Suma máxima después de dividir repetidamente N por un divisor

Dado un número entero N . La tarea es encontrar la suma máxima posible de valores intermedios (incluidos N y 1 ) obtenidos después de aplicar la siguiente operación:
 

Divide N entre cualquier divisor (>1) hasta que sea 1. 

Ejemplos: 
 

Input: N = 10
Output: 16
Initially, N=10
1st Division -> N = 10/2 = 5 
2nd Division -> N= 5/5 = 1

Input: N = 8
Output: 15
Initially, N=8
1st Division -> N = 8/2 = 4 
2nd Division -> N= 4/2 = 2
3rd Division -> N= 2/2 = 1

Enfoque: dado que la tarea es maximizar la suma de valores después de cada paso, intente maximizar los valores individuales. Entonces, reduzca el valor de N lo menos posible. Para lograr eso, dividimos N por su divisor más pequeño.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the smallest divisor
int smallestDivisor(int n)
{
    int mx = sqrt(n);
    for (int i = 2; i <= mx; i++)
        if (n % i == 0)
            return i;
    return n;
}
 
// Function to find the maximum sum
int maxSum(int n)
{
    long long res = n;
    while (n > 1) {
        int divi = smallestDivisor(n);
        n /= divi;
        res += n;
    }
    return res;
}
 
// Driver Code
int main()
{
    int n = 34;
    cout << maxSum(n);
 
    return 0;
}

// C implementation of the above approach
#include <stdio.h>
#include<math.h>
 
// Function to find the smallest divisor
int smallestDivisor(int n)
{
    int mx = sqrt(n);
    for (int i = 2; i <= mx; i++)
        if (n % i == 0)
            return i;
    return n;
}
 
// Function to find the maximum sum
int maxSum(int n)
{
    long long res = n;
    while (n > 1) {
        int divi = smallestDivisor(n);
        n /= divi;
        res += n;
    }
    return res;
}
 
// Driver Code
int main()
{
    int n = 34;
    printf("%d",maxSum(n));
 
    return 0;
}
 
// This code is contributed by allwink45.

// Java implementation of the above approach
import java.io.*;
 
class GFG
{
     
// Function to find the smallest divisor
static double smallestDivisor(int n)
{
    double mx = Math.sqrt(n);
    for (int i = 2; i <= mx; i++)
        if (n % i == 0)
            return i;
    return n;
}
 
// Function to find the maximum sum
static double maxSum(int n)
{
    long res = n;
    while (n > 1)
    {
        double divi = smallestDivisor(n);
        n /= divi;
        res += n;
    }
    return res;
}
 
    // Driver Code
    public static void main (String[] args)
    {
        int n = 34;
        System.out.println (maxSum(n));
    }
}
 
// This code is contributed by jit_t.

from math import sqrt
# Python 3 implementation of the above approach
 
# Function to find the smallest divisor
def smallestDivisor(n):
    mx = int(sqrt(n))
    for i in range(2, mx + 1, 1):
        if (n % i == 0):
            return i
    return n
 
# Function to find the maximum sum
def maxSum(n):
    res = n
    while (n > 1):
        divi = smallestDivisor(n)
        n = int(n/divi)
        res += n
     
    return res
 
# Driver Code
if __name__ == '__main__':
    n = 34
    print(maxSum(n))
 
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to find the smallest divisor
    static double smallestDivisor(int n)
    {
        double mx = Math.Sqrt(n);
        for (int i = 2; i <= mx; i++)
            if (n % i == 0)
                return i;
        return n;
    }
     
    // Function to find the maximum sum
    static double maxSum(int n)
    {
        long res = n;
        while (n > 1)
        {
            double divi = smallestDivisor(n);
            n /= (int)divi;
            res += n;
        }
        return res;
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 34;
        Console.WriteLine(maxSum(n));
    }
}
 
// This code is contributed by Ryuga.

<?php
 
// PHP implementation of the above approach
// Function to find the smallest divisor
function smallestDivisor($n)
{
    $mx = sqrt($n);
    for ($i = 2; $i <= $mx; $i++)
        if ($n % $i == 0)
            return $i;
    return $n;
}
 
// Function to find the maximum sum
function maxSum($n)
{
    $res = $n;
    while ($n > 1)
    {
        $divi = smallestDivisor($n);
        $n /= $divi;
        $res += $n;
    }
    return $res;
}
 
    // Driver Code
    $n = 34;
    echo maxSum($n);
 
#This code is contributed by akt_mit.
?>

<script>
// javascript implementation of the above approach
 
    // Function to find the smallest divisor
    function smallestDivisor(n) {
        var mx = Math.sqrt(n);
        for (i = 2; i <= mx; i++)
            if (n % i == 0)
                return i;
        return n;
    }
 
    // Function to find the maximum sum
    function maxSum(n) {
        var res = n;
        while (n > 1) {
            var divi = smallestDivisor(n);
            n /= divi;
            res += n;
        }
        return res;
    }
 
    // Driver Code
        var n = 34;
        document.write(maxSum(n));
 
// This code is contributed by Rajput-Ji
</script>

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Complejidad de tiempo: O(sqrt(n)*log(n)), donde n representa el entero dado.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.