HDU ACM 1.2.4 hide handkerchief

hide handkerchief

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.

 

Sample Input

3 2-1 -1

 

Sample Output

YES

分析题意，其实就是要求给定的N和M之间的最大公约数，如果公约数等于1，则可以遍历所有箱子，如果不等于1，则无法遍历。

代码如下：

#include <iostream>using namespace std;int Z(int a, int b){if (a %b == 0){return b;}elseZ(b, a%b);}void main(){int M, N;//cout <<INT_MAX<< endl;cin >> N >> M;while (M != -1 && N != -1){int gongyueshu = 0;gongyueshu = Z(N, M);if(1 == gongyueshu)cout << "YES" << endl;else if(gongyueshu > 1)cout << "POOR Haha" << endl;cin >> N >> M;}return;}