Given two integer N and K, the task is to find the sum of all the numbers from the range [1, N] excluding those which are powers of K.
Examples:
Input: N = 10, K = 3
Output: 42
2 + 4 + 5 + 6 + 7 + 8 + 10 = 42
1, 3 and 9 are excluded as they are powers of 3.
Input: N = 200, K = 30
Output: 20069
Approach:
Approach to solve this problem could be iterating through the range [1, N] and checking if each number is a power of K or not. If it is not, then add it to the sum. This can be achieved using a loop and checking the value of each number raised to the power of K using the log function.
- Initialize a variable sum to store the sum of all the elements from the range [1, n] excluding those which are powers of k.
- Iterate through the range [1, n] using a for loop.
- Check if the current element i is a power of k or not. This can be done by calculating log(i) / log(k) and checking if the result is not an integer, i.e., if log(i) / log(k) != floor(log(i) / log(k)).
- If the current element i is not a power of k, add it to the sum.
- After the loop, return the required sum.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;#define ll long long int// Function to return the sum of all the elements// from the range [1, n] excluding those which are// powers of kll getSum(ll n, ll k){ // Variable to store the sum ll sum = 0; // Iterate through the range [1, n] for (ll i = 1; i <= n; i++) { // Check if i is a power of k or not if (log(i) / log(k) != floor(log(i) / log(k))) { // If i is not a power of k, add it to the sum sum += i; } } // Return the required sum return sum;}// Driver codeint main(){ ll n = 10, k = 3; cout << getSum(n, k); return 0;} |
Java
import java.util.*;public class Main { public static void main(String[] args) { long n = 10, k = 3; System.out.println(getSum(n, k)); } // Function to calculate the sum of numbers that do not have an exact base-k logarithm public static long getSum(long n, long k) { long sum = 0; for (long i = 1; i <= n; i++) { // Calculate the base-k logarithm of 'i' and check if it is not an integer value if (Math.log(i) / Math.log(k) != Math.floor(Math.log(i) / Math.log(k))) { // If the logarithm is not an integer, add 'i' to the sum sum += i; } } return sum; }} |
Python3
import math# Function to return the sum of all the elements# from the range [1, n] excluding those which are# powers of kdef get_sum(n, k): # Variable to store the sum sum_val = 0 # Iterate through the range [1, n] for i in range(1, n + 1): # Check if i is a power of k or not if math.log(i, k) != int(math.log(i, k)): # If i is not a power of k, add it to the sum sum_val += i # Return the required sum return sum_val# Driver codeif __name__ == "__main__": n = 10 k = 3 print(get_sum(n, k)) |
C#
using System;namespace SumExcludingPowers{ class GFG { // Function to return the sum of all the elements // from the range [1, n] excluding those which are // powers of k static long GetSum(long n, long k) { // Variable to store the sum long sum = 0; // Iterate through the range [1, n] for (long i = 1; i <= n; i++) { // Check if i is a power of k or not if (Math.Log(i, k) != Math.Floor(Math.Log(i, k))) { // If i is not a power of k, add it to the sum sum += i; } } // Return the required sum return sum; } // Main function static void Main(string[] args) { long n = 10, k = 3; Console.WriteLine(GetSum(n, k)); } }} |
Javascript
// Function to return the sum of all the elements// from the range [1, n] excluding those which are// powers of kfunction getSum(n, k) { // Variable to store the sum let sumVal = 0; // Iterate through the range [1, n] for (let i = 1; i <= n; i++) { // Check if i is a power of k or not if (Math.log(i) / Math.log(k) !== parseInt(Math.log(i) / Math.log(k))) { // If i is not a power of k, add it to the sum sumVal += i; } } // Return the required sum return sumVal;}// Driver codeconst n = 10;const k = 3;console.log(getSum(n, k)); |
Output
42
Time Complexity: O(n log(n)) because for each element in the range [1,n], we are computing log(i) which takes O(log(i)) time.
Space Complexity: O(1) because we are only using a constant amount of extra space to store the sum variable.
Approach: Find the sum of the following series:
- pwrK: The sum of all the powers of K from [1, N] i.e. K0 + K1 + K2 + … + Kr such that Kr ≤ N
- sumAll: The sum of all the integers from the range [1, N] i.e. (N * (N + 1)) / 2.
The result will be sumAll – pwrK
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define ll long long int// Function to return the sum of all the// powers of k from the range [1, n]ll sumPowersK(ll n, ll k){ // To store the sum of the series ll sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum;}// Find to return the sum of the// elements from the range [1, n]// excluding those which are powers of kll getSum(ll n, ll k){ // Sum of all the powers of k from [1, n] ll pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] ll sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK);}// Driver codeint main(){ ll n = 10, k = 3; cout << getSum(n, k); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Function to return the sum of all the// powers of k from the range [1, n]static long sumPowersK(long n, long k){ // To store the sum of the series long sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum;}// Find to return the sum of the// elements from the range [1, n]// excluding those which are powers of kstatic long getSum(long n, long k){ // Sum of all the powers of k from [1, n] long pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] long sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK);} // Driver code public static void main (String[] args) { long n = 10, k = 3; System.out.println( getSum(n, k)); }}// This code is contributed by anuj_67.. |
Python3
# Python3 implementation of the approach # Function to return the sum of all the # powers of k from the range [1, n] def sumPowersK(n, k) : # To store the sum of the series sum = 0; num = 1; # While current power of k <= n while (num <= n) : # Add current power to the sum sum += num; # Next power of k num *= k; # Return the sum of the series return sum; # Find to return the sum of the # elements from the range [1, n] # excluding those which are powers of k def getSum(n, k) : # Sum of all the powers of k from [1, n] pwrK = sumPowersK(n, k); # Sum of all the elements from [1, n] sumAll = (n * (n + 1)) / 2; # Return the required sum return (sumAll - pwrK); # Driver code if __name__ == "__main__" : n = 10; k = 3; print(getSum(n, k)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG {// Function to return the sum of all the// powers of k from the range [1, n]static long sumPowersK(long n, long k){ // To store the sum of the series long sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum;}// Find to return the sum of the// elements from the range [1, n]// excluding those which are powers of kstatic long getSum(long n, long k){ // Sum of all the powers of k from [1, n] long pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] long sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK);}// Driver codepublic static void Main () { long n = 10, k = 3; Console.WriteLine( getSum(n, k));}}// This code is contributed by anuj_67.. |
Javascript
<script>// javascript implementation of the approach // Function to return the sum of all the // powers of k from the range [1, n] function sumPowersK(n , k) { // To store the sum of the series var sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum; } // Find to return the sum of the // elements from the range [1, n] // excluding those which are powers of k function getSum(n , k) { // Sum of all the powers of k from [1, n] var pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] var sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK); } // Driver code var n = 10, k = 3; document.write(getSum(n, k));// This code contributed by Rajput-Ji</script> |
Output
42
Time Complexity: O(logkN)
Auxiliary Space: O(1), since no extra space has been taken.
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