Wolf and Rabbit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9060 Accepted Submission(s): 4612
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).
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
1 2
2 2
Sample Output
NO
YES
YES
/*************************************************************************
> File Name: HDU1222.cpp
> Author: LyuCheng
> Created Time: 2017年10月18日 星期三 21时11分08秒
************************************************************************/
/*
* 题意：大灰狼追小白兔。小白兔可以躲起来的洞绕成一个圈，大灰狼从0这个点出发，
* 每次走m个，问这些洞有木有可以不被狼找到的。
* 思路：如果gcd(n,m)不等于1的话，那么最小公倍数不为n*m,那么用x(x<n)次，大灰
* 狼就会跑到出发位置，肯定不会遍历n个点.
* */
#include <bits/stdc++.h> using namespace std; int t;
int a,b; int main(){
scanf("%d",&t);
while(t--){
scanf("%d%d",&a,&b);
if(__gcd(a,b)==){
puts("NO");
}else{
puts("YES");
}
}
return ;
}