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++
// C++ program to find sum of multiples of a number // up to N efficiently #include <iostream> 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. |
PHP
<?php // PHP 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 $a = 7; $N = 49; echo "Sum of multiples of ". $a , " up to " . $N . " = " . calculate_sum($a, $N) ; // This code is contributed by Sam007 ?> |
Output :
Sum of multiples of 7 upto 49 = 196
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Improved By : Sam007
