MCD de factoriales de dos números

Dados dos números m y n. Encuentre el MCD de su factorial.
Ejemplos: 
 

Input : n = 3, m = 4
Output : 6
Explanation:
Factorial of n = 1 * 2 * 3 = 6
Factorial of m = 1 * 2 * 3 * 4 = 24
GCD(6, 24) = 6.

Input : n = 9, m = 5
Output : 20
Explanation:
Factorial of n = 1 * 2 * 3 *4 * 5 * 6 * 7 * 8 
* 9 = 362880
Factorial of m = 1 * 2 * 3 * 4 * 5 = 120
GCD(362880, 120) = 120

Una solución simple es encontrar factoriales de ambos números. Luego encuentre el MCD de los factoriales calculados.
Una solución eficiente se basa en el hecho de que el MCD de dos factoriales es igual al factorial más pequeño (tenga en cuenta que los factoriales tienen todos los términos en común).
A continuación se muestra la implementación del enfoque anterior. 
 

// CPP program to find GCD of factorial of two
// numbers.
#include <bits/stdc++.h>
using namespace std;
 
int factorial(int x)
{
    if (x <= 1)
        return 1;
    int res = 2;
    for (int i = 3; i <= x; i++)
        res = res * i;
    return res;
}
 
int gcdOfFactorial(int m, int n)
{
    return factorial(min(m, n));
}
 
int main()
{
    int m = 5, n = 9;
    cout << gcdOfFactorial(m, n);
    return 0;
}

// Java program to find GCD of factorial
// of two numbers.
public class FactorialGCD{
     
static int factorial(int x)
{
    if (x <= 1)
        return 1;
    int res = 2;
    for (int i = 3; i <= x; i++)
        res = res * i;
    return res;
}
 
static int gcdOfFactorial(int m, int n)
{
    int min = m < n ? m : n;
    return factorial(min);
}
 
    /* Driver program to test above functions */
    public static void main (String[] args)
    {
        int m = 5, n = 9;
         
        System.out.println(gcdOfFactorial(m, n));
    }
}
 
// This code is contributed by Prerna Saini

# Python code to find GCD of factorials of
# two numbers.
import math
 
def gcdOfFactorial(m, n) :
    return math.factorial(min(m, n))
 
# Driver code
m = 5
n = 9
print(gcdOfFactorial(m, n))

// C# program to find GCD of factorial
// of two numbers.
using System;
 
public class GFG{
      
    static int factorial(int x)
    {
        if (x <= 1)
            return 1;
             
        int res = 2;
         
        for (int i = 3; i <= x; i++)
            res = res * i;
             
        return res;
    }
      
    static int gcdOfFactorial(int m, int n)
    {
        int min = m < n ? m : n;
        return factorial(min);
    }
  
    /* Driver program to test above functions */
    public static void Main()
    {
         
        int m = 5, n = 9;
          
        Console.WriteLine(gcdOfFactorial(m, n));
    }
}
  
// This code is contributed by Anant Agarwal.

<?php
// PHP program to find GCD
// of factorial of two numbers.
 
function factorial($x)
{
    if ($x <= 1)
        return 1;
    $res = 2;
    for ($i = 3; $i <= $x; $i++)
        $res = $res * $i;
    return $res;
}
 
function gcdOfFactorial($m, $n)
{
    return factorial(min($m, $n));
}
 
// Driver Code
$m = 5; $n = 9;
echo gcdOfFactorial($m, $n);
 
// This code is contributed by ajit
?>

<script>
 
// JavaScript program to find GCD of factorial
// of two numbers.
 
    function factorial(x) {
        if (x <= 1)
            return 1;
        var res = 2;
        for (i = 3; i <= x; i++)
            res = res * i;
        return res;
    }
 
    function gcdOfFactorial(m , n) {
        var min = m < n ? m : n;
        return factorial(min);
    }
 
    /* Driver program to test above functions */
     
        var m = 5, n = 9;
 
        document.write(gcdOfFactorial(m, n));
 
// This code is contributed by aashish1995
 
</script>

Producción : 

120