Factorials Product Number

In mathematics, there is a little game. The following is how a function F(x) is defined:

• F(x) = Product of the factorials of digits in x.

ex :
   F (234) = 2! * 3! * 4!

A decimal number a is selected which has n digits. This number can have leading zeros as well. Furthermore, ‘a’ has at least one digit larger than 1.

Task

You should be able to find maximum positive number x with satisfying F(x)=F(a) and x contains neither digit 0 nor digit 1. This number may have more digits than the original number.

Input Format

First-line contains an integer n, the number of digits in a
The second line contains n digits of a

(‘a’ has at least one digit larger than 1. This number can have leading zeros as well)

Constraints

1 ≤ n ≤ 15

Output Format

An Integer, maximum possible number satisfying the conditions.

(There should be no 0's and 1's in this number's decimal representation)

Sample Input 0

4
1256

Sample Output 0

5532

Sample Input 1

3
564

Sample Output 1

553322

Solution 1

By Nadeesha Jayamanne

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
using namespace std;

int fact(int num){
    map<int,int> factorials={{2,2},{3,6},{4,24},{5,120},{6,720},{7,5040},{8,40320},{9,362880}};
    return factorials[num];
}
bool isPrime(int num){
    if(num==2 || num==3 || num==5 || num==7 )
    return true;
    else return false;
}

int Big_Prime(int digit, vector<int> &X){
    for(int i=digit;i>1;i--){
        if(isPrime(i)){
            X.push_back(i);
            int Rest=1;
            Rest=fact(digit)/fact(i);
            return Rest;
        }
    }
    return -1;
}


void div_prime(int value, vector<int> &X){
    int Rest=0;
    for(int i=value;i>1;i--){
        if(isPrime(i) && value%i==0){
            X.push_back(i);
            Rest=value/fact(i);
            break;
        }
    }
    if(Rest>2){
        div_prime(Rest, X);
    }
    else if (Rest==2){
        X.push_back(Rest);
    }
}

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int n;
    long long a;
    cin >> n;
    cin >> a;
    stack<int> digits;
    for(int i=n; i>0;i--){
        int num = a%10;
        a /= 10;
        digits.push(num);
    }

    vector<int> X;

    while(!digits.empty()){
        int digit = digits.top();
        digits.pop();

        if(digit==0 || digit ==1){
            continue;
        }
        int rest = Big_Prime(digit, X);
        div_prime(rest, X);
    }
    sort(X.begin(), X.end(), greater<int>());
    for (int i : X){
        cout<< i;
    }

    return 0;
}

Solution 2

By Chanupa Gurusinghe

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    int n;
    string a;
    cin >> n >> a;

    vector<char> result;

    for (char ch : a) {
        switch (ch) {
            case '0': case '1':
                break; // skip
            case '2':
                result.push_back('2');
                break;
            case '3':
                result.push_back('3');
                break;
            case '4':
                result.push_back('3');
                result.push_back('2');
                result.push_back('2');
                break;
            case '5':
                result.push_back('5');
                break;
            case '6':
                result.push_back('5');
                result.push_back('3');
                break;
            case '7':
                result.push_back('7');
                break;
            case '8':
                result.push_back('7');
                result.push_back('2');
                result.push_back('2');
                result.push_back('2');
                break;
            case '9':
                result.push_back('7');
                result.push_back('3');
                result.push_back('3');
                result.push_back('2');
                break;
        }
    }

    sort(result.begin(), result.end(), greater<char>());

    for (char c : result)
        cout << c;
    cout << endl;

    return 0;
}