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 the following. We will also check for the result if the result at any step becomes 1 we will just return the 1 as gcd(1,x)=1.
result = arr[0] For i = 1 to n-1 result = GCD(result, arr[i])
Below is the implementation of the 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); if(result == 1) { return 1; } } return result; } // Driver code int main() { int arr[] = { 2, 4, 6, 8, 16 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findGCD(arr, n) << endl; return 0; } |
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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 = 0; for (int element: arr){ result = gcd(result, element); if(result == 1) { return 1; } } 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 |
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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 |
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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); if(result == 1) { return 1; } } 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 |
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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], $result); if($result == 1) { return 1; } } 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. ?> |
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Output:
2
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