【杭电1222】Wolf and Rabbit

野性酷女 2022-09-25 01:18 71阅读 0赞

# ``     Wolf and Rabbit #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 7476 Accepted Submission(s): 3749**

Problem Description

There is a hill with n holes around. The holes are signed from 0 to n-1.  
  
![C9-1004-1.jpg][]  
  
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
      
       
      
       
        
       这是有趣的数学问题啊，能有效锻炼思维能力啊，
      
       
      
       
        
       两个数的最大公约数只要不等于1，就有数字遍历不到。因为无论这两个数中哪一个数成倍增大，它与另外一个不变的数的余数都能被他们的最大公约数除尽，所以只有最大公约数为一才能又到达不到的地方，这可以列几个例子试试。比如6和10，无论6的多少倍于对10取余都只能是2、4、6、8、0中的一个。
    
       
        #include<stdio.h>
    int max(int a,int b) {
    	if(a%b==0)
    		return b;
    	else
    		return max(b,a%b);
    }
    int main() {
    	int n;
    	scanf("%d",&n);
    	while(n--) {
    		int i,j;
    		scanf("%d %d",&i,&j);
    		if(max(i,j)==1)
    			printf("NO\n");
    		else
    			printf("YES\n");
    	}
    	return 0;
    }

[C9-1004-1.jpg]: /images/20220719/67a52ca6b2314441a836cc2664e26ac0.png