CodeForces 515C. Drazil and Factorial

C. 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

Note

In the first case,

真的说，这道题目是比较简单的，只是我的想法有点问题。

我做的过程太过于复杂了！在中间操作过程已经使结果超过 long long

而JS相当于在中间过程有简单的优化，是我没想到的，我把每个数还原的过于简单了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <ctype.h>
using namespace std;
#define LL long long

int main()
{
    int n;
    char num[20];
    int cnt[10] = {0};
    int bit[100000];
    while(scanf("%d%*c", &n) != EOF)
    {
        int tag = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%c", &num[i]);
            cnt[num[i] - '0']++;
        }

        for(int i = 0; i < n; i++)
        {
            if(cnt[9] && num[i] == '9')
            {
                bit[tag++] = 7;
                bit[tag++] = 3;
                bit[tag++] = 3;
                bit[tag++] = 2;
                cnt[9]--;
            }
            else if(cnt[8] && num[i] == '8')
            {
                bit[tag++] = 7;
                bit[tag++] = 2;
                bit[tag++] = 2;
                bit[tag++] = 2;
                cnt[8]--;
            }
            else if(cnt[7] && num[i] == '7')
            {
                bit[tag++] = 7;
                cnt[7]--;
            }
            else if(cnt[6] && num[i] == '6')
            {
                bit[tag++] = 5;
                bit[tag++] = 3;
                cnt[6]--;
            }
            else if(cnt[5] && num[i] == '5')
            {
                bit[tag++] = 5;
                cnt[5]--;
            }
            else if(cnt[4] && num[i] == '4')
            {
                bit[tag++] = 3;
                bit[tag++] = 2;
                bit[tag++] = 2;
                cnt[4]--;
            }
            else if(cnt[3] && num[i] == '3')
            {
                bit[tag++] = 3;
                cnt[3]--;
            }
            else if(cnt[2] && num[i] == '2')
            {
                bit[tag++] = 2;
                cnt[2]--;
            }
        }
        sort(bit, bit+tag);
        for(int i = tag-1; i >= 0; i--)
        {
            cout << bit[i];
        }
        cout << endl;
    }
    return 0;
}