Encontrar la suma de un número y su máximo factor primo

Dado un número entero N , la tarea es encontrar la suma de N y su factor primo máximo.
Ejemplos: 
 

Entrada: 19 
Salida: 38 
El factor primo máximo de 19 es 19. 
Por lo tanto, 19 + 19 = 38
Entrada: 8 
Salida: 10 
8 + 2 = 10 
 

Enfoque: encuentre el factor primo más grande del número y guárdelo en maxPrimeFact , luego imprima el valor de N + maxPrimeFact .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to find sum of n and
// it's largest prime factor
#include <cmath>
#include <iostream>
using namespace std;
 
// Function to return the sum of n and
// it's largest prime factor
int maxPrimeFactors(int n)
{
    int num = n;
 
    // Initialise maxPrime to -1.
    int maxPrime = -1;
 
    while (n % 2 == 0) {
        maxPrime = 2;
        n /= 2;
    }
 
    // n must be odd at this point, thus skip
    // the even numbers and iterate only odd numbers
    for (int i = 3; i <= sqrt(n); i += 2) {
        while (n % i == 0) {
            maxPrime = i;
            n = n / i;
        }
    }
 
    // This condition is to handle the case
    // when n is a prime number greater  than 2
    if (n > 2)
        maxPrime = n;
 
    // finally return the sum.
    int sum = maxPrime + num;
    return sum;
}
 
// Driver Program to check the above function.
int main()
{
    int n = 19;
 
    cout << maxPrimeFactors(n);
    return 0;
}

// Java program to find sum of n and
// it's largest prime factor
import java.io.*;
 
class GFG
{
 
// Function to return the sum of n
// and it's largest prime factor
static int maxPrimeFactors(int n)
{
int num = n;
 
// Initialise maxPrime to -1.
int maxPrime = -1;
 
while (n % 2 == 0)
{
maxPrime = 2;
n /= 2;
}
 
// n must be odd at this point,
// thus skip the even numbers and
// iterate only odd numbers
for (int i = 3; i <= Math.sqrt(n); i += 2) {
     
    while (n % i == 0) {
        maxPrime = i; n = n / i;
        }
     
}
        // This condition is to handle the case
        // when n is a prime number greater than 2
        if (n > 2) {
            maxPrime = n;
        }
// finally return the sum.
int sum = maxPrime + num;
return sum;
}
 
// Driver Code
public static void main (String[] args)
{
int n = 19;
 
System.out.println(maxPrimeFactors(n));
}
}
// This code is contributed by anuj_67

# Python 3 program to find sum of n and
# it's largest prime factor
from math import sqrt
 
# Function to return the sum of n and
# it's largest prime factor
def maxPrimeFactors(n):
    num = n
 
    # Initialise maxPrime to -1.
    maxPrime = -1;
 
    while (n % 2 == 0):
        maxPrime = 2
        n = n / 2
     
    # n must be odd at this point, thus skip
    # the even numbers and iterate only odd numbers
    p = int(sqrt(n) + 1)
    for i in range(3, p, 2):
        while (n % i == 0):
            maxPrime = i
            n = n / i
         
    # This condition is to handle the case
    # when n is a prime number greater than 2
    if (n > 2):
        maxPrime = n
 
    # finally return the sum.
    sum = maxPrime + num
    return sum
 
# Driver Code
if __name__ == '__main__':
    n = 19
 
    print(maxPrimeFactors(n))
 
# This code is contributed by
# Surendra_Gangwar

// C# program to find sum of n and
// it's largest prime factor
using System;
 
class GFG
{
// Function to return the sum of n
// and it's largest prime factor
static int maxPrimeFactors(int n)
{
int num = n;
 
// Initialise maxPrime to -1.
int maxPrime = -1;
 
while (n % 2 == 0)
{
    maxPrime = 2;
    n /= 2;
}
 
// n must be odd at this point,
// thus skip the even numbers and
// iterate only odd numbers
for (int i = 3;
         i <= Math.Sqrt(n); i += 2)
{
     
    while (n % i == 0)
    {
        maxPrime = i; n = n / i;
    }
     
}
 
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
{
    maxPrime = n;
}
 
// finally return the sum.
int sum = maxPrime + num;
return sum;
}
 
// Driver Code
static void Main ()
{
    int n = 19;
     
    Console.WriteLine(maxPrimeFactors(n));
}
}
 
// This code is contributed by Ryuga

<?php
// PHP program to find sum of n and
// it's largest prime factor
 
// Function to return the sum of n
// and it's largest prime factor
function maxPrimeFactors($n)
{
    $num = $n;
 
    // Initialise maxPrime to -1.
    $maxPrime = -1;
 
    while ($n % 2 == 0)
    {
        $maxPrime = 2;
        $n /= 2;
    }
 
    // n must be odd at this point,
    // thus skip the even numbers
    // and iterate only odd numbers
    for ($i = 3; $i <= sqrt($n); $i += 2)
    {
        while ($n % $i == 0)
        {
            $maxPrime = $i;
            $n = $n / $i;
        }
    }
 
    // This condition is to handle the case
    // when n is a prime number greater than 2
    if ($n > 2)
        $maxPrime = $n;
 
    // finally return the sum.
    $sum = $maxPrime + $num;
    return $sum;
}
 
// Driver Code
$n = 19;
 
echo maxPrimeFactors($n);
 
// This code is contributed
// by inder_verma
?>

<script>
 
// Javascript program to find sum of n and
// it's largest prime factor
 
// Function to return the sum of n
// and it's largest prime factor
function maxPrimeFactors(n)
{
    var num = n;
     
    // Initialise maxPrime to -1.
    var maxPrime = -1;
     
    while (n % 2 == 0)
    {
        maxPrime = 2;
        n = parseInt(n/2);
    }
     
    // n must be odd at this point,
    // thus skip the even numbers and
    // iterate only odd numbers
    for (var i = 3; i <= parseInt(Math.sqrt(n)); i += 2)
    {
         
        while (n % i == 0) {
            maxPrime = i;
            n = parseInt(n / i);
        }
         
    }
            // This condition is to handle the case
            // when n is a prime number greater than 2
            if (n > 2) {
                maxPrime = n;
            }
    // finally return the sum.
    var sum = maxPrime + num;
    return sum;
}
 
// Driver Code
 
var n = 19;
 
document.write(maxPrimeFactors(n));
 
 
// This code contributed by shikhasingrajput
 
</script>

38

Complejidad de tiempo: O(sqrtn*logn)

Espacio Auxiliar: O(1)