Codeforce 515 C . Drazil and Factorial 规律

传送门： http://codeforces.com/contest/515/problem/C

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, 

题目问你求最大的a

这样考虑  4！=3！*2！*2！

6！=5！*3！

8！=7！*2！*2！*2！

9！=7！*3！*3！*2！

于是得到了f(4)=f(322) f(6)=f(53) f(8)=f(7222) f(9)=f(7332)

AC代码

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
using namespace std;

int main()
{
	int i,j,k,l,m,n,temp;
	int a[10];
	memset(a,0,sizeof(a));
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%1d",&temp);
		switch(temp)
		{
			case 2: 
				a[2]++;
				break;
			case 3:
				a[3]++;
				break;
			case 4:
				a[3]++;
				a[2]+=2;
				break;
			case 5:
				a[5]++;
				break;
			case 6:
				a[3]++;
				a[5]++;
				break;
			case 7:
				a[7]++;
				break;
			case 8:
				a[2]+=3;
				a[7]++;
				break;
			case 9:
				a[2]++;
				a[3]+=2;
				a[7]++;
				break;
				
			
		}
		
	}
	for(i=9;i>=2;i--)
	{
		for(j=0;j<a[i];j++)
		{
			printf("%d",i);
		}
	}
	printf("\n");

}