题目链接：http://codeforces.com/problemset/problem/472/A
472A. Design Tutorial: Learn from Math
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
题目描述
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.
输入描述
Input
The only line contains an integer n (12 ≤ n ≤ 106).
输出描述
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.
样例
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.
题意：将一个数拆为两个合数之和。
思路：输入偶数最小为4，可拆为合数4与0。输入奇数最小为13，可拆为合数4与9。所以拆分方法用4或9进行。
代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
if(n%2==0) //偶数拆为4和n-4
cout<<"4 "<<n-4<<endl;
else //奇数拆为9和n-9
cout<<"9 "<<n-9<<endl;
return 0;
}