CodeForces 4A

A

A - Water~melon
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 4A
Description
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.

Sample Input
Input
8
Output
YES
Hint
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).
AC代码：

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

题意：如果一个数可以被拆成两个偶数相加，即输出YES，等价于输入一个数，如果这个数是偶数，输出YES，否则输出NO
注意：2 要单独考虑输出NO
zsy:
题意：给出一个1-100的数，判断能否拆成两个偶数之和。
思路：所有除二以外的偶数都可拆成两个偶数之和。

//#define LOCAL
#include <stdio.h>

int main(){
    int w;
    #ifdef LOCAL
    freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
    #endif
    while(~scanf("%d",&w)){
        if(w==2) {          //注意对w==2的情况 
            printf("NO\n");
            continue;
        }
        if(w%2==0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}