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 <bits/stdc++.h>
#include <unordered_map>
#include <queue>
#include <vector>
#define MAX 10001

using namespace std;

int mp[]={0,2,5,5,4,5,6,3,7,6};
vector<int> sticks;
int n,extra,m,firstMatch,firstCnt;
int d[MAX];

// string maxStr(string a,string b){
//     if(a[0]=='-')return b;
//     if(b[0]=='-')return a;
//     if(a.size()==b.size())return max(a,b);
//     return a.size()>b.size()?a:b;
// }

void dp(int extra) {
    int i,j;
    d[0]=0;
    for(i=0;i<n-2;i++){
        for(j=0;j<m;j++){
            d[i+mp[sticks[j]]]=max(d[i+mp[sticks[j]]],d[i]+1);
        }
    }
    int r=n;
    sort(sticks.rbegin(),sticks.rend());

    while(r>0){
        for(i=0;i<m;i++){
            if(r-mp[sticks[i]]>=0&&d[r-mp[sticks[i]]]==d[r]-1){
                cout<<sticks[i];
                r-=mp[sticks[i]];
                break;
            }
        }
    }
}
int main() {
    int i, t;
    cin>>n>>m;
    for(i=0;i<MAX;i++){
        d[i]=-100;
    }
    int first = 999;
    for(i=0;i<m;i++){
        cin >> t;
        sticks.push_back(t);
    }
    dp(n);
    // cout<<;
    return 0;
}