HDU 2104 hide handkerchief

题目HDU 2104：http://acm.hdu.edu.cn/showproblem.php?pid=2104

Problem Description

The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.
Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .
Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A.
So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha".

Input

There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.

Output

For each input case, you should only the result that Haha can find the handkerchief or not.

解题思路

开始时将题目理解错了，一位是自己随机产生一个随机数用来存放handkerchief位置，这样的话只需要随机产生一个随机数，并判断初始位置加上M-1并对N取模后能否等于这个随机数就可以了。考虑之后，发现初始位置不能确定，而初始位置对最终的结果是十分重要的，所以题目的意思应该为Haha这个人遍历了所有的箱子。
若M与N不互质，则一定会出现重复的数字进行循环。所以只有M与N互质是才能将所有的箱子循环完。直接放出代码：

#include <iostream>  
using namespace std;  
int gcd(int a,int b)  
{  
    return (b?gcd(b,a%b):a);  
}  
int main()  
{  
    int m,n;  
    while(cin>>m>>n&&m+n!=-2)  
    {  
        if(gcd(m,n)==1)  cout<<"YES
";  
        else cout<<"POOR Haha
";  
    }  
    return 0;  
}