Sgu 116

116. 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

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Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff 
#define mod 1000000007
#define ll long long
#define maxn 20025
#define pi acos(-1.0)  
int n, m, k, c, ans=10000;
int prv[maxn], pri[maxn],pr2[maxn];
int dp[maxn],p[maxn];
bool cmp(int a, int b){ return a > b; }
int main(){
    for (int i = 2; i <= 10000;i++)
    if (!prv[i])for (int j = i*i; j <= 10000; j += i)
    if (!prv[j])prv[j] = 1;
    for (int j = 0; j < maxn; j++)dp[j] = 10000;
    for (int i = 2; i <= 10000; i++)if (!prv[i])pri[k++] = i;
    for (int i = 2; i < k; i++)if (!prv[i])pr2[c++] = pri[i-1];  
    scanf("%d", &n);                         
    dp[0] = 0;
    for (int i = 0; i < c;i++)
    for (int j = pr2[i]; j <= 10000; j++)
        dp[j] = min(dp[j], dp[j - pr2[i]] + 1);
    if (dp[n] >= 10000)dp[n] = 0;
    printf("%d\n", dp[n]);
    int t=n,x = 0;
    for (int i = dp[n]-1; i >=0; i--)
        for (int j = 0; j < c; j++)
            if (dp[t - pr2[j]]==i){
                p[x++] = pr2[j];
                t -= pr2[j];
                break;
            }
    sort(p, p+x, cmp);
    for (int i = 0; i < x; i++)printf("%d ", p[i]);
    return 0;
}