Given an integer N, the task is to find the sum of the digits of the number N written in all the bases from 2 to N / 2.
Examples:
Input: N = 6
Output: 4
In base 2, 6 is represented as 110.
In base 3, 6 is represented as 20.
Sum = 1 + 1 + 0 + 2 + 0 = 4
Input: N = 8
Output: 7
Approach:
- For every base from 2 to (n / 2) calculate the digits of n in the particular base with the following:
- Calculate the remainder on dividing n by base and the remainder is one of the digits of n in that base.
- Add the digit to the sum and update n as (n = n / base).
- Repeat the above steps while n > 0
- Print the sum calculated in the previous steps.
Below is the implementation of the above approach:
C++
// 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 baseint 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/2void 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 programint main(){ int n = 8; SumsOfDigits(n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to calculate the sum of the digits of// n in the given basestatic 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/2static 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.. |
Python3
# Python 3 implementation of the approachfrom math import floor# Function to calculate the sum of the digits of# n in the given basedef 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/2def 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 programif __name__ == '__main__': n = 8 SumsOfDigits(n) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG{ // Function to calculate the sum of // the digits of n in the given basestatic 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 Codepublic static void Main (String []args){ int n = 8; SumsOfDigits(n); } }// This code is contributed by Arnab Kundu |
PHP
<?php// PHP implementation of the approach// Function to calculate the sum of// the digits of n in the given basefunction 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 $sum;}// Function to calculate the sum of// digits of n in bases from 2 to n/2function SumsOfDigits($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?> |
Javascript
<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> |
Output:
7
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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