一个数字的倍数之和，最多N

给定数字a并限制N。找出最多N的倍数之和。

例子 ：

Input : a = 4, N = 23
Output : sum = 60
[Multiples : 4, 8, 12, 16, 20]

Input :a = 7, N = 49
Output :sum = 196
[Multiples: 7, 14, 21, 28, 35, 42, 49]

基本思想是从i = a迭代到i = n，i ++并检查i％a == 0是否为零，如果为零，则将i加到sum(最初为sum = 0)，这样我们就得到了总和。将花费O(n)时间。
我们可以将循环修改为i = a，i <= n，i = i + a以减少迭代次数，但是如果a的m倍，也将花费O(m)时间。
为了得到O(1)时间的结果，我们可以使用n个自然数之和的公式。
a = 4且N = 23，即a的倍数，m = N / a(整数除法)。倍数是4、8、12、16、20。
我们可以将其写为4 X [1、2、3、4、5]。这样我们就可以得到倍数之和：

sum = a * (Summation of 1 to m [natural numbers from 1 to m]) 
 sum = 4 * (m*(m+1) / 2)
 sum = 4 * (5*6 / 2) = 4 * 15 = 60 
 

C++

// C++ program to find sum of multiples of a number
// up to N efficiently
#include 
using namespace std;
 
// Function for calculating sum of multiples of
// a upto N
int calculate_sum(int a, int N)
{
    // Number of multiples
    int m = N / a;
 
    // sum of first m natural numbers
    int sum = m * (m + 1) / 2;
 
    // sum of multiples
    int ans = a * sum;
 
    return ans;
}
 
// Driver code
int main()
{
    int a = 7, N = 49;
    cout << "Sum of multiples of "
         << a << " up to " << N << " = "
         << calculate_sum(a, N) << endl;
    return 0;
}

Java

// Java program to find sum of multiples
// of a number up to N efficiently
 
class GFG {
     
// Function for calculating sum
// of multiples of a upto N
static int calculate_sum(int a, int N) {
     
    // Number of multiples
    int m = N / a;
 
    // sum of first m natural numbers
    int sum = m * (m + 1) / 2;
 
    // sum of multiples
    int ans = a * sum;
 
    return ans;
}
 
// Driver code
public static void main(String[] args) {
     
    int a = 7, N = 49;
    System.out.println("Sum of multiples of " + a +
                       " up to " + N + " = " +
                               calculate_sum(a, N));
}
}
 
// This code is contributed by Anant Agarwal.

Python3

"""Python program to find sum of
multiples of a number up to N"""
 
# Calculates sum of multiples of
# a number upto N
def calculate_sum(a, N):
 
    # Number of multiples
    m = N / a
 
    # sum of first m natural numbers
    sum = m * (m + 1) / 2
 
    # sum of multiples
    ans = a * sum
 
    print("Sum of multiples of ", a,
          " up to ", N, " = ", ans)
 
# Driver Code
calculate_sum(7, 49)
     
# This code is contributed by Abhishek Agrawal.

C#

// C# program to find sum of multiples
// of a number up to N efficiently
using System;
 
class GFG {
 
    // Function for calculating sum
    // of multiples of a upto N
    static int calculate_sum(int a, int N)
    {
 
        // Number of multiples
        int m = N / a;
 
        // sum of first m natural numbers
        int sum = m * (m + 1) / 2;
 
        // sum of multiples
        int ans = a * sum;
 
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
 
        int a = 7, N = 49;
        Console.WriteLine("Sum of multiples of " + a +
         " up to " + N + " = " + calculate_sum(a, N));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript program to find sum
// of multiples of a number
// up to N efficiently
 
// Function for calculating sum
// of multiples of a upto N
function calculate_sum(a, N)
{
     
    // Number of multiples
    m = N / a;
 
    // Sum of first m
    // natural numbers
    sum = m * (m + 1) / 2;
 
    // Sum of multiples
    ans = a * sum;
 
    return ans;
}
 
// Driver code
let a = 7;
let N = 49;
 
document.write("Sum of multiples of "+ a +
               " up to " + N + " = " +
               calculate_sum(a, N));
 
// This code is contributed by mohan1240760
 

输出 ：

Sum of multiples of 7 upto 49 = 196