HDoj 2014 判断互质

hide handkerchief

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4629    Accepted Submission(s): 2162

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

 

output

YES

这个题就是判断一下N和M是否互质就行，互质就是指最大公约数为1：

找最大公约数是：

辗转相除法：

有两整数a和b：

① a%b得余数c

② 若c=0，则b即为两数的最大公约数

③ 若c≠0，则a=b，b=c，再回去执行①

#include<stdio.h>
int main()
{
int a=3,b;
int c;
c=a%b;
while(c!=0)
{
a=b;
b=c;
c=a%b;
}
printf("最大公约数是:%d\n",b);
return 0;
}

本题解：

#include<stdio.h> 


int get(int a,int b);
  
int main()  
{  
    int n,m;  
    while(scanf("%d%d",&n,&m)&&n!=-1&&m!=-1)  
    {  
        if(get(n,m)==1)  
           printf("YES\n");  
        else  
           printf("POOR Haha\n");  
    }  
    return 0;  
}  


int get(int a,int b)
{

int c;
c=a%b;
while(c!=0)
{
a=b;
b=c;
c=a%b;
}
return b;
}