Given two integers N and K, the task is to find the sum of first N natural numbers then update N as the previously calculated sum. Repeat these steps K times and finally print the value of N.
Examples:
Input: N = 2, K = 2
Output: 6
Operation 1: n = sum(n) = sum(2) = 1 + 2 = 3
Operation 2: n = sum(n) = sum(3) = 1 + 2 + 3 = 6
Input: N = 3, K = 2
Output: 21
Approach: Find the sum of first N natural numbers using the formula (N * (N + 1)) / 2 then update N with the calculated sum. Repeat these steps exactly K times and print the final value of N.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the sum of// the first n natural numbersint sum(int n){ int sum = (n * (n + 1)) / 2; return sum;}// Function to return the repeated sumint repeatedSum(int n, int k){ // Perform the operation exactly k times for (int i = 0; i < k; i++) { // Update n with the sum of // first n natural numbers n = sum(n); } return n;}// Driver codeint main(){ int n = 2, k = 2; cout << repeatedSum(n, k); return 0;} |
Java
// Java implementation of the approach class GFG{ // Function to return the sum of // the first n natural numbers static int sum(int n) { int sum = (n * (n + 1)) / 2; return sum; } // Function to return the repeated sum static int repeatedSum(int n, int k) { // Perform the operation exactly k times for (int i = 0; i < k; i++) { // Update n with the sum of // first n natural numbers n = sum(n); } return n; } // Driver code public static void main (String[] args) { int n = 2, k = 2; System.out.println(repeatedSum(n, k)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the sum of# the first n natural numbersdef sum(n): sum = (n * (n + 1)) // 2 return sum# Function to return the repeated sumdef repeatedSum(n, k): # Perform the operation exactly k times for i in range(k): # Update n with the sum of # first n natural numbers n = sum(n) return n# Driver coden = 2k = 2print(repeatedSum(n, k))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG{ // Function to return the sum of // the first n natural numbers static int sum(int n) { int sum = (n * (n + 1)) / 2; return sum; } // Function to return the repeated sum static int repeatedSum(int n, int k) { // Perform the operation exactly k times for (int i = 0; i < k; i++) { // Update n with the sum of // first n natural numbers n = sum(n); } return n; } // Driver code public static void Main (String[] args) { int n = 2, k = 2; Console.WriteLine(repeatedSum(n, k)); } }// This code is contributed by PrinciRaj1992 |
Javascript
<script>// javascript implementation of the approach // Function to return the sum of// the first n natural numbers function sum(n) { var sum = (n * (n + 1)) / 2; return sum; } // Function to return the repeated sum function repeatedSum(n , k) { // Perform the operation exactly k times for (i = 0; i < k; i++) { // Update n with the sum of // first n natural numbers n = sum(n); } return n; } // Driver code var n = 2, k = 2; document.write(repeatedSum(n, k));// This code contributed by Rajput-Ji </script> |
Output:
6
Time Complexity: O(k)
Auxiliary Space: O(1)
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