# Count of multiplicative partitions of N

Given an integer N, the task is to find the total number of multiplicative partition for N.

Multiplicative Partition: Number of ways of factoring of an integer with all factors greater than 1.

Examples:

Input: N = 20
Output:
Explanation:
Multiplicative partitions of 20 are:
2 × 2 × 5 = 2 × 10 = 4 × 5 = 20.
Input: N = 30
Output:
Explanation:
Multiplicative partitions of 30 are:
2 × 3 × 5 = 2 × 15 = 6 × 5 = 3 × 10 = 30

Approach: The idea is to try for every divisor of the N and then recursively break the dividend to get the multiplicative partitions. Below are the illustrations of the steps of approach:

• Initialize minimum factor as 2. Since it is the minimum factor other than 1.
• Start a loop from i = minimum to N – 1, and check if the number divides N and N/i > i, then increment the counter by 1 and again call the same function. Since, i divides n so it means i and N/i can be factorized some more times.

For Example:

If N = 30, let i = min = 2
30 % 2 = 0, so again recur with (2, 15)
15 % 3 = 0, so again recur with (3, 5)

and so on.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find ` `// the multiplicative partitions of ` `// the given number N ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return number of ways ` `// of factoring N with all ` `// factors greater than 1 ` `static` `int` `getDivisors(``int` `min, ``int` `n) ` `{ ` `     `  `    ``// Variable to store number of ways ` `    ``// of factoring n with all ` `    ``// factors greater than 1 ` `    ``int` `total = 0; ` `     `  `    ``for``(``int` `i = min; i < n; ++i) ` `    ``{ ` `        ``if` `(n % i == 0 && n / i >= i) ` `        ``{ ` `            ``++total; ` `            ``if` `(n / i > i) ` `                ``total += getDivisors(i, n / i); ` `        ``} ` `    ``} ` `    ``return` `total; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 30; ` `     `  `    ``// 2 is the minimum factor of ` `    ``// number other than 1. ` `    ``// So calling recursive ` `    ``// function to find ` `    ``// number of ways of factoring N ` `    ``// with all factors greater than 1 ` `    ``cout << 1 + getDivisors(2, n); ` `     `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by rutvik_56 `

## Java

 `// Java implementation to find ` `// the multiplicative partitions of ` `// the given number N ` ` `  `class` `MultiPart { ` ` `  `    ``// Function to return number of ways ` `    ``// of factoring N with all ` `    ``// factors greater than 1 ` `    ``static` `int` `getDivisors(``int` `min, ``int` `n) ` `    ``{ ` ` `  `        ``// Variable to store number of ways ` `        ``// of factoring n with all ` `        ``// factors greater than 1 ` `        ``int` `total = ``0``; ` ` `  `        ``for` `(``int` `i = min; i < n; ++i) ` ` `  `            ``if` `(n % i == ``0` `&& n / i >= i) { ` `                ``++total; ` `                ``if` `(n / i > i) ` `                    ``total ` `                        ``+= getDivisors( ` `                            ``i, n / i); ` `            ``} ` ` `  `        ``return` `total; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``30``; ` ` `  `        ``// 2 is the minimum factor of ` `        ``// number other than 1. ` `        ``// So calling recursive ` `        ``// function to find ` `        ``// number of ways of factoring N ` `        ``// with all factors greater than 1 ` `        ``System.out.println( ` `            ``1` `+ getDivisors(``2``, n)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation to find ` `# the multiplicative partitions of ` `# the given number N ` ` `  `# Function to return number of ways ` `# of factoring N with all ` `# factors greater than 1 ` `def` `getDivisors(``min``, n): ` `     `  `    ``# Variable to store number of ways ` `    ``# of factoring n with all ` `    ``# factors greater than 1 ` `    ``total ``=` `0` ` `  `    ``for` `i ``in` `range``(``min``, n): ` `        ``if` `(n ``%` `i ``=``=` `0` `and` `n ``/``/` `i >``=` `i): ` `            ``total ``+``=` `1` `            ``if` `(n ``/``/` `i > i): ` `                ``total ``+``=` `getDivisors(i, n ``/``/` `i) ` `                 `  `    ``return` `total ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `   `  `    ``n ``=` `30` ` `  `    ``# 2 is the minimum factor of ` `    ``# number other than 1. ` `    ``# So calling recursive ` `    ``# function to find ` `    ``# number of ways of factoring N ` `    ``# with all factors greater than 1 ` `    ``print``(``1` `+` `getDivisors(``2``, n)) ` ` `  `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation to find ` `// the multiplicative partitions of ` `// the given number N ` `using` `System; ` ` `  `class` `GFG{  ` `     `  `// Function to return number of ways ` `// of factoring N with all ` `// factors greater than 1  ` `static` `int` `getDivisors(``int` `min, ``int` `n) ` `{ ` ` `  `    ``// Variable to store number of ways ` `    ``// of factoring n with all ` `    ``// factors greater than 1 ` `    ``int` `total = 0; ` ` `  `    ``for``(``int` `i = min; i < n; ++i) ` `        ``if` `(n % i == 0 && n / i >= i) ` `        ``{ ` `            ``++total; ` `            ``if` `(n / i > i) ` `                ``total+= getDivisors(i, n / i); ` `        ``} ` ` `  `    ``return` `total; ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `n = 30; ` ` `  `    ``// 2 is the minimum factor of ` `    ``// number other than 1. ` `    ``// So calling recursive ` `    ``// function to find ` `    ``// number of ways of factoring N ` `    ``// with all factors greater than 1 ` `    ``Console.Write(1 + getDivisors(2, n)); ` `}  ` `}  ` ` `  `// This code is contributed by adityakumar27200 `

Output:

```5
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.