C++ Program for GCD of more than two (or array) numbers

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)

CPP

// 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;
}

C#

using System;

public class Program {
    // Function to return gcd of a and b
    public 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
    public 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;
    }

    // Driven code
    public static void Main()
    {
        int[] arr = { 2, 4, 6, 8, 16 };
        int n = arr.Length;
        Console.WriteLine(findGCD(arr, n));
    }
}
//this code is added by snehalsalokhe

Output:

2

Time Complexity: O(N * log(N)), where N is the largest element of the array
Auxiliary Space: O(1), ignoring the stack space used in recursion

Implementation of the same code with Iterative GCD Function

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)
{
    int result = min(a, b); // Find Minimum of a nd b
    while (result > 0) {
        if (a % result == 0 && b % result == 0) {
            break;
        }
        result--;
    }
    return result; // return gcd of a nd b
}

// 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;
}

Output

2

Time Complexity: O(N * log(N)), where N is the largest element of the array
Auxiliary Space: O(1)

Please refer complete article on GCD of more than two (or array) numbers for more details!