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  • Wolf and Rabbit

    Wolf and Rabbit

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7338    Accepted Submission(s): 3672


    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
    这道题题解看这里点击打开链接
    #include<cstdio>
    int yue(int a,int b)
    {
          if(a%b==0)
    	  return b;
    	  else
    	  return  yue(b,a%b);
    }
    int main()
    {
    
    	int n,m,t;
    	scanf("%d",&t);
    	
    	while(t--)
    	{
    		scanf("%d%d",&n,&m);
    		if(yue(n,m)==1)
    		{
    			printf("NO
    ");
    		}
    		else
    		printf("YES
    ");
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027209.html
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