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>

typedef long long ll;
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int def[10] = {-100000, 2, 5, 5, 4, 5, 6, 3, 7, 6};

    int n, m; cin >> n >> m;
    int a[m];
    vector<pair<int, int>> hand(m); // num, cost
    for (int i = 0; i < m; i++) {
        cin >> a[i];
        hand[i] = {a[i], def[a[i]]};
    }
    sort(hand.begin(), hand.end(), greater<>());
    int mnv = hand[0].first, mnc = hand[0].second;
    for (int i = 1; i < m; i++) {
        int cost = hand[i].second;
        int val = hand[i].first;
        if (mnc > cost) {
            mnc = cost;
            mnv = val;
        }
    }
    int order = n/mnc;

    string ans = "";
    for (int ord = order; ord >= 0; ord--) {
        ans = "";
        for (int i = 0; i < ord; i++) {
            ans += to_string(mnv);
        }
        int left = n - mnc * ord;
        queue<pair<int, string>> q;
        q.push({left, ans});
        bool flag = false;
        while (!q.empty()) {
            int fl = q.front().first;
            string fs = q.front().second;
            q.pop();
            if (fl == 0) {
                flag = true;
                ans = fs;
                break;
            }
            if (fl < 0) {
                continue;
            }
            // find included unique number
            set<char> uniH;
            for (auto &c : fs) {
                uniH.insert(c);
            }
            for (auto it = uniH.begin(); it != uniH.end(); it++) {
                for (int i = 0; i < m; i++) {
                    int v = fl;
                    string ns(fs);
                    ns[ns.find(*it)] = to_string(hand[i].first)[0];
                    sort(ns.begin(), ns.end(), greater<>());
                    int diff = hand[i].second - def[*it - '0'];
                    // if (diff >= 0 && *it - '0' != hand[i].first) {
                    //     q.push({v - diff, ns});
                    // }
                    if (diff > 0) {
                        q.push({v - diff, ns});
                    }
                }
            }
        }
        if (flag) break;
    }
    sort(ans.begin(), ans.end(), greater<>());
    cout << ans << endl;
    return 0;
}