Given a number a and limit N. Find the sum of multiple of a upto N.
Examples :
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]
The basic idea is to iterate from i = a to i = n, i++ and check whether i % a == 0 or not.If zero then add i to sum(initially sum = 0).Thus we will get the sum.It will take O(n) time.
We can modify the loop as i = a, i <= n, i = i + a to reduce the number of iterations.But it will also take O(m) time if there is m multiples of a.
To get the result in O(1) time we can use the formula of summation of n natural numbers.For the above example,
a = 4 and N = 23, number of multiples of a, m = N/a(integer division). The multiples are 4, 8, 12, 16, 20.
We can write it as 4 X [1, 2, 3, 4, 5]. So we can get the sum of multiples as:
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++
#include <iostream>
using namespace std;
int calculate_sum(int a, int N)
{
int m = N / a;
int sum = m * (m + 1) / 2;
int ans = a * sum;
return ans;
}
int main()
{
int a = 7, N = 49;
cout << "Sum of multiples of "
<< a << " up to " << N << " = "
<< calculate_sum(a, N) << endl;
return 0;
}
|
Java
class GFG {
static int calculate_sum(int a, int N) {
int m = N / a;
int sum = m * (m + 1) / 2;
int ans = a * sum;
return ans;
}
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));
}
}
|
Python3
def calculate_sum(a, N):
m = N / a
sum = m * (m + 1) / 2
ans = a * sum
print("Sum of multiples of ", a,
" up to ", N, " = ", ans)
calculate_sum(7, 49)
|
C#
using System;
class GFG {
static int calculate_sum(int a, int N)
{
int m = N / a;
int sum = m * (m + 1) / 2;
int ans = a * sum;
return ans;
}
public static void Main()
{
int a = 7, N = 49;
Console.WriteLine("Sum of multiples of " + a +
" up to " + N + " = " + calculate_sum(a, N));
}
}
|
PHP
<?php
function calculate_sum($a, $N)
{
$m = $N / $a;
$sum = $m * ($m + 1) / 2;
$ans = $a * $sum;
return $ans;
}
$a = 7;
$N = 49;
echo "Sum of multiples of ". $a ,
" up to " . $N . " = " .
calculate_sum($a, $N) ;
?>
|
Javascript
<script>
function calculate_sum(a, N)
{
m = N / a;
sum = m * (m + 1) / 2;
ans = a * sum;
return ans;
}
let a = 7;
let N = 49;
document.write("Sum of multiples of "+ a +
" up to " + N + " = " +
calculate_sum(a, N));
</script>
|
Output :
Sum of multiples of 7 upto 49 = 196
This article is contributed by Sukanta Nandi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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