Suma de los dígitos de un número N escrito en todas las bases de 2 a N/2

Dado un número entero N , la tarea es encontrar la suma de los dígitos del número N escrito en todas las bases de 2 a N/2 .
Ejemplos: 
 

Entrada: N = 6 
Salida: 4 
En base 2, 6 se representa como 110. 
En base 3, 6 se representa como 20. 
Suma = 1 + 1 + 0 + 2 + 0 = 4
Entrada: N = 8 
Salida: 7 
 

Acercarse: 
 

• Para cada base de 2 a (n/2) calcule los dígitos de n en la base particular con lo siguiente: 
  • Calcula el resto de dividir n por base y el resto es una de las cifras de n en esa base.
  • Agregue el dígito a la suma y actualice n como (n = n / base) .
  • Repita los pasos anteriores mientras n > 0
• Imprime la suma calculada en los pasos anteriores.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the sum of the digits of
// n in the given base
int solve(int n, int base)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0) {
 
        // Digit of n in the given base
        int remainder = n % base;
 
        // Add the digit
        sum += remainder;
        n = n / base;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base = 2; base <= n / 2; ++base)
        sum += solve(n, base);
 
    cout << sum;
}
 
// Driver program
int main()
{
    int n = 8;
    SumsOfDigits(n);
    return 0;
}

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
     
 
// Function to calculate the sum of the digits of
// n in the given base
static int solve(int n, int base)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0) {
 
        // Digit of n in the given base
        int remainder = n % base;
 
        // Add the digit
        sum += remainder;
        n = n / base;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
static void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base = 2; base <= n / 2; ++base)
        sum += solve(n, base);
 
    System.out.println(sum);
}
 
// Driver program
 
 
    public static void main (String[] args) {
        int n = 8;
    SumsOfDigits(n);
    }
     
}
// This code is contributed by anuj_67..

# Python 3 implementation of the approach
from math import floor
# Function to calculate the sum of the digits of
# n in the given base
def solve(n, base):
    # Sum of digits
    sum = 0
 
    while (n > 0):
        # Digit of n in the given base
        remainder = n % base
 
        # Add the digit
        sum = sum + remainder
        n = int(n / base)
     
    return sum
 
# Function to calculate the sum of
# digits of n in bases from 2 to n/2
def SumsOfDigits(n):
     
    # to store digit sum in all base
    sum = 0
    N = floor(n/2)
    # function call for multiple bases
    for base in range(2,N+1,1):
        sum = sum + solve(n, base)
 
    print(sum)
 
# Driver program
if __name__ == '__main__':
    n = 8
    SumsOfDigits(n)
     
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to calculate the sum of
// the digits of n in the given base
static int solve(int n, int base1)
{
    // Sum of digits
    int sum = 0;
 
    while (n > 0)
    {
 
        // Digit of n in the given base
        int remainder1 = n % base1;
 
        // Add the digit
        sum += remainder1;
        n = n / base1;
    }
 
    return sum;
}
 
// Function to calculate the sum of
// digits of n in base1s from 2 to n/2
static void SumsOfDigits(int n)
{
    // to store digit sum in all bases
    int sum = 0;
 
    // function call for multiple bases
    for (int base1 = 2;
             base1 <= n / 2; ++base1)
        sum += solve(n, base1);
 
    Console.WriteLine(sum);
}
 
// Driver Code
public static void Main (String []args)
{
    int n = 8;
    SumsOfDigits(n);
}
}
 
// This code is contributed by Arnab Kundu

<?php
// PHP implementation of the approach
 
// Function to calculate the sum of
// the digits of n in the given base
function solve($n, $base)
{
    // Sum of digits
    $sum = 0;
 
    while ($n > 0)
    {
 
        // Digit of n in the given base
        $remainder = $n % $base;
 
        // Add the digit
        $sum += $remainder;
        $n = $n / $base;
    }
 
    return $sum;
}
 
// Function to calculate the sum of
// digits of n in bases from 2 to n/2
function SumsOfDigits($n)
{
    // to store digit sum in all bases
    $sum = 0;
 
    // function call for multiple bases
    for ($base = 2;
         $base <= $n / 2; ++$base)
        $sum += solve($n, $base);
 
    echo $sum;
}
 
// Driver Code
$n = 8;
SumsOfDigits($n);
 
// This code is contributed
// by Akanksha Rai
?>

<script>
// javascript implementation of the approach   
// Function to calculate the sum of the digits of
    // n in the given base
    function solve(n , base)
    {
     
        // Sum of digits
        var sum = 0;
 
        while (n > 0) {
 
            // Digit of n in the given base
            var remainder = n % base;
 
            // Add the digit
            sum += remainder;
            n = parseInt(n / base);
        }
 
        return sum;
    }
 
    // Function to calculate the sum of
    // digits of n in bases from 2 to n/2
    function SumsOfDigits(n)
    {
     
        // to store digit sum in all bases
        var sum = 0;
 
        // function call for multiple bases
        for (base = 2; base <= n / 2; ++base)
            sum += solve(n, base);
 
        document.write(sum);
    }
 
    // Driver program
        var n = 8;
        SumsOfDigits(n);
 
// This code is contributed by gauravrajput1
</script>

7

Complejidad de tiempo: O(n * log n)

Espacio Auxiliar: O(1)