C++ Program for GCD of more than two (or array) numbers
Last Updated :
20 Mar, 2023
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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)
// 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++ program to find GCD of two or// more numbersusing namespace std;// Function to return gcd of a and bint gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);}// Function to find gcd of array of// numbersint findGCD(int arr[], int n){ int result = arr[0]; for (int i = 1; i < n; i++) result = gcd(arr[i], result); return result;}// Driven codeint main(){ int arr[] = { 2, 4, 6, 8, 16 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findGCD(arr, n) << endl; return 0;}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++ 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;
}
// C++ program to find GCD of two or// more numbersusing namespace std;// Function to return gcd of a and bint 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// numbersint findGCD(int arr[], int n){ int result = arr[0]; for (int i = 1; i < n; i++) result = gcd(arr[i], result); return result;}// Driven codeint 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!
