HDOJ 1222 Wolf and Rabbit
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Wolf and Rabbit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7212 Accepted Submission(s): 3607
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.
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
21 22 2
Sample Output
NOYES
Author
weigang Lee
Source
杭州电子科技大学第三届程序设计大赛
应该算是数学题的范围。
只有n,m最大公约数是1的时候才能走到n=1这个洞。其余应该都是最大公约数是几过几个洞查看一下。
代码如下:
#include<stdio.h>int GCD(int a,int b){if(a%b==0)return b;elsereturn GCD(b,a%b);}int main(){int T;int m,n;scanf("%d",&T);while(T--){scanf("%d%d",&m,&n);if(GCD(m,n)==1)printf("NO\n");elseprintf("YES\n");}return 0;}
0 0
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