4A - Watermelon

本文介绍了一个简单的编程问题——切西瓜问题。两个朋友买了一个西瓜,希望将其切成两半且每半重量均为偶数公斤。文章提供了判断是否能按要求切分西瓜的C++代码实现。

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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.
Examples
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).

本题纯种水题一个,题意就是两个强迫症患者切西瓜,一个西瓜切两半,要求两半都是,等同于买的西瓜是除了2之外的偶数。

code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n&1||(n==2))
        {
            printf("NO\n");
        }
        else
        {
            printf("YES\n");
        }
    }
    return 0;
}
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