Problem:
Score : 400 points

Problem Statement

Find the largest integer that can be formed with exactly N matchsticks, under the following conditions:

Every digit in the integer must be one of the digits A_1, A_2, ..., A_M (1 \leq A_i \leq 9).

The number of matchsticks used to form digits 1, 2, 3, 4, 5, 6, 7, 8, 9 should be 2, 5, 5, 4, 5, 6, 3, 7, 6, respectively.

Constraints

All values in input are integers.

2 \leq N \leq 10^4

1 \leq M \leq 9

1 \leq A_i \leq 9

A_i are all different.

There exists an integer that can be formed by exactly N matchsticks under the conditions.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_M

Output

Print the largest integer that can be formed with exactly N matchsticks under the conditions in the problem statement.

Sample Input 1

20 4
3 7 8 4

Sample Output 1

777773

The integer 777773 can be formed with 3 + 3 + 3 + 3 + 3 + 5 = 20 matchsticks, and this is the largest integer that can be formed by 20 matchsticks under the conditions.

Sample Input 2

101 9
9 8 7 6 5 4 3 2 1

Sample Output 2

71111111111111111111111111111111111111111111111111

The output may not fit into a 64-bit integer type.

Sample Input 3

15 3
5 4 6

Sample Output 3

654

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

int main(){
  int N,M;
  std::cin>>N>>M;
  std::vector<long long> m{2,5,5,4,5,6,3,7,6};
  std::vector<long long> a(M),b(M);
  int b_min = 10;
  for (int i = 0; i < M; i++){
    std::cin>> a[i];
    b[i] = m[a[i]-1];
    if(b[i] < b_min){
      b_min = b[i];
    }
  }

  /*std::sort(a.begin(),a.end());
  std::sort(b.begin(),b.end());*/

  int k = N / b_min;
  std::vector<std::vector<long long>> DP(k+1,std::vector<long long>(N));

  for (int i = k-1; i >= 0; i--){
    for (int j = N-1; j >= 0; j--){
      long long temp = 0;
      for (int h = 0; h < M; h++){
        if(j+b[h] == N){
          temp = std::max(temp,a[h]*static_cast<long long>(std::pow(10,i)));
        }else if(j+b[h] < N && DP[i+1][j+b[h]] != 0){
          temp = std::max(temp,DP[i+1][j+b[h]] + a[h]*static_cast<long long>(std::pow(10,i)));
        }
      }
      DP[i][j] = temp;

    }
  }
  std::cout<<DP[0][0]<<std::endl;

  return 0;
}