Watermelon_Codeforces_4A

最新推荐文章于 2021-10-17 23:33:05 发布

MoogleAndChocobo

最新推荐文章于 2021-10-17 23:33:05 发布

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分类专栏： acm codeforces 文章标签： codeforces acm

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在一个炎热的夏日，Pete和Billy买了一个大西瓜，重量为w公斤。他们希望将西瓜分成两部分，每部分重量都是偶数，但不一定要相等。如果可以这样分割，输出'YES'，否则输出'NO'。例如，可以分割成2公斤和6公斤的两部分。

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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).

重点：并非简单地判断奇偶数，2为特殊情况

#include<cstdio>

int main()
{
    int n;
    scanf("%d", &n);
    if(n >= 4 && n % 2 == 0)
        printf("YES\n");
    else
        printf("No\n");
    return 0;
}