Recuento de potencia distinta del factor primo de N

Dado un entero positivo N , la tarea es encontrar el número total de potencias distintas del factor primo del número dado N .
Ejemplos: 
 

Entrada: N = 216 
Salida: 4 
Explicación: 
216 se puede expresar como 2 * 2 ²  * 3 * 3 ² . 
Los factores que satisfacen las condiciones son 2, 2 ² , 3 y 3 ² ya que todos ellos se escriben como distintas potencias positivas de factores primos.
Entrada: N = 24 
Salida: 3 
Explicación: 
24 se puede expresar como 2 * 2 ² * 3 
 

Enfoque: La idea es encontrar todos los factores primos de N y cuántas veces cada factor primo divide a N . 
Supongamos que el factor primo ‘p’ divide N ‘z’ veces, entonces los factores primos distintos requeridos son p, p ² , …, p ⁱ . 
Para encontrar el número de factores primos distintos para el número primo p, encuentre el valor mínimo de i tal que (1 + 2 + …. + i) ≤ z .
Por lo tanto, para cada número primo que divide N K número de veces, encuentre el valor mínimo de i tal que (1 + 2 + …. + i) ≤ K .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
int countFac(int n)
{
    int m = n;
    int count = 0;
 
    // Iterate for all prime factor
    for (int i = 2; (i * i) <= m; ++i) {
 
        int total = 0;
 
        // If it is a prime factor,
        // count the total number
        // of times it divides n.
        while (n % i == 0) {
            n /= i;
            ++total;
        }
 
        int temp = 0;
 
        // Find the Number of distinct
        // possible positive numbers
        for (int j = 1;
             (temp + j) <= total;
             ++j) {
            temp += j;
            ++count;
        }
    }
    if (n != 1)
        ++count;
 
    // Return the final count
    return count;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 24;
 
    // Function Call
    cout << countFac(N);
    return 0;
}

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
static int countFac(int n)
{
    int m = n;
    int count = 0;
 
    // Iterate for all prime factor
    for(int i = 2; (i * i) <= m; ++i)
    {
       int total = 0;
        
       // If it is a prime factor,
       // count the total number
       // of times it divides n.
       while (n % i == 0)
       {
           n /= i;
           ++total;
       }
       int temp = 0;
        
       // Find the Number of distinct
       // possible positive numbers
       for(int j = 1; (temp + j) <= total; ++j)
       {
          temp += j;
          ++count;
       }
    }
    if (n != 1)
        ++count;
         
    // Return the final count
    return count;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number N
    int N = 24;
 
    // Function Call
    System.out.println(countFac(N));
}
}
 
// This code is contributed by Pratima Pandey

# Python3 program for the above approach
 
# Function to count the number
# of distinct positive power of
# prime factor of integer N
def countFac(n):
     
    m = n
    count = 0
 
    # Iterate for all prime factor
    i = 2
    while((i * i) <= m):
        total = 0
 
        # If it is a prime factor,
        # count the total number
        # of times it divides n.
        while (n % i == 0):
            n /= i
            total += 1
         
        temp = 0
 
        # Find the Number of distinct
        # possible positive numbers
        j = 1
        while((temp + j) <= total):
            temp += j
            count += 1
            j += 1
         
        i += 1
     
    if (n != 1):
        count += 1
 
    # Return the final count
    return count
 
# Driver Code
 
# Given number N
N = 24
 
# Function call
print(countFac(N))
 
# This code is contributed by sanjoy_62

// C# program for the above approach
using System;
class GFG{
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
static int countFac(int n)
{
    int m = n;
    int count = 0;
 
    // Iterate for all prime factor
    for(int i = 2; (i * i) <= m; ++i)
    {
       int total = 0;
        
       // If it is a prime factor,
       // count the total number
       // of times it divides n.
       while (n % i == 0)
       {
           n /= i;
           ++total;
       }
       int temp = 0;
        
       // Find the Number of distinct
       // possible positive numbers
       for(int j = 1; (temp + j) <= total; ++j)
       {
          temp += j;
          ++count;
       }
    }
    if (n != 1)
        ++count;
         
    // Return the final count
    return count;
}
 
// Driver code
public static void Main()
{
     
    // Given Number N
    int N = 24;
 
    // Function Call
    Console.Write(countFac(N));
}
}
 
// This code is contributed by Code_Mech

<script>
 
// Javascript program for the above approach
 
// Function to count the number
// of distinct positive power of
// prime factor of integer N
function countFac(n)
{
    var m = n;
    var count = 0;
 
    // Iterate for all prime factor
    for(var i = 2; (i * i) <= m; ++i)
    {
       var total = 0;
        
       // If it is a prime factor,
       // count the total number
       // of times it divides n.
       while (n % i == 0)
       {
           n /= i;
           ++total;
       }
       var temp = 0;
        
       // Find the Number of distinct
       // possible positive numbers
       for(var j = 1; (temp + j) <= total; ++j)
       {
          temp += j;
          ++count;
       }
    }
    if (n != 1)
        ++count;
         
    // Return the final count
    return count;
}
 
// Driver code
 
// Given Number N
var N = 24;
 
// Function Call
document.write(countFac(N));
 
// This code is contributed by Khushboogoyal499
    
</script>

3

Complejidad del tiempo: O(sqrt(N))