整数不少于12可以表示为两个复合数字的和

One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.

For example, there is a statement called the “Goldbach’s conjecture”. It says: “each even number no less than four can be expressed as the sum of two primes”. Let’s modify it. How about a statement like that: “each integer no less than 12 can be expressed as the sum of two composite numbers.” Not like the Goldbach’s conjecture, I can prove this theorem.

You are given an integer n no less than 12, express it as a sum of two composite numbers.

Input
The only line contains an integer n (12 ≤ n ≤ 106).

Output
Output two composite integers x and y (1 < x, y < n) such that x + y = n. If there are multiple solutions, you can output any of them.

Examples
Input
12
Output
4 8
Input
15
Output
6 9
Input
23
Output
8 15
Input
1000000
Output
500000 500000
Note
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output “6 6” or “8 4” as well.

In the second example, 15 = 6 + 9. Note that you can’t output “1 14” because 1 is not a composite number.
题目大意：每个整数不少于12可以表示为两个复合数字的和。
代码如下（已AC）：

#include<iostream>
using namespace std;
bool isPrim(int x)
{
	for (int i = 2; i <= x / 2; i++)
	{
		if (x%i == 0)
			return false;
	}
	return true;
}//判断一个数是不是素数；
void gotbahe(int x)
{
	for (int i = 2; i <= x / 2; i++)
		if (!isPrim(i) && !isPrim(x - i))
		{
			cout <<  i << " " << x - i; break;
		}
}//把一个大于12的数分成两复数；
int main()
{
	int x;
	cin >> x;
	if(x<12||x>1000000)	
		return 0;
	gotbahe(x);
}