CF Drazil and Factorial (打表)

Drazil and Factorial

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Drazil is playing a math game with Varda.

Let's define  for positive integer x as a product of factorials of its digits. For example, .

First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:

1. x doesn't contain neither digit 0 nor digit 1.

2.  = .

Help friends find such number.

Input

The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.

The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.

Output

Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.

Sample test(s)

input

4
1234

output

33222

input

3
555

output

555



把2-9每个数打表，看最多能分解成哪几个的阶乘，然后直接替换，最后排序输出。

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cctype>
#include <queue>
#include <map>
using    namespace    std;

const    int    SIZE = 1000;

bool    comp(const int a,const int b);

char    RULE[][10] = {{"0"},{"1"},{"2"},{"3"},{"322"},{"5"},{"53"},{"7"},{"7222"},{"7332"}};
int    main(void)
{
    int    n;
    char    s[SIZE];
    char    ans[SIZE] = {'\0'};
    
    cin >> n;
    cin >> s;
    for(int i = 0;s[i];i ++)
        if(s[i] != '1' && s[i] != '0')
            strcat(ans,RULE[s[i] - '0']);
    sort(ans,ans + strlen(ans),comp);
    cout << ans << endl;

    return    0;
}

bool    comp(const int a,const int b)
{
    return    a > b;
}