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 Scala (2.11.7) (judge verdict: Accepted)

import scala.io.StdIn.readLine

object Main{
  def main(args : Array[String]): Unit ={
    val Array(n, m) : Array[Int] = readLine().split(' ').map(_.toInt)
    val can : Array[Int] = readLine().split(' ').map(_.toInt)
    val SLV = new SOLVE(n, can)
    println(SLV.solve(n))
  }
}

class SOLVE(private val n : Int, private val can : Array[Int]){
  val caned : Array[Int] = can.sorted
  val COST : Array[Int] = Array(0, 2, 5, 5, 4, 5, 6, 3, 7, 6)
  val DP : Array[Int] = Array.fill(n + 10)(-1)
  DP(0) = 0

  def MakeReversing(k : Int): String ={
    def getNextString(u : Int): (Char, Int) ={
      var res1 : Char = '\n'
      var res2 : Int = 114514
      for(ele <- caned;if(u >= COST(ele))){
        if(DP(u) - DP(u - COST(ele)) == 1) {
          res1 = ('0' + ele).toChar
          res2 = COST(ele)
        }
      }
      (res1, res2)
    }
    var search : Int = k
    var res : String = ""
    while(search > 0){
      val (a, b) : (Char, Int) = getNextString(search)
      search -= b
      res = res + a
    }
    res
  }
  def solve(k : Int): String ={
    for(i <- 0 until k; j <- can){
      val idx : Int = i + COST(j)
      if(DP(idx) < DP(i) + 1) DP(idx) = DP(i) + 1
    }
    MakeReversing(k)
  }
}