HDU 2104：丢手绢【判断是否互质】

Discription
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.

Sample Input

3 2
-1 -1

Sample Output

YES

题意
n个人人围成一圈，每个人有一个盒子。只有一个盒子里有手绢。
从第一个盒子开始找，每次隔m-1和查找。
问能否找到有手绢的盒子。

思路
根据题意可以知道，能够找到手绢的前提是能够遍历所有的盒子。
能够便利所有的盒子的条件是，n和m互质，直接判断就可以了。

AC代码

#include<bits/stdc++.h>
using namespace std;
int n,m;
int main()
{
    while(scanf("%d %d",&n,&m))
    {
        if(n==-1&&m==-1)
            break;
        int tmp=__gcd(n,m);
        if(tmp==1)
            cout<<"YES"<<endl;
        else
            cout<<"POOR Haha"<<endl;
    }
    return 0;
}