Given three integers N, K, and R. The task is to calculate the sum of all those numbers from 1 to N which yields remainder R upon division by K.
Examples:
Input: N = 20, K = 4, R = 3
Output: 55
3, 7, 11, 15 and 19 are the only numbers that give 3 as the remainder on division with 4.
3 + 7 + 11 + 15 + 19 = 55
Input: N = 15, K = 13, R = 2
Output: 17
Approach:
- Initialize sum = 0 and take the modulo of each element from 1 to N with K.
- If the remainder is equal to R, then update sum = sum + i where i is the current number that gave R as the remainder on dividing by K.
- Print the value of sum in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the sumlong long int count(int N, int K, int R){ long long int sum = 0; for (int i = 1; i <= N; i++) { // If current number gives R as the // remainder on dividing by K if (i % K == R) // Update the sum sum += i; } // Return the sum return sum;}// Driver codeint main(){ int N = 20, K = 4, R = 3; cout << count(N, K, R); return 0;} |
Java
// Java implementation of the approach class GfG { // Function to return the sum static long count(int N, int K, int R) { long sum = 0; for (int i = 1; i <= N; i++) { // If current number gives R as the // remainder on dividing by K if (i % K == R) // Update the sum sum += i; } // Return the sum return sum; } // Driver code public static void main(String[] args) { int N = 20, K = 4, R = 3; System.out.println(count(N, K, R)); }} // This code is contributed by// prerna saini. |
Python3
# Python 3 implementation of the approach# Function to return the sumdef count(N, K, R): sum = 0 for i in range(1, N + 1): # If current number gives R as the # remainder on dividing by K if (i % K == R): # Update the sum sum += i # Return the sum return sum# Driver codeif __name__ == '__main__': N = 20 K = 4 R = 3 print(count(N, K, R))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG { // Function to return the sum static long count(int N, int K, int R) { long sum = 0; for (int i = 1; i <= N; i++) { // If current number gives R as the // remainder on dividing by K if (i % K == R) // Update the sum sum += i; } // Return the sum return sum; } // Driver code public static void Main() { int N = 20, K = 4, R = 3; Console.Write(count(N, K, R)); }} // This code is contributed by// Akanksha Rai |
PHP
<?php// PHP implementation of the approach// Function to return the sumfunction count1($N, $K, $R){ $sum = 0; for ($i = 1; $i <= $N; $i++) { // If current number gives R as the // remainder on dividing by K if ($i % $K == $R) // Update the sum $sum += $i; } // Return the sum return $sum;}// Driver code$N = 20; $K = 4; $R = 3;echo count1($N, $K, $R);// This code is contributed// by Akanksha Rai?> |
Javascript
<script>// Javascript implementation of the approach // Function to return the sum function count(N , K , R) { var sum = 0; for (i = 1; i <= N; i++) { // If current number gives R as the // remainder on dividing by K if (i % K == R) // Update the sum sum += i; } // Return the sum return sum; } // Driver code var N = 20, K = 4, R = 3; document.write(count(N, K, R));// This code contributed by aashish1995</script> |
Output:
55
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.
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