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.
InputThe 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).
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.
OutputFor 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 2Sample 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"); } }