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<stdio.h>
#define ONE 2
#define TWO 5
#define THR 5
#define FOU 4
#define FIV 5
#define SIX 6
#define SEV 3
#define EIG 7
#define NIN 6
#define max(a,b) (a)>(b)?(a):(b)
/*#define dpro(a) {
  switch(a){
    case 1:;
  }
}
i-7>=0?
dp[i]=max(dp[i-ONE]*10+1
      ,max(dp[i-TWO]*10+2
      ,max(dp[i-THR]*10+3
      ,max(dp[i-FOU]*10+4
      ,max(dp[i-FIV]*10+5
      ,max(dp[i-SIX]*10+6
      ,max(dp[i-SEV]*10+7
      ,max(dp[i-EIG]*10+8
      ,max(dp[i-NIN]*10+9
      ,dp[i-1])))))))))
:(i-6>=0?
  dp[i]=max(dp[i-ONE]*10+1
        ,max(dp[i-TWO]*10+2
        ,max(dp[i-THR]*10+3
        ,max(dp[i-FOU]*10+4
        ,max(dp[i-FIV]*10+5
        ,max(dp[i-SIX]*10+6
        ,max(dp[i-EIG]*10+8
        ,max(dp[i-NIN]*10+9
        ,dp[i-1]))))))))
)*/
int main(void)
{
  int n,m,a[10];
  long long dp[10010]={};
  scanf("%d%d",&n,&m);
  for(int i=0;i<m;i++){
    scanf("%d",&a[i]);
  }
  for(int i=1;i<=n;i++){
    for(int j=0;j<m;j++){
      switch(a[j]){
        case 1:if(i-ONE>=0)dp[i]=max(dp[i-ONE]*10+1,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 2:if(i-TWO>=0)dp[i]=max(dp[i-TWO]*10+2,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 3:if(i-THR>=0)dp[i]=max(dp[i-THR]*10+3,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 4:if(i-FOU>=0)dp[i]=max(dp[i-FOU]*10+4,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 5:if(i-FIV>=0)dp[i]=max(dp[i-FIV]*10+5,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 6:if(i-SIX>=0)dp[i]=max(dp[i-SIX]*10+6,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 7:if(i-SEV>=0)dp[i]=max(dp[i-SEV]*10+7,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 8:if(i-EIG>=0)dp[i]=max(dp[i-EIG]*10+8,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
        case 9:if(i-NIN>=0)dp[i]=max(dp[i-NIN]*10+9,max(dp[i],dp[i-1]));
               else{dp[i]=max(dp[i],dp[i-1]);}break;
      }
    }
  }

  printf("%lld",dp[n]);
  return 0;
}