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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". |
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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 |
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Sample Output
YES |
当从m开始找时，如果m与n有公约数时，在找盒子时，就只会在这些约数的倍数之间找，而不是约数倍数的盒子就永远也不会被找到，所以当m与n的最大公约数为1，即m与n互质时才能找遍所有的盒子。
(一)
#include<stdio.h>
#include<math.h>
int Zhisu(int a,int b)
{
if((a-b)==0)
return b;
else
Zhisu(b,abs(a-b));
}
void main()
{
int a,b;
scanf("%d %d",&a,&b);
while(a!=-1 && b!=-1)
{
if(Zhisu(a,b)==1)
printf("YES\n");
else
printf("POOR Haha\n");
scanf("%d %d",&a,&b);
}
}
(二)
#include<stdio.h>
void main()
{
int a,b,c;
while(1)
{
scanf("%d %d",&a,&b);
if(a==-1 && b==-1)
return;
while(b!=0)
{
c=a%b;
a=b;
b=c;
}
if(a==1)
printf("YES\n");
else
printf("POOR Haha\n");
}
}
