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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class Main {
    public static void main(String [] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int[] num = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
        int [] a = new int[m];
        for (int i = 0; i < m; ++i) {
            a[i] = scanner.nextInt();
        }
        String [][] dp = new String[m][n + 1];
        String res = "";

        for (int i = 0; i < m; ++i) {
            dp[i][0] = "";
            for (int j = 1; j <= n; ++j) {
                dp[i][j] = "";
                if (i == 0) {
                    if (j % num[a[0]] == 0) {
                        dp[i][j] = String.join("", Collections.nCopies(j / num[a[0]], String.valueOf(a[0])));
                    }
                } else {
                    dp[i][j] = dp[i - 1][j];
                    for (int k = 1; k * num[a[i]] <= j; ++k) {
                        if (!dp[i - 1][j - num[a[i]] * k].isEmpty()
                        || j == num[a[i]] * k) {
                            String newStr = dp[i - 1][j - num[a[i]] * k] +
                                    String.join("", Collections.nCopies(k, String.valueOf(a[i])));

                            dp[i][j] = getMax(dp[i][j], newStr);
                        }

                    }
                }
                if (j == n && !dp[i][j].isEmpty()) {
                    res = getMax(res, dp[i][j]);
                }
            }
        }
        System.out.println(res);

    }

    static String getMax(final String a, final String b) {
        if (a.length() > b.length()) {
            return a;
        } else if (a.length() < b.length()) {
            return b;
        } else {
            int [] numA = new int[10];
            int [] numB = new int[10];
            for (int i = 0; i < a.length(); ++i) {
                numA[a.charAt(i) - '0']++;
                numB[b.charAt(i) - '0']++;
            }
            for (int i = 9; i >= 1; --i) {
                if (numA[i] > numB[i]) {
                    return a;
                } else if (numA[i] < numB[i]) {
                    return b;
                }
            }
            return a;
        }
    }
}