Encuentra la suma de todos los múltiplos de 2 y 5 debajo de N

Dado un número N. y la tarea es encontrar la suma de todos los múltiplos de 2 y 5 por debajo de N (N puede ser hasta 10^10). 

Ejemplos :  

Input : N = 10
Output : 25
Explanation : 2 + 4 + 6 + 8 + 5

Input : N = 20
Output : 110

Acercarse : 

Sabemos que múltiplos de 2 forman un AP como: 

2, 4, 6, 8, 10, 12, 14 ….(1) 
 

De manera similar, los múltiplos de 5 forman un AP como: 

5, 10, 15 ……(2) 
 

Ahora, Suma(1) + Suma(2) = 2, 4, 5, 6, 8, 10, 10, 12, 14, 15….
Aquí, 10 se repite. De hecho, todos los múltiplos de 10 o 2*5 se repiten porque se cuenta dos veces, una vez en la serie de 2 y otra vez en la serie de 5. Por lo tanto, restaremos la suma de la serie de 10 de Sum( 1) + Suma(2).

La fórmula para la suma de un AP es:  

n * ( a + l ) / 2
Donde  es el número de términos,  es el término inicial y  es el último término. 
 

Entonces, la respuesta final es:  

S ₂ + S ₅ – S ₁₀ 
 

A continuación se muestra la implementación del enfoque anterior:  

// CPP program to find the sum of all
// multiples of 2 and 5 below N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of AP series
long long sumAP(long long n, long long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
int main()
{
    long long n = 20;
 
    cout << sumMultiples(n);
 
    return 0;
}

// C program to find the sum of all
// multiples of 2 and 5 below N
#include <stdio.h>
 
// Function to find sum of AP series
long long sumAP(long long n, long long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
int main()
{
    long long n = 20;
 
    printf("%lld",sumMultiples(n));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

// Java program to find the sum of all
// multiples of 2 and 5 below N
 
class GFG{
// Function to find sum of AP series
static long sumAP(long n, long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
public static void main(String[] args)
{
    long n = 20;
 
    System.out.println(sumMultiples(n));
}
}
// This code is contributed by mits

# Python3 program to find the sum of
# all multiples of 2 and 5 below N
 
# Function to find sum of AP series
def sumAP(n, d):
 
    # Number of terms
    n = int(n / d);
 
    return (n) * (1 + n) * (d / 2);
 
# Function to find the sum of all
# multiples of 2 and 5 below N
def sumMultiples(n):
 
    # Since, we need the sum of
    # multiples less than N
    n -= 1;
 
    return (int(sumAP(n, 2) + sumAP(n, 5) -
                              sumAP(n, 10)));
 
# Driver code
n = 20;
 
print(sumMultiples(n));
     
# This code is contributed by mits

// C# program to find the sum of all
// multiples of 2 and 5 below N
 
using System;
 
public class GFG{
     
    // Function to find sum of AP series
static long sumAP(long n, long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
     
    static public void Main (){
            long n = 20;
 
        Console.WriteLine(sumMultiples(n));
    }
}

<?php
// PHP program to find the sum of all
// multiples of 2 and 5 below N
// Function to find sum of AP series
function sumAP($n, $d)
{
    // Number of terms
    $n = (int)($n /$d);
 
    return ($n) * ((1 + $n) *
                   (int)$d / 2);
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
function sumMultiples($n)
{
    // Since, we need the sum of
    // multiples less than N
    $n--;
 
    return sumAP($n, 2) + sumAP($n, 5) -
                          sumAP($n, 10);
}
 
// Driver code
$n = 20;
 
echo sumMultiples($n);
 
// This code is contributed
// by Sach_Code
?>

<script>
 
// Java script program to find the sum of all
// multiples of 2 and 5 below N
 
// Function to find sum of AP series
function sumAP(n, d)
{
     
    // Number of terms
    n = parseInt(n / d);
 
    return (n) * ((1 + n) *
         parseInt(d) / 2);
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
function sumMultiples(n)
{
     
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) -
           sumAP(n, 10);
}
 
// Driver code
n = 20;
 
document.write( sumMultiples(n));
 
// This code is contributed by bobby
 
</script>

110

Complejidad temporal: O(1), ya que no hay bucle ni recursividad.

Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.