# Count of the non-prime divisors of a given number

• Difficulty Level : Medium
• Last Updated : 19 May, 2021

Given a number N, the task is to find the count of non-prime divisors of the given number N.

Examples:

Input: N = 8
Output:
Explanation:
Divisors of 8 are – {1, 2, 4, 8}
Non-Prime Divisors – {1, 4, 8}

Input: N = 20
Output:
Explanation:
Divisors of 20 are – {1, 2, 4, 5, 10, 20}
Non-Prime Divisors – {1, 4, 10, 20}

Approach: The key observation in the problem is that any number can be written as a product of its prime factors as

where K is the count of the prime factors of the given number. Using permutation and combination we can find that the total count of the
factors that are

Therefore, the count of the non-prime factors will be:

Below is the implementation of the above approach:

## C++

 // C++ program to find count of // non-prime divisors of given number   #include  using namespace std;   // Function to factors of the given // number vector<int> getFactorization(int x) {     int count = 0;     vector<int> v;       // Loop to find the divisors of     // the number 2     while (x % 2 == 0) {         count++;         x = x / 2;     }     if (count != 0)         v.push_back(count);       // Loop to find the divisors of the     // given number upto SQRT(N)     for (int i = 3; i <= sqrt(x); i += 2) {         count = 0;         while (x % i == 0) {             count++;             x /= i;         }         if (count != 0)             v.push_back(count);     }       // Condition to check if the rest     // number is also a prime number     if (x > 1) {         v.push_back(1);     }     return v; }   // Function to find the non-prime // divisors of the given number int nonPrimeDivisors(int N) {     vector<int> v = getFactorization(N);     int ret = 1;       // Loop to count the number of     // the total divisors of given number     for (int i = 0; i < v.size(); i++)         ret = ret * (v[i] + 1);       ret = ret - v.size();     return ret; }   // Driver Code int main() {     int N = 8;       // Function Call     cout << nonPrimeDivisors(N) << endl;     return 0; }

## Java

 // Java program to find  // count of non-prime  // divisors of given number import java.util.*; class GFG{   // Function to factors // of the given number static Vector getFactorization(int x) {   int count = 0;   Vector v = new Vector<>();     // Loop to find the    // divisors of the number 2   while (x % 2 == 0)    {     count++;     x = x / 2;   }       if (count != 0)     v.add(count);     // Loop to find the divisors    // of the given number upto SQRT(N)   for (int i = 3;             i <= Math.sqrt(x); i += 2)    {     count = 0;     while (x % i == 0)      {       count++;       x /= i;     }           if (count != 0)       v.add(count);   }     // Condition to check if    // the rest number is also    // a prime number   if (x > 1)    {     v.add(1);   }   return v; }   // Function to find the non-prime // divisors of the given number static int nonPrimeDivisors(int N) {   Vector v = getFactorization(N);   int ret = 1;     // Loop to count the number of   // the total divisors of given number   for (int i = 0; i < v.size(); i++)     ret = ret * (v.get(i) + 1);     ret = ret - v.size();   return ret; }   // Driver Code public static void main(String[] args) {   int N = 8;     // Function Call   System.out.println(nonPrimeDivisors(N)); } }   // This code is contributed by shikhasingrajput

## Python3

 # Python3 program to find count of # non-prime divisors of given number from math import sqrt   # Function to factors of the given # number def getFactorization(x):           count = 0     v = []       # Loop to find the divisors of     # the number 2     while (x % 2 == 0):         count += 1         x = x // 2       if (count != 0):         v.append(count)       # Loop to find the divisors of the     # given number upto SQRT(N)     for i in range(3, int(sqrt(x)) + 12):         count = 0                   while (x % i == 0):             count += 1             x //= i                       if (count != 0):             v.append(count)       # Condition to check if the rest     # number is also a prime number     if (x > 1):         v.append(1)               return v   # Function to find the non-prime # divisors of the given number def nonPrimeDivisors(N):           v = getFactorization(N)     ret = 1       # Loop to count the number of     # the total divisors of given number     for i in range(len(v)):         ret = ret * (v[i] + 1)     ret = ret - len(v)           return ret   # Driver Code if __name__ == '__main__':           N = 8       # Function Call     print(nonPrimeDivisors(N))   # This code is contributed by Samarth

## C#

 // C# program to find  // count of non-prime  // divisors of given number using System; using System.Collections.Generic; class GFG{   // Function to factors // of the given number static List<int> getFactorization(int x) {   int count = 0;   List<int> v = new List<int>();     // Loop to find the    // divisors of the number 2   while (x % 2 == 0)    {     count++;     x = x / 2;   }       if (count != 0)     v.Add(count);     // Loop to find the divisors    // of the given number upto    // SQRT(N)   for (int i = 3;             i <= Math.Sqrt(x); i += 2)    {     count = 0;     while (x % i == 0)      {       count++;       x /= i;     }           if (count != 0)       v.Add(count);   }     // Condition to check if    // the rest number is also    // a prime number   if (x > 1)    {     v.Add(1);   }   return v; }   // Function to find the non-prime // divisors of the given number static int nonPrimeDivisors(int N) {   List<int> v = getFactorization(N);   int ret = 1;     // Loop to count the number of   // the total divisors of given number   for (int i = 0; i < v.Count; i++)     ret = ret * (v[i] + 1);     ret = ret - v.Count;   return ret; }   // Driver Code public static void Main(String[] args) {   int N = 8;     // Function Call   Console.WriteLine(nonPrimeDivisors(N)); } }   // This code is contributed by gauravrajput1

## Javascript

 

Output:

3

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