Wolf and Rabbit||HDU1222

link:http://acm.hdu.edu.cn/showproblem.php?pid=1222
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

其实就是判断最大公约数是不是1

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int gcd(int a,int b)
{
    if(b==0)
        return a;
    return 
        gcd(b,a%b);
}
int main()
{
    int t,a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&a,&b);
        if(gcd(a,b)==1)
          printf("NO\n");
        else
          printf("YES\n");
    }
return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

三更鬼

谢谢老板!

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值