Given two positive integers N and K. The task is to evaluate the value of 1K + 2 K + 3K + … + NK.
Examples:
Input: N = 3, K = 4
Output: 98
Explanation:
?(x4) = 14 + 24 + 34, where 1 ? x ? N
?(x4) = 1 + 16 + 81
?(x4) = 98Input: N = 8, K = 4
Output: 8772
Approach:
- The idea is to find the value of xK using pow() function. (where x is from 1 to N).
- The required sum is the summation of all values calculated above.
Below is the implementation of the above approach:
C++
// C++ Program to find the value// 1^K + 2^K + 3^K + .. + N^K#include <bits/stdc++.h>using namespace std;// Function to find value of// 1^K + 2^K + 3^K + .. + N^Kint findSum(int N, int k){ // Initialise sum to 0 int sum = 0; for (int i = 1; i <= N; i++) { // Find the value of // pow(i, 4) and then // add it to the sum sum += pow(i, k); } // Return the sum return sum;}// Drivers Codeint main(){ int N = 8, k = 4; // Function call to // find the sum cout << findSum(N, k) << endl; return 0;} |
Java
// Java Program to find the value// 1^K + 2^K + 3^K + .. + N^Kclass GFG { // Function to find value of // 1^K + 2^K + 3^K + .. + N^K static int findSum(int N, int k) { // Initialise sum to 0 int sum = 0; for (int i = 1; i <= N; i++) { // Find the value of // pow(i, 4) and then // add it to the sum sum += (int)Math.pow(i, k); } // Return the sum return sum; } // Drivers Code public static void main (String[] args) { int N = 8, k = 4; // Function call to // find the sum System.out.println(findSum(N, k)); }}// This code is contributed by AnkitRai01 |
Python 3
# Python 3 Program to find the value# 1^K + 2^K + 3^K + .. + N^Kfrom math import pow# Function to find value of# 1^K + 2^K + 3^K + .. + N^Kdef findSum(N, k): # Initialise sum to 0 sum = 0 for i in range(1, N + 1, 1): # Find the value of # pow(i, 4) and then # add it to the sum sum += pow(i, k) # Return the sum return sum# Drives Codeif __name__ == '__main__': N = 8 k = 4 # Function call to # find the sum print(int(findSum(N, k)))# This code is contributed by Surendra_Gangwar |
C#
// C# Program to find the value// 1^K + 2^K + 3^K + .. + N^Kusing System;public class GFG { // Function to find value of // 1^K + 2^K + 3^K + .. + N^K static int findSum(int N, int k) { // Initialise sum to 0 int sum = 0; for (int i = 1; i <= N; i++) { // Find the value of // pow(i, 4) and then // add it to the sum sum += (int)Math.Pow(i, k); } // Return the sum return sum; } // Drivers Code public static void Main (string[] args) { int N = 8, k = 4; // Function call to // find the sum Console.WriteLine(findSum(N, k)); }}// This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript Program to find the value // 1^K + 2^K + 3^K + .. + N^K // Function to find value of // 1^K + 2^K + 3^K + .. + N^K function findSum(N, k) { // Initialise sum to 0 let sum = 0; for (let i = 1; i <= N; i++) { // Find the value of // pow(i, 4) and then // add it to the sum sum += Math.pow(i, k); } // Return the sum return sum; } let N = 8, k = 4; // Function call to // find the sum document.write(findSum(N, k));// This code is contributed by divyesh072019.</script> |
Output:
8772
Time Complexity: O(N * log(k)) (here we iterate for loop from i = 1 to N and we need to calculate pow (i,k) which required log(k) time so overall time complexity will be O(N * log(k)) )
Auxiliary space: O(1)
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