cf472A Design Tutorial: Learn from Math

A. Design Tutorial: Learn from Math

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

给定n，要求找出a,b,使得a、b是合数，并且n=a+b

3分钟打完……感觉手速还是慢

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
LL n;
LL a[100010];
bool mark[1000010];
int main()
{
	n=read();
	memset(mark,1,sizeof(mark));
	mark[1]=0;
	for (int i=2;i<=n;i++)
	  if (mark[i])
	    for (int j=2*i;j<=n;j+=i)
	      mark[j]=0;
	for (int i=2;i<=n;i++)
	  if (!mark[i]&&!mark[n-i]&&i<n-i)
	  {
	  	printf("%d %d\n",i,n-i);
	  	return 0;
	  }
}