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GCD of more than two (or array) numbers

Given an array of numbers, find GCD of the array elements. In a previous post we find GCD of two number.

Examples:

Input  : arr[] = {1, 2, 3}
Output : 1

Input  : arr[] = {2, 4, 6, 8}
Output : 2


The GCD of three or more numbers equals the product of the prime factors common to all the numbers, but it can also be calculated by repeatedly taking the GCDs of pairs of numbers.

gcd(a, b, c) = gcd(a, gcd(b, c)) 
             = gcd(gcd(a, b), c) 
             = gcd(gcd(a, c), b)

For an array of elements, we do following.



result = arr[0]
For i = 1 to n-1
   result = GCD(result, arr[i])

Below is the implementation of above idea.

C++

// C++ program to find GCD of two or
// more numbers
#include <bits/stdc++.h>
using namespace std;
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to find gcd of array of
// numbers
int findGCD(int arr[], int n)
{
    int result = arr[0];
    for (int i = 1; i < n; i++)
        result = gcd(arr[i], result);
  
    return result;
}
  
// Driven code
int main()
{
    int arr[] = { 2, 4, 6, 8, 16 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findGCD(arr, n) << endl;
    return 0;
}

Java

// Java program to find GCD of two or
// more numbers
  
public class GCD {
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // Function to find gcd of array of
    // numbers
    static int findGCD(int arr[], int n)
    {
        int result = arr[0];
        for (int i = 1; i < n; i++)
            result = gcd(arr[i], result);
  
        return result;
    }
  
    public static void main(String[] args)
    {
        int arr[] = { 2, 4, 6, 8, 16 };
        int n = arr.length;
        System.out.println(findGCD(arr, n));
    }
}
  
// This code is contributed by Saket Kumar

Python

# GCD of more than two (or array) numbers
  
# Function implements the Euclidian 
# algorithm to find H.C.F. of two number
def find_gcd(x, y):
      
    while(y):
        x, y = y, x % y
      
    return x
          
# Driver Code        
l = [2, 4, 6, 8, 16]
  
num1 = l[0]
num2 = l[1]
gcd = find_gcd(num1, num2)
  
for i in range(2, len(l)):
    gcd = find_gcd(gcd, l[i])
      
print(gcd)
  
# Code contributed by Mohit Gupta_OMG

C#

// C# program to find GCD of 
// two or more numbers
using System;
  
public class GCD {
      
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // Function to find gcd of 
    // array of numbers
    static int findGCD(int[] arr, int n)
    {
        int result = arr[0];
        for (int i = 1; i < n; i++)
            result = gcd(arr[i], result);
  
        return result;
    }
      
    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, 4, 6, 8, 16 };
        int n = arr.Length;
        Console.Write(findGCD(arr, n));
    }
}
  
// This code is contributed by nitin mittal

PHP

<?php
// PHP program to find GCD of two or
// more numbers
  
// Function to return gcd of a and b
function gcd( $a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// Function to find gcd of array of
// numbers
function findGCD($arr, $n)
{
    $result = $arr[0];
    for ($i = 1; $i < $n; $i++)
        $result = gcd($arr[$i], $result);
   
    return $result;
}
  
// Driver code
$arr = array( 2, 4, 6, 8, 16 );
$n = sizeof($arr);
echo (findGCD($arr, $n));
  
// This code is contributed by
// Prasad Kshirsagar.
?>


Output:

2

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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