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.
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Output
For each input case, you should only the result that Haha can find the handkerchief or not.
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Sample Input
3 2 -1 -1 |
Sample Output
YES |
代码:
#include<iostream> #include<fstream> using namespace std; int main() { int n,m; while(cin >> n >> m && n != -1 && m != -1) { int r; while( n % m != 0) { r = n % m; n = m; m = r; } if( m == 1) cout << "YES" << endl; else cout << "POOR Haha" << endl; } return 0; }