杭电 1.2.2 hide handkerchief

原题地址： http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=2&problemid=6

hide handkerchief

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1708 Accepted Submission(s): 609 

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

 

首先这题开始没看出来什么道道，就打算用遍历枚举的方法做，结果是堆栈溢出，数组开得太大了，于是乎不知道怎么办，所以我百度了一下，的值这题的规律是M与N互质，所以就写了这个代码~~贴上

#include<iostream>
using namespace std;
int main()
{
    int m, n, temp;
    while (cin>>m>>n)
    {
          if ((m == -1)&&(n == -1))
          break;     
          while (n != 0)
          {
                temp = n;
                n = m % n;
                m = temp;    
          }
          if (m == 1) cout<<"YES"<<endl;
          else cout<<"POOR Haha"<<endl;
    }
    system ("pause");
    return 0;
}

其实这题本身没有什么讲的，关键是我又踏回了编程之路……还有决定这段时间学习数论，感觉挺有意思的。