Programa C/C++ para contar ceros finales en factorial de un número

Dado un entero n, escriba una función que devuelva el recuento de ceros finales en n!.  Ejemplos:

Input: n = 5
Output: 1 
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24

Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....

// C++ program to count
// trailing 0s in n!
#include <iostream>
using namespace std;
 
// Function to return trailing
// 0s in factorial of n
int findTrailingZeros(int n)
{
    // Initialize result
    int count = 0;
 
    // Keep dividing n by powers of
    // 5 and update count
    for (int i = 5; n / i >= 1; i *= 5)
        count += n / i;
 
    return count;
}
 
// Driver Code
int main()
{
    int n = 100;
    cout << "Count of trailing 0s in " << 100
         << "! is " << findTrailingZeros(n);
    return 0;
}

Count of trailing 0s in 100! is 24

Complejidad de tiempo: O (log ₅ n)

Espacio Auxiliar: O(1)

Consulte el artículo completo sobre Contar ceros finales en factorial de un número para obtener más detalles.