Codeforces Round #292 (Div. 2) -- C. 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, 

思路：就是尽可能的将数字化简拆分，0，1舍弃，2，3，5，7不动，4拆分成3，2和2，6拆分成5和3,8拆分成7,2,2,和2,9拆分成7,3,3和2，然后从大到小输出即可

比赛时写完才发现自己有个小bug，哎，这时已经锁了(๑´ㅂ`๑)

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main() {
	int n, num[11];
	char a[20];
	while(scanf("%d %s", &n, a) != EOF) {
		memset(num, 0, sizeof(num));
		for(int i = 0; i < n; i++) {
			if(a[i] != '0' || a[i] != '1') {
				if(a[i] == '6') {
					num[5]++; 
					num[3]++;
				} 
				else if(a[i] == '8') {
					num[2] += 3;
					num[7] ++;
				}
				else if(a[i] == '4') {
					num[2] += 2;
					num[3]++;
				}
				else if(a[i] == '9') {
					num[7]++;
					num[2]++;
					num[3]+=2;
				}
				else num[a[i] - '0']++;
			}
		}
		
		for(int i = 9; i >= 2; i--) {
			while(num[i]){
				printf("%d", i);
				num[i]--;
			}
		}
		printf("\n");
	}
	return 0;
}