# SGU 116 - Index of super-prime

## Description

Let $P_1, P_2,\cdots ,P_N,\cdots$ 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

 1  6 

## Sample Output

 1 2  2 3 3 

## Solution

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66  #include #include #include #include using namespace std; const int MAX = 10240; vector P, SP; int pP[MAX], pSP[MAX], nCnt; int f[MAX], pPath[MAX]; int main() { int N; P.push_back(0); SP.push_back(0); memset(pP, 0, sizeof(pP)); memset(pSP, 0, sizeof(pSP)); for(int i = 2; i < MAX; i++) { if(pP[i] == 0) { P.push_back(i); for(int j = i + i; j < MAX; j += i) { pP[j] = 1; } } } for(int i = 2; i < P.size(); i++) { if(pSP[i] == 0) { SP.push_back(P[i]); for(int j = i + i; j < P.size(); j += i) { pSP[j] = 1; } } } while(cin >> N) { memset(f, 0, sizeof(f)); memset(pPath, 0, sizeof(pPath)); f[0] = 0; for(int i = 1; i <= N; i++) { f[i] = 214748364; } for(int i = 1; i < SP.size(); i++) { for(int j = SP[i]; j <= N; j++) { if(f[j - SP[i]] + 1 < f[j]) { f[j] = f[j - SP[i]] + 1; pPath[j] = j - SP[i]; } } } if(f[N] == 214748364) { cout << 0 << endl; } else { cout << f[N] << endl << N - pPath[N]; for(int i = pPath[N]; i; i = pPath[i]) { cout << " " << i - pPath[i]; } cout << endl; } } return 0;