Wolf and Rabbit（gcd）

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Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5298    Accepted Submission(s): 2658

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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HDU 1222 Wolf and Rabbit

该题是一题找规律题，当n与m都是偶数或是倍数是就存在这样的洞，

方法一：

#include<stdio.h>
#include<stdlib.h>
int  main()
{
    int  n,m,N;
    scanf ( "%d" ,&N );
    for ( int  i=1; i<=N; i++ )
    {
         scanf ( "%d%d" ,&n,&m );
         if ( n==1 || m==1)
           printf ( "NO\n"  );
         else
         {
                 if ( (n%2==0) && (m%2==0) )
                         printf ( "YES\n"  );
                 else
                 {
                     if ( (n%m==0)||(m%n==0) )
                             printf ( "YES\n"  );
                     else  printf ( "NO\n"  );
                 }
         }
    }
    return  0;
}

　　由第一种方法得到，我们可用Gcd()函数，当公约数大于1时就代表安全。

#include<stdio.h>
int  Gcd( int  a, int  b )
{
    return  b==0?a:Gcd( b,a%b );
}
int  main()
{
   int  T,n,m;
   scanf ( "%d" ,&T );
   while ( T-- )
   {
      scanf ( "%d%d" ,&n,&m );
      if ( Gcd( n,m )>1 )
      printf ( "YES\n"  );
      else  printf ( "NO\n"  );
   }
   return  0;
}