sgu116

 Index of super-prime

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

Let P₁, P₂, … ,P_N, … be a sequence of prime numbers. Super-prime number is such a prime number that its current number in prime numbers sequence is a prime number too. For example, 3 is a super-prime number, but 7 is not. Index of super-prime for number is 0 iff it is impossible to present it as a sum of few (maybe one) super-prime numbers, and if such presentation exists, index is equal to minimal number of items in such presentation. Your task is to find index of super-prime for given numbers and find optimal presentation as a sum of super-primes.

 

Input

There is a positive integer number in input. Number is not more than 10000.

 

Output

Write index I for given number as the first number in line. Write I super-primes numbers that are items in optimal presentation for given number. Write these I numbers in order of non-increasing.

 

Sample Input

6

Sample Output

2
3 3

#include<iostream>
#include<algorithm>
#include<string.h>

using namespace std;
const int maxn = 1000;

int Q, Ps[maxn] = {0}, cnt = 0, snt = 0;

bool isprime(int p)
{
	if(p == 2) return true;
	if(p % 2 == 0)
		return false;
	for(int i = 3, j= 9; j <= p; i+=2, j = i*i)
	{
		if(p %i == 0) return false;
	}
	return true;
}

void init()
{
	cin >> Q;
	cnt = 1;
	for(int i = 3; i <= Q; i++)
	{
		if(isprime(i))
		{
			cnt++;
			if(isprime(cnt))
			{
				Ps[snt++] = i;
			}
		}
	}
}
bool G[maxn * 10] = {0};
int F[maxn * 10] = {0};
int P[maxn * 10] = {0};

int DP(int x)
{
	if(G[x]) return F[x];
	G[x] = true;
	for(int i = 0; i < snt; i++)
	{
		if(x >= Ps[i])
		{
			if(DP(x-Ps[i]) < F[x])
			{
				F[x] = F[x-Ps[i]];
				P[x] = Ps[i];
			}
		}
	}
	F[x]++; 
	return F[x];
}
bool cmp(const int & x, const int & y)
{
	return x > y;
}

void print(int x)
{
	int ans[maxn], cnt = 0;
	if(P[x] == 0) return;
	while(x)
	{
		ans[cnt++] = P[x];
		x -= P[x];
	}
	sort(ans, ans+cnt, cmp);
	cout << ans[0];
	for(int i = 1; i < cnt; i++)
		cout << " " << ans[i];
}

int main()
{
	init();
	for(int i = 0; i <= Q; i++)
		F[i]=~0U >> 3;
	F[0] = 0; G[0] = true;
	if(DP(Q) > 10000)
		cout<< 0 << endl;
	else
	{
		cout << DP(Q) << endl;
		print(Q);
	}
	return 0;
}