A. Watermelon
time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed w kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.

Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

Input

The first (and the only) input line contains integer number w (1 ≤ w ≤ 100) — the weight of the watermelon bought by the boys.

Output

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

Input
8
Output 
YES
Note

For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

题意:分开的两部分都是正偶数输出YES,否则输出NO

         分析:两部分不一定相等
         代码如下:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int w;
cin>>w; if(w%==&&w!=)
{
cout<<"Yes"<<endl;
}
else
{
cout<<"No"<<endl;
}
return ;
}

Codeforces - Watermelon的更多相关文章

  1. Watermelon -- codeforces

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93241#problem/A  (654123) http://codeforces.co ...

  2. Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon 水题

    A. Watermelon 题目连接: http://www.codeforces.com/contest/4/problem/A Description One hot summer day Pet ...

  3. codeforces水题100道 第二题 Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon (math)

    题目链接:http://www.codeforces.com/problemset/problem/4/A题意:一个整数能否表示成两个正偶数的和.C++代码: #include <cstdio& ...

  4. Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon【暴力/数学/只有偶数才能分解为两个偶数】

    time limit per test 1 second memory limit per test 64 megabytes input standard input output standard ...

  5. Codeforces - A. Watermelon

    A. Watermelon time limit per test 1 second memory limit per test 64 megabytes input standard input o ...

  6. CodeForces 4A

    A A - Water~melon Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit St ...

  7. Codeforces初体验

    Codeforces印象 这两天抽时间去codeforces体验了一把. 首先,果然有众多大牛存在.非常多名人一直參加每周一次的比赛.积分2000+,并參与出题. 另外.上面题目非常多.预计至少一千题 ...

  8. python爬虫学习(5) —— 扒一下codeforces题面

    上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题 ...

  9. 【Codeforces 738D】Sea Battle(贪心)

    http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n g ...

随机推荐

  1. 自动控制理论的MATLAB仿真实例(一)

    拉普拉斯变换及其反变换 Laplace变换及其反变换的定义为:

  2. HTTP中主要的头字段

    HTTP中主要的头字段 头字段类型 含义 备注 通用头:适用于请求和响应消息的头字段 Date 表示请求和响应生成的日期   Pragma 表示数据是否允许缓存的通信选项   Cache-Contro ...

  3. Nacos作为微服务注册中心,爱不释手的感觉

    我觉得Nacos用起来还不错 在使用SpringCloud做分布式微服务架构时,注册中心是必不可少的一个组件.目前可以用的主要有:Eureka.Consul.Zookeeper.今天,我们就来说一下A ...

  4. Alterations of brain quantitative proteomics profiling revealed the molecular mechanisms of diosgenin against cerebral ischemia reperfusion effects(大脑的定量蛋白质组学揭示了薯蓣皂苷元对脑缺血再灌注效应的分子机制)

    文献名:Alterations of brain quantitative proteomics profiling revealed the molecular mechanisms of dios ...

  5. Web的服务器和Javaweb结构

    上一节介绍了Eclipse中集成Tomcat环境搭建及javaweb项目的创建,下面说说什么是web服务器及javaweb的结构. 1.web应用的演变 1.1 b/s与c/s模式 B/S:Brows ...

  6. JWT校验流程源码

    目录 一. jwt token校验源码简析 1.1 前言 1.2 jwt的authenticate方法 二. jwt的RefreshJSONWebToken 返回目录 一. jwt token校验源码 ...

  7. 接口自动化框架pyface详细介绍

    版权说明 本框架系本人结合一些实践经验和开源框架设计思想,在家基于兴趣爱好独立完成的代码开发. 源码只保存在私人电脑,办公电脑上无.github开源与公司无关,先把关系撇清,不涉及侵权. 嘘. 框架定 ...

  8. 题解 P5663 【加工零件【民间数据】】

    博客园体验更佳 讲讲我的做法 确定做法 首先,看到这道题,我直接想到的是递归,于是复杂度就上天了,考虑最短路. 如何用最短路 首先,看一张图 我们该如何解决问题? 问题:\(3\)做\(5\)阶段的零 ...

  9. 【2019HDU多校】第九场1006/HDU6685-Rikka with Coin——位运算打表

    题目链接 题目大意 使用10.20.50.100元面额的硬币能分别组成题目给出的面额,需要最少的硬币个数 分析 一开始队友想用一堆if-else解决问题,然后WA了无数发-- 我想到了一种比较简单的打 ...

  10. 贪心-最大相容区间-Maximum Number of Events That Can Be Attended

    2020-02-16 16:24:19 问题描述: 问题求解: 看起来就像是sort + 贪心,但是具体如何做呢? 实际上本题是最大相容区间的变种题,在最大相容区间里,我们按照结束时间对interva ...