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# 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// numbervector<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 numberint 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 Codeint main(){    int N = 8;     // Function Call    cout << nonPrimeDivisors(N) << endl;    return 0;}

## Java

 // Java program to find// count of non-prime// divisors of given numberimport java.util.*;class GFG{ // Function to factors// of the given numberstatic 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 numberstatic 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 Codepublic 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 numberfrom math import sqrt # Function to factors of the given# numberdef 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 numberdef 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 Codeif __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 numberusing System;using System.Collections.Generic;class GFG{ // Function to factors// of the given numberstatic 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 numberstatic 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 Codepublic 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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