4A. Watermelon

A. Watermelon

  • time limit per test1 second
  • memory limit per test64 megabytes
  • inputstandard input
  • outputstandard 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.

一个炎热的夏日，皮特和他的朋友比利决定买一个西瓜。在他们看来，他们选择了最大最成熟的。然后称西瓜的重量，秤上显示重量为w公斤。他们迫不及待地赶回了家，渴死了，决定把浆果分开，但他们面临着一个棘手的问题。

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.

皮特和比利非常喜欢偶数，这就是为什么他们想要把西瓜分成两个部分，每个部分的重量都是偶数公斤，同时这两个部分并不一定是相等的。男孩们非常累，想尽快开始吃饭，这就是为什么你应该帮助他们，看看他们是否能按照他们想要的方式分西瓜。当然，他们每个人都应该得到一部分正权重。

Input

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

第一个（也是唯一一个）输入行包含整数w（1） ≤ W ≤ 100）男孩们买的西瓜的重量。

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.

如果男孩们能把西瓜分成两部分，每一部分的重量都是偶数公斤，那么请打印“是”；在相反的情况下没有。

Examples
input

8

output

YES

Note 笔记

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).

例如，男孩们可以将西瓜分成两部分，分别为2公斤和6公斤（另一种变体是4公斤和4公斤的两部分）。

Solution

将西瓜分成两部分，每一部分的重量都是偶数公斤，意味着西瓜的重量是偶数（w % 2 == 0） 考虑边界条件，每部分最少是2公斤，西瓜的重量最小是4公斤（w >= 4）

Code
#include <iostream>
using namespace std;

//A. Watermelon
int main() {
    int w = 0;//w：男孩们买的西瓜的重量
    cin >> w;
    if(w % 2 == 0 && w >= 4){//如果男孩们能把西瓜分成两部分
        cout << "YES" << endl;
    }else{
        cout << "NO" << endl;
    }
    return 0;
}