acm寒假特辑 1月19日 CodeForces

D - 4 CodeForces - 472A

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：一个数拆成两个不相同的合数（合数指：自然数中除了能被1和本身整除外，还能被其他数（0除外）整除的数。）。

题目思想：①因为大部分偶数都为合数所以 i（偶数） 从4开始试探，判断n-i是否为合数。
故可两者都为合数。
②还有一种偷懒的办法，4，9为合数，偶数则用i=4，奇数则用i=9，则n-i始终为偶数，除了2以外的偶数都为合数。输出 i 与 n-i。

//miku saiko依然是旧题
#include<iostream>
using namespace std;
int x(int a)//此处用于判断是否为合数
{
	for (int j = 2; j*j <= a;j++)
	{
		if (a % j == 0)
			return 1;
	}
	return 0;
};

int main()
{
	int n;
	cin >> n;
	for (int i = 4;; i += 2)
	{
		if (x(n-i))
		{
			cout << i << ' ' << (n - i) << endl; 
			break;
		}
	}
    return 0;
}