Suma de los primeros K números que no son divisibles por N

Dados dos números N y K , la tarea es encontrar la suma de los primeros K números que no son divisibles por N.

Ejemplos: 

Entrada: N = 5, K = 10 
Salida: 63 
Explicación: La suma de { 1, 2, 3, 4, 6, 7, 8, 9, 11, 12 } es 63.

Entrada: N = 3, k = 13 
Salida: 127 
Explicación: La suma de { 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19 } es 127. 
 

Planteamiento: Para resolver este problema, debemos seguir los siguientes pasos: 
 

• Calcular el último múltiplo de N por (K / (N – 1)) * N
• Calcular K%(N – 1) . Si el resto es 0, (K / (N – 1)) * N – 1 da el último valor que no es divisible por N. De lo contrario, debemos sumar el resto a (K / (N – 1)) * N .
• Calcule la suma hasta el último valor que no es divisible por N obtenido en el paso anterior.
• Calcule la suma de múltiplos de N y reste de la suma calculada en el paso anterior para obtener el resultado deseado.

El siguiente código es la implementación del enfoque anterior: 

// C++ Program to calculate
// the sum of first K
// numbers not divisible by N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum
int findSum(int n, int k)
{
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0) {
        val = val - 1;
    }
    else {
        val = val + rem;
    }
 
    // Calculate the  sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples
        = (x
           * (x + 1) * n)
          / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 7, k = 13;
    cout << findSum(n, k)
         << endl;
}

// Java program to calculate
// the sum of first K numbers
// not divisible by N
import java.util.*;
 
class GFG {
 
// Function to find the sum
static int findSum(int n, int k)
{
     
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
 
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 7, k = 13;
 
    System.out.println(findSum(n, k));
}
}
 
// This code is contributed by offbeat

# Python3 Program to calculate
# the sum of first K
# numbers not divisible by N
 
# Function to find the sum
def findSum(n, k):
 
    # Find the last multiple of N
    val = (k // (n - 1)) * n;
 
    rem = k % (n - 1);
 
    # Find the K-th non-multiple of N
    if (rem == 0):
        val = val - 1;
     
    else:
        val = val + rem;
     
    # Calculate the sum of
    # all elements from 1 to val
    sum = (val * (val + 1)) // 2;
 
    # Calculate the sum of
    # all multiples of N
    # between 1 to val
    x = k // (n - 1);
    sum_of_multiples = (x * (x + 1) * n) // 2;
    sum -= sum_of_multiples;
 
    return sum;
 
# Driver code
n = 7; k = 13;
print(findSum(n, k))
 
# This code is contributed by Code_Mech

// C# program to calculate
// the sum of first K numbers
// not divisible by N
using System;
 
class GFG{
 
// Function to find the sum
static int findSum(int n, int k)
{
     
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
 
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 7, k = 13;
 
    Console.WriteLine(findSum(n, k));
}
}
 
// This code is contributed by 29AjayKumar

<script>
 
// Javascript Program to calculate
// the sum of first K
// numbers not divisible by N
 
// Function to find the sum
function findSum(n, k)
{
     
    // Find the last multiple of N
    var val = parseInt(k / (n - 1)) * n;
 
    var rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
 
    // Calculate the  sum of
    // all elements from 1 to val
    var sum = parseInt((val * (val + 1)) / 2);
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    var x = parseInt(k / (n - 1));
    var sum_of_multiples = parseInt(
        (x * (x + 1) * n) / 2);
         
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
var n = 7, k = 13;
 
document.write(findSum(n, k))
 
// This code is contributed by rrrtnx
 
</script>

99

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