y;
		y=x%y;
		x=t;
	}
	return x;
}
int main()
{
	int m,a,b;
	scanf("%d",&m);
	while(m--)
	{
		scanf("%d%d",&a,&b);
		if(gcd(a,b)==1)
		printf("NO\n");
		else
		printf("YES\n");
	}
	return 0;
}