Codeforces Round #292 (Div. 1)---A. Drazil and Factorial

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,

水题,
将阶乘分解成素数的阶乘即可

/*************************************************************************
    > File Name: cf292-A.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月18日 星期三 00时26分55秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

char str[20];
int num[10];

int main ()
{
    int n;
    while (~scanf("%d", &n))
    {
        scanf("%s", str);
        memset (num, 0, sizeof(num));
        int len = strlen (str);
        for (int i = 0; i < len; ++i)
        {
            if (str[i] == ‘0‘ || str[i] == ‘1‘)
            {
                continue;
            }
            else if (str[i] == ‘2‘ || str[i] == ‘3‘ || str[i] == ‘5‘ || str[i] == ‘7‘)
            {
                num[str[i] - ‘0‘]++;
            }
            else if (str[i] == ‘4‘)
            {
                num[2] += 2;
                num[3]++;
            }
            else if (str[i] == ‘6‘)
            {
                num[5]++;
                num[3]++;
            }
            else if (str[i] == ‘8‘)
            {
                num[7]++;
                num[2] += 3;
            }
            else
            {
                num[7]++;
                num[3] += 2;
                num[2]++;
            }
        }
        for (int i = 9; i >= 2; --i)
        {
            for (int j = 1; j <= num[i]; ++j)
            {
                printf("%d", i);
            }
        }
        printf("\n");
    }
    return 0;
}