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^K 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 += pow (i, k);
}
// Return the sum
return sum;
} // Drivers Code int 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^K 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
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^K from math import pow
# Function to find value of # 1^K + 2^K + 3^K + .. + N^K def 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 Code if __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^K using 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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