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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <malloc.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

//ABC118D
int compare_int(const void *a, const void *b) {//sort
  return *(int*) a - *(int*) b;
}

int main(void) {
  int minb=10, n, m, k, i, j, cou=0, tmp, b[]={2, 5, 5, 4, 5, 6, 3, 7, 6};
  scanf(" %d %d", &n, &m);
  int a[m], dp[n+1];
  for(i=0; i<m; i++){
    scanf(" %d", &a[i]);
    minb=minb>b[a[i]-1] ? b[a[i]-1] : minb;
    //printf("minb=%d\n", minb);
  }
  qsort(a, m, sizeof(int), compare_int);//sort
  dp[0]=0;

  //printf("dp[%d]=%d\n", 0, dp[0]);

  for(i=1; i<=n; i++){
    dp[i]=-1e9;
    for(j=0; j<m; j++){
      if(i-b[a[j]-1]>=0&&dp[i-b[a[j]-1]]!=-1e9&&dp[i]<dp[i-b[a[j]-1]]+1){
        dp[i]=dp[i-b[a[j]-1]]+1;
      }
    }
    //printf("dp[%d]=%d\n", i, dp[i]);
  }
  //printf("%d\n\n", dp[n-1]);

  tmp=n;
  for(i=0;i<m;i++){
    while(dp[n]>0){
      if(dp[n]-dp[tmp-b[a[m-i-1]-1]]==1){
        dp[n]--;
        printf("%d",a[m-i-1]);
        tmp-=b[a[m-i-1]-1];
      }else break;
    }
  }
}