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>
using namespace std;
int a1[10]={0,2,5,5,4,5,6,3,7,6},n,m,sum;
int a2[10]={0,1,7,4,5,3,2,9,6,8};
int ans[1000005],ans1[1000005],ans2[1000005],l1,l2,l3;
int ans3[1000005],ans4[1000005];
bool f1[10];
inline bool X(int *x1,int *x2,int ll1,int ll2)
{
	if(ll1!=ll2)
		return ll1>ll2;
	for(int i=1;i<=ll1;i++)
		if(x1[i]!=x2[i])
			return x1[i]>x2[i];
	return 0;
}
inline void dfs(int x,int tot)
{
	if(x==0)
	{
		if(X(ans2,ans1,tot,l2))
			memcpy(ans1,ans2,sizeof(ans2)),l2=tot;
		return;
	}
	for(int i=1;i<=9;i++)
		if(f1[i]&&x>=a1[i])
		{
			ans2[tot+1]=i;
			dfs(x-a1[i],tot+1);
		}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x;
		scanf("%d",&x);
		f1[x]=1;
	}
	for(int i=1;i<=9;i++)
		if(f1[a2[i]])
			while(n-a1[a2[i]]>=9)
				ans[++l1]=a2[i],n-=a1[a2[i]];
	dfs(n,0);
	l3=l1+l2;
	for(int i=1;i<=l1;i++)
		ans3[i]=ans[i],ans4[l2+i]=ans[i];
	for(int i=1;i<=l2;i++)
		ans3[i+l1]=ans1[i],ans4[i]=ans1[i];
	if(X(ans3,ans4,l3,l3))
	{
		for(int i=1;i<=l3;i++)
			printf("%d",ans3[i]);
	}
	else
	{
		for(int i=1;i<=l3;i++)
			printf("%d",ans4[i]);
	}
}