【Codeforces Round 324 (Div 2)D】【miller-rabin素数检验 猜想】Dima and Lisa 奇数拆分成三素数

D. Dima and Lisa

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers p_i (1 ≤ i ≤ k), such that

1. 1 ≤ k ≤ 3
2. p_i is a prime
3. 

The numbers p_i do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 10⁹).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers p_i in any order. If there are multiple possible solutions, you can print any of them.

Sample test(s)

input

27

output

3
5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
LL MUL(LL x,int p,int Z)
{
    LL y=1;
    while(p)
    {
        if(p&1)y=y*x%Z;
        x=x*x%Z;
        p>>=1;
    }
    return y;
}
bool miller_rabin(int n)
{
    if(n<=1)return 0;
    if(n==2)return 1;
    if(n%2==0)return 0;
    int p=n-1;
    srand(time(NULL));
    int TIMES=9;
    for(int i=1;i<=TIMES;i++)
    {
        int x=rand()%(n-1)+1;
        if(MUL(x,p,n)!=1)return 0;
    }
    return 1;
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(miller_rabin(n))printf("1\n%d\n",n);
		else
		{
			int top=n/2-1;
			if(top%2==0)--top;
			for(int i=top;i>=3;i-=2)
			{
				if(miller_rabin(i)&&miller_rabin(n-i*2))
				{
					printf("3\n%d %d %d\n",i,i,n-i*2);
					break;
				}
			}
		}
	}
	return 0;
}
/*
【题意】
让你把一个奇数n（[3,1e9]），拆分成1~3个奇数之和的形式

【类型】
猜想 暴力 miller_rabin素数检验

【分析】
方法1，哥德巴赫猜想——任意一个大于2的偶数都可以写成2个质数之和。
于是我们对于>=9的数，先拆出一个3，剩下的暴力枚举=x+y。

方法2，wkc猜想>_<——我感觉任意一个>=9的奇数都可以写成素数x+素数x+素数y
于是就这么写，然后就过了QwQ

然而如何判定拆分的数是否为素数呢？素数检验！

【时间复杂度&&优化】
复杂度为玄学！

*/