【HD 1222】Wolf and Rabbit

最新推荐文章于 2021-01-12 23:48:19 发布

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本文通过狼追兔问题，探讨了如何判断是否存在安全位置的方法。采用最大公约数(GCD)来确定狼是否能覆盖所有洞穴，从而决定兔子是否安全。提供了一段使用C++实现的示例代码。

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

——————————————————————————————————
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
——————————————————————————————————

#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");
    }


}