hdoj 1222 Wolf and Rabbit
来源:互联网 发布:重庆大学网络教育 编辑:程序博客网 时间:2024/06/05 15:29
Wolf and Rabbit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7065 Accepted Submission(s): 3532
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
解题思路:只要n和m互质时-.-才可遍历所有的洞。。
代码:
#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int gcd(int nn,int mm){ if (nn%mm==0) return mm; else return gcd(mm,nn%mm);}int main(){ int n,m,lp,t;scanf("%d",&t); while (t--) { scanf("%d%d",&n,&m); if (n>m) lp=gcd(n,m); else lp=gcd(m,n); if (lp==1) printf("NO\n"); else printf("YES\n"); } return 0;}
0 0
- hdoj 1222 Wolf and Rabbit
- HDOJ 1222 Wolf and Rabbit
- HDOJ -- 1222 Wolf and Rabbit
- 【杂题】 HDOJ 1222 Wolf and Rabbit
- hdoj 1222 Wolf and Rabbit 【GCD】
- 1222 Wolf and Rabbit
- 1222 Wolf and Rabbit
- [数论]HDOJ 1222 Wolf and Rabbit 欧几里得算法
- HDOJ 1222 Wolf and Rabbit(数学解题技巧)
- HDOJ 1222 Wolf and Rabbit(判断互质,水)
- HDOJ 1222 Wolf and Rabbit ( GCD 和思维 )
- hdu 1222 Wolf and Rabbit
- HDU 1222 Wolf and Rabbit
- hdu 1222 Wolf and Rabbit
- hdu 1222 Wolf and Rabbit
- HDU 1222 Wolf and Rabbit
- hdu--1222 Wolf and Rabbit
- hdu 1222 Wolf and Rabbit
- Java常用类库之时间操作类——Date、Calendar、DateFormat、SimpleDateFormat及实例操作
- AbstractRoutingDataSource源码分析
- Java 8新特性:Stream API
- Java 8为什么需要Lambda表达式
- Struts2 拦截器interceptor控制登录和访问权限
- hdoj 1222 Wolf and Rabbit
- 人工智能----TensorFlow开篇简介
- log4j.properties 详解与配置步骤
- Matlab 字符与字符串
- 网友总结了26个面试问题解答
- linux中使用cifs挂载windows 7的共享文件提示mount error(5): Input/output error
- 用过滤器实现登录和访问权限
- ActiveMQ中Consumer特性详解与优化
- Leetcode-104. Maximum Depth of Binary Tree c语言