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

• Difficulty Level : Expert
• Last Updated : 17 Jan, 2023

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 ``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;``    ``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);``    ``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), ignoring the stack space used in recursion

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

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