在N以下找到2和5的所有倍数的总和

给定数字N。任务是找到N以下2和5的所有倍数的总和(N可能高达10 ^ 10)。

例子：

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

Input : N = 20
Output : 110

方法 ：
我们知道2的倍数形成AP的原因是：

2, 4, 6, 8, 10, 12, 14….(1) 
 

类似地，5的倍数形成一个AP，如下所示：

5, 10, 15……(2) 
 

现在， Sum(1)+ Sum(2) = 2、4、5、6、8、10、10、12、14、15 …。
在此，重复10。实际上，所有10或2 * 5的倍数都被重复，因为它被计数了两次，一次在2系列中，一次在5系列中。因此，我们将从Sum()中减去10系列的总和。 1)+和(2)。

AP的总和的公式为：

n * ( a + l ) / 2
Where $n$is the number of terms, $a$is the starting term, and $l$is the last term. 
 

因此，最终答案是：

S₂ + S₅ – S₁₀ 
 

下面是上述方法的实现：

C++

// CPP program to find the sum of all
// multiples of 2 and 5 below N
 
#include 
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;
}

Java

// 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

# 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#

// 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));
    }
}

Javascript

 
// 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
 

输出：

110