# Count of integers that divide all the elements of the given array

Given an array arr[] of N elements. The task is to find the count of positive integers that divide all the array elements.

Examples:

Input: arr[] = {2, 8, 10, 6}
Output: 2
1 and 2 are the only integers that divide
all the elements of the given array.

Input:arr[] = {6, 12, 18, 12, 6}
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We know that the maximum integer that will divide all the array elements will be the gcd of the array and all the other integers that will divide all the elements of the array will have to be the factors of this gcd. Hence, the count of valid integers will be equal to the count of factors of the gcd of all the array elements.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of  ` `// the required integers ` `int` `getCount(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// To store the gcd of the array elements ` `    ``int` `gcd = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``gcd = __gcd(gcd, a[i]); ` ` `  `    ``// To store the count of factors ` `    ``// of the found gcd ` `    ``int` `cnt = 0; ` ` `  `    ``for` `(``int` `i = 1; i * i <= gcd; i++) { ` `        ``if` `(gcd % i == 0) { ` ` `  `            ``// If g is a perfect square ` `            ``if` `(i * i == gcd) ` `                ``cnt++; ` ` `  `            ``// Factors appear in pairs ` `            ``else` `                ``cnt += 2; ` `        ``} ` `    ``} ` ` `  `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 4, 16, 1024, 48 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``cout << getCount(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `    ``// Recursive function to return gcd  ` `    ``static` `int` `calgcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``if` `(b == ``0``)  ` `            ``return` `a;  ` `        ``return` `calgcd(b, a % b);  ` `    ``} ` `     `  `    ``// Function to return the count of  ` `    ``// the required integers ` `    ``static` `int` `getCount(``int` `[] a, ``int` `n) ` `    ``{ ` `     `  `        ``// To store the gcd of the array elements ` `        ``int` `gcd = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``gcd = calgcd(gcd, a[i]); ` `     `  `        ``// To store the count of factors ` `        ``// of the found gcd ` `        ``int` `cnt = ``0``; ` `     `  `        ``for` `(``int` `i = ``1``; i * i <= gcd; i++)  ` `        ``{ ` `            ``if` `(gcd % i == ``0``)  ` `            ``{ ` `     `  `                ``// If g is a perfect square ` `                ``if` `(i * i == gcd) ` `                    ``cnt++; ` `     `  `                ``// Factors appear in pairs ` `                ``else` `                    ``cnt += ``2``; ` `            ``} ` `        ``} ` `        ``return` `cnt; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `[] a = { ``4``, ``16``, ``1024``, ``48` `}; ` `        ``int` `n = a.length; ` `     `  `        ``System.out.println(getCount(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of ` `# the required integers ` `from` `math ``import` `gcd as __gcd ` `def` `getCount(a, n): ` ` `  `    ``# To store the gcd of the array elements ` `    ``gcd ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``gcd ``=` `__gcd(gcd, a[i]) ` ` `  `    ``# To store the count of factors ` `    ``# of the found gcd ` `    ``cnt ``=` `0` ` `  `    ``for` `i ``in` `range``(``1``, gcd ``+` `1``): ` `        ``if` `i ``*` `i > gcd: ` `            ``break` `        ``if` `(gcd ``%` `i ``=``=` `0``): ` ` `  `            ``# If g is a perfect square ` `            ``if` `(i ``*` `i ``=``=` `gcd): ` `                ``cnt ``+``=` `1` ` `  `            ``# Factors appear in pairs ` `            ``else``: ` `                ``cnt ``+``=` `2` ` `  `    ``return` `cnt ` ` `  `# Driver code ` `a ``=` `[``4``, ``16``, ``1024``, ``48``] ` `n ``=` `len``(a) ` ` `  `print``(getCount(a, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Recursive function to return gcd  ` `    ``static` `int` `calgcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``if` `(b == 0)  ` `            ``return` `a;  ` `        ``return` `calgcd(b, a % b);  ` `    ``} ` `     `  `    ``// Function to return the count of  ` `    ``// the required integers ` `    ``static` `int` `getCount(``int` `[] a, ``int` `n) ` `    ``{ ` `     `  `        ``// To store the gcd of the array elements ` `        ``int` `gcd = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``gcd = calgcd(gcd, a[i]); ` `     `  `        ``// To store the count of factors ` `        ``// of the found gcd ` `        ``int` `cnt = 0; ` `     `  `        ``for` `(``int` `i = 1; i * i <= gcd; i++)  ` `        ``{ ` `            ``if` `(gcd % i == 0)  ` `            ``{ ` `     `  `                ``// If g is a perfect square ` `                ``if` `(i * i == gcd) ` `                    ``cnt++; ` `     `  `                ``// Factors appear in pairs ` `                ``else` `                    ``cnt += 2; ` `            ``} ` `        ``} ` `        ``return` `cnt; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `[] a = { 4, 16, 1024, 48 }; ` `        ``int` `n = a.Length; ` `     `  `        ``Console.WriteLine(getCount(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

Output:

```3
```

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Improved By : mohit kumar 29, ihritik