【HD 1222】Wolf and Rabbit

第一下看到的时候懵了，想了半天才明白个大概为什么是GCD。

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7349 Accepted Submission(s): 3678

Problem Description

There is a hill with n holes around. The holes are signed from 0 to n-1.

A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n( 0 < m,n<2147483648).

Output

For each input m n, if safe holes exist, you should output “YES”, else output “NO” in a single line.

Sample Input

2

1 2

2 2

Sample Output

NO

YES

Author

weigang Lee

Source

杭州电子科技大学第三届程序设计大赛

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思路： m个洞，每隔n个进一次。

m/n …… a

( a+m )/n ……. b

( b+m )/n ……. c

相当于是在交错的走，如果没有一点使得和前面的轨迹重合的话所有洞都会走一遍，所以，就是看是不是有GCD

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#include<cstdio>
int gcd( int a,int b )
{
if( a%b == 0) return b;
else return gcd( b,a%b );
}

int main()
{
int x,y,t;
scanf("%d",&t);
while( t-- )
{
scanf("%d%d",&x,&y);
if( gcd(x,y) != 1 )
{
printf("YES\n");
}
else
printf("NO\n");
}

}