A - Problem A
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.
问题简述：两个小男孩是偶数的粉丝，他们分西瓜要分成重量为偶数的两部分。第一行输入重量w，然后判断能否分成重量为偶数的两部分，如果可以输出yes，否则输出no。
程序简述：只有偶数才能分成两个偶数，但是2不行。于是用if语句对2求余判断是否为偶数，然后&&w!=2。
#include <iostream>
using namespace std;
int main()
{
int w;
cin >> w;
if (w % 2 == 0&&w!=2)
cout << "YES" << endl;
else cout << "NO" << endl;
}