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 Rust (1.15.1) (judge verdict: Wrong Answer)

// -----------------------------------------------
// Framework <https://github.com/vain0x/procon>
//
// See the bottom of file for solution.
// -----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

static W: &'static [usize] = &[0, 2, 5, 5, 4, 5, 6, 3, 7, 6];

fn dfs(i: usize, w: usize, v: usize, A: &[usize], bag: &mut [usize]) -> (bool, usize, Vec<usize>) {
    if i >= A.len() {
        if w == 0 {
            return (true, v, bag.to_owned());
        }
        return (false, 0, vec![]);
    }

    let mut ok = false;
    let mut max_v = 0;
    let mut max_bag = vec![];

    let mut w = w;
    let mut v = v;
    let mut k = 0;
    loop {
        bag[A[i]] = k;
        let (z_ok, z_v, z_bag) = dfs(i + 1, w, v, A, bag);
        if z_ok && max_v < z_v {
            ok = true;
            max_v = z_v;
            max_bag = z_bag;
        }

        if w < W[A[i]] {
            bag[A[i]] = 0;
            break;
        }

        w -= W[A[i]];
        v = v * 10 + A[i];
        k += 1;
    }

    (ok, max_v, max_bag)
}

fn main() {
    let (N, _) = read!(usize, usize);
    let mut A = read![[usize]];

    A.sort();
    A.reverse();

    let mut min_d = 0;
    let mut min_w = 99;
    for &d in &A {
        if min_w > W[d] {
            min_w = W[d];
            min_d = d;
        }
    }
    debug!(N, A, (min_d, min_w));

    let d = min_d;
    let mut l = N / W[d];
    loop {
        let mut bag = vec![0; 10];

        let w = N - l * W[d];
        let (ok, _, mut bag) = dfs(0, w, 0, &A, &mut bag);
        debug!((l, w), bag);

        if !ok {
            l -= 1;
            continue;
        }

        bag[d] += l;

        let mut s = vec![];
        for d in (1..10).rev() {
            for _ in 0..bag[d] {
                s.push(b'0' + d as u8);
            }
        }
        println!("{}", String::from_utf8(s).unwrap());
        break;
    }
}