Codeforces 516A Drazil and Factorial【暴搜找规律+贪心】

A. 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 integerx as a product of factorials of its digits. For example,.

First, they choose a decimal number a consisting ofn digits that contains at least one digit larger than1. This number may possibly start with leading zeroes. Then they should find maximum positive numberx 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 ina.

The second line contains n digits of a. There is at least one digit in a that is larger than1. 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.

Examples

Input

4
1234

Output

33222

Input

3
555

Output

555

Note

In the first case,

题目大意：

定义一个F（a）=a的每个位子上的数的阶乘相乘。

给你一个数a，让你找一个数x，使得x中没有0也没有1.并且x尽可能的大。

思路：

1、肯定我们是将a中能够拆分的数进行拆分，比如4！=2！*2！*3！

那么我们肯定第一个考虑的问题就是从2~9这些数怎样拆分才能使得拆分出来的数字量最大，因为拆分出来的数字量越大，我们能够组成的数x长度也就越大，长度也大的数，数也就越大。

2、那么我们暴搜判断每个阶乘数最多能够拆分出来几个数，这几个数分别都是多少？

暴力处理代码：

运行结果对应每一行的含义就是对应当前这个阶乘数能够拆分成哪些接成数（包括几个）

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
int vis[15];
char a[20];
ll jiecheng[20];
void init()
{
    for(int i=2;i<=9;i++)
    {
        ll sum=1;
        for(int j=1;j<=i;j++)
        {
            sum*=j;
        }
        jiecheng[i]=sum;
    }
}
void Dfs(long long int now,long long int n)
{
    if(now>n)return ;
    else if(now==n)
    {
        for(int i=2;i<=9;i++)
        {
            printf("%d ",vis[i]);
        }
        printf("\n");
        return ;
    }
    for(int i=2;i<=9;i++)
    {
        vis[i]++;
        Dfs(now*jiecheng[i],n);
        vis[i]--;
    }
}
int main()
{
    int n;
    memset(vis,0,sizeof(vis));
    init();
    for(int i=2;i<=9;i++)
    {
        printf("%d\n",i);
        Dfs(1,jiecheng[i]);
    }
}

3、那么根据暴力处理出来的结果，我们对应进行处理每个数能够拆分出哪些数，然后此时拆分出来的数一共有多长也就能够确定下来了。最终将每个数从大到小输出即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int vis[15];
char a[20];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        scanf("%s",a);
        for(int i=0;i<n;i++)
        {
            if(a[i]=='0'||a[i]=='1')continue;
            if(a[i]=='2')vis[2]++;
            if(a[i]=='3')vis[3]++;
            if(a[i]=='4')vis[3]++,vis[2]+=2;
            if(a[i]=='5')vis[5]++;
            if(a[i]=='6')vis[5]++,vis[3]++;
            if(a[i]=='7')vis[7]++;
            if(a[i]=='8')vis[7]++,vis[2]+=3;
            if(a[i]=='9')vis[2]++,vis[3]+=2,vis[7]++;
        }
        for(int i=9;i>=2;i--)
        {
            for(int j=0;j<vis[i];j++)
            {
                printf("%d",i);
            }
        }
        printf("\n");
    }
}