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 <fstream>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <list>
#include <string>
#include <math.h>
#include <set>
#include <string>
#include <numeric>
#include <malloc.h>
using namespace std;

bool StringCompare(string A,string B)
{
	if(A.length()>B.length())
	{
		return true;
	}
	else if(A.length()<B.length())
	{
		return false;
	}
	else
	{
		if(A.compare(B)>=0)
		{
			return true;
		}
		else
		{
			return false;
		}
	}
}
int main()
{
	int N,M;
	cin>>N>>M;
	int cost[] = {2,5,5,4,5,6,3,7,6};
	vector<int> A;
	A.resize(M);
	for(int i=0; i<M; i++)
	{
		cin>>A[i];
	}

	vector<string> result;
	result.resize(N+1);
	for(int i=0; i<=N; i++)
	{
		result[i] = "";
	}
	//配るDPを構築
	for(int i=0; i<N; i++)
	{
		for(int j=0; j<M; j++)
		{
			if(result[i]=="" && i>0)
			{
				break;
			}
			if(i+cost[A[j]-1]>N)
			{
				continue;
			}
			if(StringCompare(result[i]+to_string(A[j]),result[i+cost[A[j]-1]]))
			{
				result[i+cost[A[j]-1]]=result[i]+to_string(A[j]);
			}
			
		}
	}
	cout<<endl;
	for(int i=0; i<=N; i++)
	{
		cout<<result[i]<<endl;
	}
	return 0;
}