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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.Comparator;

public class Main implements Runnable {
    
    private int n;
    private int m;
    private int[] a;
    private int[] use;
    private int[] mat = {2,5,5,4,5,6,3,7,6};

    public static void main(String[] args) throws IOException {
        Main m = new Main();
		m.run();
    }

    private void set() {
		try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
	  		String[] parts = br.readLine().split("[\\s]+");
	 	   	n = Integer.parseInt(parts[0]);
	 	   	m = Integer.parseInt(parts[1]);
	 	   	a = Arrays.stream(br.readLine().split("[\\s]+"))
				.mapToInt(Integer::parseInt).toArray();
		    use = new int[m];
		    sort();
		} catch (IOException e) {
		    e.printStackTrace();
		}
    }

    private void sort() {
		for (int i=m-1;i>0;i--) {
		    for (int j=0;j<i;j++) {
				if ((mat[a[j]-1] > mat[a[j+1]-1]) ||
				    (mat[a[j]-1] == mat[a[j+1]-1] && a[j] < a[j+1])) {
				    int b = a[j];
		    		a[j] = a[j+1];
		   		 	a[j+1] = b;
				}
	    	}
		}
    }

    private int sum() {
		int s = 0;
		for (int i=0;i<m;i++) {
	  		s += use[i]*mat[a[i]-1];
		}
		return s;
    }

    @Override
    public void run() {
		set();
		use[0] = n/mat[a[0]-1];
		int idx = 0;
		while (sum() != n) {
	    	if (use[idx] > 0) {
				use[idx]--;
	    	} else {
				idx++;
				if (idx == m) idx = 0;
	    	}
	    	int left = n - sum();
	    	for (int i=idx+1;i<m;i++) {
				if (mat[a[i]-1] <= left) {
		    		use[i] += left/mat[a[i]-1];
		    		left = n - sum();
				}
	    	}
		}
		StringBuilder sb = new StringBuilder();
		while (sum() != 0) {
	    	int num = 0;
	    	idx = 0;
	    	for (int i=0;i<m;i++) {
				if (num < a[i] && use[i] != 0) {
		    		num = a[i];
		    		idx = i;
				}
	    	}
	    	for (int i=0;i<use[idx];i++) {
				sb.append(num);
	    	}
	    	use[idx] = 0;
		}
		System.out.println(sb.toString());
    }

}