http://acm.hdu.edu.cn/showproblem.php?pid=1222

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6052    Accepted Submission(s): 3032

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
 

题解:就是求n和m 是否互质,求gcd(n,m)是否等于1即可

代码:

 #include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int gcd (int a , int b)
{
return b==? a :gcd(b,a%b);
}
int main()
{
int n , m ;
int t ;
cin>>t;
while(t--)
{
scanf("%d%d",&n,&m);
if(gcd(n,m)==)
cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return ;
}

Wolf and Rabbit的更多相关文章

  1. Wolf and Rabbit(gcd)

    Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  2. HDU 1222 Wolf and Rabbit(gcd)

    HDU 1222   Wolf and Rabbit   (最大公约数)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...

  3. HDU 1222 Wolf and Rabbit(欧几里得)

    Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  4. gcd(最大公约数)lcm(最小公倍数)E - Wolf and Rabbit

    1.gcd 递归实现 int gcd(int i,int j){ if(j==0) return i; else return gcd(j,i%j);} 2.lcm int gcd(int i,int ...

  5. hdu 1222 Wolf and Rabbit

    Problem Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbi ...

  6. 转化为用欧几里得算法判断互质的问题D - Wolf and Rabbit

    Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbit must h ...

  7. HDU Wolf and Rabbit

    Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbit must h ...

  8. GCD(1222)Wolf and Rabbit

    Problem Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbi ...

  9. HDU 1222 Wolf and Rabbit(数学,找规律)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

随机推荐

  1. Visual Studio Code 快捷键大全(Windows)

    Visual Studio Code 是一款优秀的编辑器,非常适合编写 TS 以及 React .最近在学习 AngularJs 2,开始使用 VSCode,特意整理翻译了一下官网的快捷键.因为已经习 ...

  2. ArcGIS API for JavaScript 4.2学习笔记[14] 弹窗的位置、为弹窗添加元素

    这一节我们来看看弹窗的位置和弹窗上能放什么. 先一句话总结: 位置:可以随便(点击时出现或者一直固定在某个位置),也可以指定位置 能放什么:四种,文字.媒体(图片等).表格.附件. [Part I 位 ...

  3. HTML5学习知识点

    一.文档问题 1.html5新标签:section.header.footer.nav.aside.blockquote.q.fieldest.figure.address.article.detai ...

  4. Less is exponentially more

    Less is exponentially more  (原文出处:rob pike 博客,https://commandcenter.blogspot.jp/2012/06/less-is-expo ...

  5. Django2中文文档--目录及介绍部分

    Django2文档-文档结构 我是按照官方文档的格式进行翻译,所以格式根官方格式一致 如果大家发现哪些地方有问题可以联系我 2426525089@qq.com 或者加入QQ群跟我一起翻译,群号码: 2 ...

  6. 安装MySQL容易出现的问题

    mysql 安装到最后一步时,start service 为失败状态. 解决方法: 方 式1  MySQL安装是出现could not start the service mysql error:0 ...

  7. K:java中枚举的常见用法

    用法一:常量   在JDK1.5 之前,我们定义常量都是: public static fianl.....现在好了,有了枚举,可以把相关的常量分组到一个枚举类型里,而且枚举提供了比常量更多的方法. ...

  8. centos6环境下使用yum安装Ambari

    前言: Ambari是apache下面的开源项目,主要通过web UI方式对Hadoop集群进行统一创建和管理,以节省Hadoop集群的运维成本.本文通过安装过程中的截图简要介绍一下相关步骤供需要的朋 ...

  9. display:box;display:flex;弹性盒模型

    display:box:display:flex:弹性盒模型 非常适用于移动端.PC端高级浏览器,效果也很好. display: -webkit-box; display: -moz-box; dis ...

  10. linux下分割文件

    split -l 115 XSMD.txt -d -a 2 XSMD.txt._   注:将一个文件XSMD.txt分割成两个文件,每个大小115