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

         for(int i = ; i < n; i++)
         {
             if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 bit[tag++] = ;
                 bit[tag++] = ;
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 bit[tag++] = ;
                 bit[tag++] = ;
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 bit[tag++] = ;
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 cnt[]--;
             }
             else if(cnt[] && num[i] == '')
             {
                 bit[tag++] = ;
                 cnt[]--;
             }
         }
         sort(bit, bit+tag);
         for(int i = tag-; i >= ; i--)
         {
             cout << bit[i];
         }
         cout << endl;
     }
     return ;
 }