Common divisors of N numbers
Given an array arr[] of N integers. The task is to find all the common divisors of all N integers.
Examples
Input: arr[] = {6, 90, 12, 18, 30, 18}
Output: 1 2 3 6
Explanation:
GCD of all the numbers is 6.
Now to find all the divisors of 6, we have
6 = 1 * 6
6 = 2 * 3
Hence 1, 2, 3 and 6 the common divisors of {6, 90, 12, 18, 20, 18}.Input: arr[] = {1, 2, 3, 4, 5}
Output: 1
Explanation:
GCD of all the numbers is 1.
Hence there is only one common divisor of all the numbers i.e., 1.
Approach:
- To find the common divisors of all the N integers in the given array arr[] find the greatest common divisors (gcd) of all the integers in arr[].
- Find all the divisors of greatest common divisors (gcd) obtained in the above step using the approach discussed in this article.
Below is the implementation of the above approach:
C++
// C++ program to find all common // divisors of N numbers #include <bits/stdc++.h> using namespace std; // Function to calculate gcd of // two numbers int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to print all the // common divisors void printAllDivisors(int arr[], int N) { // Variable to find the gcd // of N numbers int g = arr[0]; // Set to store all the // common divisors set<int> divisors; // Finding GCD of the given // N numbers for (int i = 1; i < N; i++) { g = gcd(arr[i], g); } // Finding divisors of the // HCF of n numbers for (int i = 1; i * i <= g; i++) { if (g % i == 0) { divisors.insert(i); if (g / i != i) divisors.insert(g / i); } } // Print all the divisors for (auto& it : divisors) cout << it << " "; } // Driver's Code int main() { int arr[] = { 6, 90, 12, 18, 30, 24 }; int n = sizeof(arr) / sizeof(arr[0]); // Function to print all the // common divisors printAllDivisors(arr, n); return 0; } |
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Java
// Java program to find all common // divisors of N numbers import java.util.*; class GFG { // Function to calculate gcd of // two numbers static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to print all the // common divisors static void printAllDivisors(int arr[], int N) { // Variable to find the gcd // of N numbers int g = arr[0]; // Set to store all the // common divisors HashSet<Integer> divisors = new HashSet<Integer>(); // Finding GCD of the given // N numbers for (int i = 1; i < N; i++) { g = gcd(arr[i], g); } // Finding divisors of the // HCF of n numbers for (int i = 1; i * i <= g; i++) { if (g % i == 0) { divisors.add(i); if (g / i != i) divisors.add(g / i); } } // Print all the divisors for (int it : divisors) System.out.print(it+ " "); } // Driver's Code public static void main(String[] args) { int arr[] = { 6, 90, 12, 18, 30, 24 }; int n = arr.length; // Function to print all the // common divisors printAllDivisors(arr, n); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 program to find all common # divisors of N numbers # Function to calculate gcd of # two numbers def gcd(a, b): if (a == 0): return b return gcd(b % a, a) # Function to prall the # common divisors def printAllDivisors(arr, N): # Variable to find the gcd # of N numbers g = arr[0] # Set to store all the # common divisors divisors = dict() # Finding GCD of the given # N numbers for i in range(1, N): g = gcd(arr[i], g) # Finding divisors of the # HCF of n numbers for i in range(1, g + 1): if i*i > g: break if (g % i == 0): divisors[i] = 1 if (g // i != i): divisors[g // i] = 1 # Prall the divisors for it in sorted(divisors): print(it, end=" ") # Driver's Code if __name__ == '__main__': arr= [6, 90, 12, 18, 30, 24] n = len(arr) # Function to prall the # common divisors printAllDivisors(arr, n) # This code is contributed by mohit kumar 29 |
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C#
// C# program to find all common // divisors of N numbers using System; using System.Collections.Generic; class GFG { // Function to calculate gcd of // two numbers static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to print all the // common divisors static void printAllDivisors(int []arr, int N) { // Variable to find the gcd // of N numbers int g = arr[0]; // Set to store all the // common divisors HashSet<int> divisors = new HashSet<int>(); // Finding GCD of the given // N numbers for (int i = 1; i < N; i++) { g = gcd(arr[i], g); } // Finding divisors of the // HCF of n numbers for (int i = 1; i * i <= g; i++) { if (g % i == 0) { divisors.Add(i); if (g / i != i) divisors.Add(g / i); } } // Print all the divisors foreach (int it in divisors) Console.Write(it+ " "); } // Driver's Code public static void Main(String[] args) { int []arr = { 6, 90, 12, 18, 30, 24 }; int n = arr.Length; // Function to print all the // common divisors printAllDivisors(arr, n); } } // This code is contributed by PrinciRaj1992 |
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Output:
1 2 3 6
Time Complexity: O(N*log(M)) where N is the length of the given array and M is the maximum element in the array.
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