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>
#include <string>
using namespace std;

int cmp(string a, string b) {
	int a_ = atoi(a.c_str());
	int b_ = atoi(b.c_str());
	return a_ > b_;
}
int main()
{
	int n, m;
	int count[9] = { 2,5,5,4,5,6,3,7,6 };
	cin >> n >> m;
	vector<int> arr(m, 0);
	for (int i = 0; i < m; i++) {
		cin >> arr[i];
	}
	vector<string> dp(n + 1, "0");
	for (int i = 2; i <= n; i++) {
		for (int j = 0; j < m; j++) {
			char first = arr[j] + '0';
			int sticks = count[arr[j] - 1];
			if (i >= sticks) {
				string cur;
				if (dp[i-sticks] != "0") cur = first + dp[i - sticks];
				else if (dp[i - sticks] == "0" && i == sticks) {
					cur = first;
				}
				if (cmp(cur, dp[i])) {
					dp[i] = cur;
				}
			}
		}
	}
	cout << dp[n] << endl;
	//for (int i = 0; i <= n; i++) cout << i << ": " << dp[i] << " ";
	//cout << endl;
	// dp[i] = max{num[j]*10^wei[j] + dp[i-count[j]]} | j=arr_0->arr_m ;
	// wei[i] = wei[j] + 1;
	return 0;
}