Suma de los números naturales (hasta N) cuyo módulo con K da R

Dados tres números enteros N , K y R . La tarea es calcular la suma de todos los números de 1 a N que produce el resto R al dividir por K.
Ejemplos: 
 

Entrada: N = 20, K = 4, R = 3 
Salida: 55 
3, 7, 11, 15 y 19 son los únicos números que dan 3 como resto de la división con 4. 
3 + 7 + 11 + 15 + 19 = 55
Entrada: N = 15, K = 13, R = 2 
Salida: 17 
 

Acercarse: 
 

• Inicialice sum = 0 y tome el módulo de cada elemento de 1 a N con K .
• Si el resto es igual a R , actualice sum = sum + i donde i es el número actual que dio a R como el resto al dividir por K .
• Imprime el valor de sum al final.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the sum
long 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 code
int main()
{
    int N = 20, K = 4, R = 3;
    cout << count(N, K, R);
 
    return 0;
}

// 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.

# Python 3 implementation of the approach
 
# Function to return the sum
def 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 code
if __name__ == '__main__':
    N = 20
    K = 4
    R = 3
    print(count(N, K, R))
 
# This code is contributed by
# Surendra_Gangwar

// 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 implementation of the approach
 
// Function to return the sum
function 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
?>

<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>

55

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)