点击这里先试一下手
Time limit : 1000ms
Memory limit 65536 KB
原题：
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).
问题简述：
第一行输入一个西瓜的总重量，把西瓜分为互为偶数的两份，如果可以分为互为偶数的两份即输出YES，否则输出NO。
问题分析：
分成的两份不需要均等，且要能够分成两份互为偶数的两份则这个总量必为偶数，但注意当总重量为大于2的偶数时才能实现，故要考虑当总重量为2的情况。
VS通过的代码如下：
#include <iostream>
using namespace std;
int main()
{
int a;
cin >> a;
if (a == 2)
{
cout << "NO" << endl;
}
else if (a % 2 == 0)
{
cout << "YES" << endl;
}
else
cout << "NO" << endl;
}