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 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.
Sample Input
8
Sample Output
YES
Hint
题意
给你一个数,问你能不能把这个数分成两个偶数之和。
题解:
先分出一个2,然后看剩下的是不是偶数就好了。
水题。
代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
if(n<=2)return puts("NO"),0;
if(n%2==0&&(n-2)%2==0)
return puts("YES"),0;
return puts("NO"),0;
}
Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon 水题的更多相关文章
- Codeforces Beta Round #14 (Div. 2) C. Four Segments 水题
C. Four Segments 题目连接: http://codeforces.com/contest/14/problem/C Description Several months later A ...
- Codeforces Beta Round #14 (Div. 2) B. Young Photographer 水题
B. Young Photographer 题目连接: http://codeforces.com/contest/14/problem/B Description Among other thing ...
- Codeforces Beta Round #6 (Div. 2 Only) A. Triangle 水题
A. Triangle 题目连接: http://codeforces.com/contest/6/problem/A Description Johnny has a younger sister ...
- 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& ...
- 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 ...
- Codeforces Beta Round #80 (Div. 2 Only)【ABCD】
Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...
- Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】
Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...
- Codeforces Beta Round #79 (Div. 2 Only)
Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...
- Codeforces Beta Round #77 (Div. 2 Only)
Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...
随机推荐
- Struts访问servletAPI方式
1.原理
- 使用JSON语法创建JS对象(重要)
JS对象的键值可以加单引号或者不加或者加双引号 JSON语法提供了一种更简单的方式来创建对象,可以避免书写函数,也可避免用new关键字,可以直接创建一个JS对象,使用一个花括号,然后将每个属性写成&q ...
- Shell-history命令加记录用户IP
记录输入的命令 history命令可以查看用户输入过的命令,一个典型history命令输出如下: 980 2017-05-29 20:17:37 cd - 981 2017-05-29 20:17:4 ...
- python基础===python自带idle的快捷键
Ctrl + [ Ctrl + ] 缩进代码Alt+3 Alt+4 注释.取消注释代码行Alt+5 Alt+6 切换缩进方式 空格<=>TabAlt+/ 单词完成,只要文中出现过,就可 ...
- JavaSE项目之员工收录系统
在Java SE中,对IO流与集合的操作在应用中比较重要.接下来,我以一个小型项目的形式,演示IO流.集合等知识点在实践中的运用. 该项目名称为“员工收录系统”,主要是通过输入员工的id.姓名信息,实 ...
- java基础9 main函数、this、static、super、final、instanceof 关键字
一.main函数详解 1.public:公共的.权限是最大的,在任何情况都可以访问 原因:为了保证jvm在任何情况下都可以访问到main法2.static:静态,静态可以让jvm调用更方便,不需要用 ...
- meta标签的使用方法(PC端)
<!DOCTYPE html> <html lang="en"> <head> <!--设定页面使用的字符集--> <meta ...
- hdu 1850(尼姆博弈)
Being a Good Boy in Spring Festival Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32 ...
- Python 中for...esle和while...else语法
Python的for...else和while...else语法,这是Python中最不常用,最为误解的语法特性之一. Python中的for.while循环都有一个可选的else分支(类似if语句和 ...
- LoadRunner的Capture Level说明
LoadRunner的Capture Level说明 Capture Level的设置说明: 1.Socket level data. Capture data using trapping on t ...