hide handkerchief （gcd）

hide handkerchief

 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个人围成一圈，每次走M步，问能否遍历到每个人

分析：

只需要判断gcd（N，M）是否等于1即可

为什么呢？

我们举个例子比如
n = 4
那么我们有1 2 3 4
如果
m = 2
我们发现每次只能走1和3并无限循环下去
如果
m = 6(gcd(m,n) = 2)
我们发现每次仍然只能走1和3并且无限循环
因此我们发现实际上每次走的步数和最大公因数有关
当最大公因数大于1时，一定能够形成间隔大于1的循环，而只有当最大公因数为1时，才能保证所有数遍历到

code：

#include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b){
    return b ? gcd(b,a%b) : a;
}
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        if(n == -1 && m == -1) break;
        if(gcd(n,m) == 1) printf("YES\n");
        else printf("POOR Haha\n");
    }
    return 0;
}