codeforces 515C Drazil and Factorial

最新推荐文章于 2018-10-20 15:45:27 发布

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本文介绍了一个基于阶乘的数学游戏——Draziland Factorial，并提供了算法解决方案。该游戏的目标是在给定条件下找到最大的不含0和1的整数。通过贪心策略和排序技巧，文章详细展示了如何将大阶乘分解为小阶乘，最终得出最优解。

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Drazil is playing a math game with Varda.

Let's define $\text{[math]}$ for positive integer x as a product of factorials of its digits. For example, $\text{[math]}$ .

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. $\text{[math]}$ = $\text{[math]}$ .

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.

Note

In the first case, $\text{[math]}$

贪心+sort，把大的阶乘拆成小的阶乘

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    int a[20],b[200];
    char x[20];
    cin>>n;
    cin>>x;
    int t=0;
    for(int i=0; i<n; i++)
    {
        a[i]=x[i]-'0';
        if(a[i]==2||a[i]==3||a[i]==5||a[i]==7)
            b[t++]=a[i];
        else if(a[i]==4)
        {
            b[t++]=2;
            b[t++]=2;
            b[t++]=3;
        }
        else if(a[i]==6)
        {
            b[t++]=5;
            //b[t++]=2;
            b[t++]=3;
        }
        else if(a[i]==8)
        {
            b[t++]=7;
            b[t++]=2;
            b[t++]=2;
            b[t++]=2;
        }
        else if(a[i]==9)
        {
            b[t++]=3;
            b[t++]=3;
            b[t++]=7;
            b[t++]=2;
            // b[t++]=2;
            //b[t++]=2;
        }
    }
    sort(b,b+t);
    for(int i=t-1; i>=0; i--)
    {
        if(b[i]>1)
            cout<<b[i];
    }
    cout<<endl;
    return 0;
}