2013
12-04

# Wolf and Rabbit

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.

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).

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

2
1 2
2 2

NO
YES

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

#include <stdio.h>
int gcd(int a,int b)
{
return a%b==0?b:gcd(b,a%b);
}
int main()
{
int p,m,n;
scanf("%d",&p);
while(p--)
{
scanf("%d%d",&m,&n);
if(gcd(m,n)==1)puts("NO");
else puts("YES");
}
return 0;
}

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2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？