codeforces_A.Watermelon

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.

#include<stdio.h>
int main()
{
    int w;
    scanf("%d",&w);
    if(w%2==0&&w>=4)
    printf("YES\n");
    else
    printf("NO\n");
    return 0;
}

?PS:

主要来判断w的一半是否是一个偶数：

思路一：使w/2是偶数，但是可能出现由于结果不为整数，导致在int限制下小数去尾，可能会出现错误。比如w=9，此时w的一半是4.5，那么在int情况下会变成4，那么此时会返回YES，会出错；

思路二：在w>=4且其为偶数情况下，w一半均为偶数。

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