A. Watermelon
time limit per test
1 second
memory limit per test
64 megabytes
input
standard input
output
standard 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.
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 test(s)
input
8
output
YES
恩 看懂题意就能做。
题意：把一个分为两个偶数之和。假设能够分。输出YES 否则输出NO
值得注意的这道题的背景是两个人分西瓜。所以每一个人至少分到一磅，so 0的情况就要排除掉，尽管0也是偶数
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
#define LL long long
int main()
{
int n;
while(cin>>n)
{
int flag=0;
for(int i=1;i<=n-1;i++)
if(i%2==0&&(n-i)%2==0)
{
flag=1;
break;
}
if(flag)
puts("YES");
else
puts("NO");
}
return 0;
}
