HDU 2104 hide handkerchief （水）

最新推荐文章于 2020-12-24 20:18:09 发布

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#search #input #each #c

本文介绍了一个寻找隐藏手帕的游戏算法，玩家需要根据特定的搜索模式来确定是否能找到隐藏的手帕。通过计算最大公约数判断玩家能否遍历所有位置。

hide handkerchief

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

#include <stdio.h>

int gcd(int m,int n){
    int r;
    while(r=m%n){
        m=n;
        n=r;
    }
    return n;
}

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m),m!=-1||n!=-1){
        if(gcd(m,n)==1)
            puts("YES");
        else
            puts("POOR Haha");
    }
    return 0;
}