Given a number n, find the sum of digits of n when represented in different bases from 2 to n-1.
Examples:
Input : 5 Output : 2 3 2 Representation of 5 is 101, 12, 11 in bases 2 , 3 , 4 . Input : 7 Output : 3 3 4 3 2
- As the given question wants the sum of digits in different bases, first we have to calculate the given number of different bases and add each digit to the number of different bases.
- So, to calculate each number’s representation we will take the mod of given number by the base in which we want to represent that number.
- Then, we have to add all those mod values as the mod values obtained will represent that number in that base.
- Finally, the sum of those mod values gives the sum of digits of that number.
Below are implementations of this approach
C++
// CPP program to find sum of digits of// n in different bases from 2 to n-1.#include <bits/stdc++.h>using namespace std;// function to calculate sum of // digit for a given baseint solve(int n, int base){ // Sum of digits int result = 0 ; // Calculating the number (n) by // taking mod with the base and adding // remainder to the result and // parallelly reducing the num value . while (n > 0) { int remainder = n % base ; result = result + remainder ; n = n / base; } // returning the result return result ;}void printSumsOfDigits(int n){ // function calling for multiple bases for (int base = 2 ; base < n ; ++base) cout << solve(n, base) <<" "; }// Driver programint main(){ int n = 8; printSumsOfDigits(n); return 0;} |
Java
// Java program to find sum of digits of// n in different bases from 2 to n-1.import java.io.*;public class GFG{// function to calculate sum of // digit for a given basestatic int solve(int n, int base){ // Sum of digits int result = 0 ; // Calculating the number (n) by // taking mod with the base and adding // remainder to the result and // parallelly reducing the num value . while (n > 0) { int remainder = n % base ; result = result + remainder ; n = n / base; } // returning the result return result ;}static void printSumsOfDigits(int n){ // function calling for multiple bases for (int base = 2 ; base < n ; ++base) System.out.print(solve(n, base)+" "); }// Driver Codepublic static void main(String[] args){ int n = 8; printSumsOfDigits(n);}}// This code is contributed by Smitha |
Python3
# Python program to find sum of digits of# n in different bases from 2 to n-1. # def to calculate sum of # digit for a given basedef solve(n, base) : # Sum of digits result = 0 # Calculating the number (n) by # taking mod with the base and adding # remainder to the result and # parallelly reducing the num value . while (n > 0) : remainder = n % base result = result + remainder n = int(n / base) # returning the result return result def printSumsOfDigits(n) : # def calling for # multiple bases for base in range(2, n) : print (solve(n, base), end=" ")# Driver coden = 8printSumsOfDigits(n) # This code is contributed by Manish Shaw# (manishshaw1) |
C#
// C# program to find the sum of digits of// n in different base1s from 2 to n-1.using System;class GFG{// function to calculate sum of // digit for a given base1static int solve(int n, int base1){ // Sum of digits int result = 0 ; // Calculating the number (n) by // taking mod with the base1 and adding // remainder to the result and // parallelly reducing the num value . while (n > 0) { int remainder = n % base1 ; result = result + remainder ; n = n / base1; } // returning the result return result ;}static void printSumsOfDigits(int n){ // function calling for multiple base1s for (int base1 = 2 ; base1 < n ; ++base1) Console.Write(solve(n, base1)+" "); }// Driver Codepublic static void Main(){ int n = 8; printSumsOfDigits(n);}}// This code is contributed by Smitha |
PHP
<?php// PHP program to find sum of digits of// n in different bases from 2 to n-1.// function to calculate sum of // digit for a given basefunction solve($n, $base){ // Sum of digits $result = 0 ; // Calculating the number (n) by // taking mod with the base and adding // remainder to the result and // parallelly reducing the num value . while ($n > 0) { $remainder = $n % $base ; $result = $result + $remainder ; $n = $n / $base; } // returning the result return $result ;}function printSumsOfDigits($n){ // function calling for // multiple bases for ($base = 2 ; $base < $n ; ++$base) { echo(solve($n, $base)); echo(" "); }}// Driver code$n = 8;printSumsOfDigits($n);// This code is contributed by Ajit.?> |
Javascript
<script>// JavaScript program to find sum of digits of// n in different bases from 2 to n-1. // function to calculate sum of // digit for a given base function solve(n , base) { // Sum of digits var result = 0; // Calculating the number (n) by // taking mod with the base and adding // remainder to the result and // parallelly reducing the num value . while (n > 0) { var remainder = n % base; result = result + remainder; n = parseInt(n / base); } // returning the result return result; } function printSumsOfDigits(n) { // function calling for multiple bases for (base = 2; base < n; ++base) document.write(solve(n, base) + " "); } // Driver Code var n = 8; printSumsOfDigits(n);// This code contributed by Rajput-Ji</script> |
Output
1 4 2 4 3 2
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
As constant extra space is used.
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