N以下3和7的所有倍数的总和

给定数字N ，任务是找到N之下3和7的所有倍数的和。
注意：总和中不得重复数字。

例子：

Input: N = 10 
Output: 25 
3 + 6 + 7 + 9 = 25

Input: N = 24 
Output: 105 
3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 = 105 

方法：

• 我们知道3的倍数形成AP，因为S ₃ = 3 + 6 + 9 + 12 + 15 + 18 + 21 +…
• 并且7的倍数形成AP，因为S ₇ = 7 + 14 + 21 + 28 +…
• 现在，总和= S ₃ + S ₇即3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 + 21 +…
• 从上一步开始，将21重复两次。实际上，所有21的倍数(或3 * 7)都将被重复，因为它们被计数了两次，一次在系列S _3中，一次在系列S _7中。因此，需要从结果中舍弃21的倍数。
• 因此，最终结果将为S ₃ + S ₇ – S ₂₁

The formula for the sum of an AP series is : 
n * ( a + l ) / 2 
Where n is the number of terms, a is the starting term, and l is the last term. 
 

下面是上述方法的实现：

C++

// C++ program to find the sum of all
// multiples of 3 and 7 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 3 and 7 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21);
}
 
// Driver code
int main()
{
    long long n = 24;
 
    cout << sumMultiples(n);
 
    return 0;
}

Java

// Java program to find the sum of all
// multiples of 3 and 7 below N
import java.util.*;
 
class solution
{
 
// 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 3 and 7 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21);
}
 
// Driver code
public static void main(String args[])
{
    long n = 24;
 
    System.out.println(sumMultiples(n));
 
 }
}
 
//This code is contributed by Surendra_Gangwar

Python3

# Python3 program to find the sum of
# all multiples of 3 and 7 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 3 and 7 below N
def sumMultiples(n):
 
    # Since, we need the sum of
    # multiples less than N
    n -= 1;
 
    return int(sumAP(n, 3) +
               sumAP(n, 7) -
               sumAP(n, 21));
 
# Driver code
n = 24;
 
print(sumMultiples(n));
 
# This code is contributed
# by mits

C#

// C# program to find the sum of all
// multiples of 3 and 7 below N
using System;
 
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 3 and 7 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 3) + sumAP(n, 7) -
                         sumAP(n, 21);
}
 
// Driver code
static public void Main(String []args)
{
    long n = 24;
 
    Console.WriteLine(sumMultiples(n));
}
}
 
// This code is contributed
// by Arnab Kundu

Javascript

 
// JavaScript program to find the sum of all
// multiples of 3 and 7 below N
 
// Function to find sum of AP series
function sumAP(n, d)
{
     
    // Number of terms
    n = parseInt(n / d);
 
    return (n) * (1 + n) * (d / 2);
}
 
// Function to find the sum of all
// multiples of 3 and 7 below N
function sumMultiples(n)
{
     
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 3) +
           sumAP(n, 7) -
           sumAP(n, 21);
}
 
// Driver code
let n = 24;
 
document.write(sumMultiples(n));
 
// This code is contributed by mohan1240760
 

输出：

105