hdu 1222 Wolf and Rabbit(递归)

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5684    Accepted Submission(s): 2848

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

21 22 2

 

Sample Output

NOYES

 

Author

weigang Lee

 

Source

杭州电子科技大学第三届程序设计大赛
题目大意：一狼每隔m个洞刨一个，问最后是否能刨完
题目分析：狼转一圈之后只会将i%m==0的洞刨掉，所以之后我们可以将每个小区间也就是1~m-1看成整个区间，因为再转一圈，它正好能够将每个区间对应此位置的洞刨开，然乎每次的步数就相当于是n%m的余数，也就是转一圈之后的偏移量，那么我们递归地考虑这个问题，很容易得到这个问题的递归解法

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;int t,n,m;bool check ( int n , int m ){    if ( m == 1 ) return true;    if ( m == 0 ) return false;    return check ( m , n%m );}int main  ( ){    scanf ( "%d" , &t );    while ( t-- )    {        scanf ( "%d%d" , &m , &n );        if ( check ( n , m ))            puts ("NO");        else puts ("YES");    }}