Count Divisors of n in O(n^1/3)

Given a number n, count all distinct divisors of it.

Examples:

Input : 18
Output : 6
Divisors of 18 are 1, 2, 3, 6, 9 and 18.

Input : 100
Output : 9
Divisors of 100 are 1, 2, 4, 5, 10, 20,
25, 50 and 100



A Naive Solution would be to iterate all the numbers from 1 to sqrt(n), checking if that number divides n and incrementing number of divisors. This approach takes O(sqrt(n)) time.

C++

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// C implementation of Naive method to count all
// divisors
#include <bits/stdc++.h>
using namespace std;
  
// function to count the divisors
int countDivisors(int n)
{
    int cnt = 0;
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            // If divisors are equal,
            // count only one
            if (n / i == i)
                cnt++;
  
            else // Otherwise count both
                cnt = cnt + 2;
        }
    }
    return cnt;
}
  
/* Driver program to test above function */
int main()
{
    printf("Total distinct divisors of 100 are : %d",
           countDivisors(100));
    return 0;
}

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Java

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// JAVA implementation of Naive method
// to count all divisors
import java.io.*;
import java.math.*;
  
class GFG {
  
    // function to count the divisors
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i <= Math.sqrt(n); i++)
        {
            if (n % i == 0) {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                    cnt++;
  
                else // Otherwise count both
                    cnt = cnt + 2;
            }
        }
        return cnt;
    }
  
    /* Driver program to test above function */
    public static void main(String args[])
    {
        System.out.println("Total distinct "
                  + "divisors of 100 are : "  
                       + countDivisors(100));
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

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Python3

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# Python3 implementation of Naive method 
# to count all divisors
  
import math
  
# function to count the divisors
def countDivisors(n) :
    cnt = 0
    for i in range(1, (int)(math.sqrt(n)) + 1) :
        if (n % i == 0) :
              
            # If divisors are equal,
            # count only one
            if (n / i == i) :
                cnt = cnt + 1
            else : # Otherwise count both
                cnt = cnt + 2
                  
    return cnt
      
# Driver program to test above function */
  
print("Total distinct divisors of 100 are : ",
      countDivisors(100))
  
# This code is contributed by Nikita Tiwari.

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C#

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// C# implementation of Naive method
// to count all divisors
using System;
  
class GFG {
  
    // function to count the divisors
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i <= Math.Sqrt(n);
                                      i++)
        {
            if (n % i == 0) {
                  
                // If divisors are equal,
                // count only one
                if (n / i == i)
                    cnt++;
  
                // Otherwise count both
                else
                    cnt = cnt + 2;
            }
        }
          
        return cnt;
    }
  
    // Driver program
    public static void Main()
    {
        Console.WriteLine("Total distinct"
               + " divisors of 100 are : "
                    + countDivisors(100));
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP implementation of Naive 
// method to count all divisors
  
// function to count the divisors
  
function countDivisors($n)
{
    $cnt = 0;
    for ($i = 1; $i <= sqrt($n); $i++)
    {
        if ($n % $i == 0)
        {
            // If divisors are equal,
            // count only one
            if ($n / $i == $i)
            $cnt++;
  
            // Otherwise count both
            else 
                $cnt = $cnt + 2;
        }
    }
    return $cnt;
}
  
// Driver Code
echo "Total distinct divisors of 100 are : ",
        countDivisors(100);
  
// This code is contributed by Ajit
?>

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Output :

Total distinct divisors of 100 are : 9

Optimized Solution (O(n^1/3))

  1. Split number n in two numbers x and y such that n=x*y where x contains only prime factors in range 2 <= x <= n(1/3) and y deals with higher prime factors greater than n(1/3).
  2. Count total factors of x using the naive trial division method. Let this count be F(x).
  3. Count total factors of y using the following three cases. Let this count be F(y).
    • If y is a prime number then factors will be 1 and y itself. That implies, F(y) = 2.
    • If y is square of a prime number, then factors will be 1, sqrt(y) and y itself. That implies, F(y) = 3.
    • If y is the product of two distinct prime numbers, then factors will be 1, both prime numbers and number y itself. That implies, F(y) = 4.
  4. Since F(x*y) is a multiplicative function and gcd(x, y) = 1, that implies, F(x*y) = F(x)*F(y) which gives the count of total distinct divisors of n.

Note that, we have only these three cases for calculating factors of y since there can be at max two prime factors of y. If it would have had more than two prime factors, one of them would surely have been less than equal to n(1/3), and hence it would be included in x and not in y.

C++

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// C++ program to count distinct divisors
// of a given number n
#include <bits/stdc++.h>
using namespace std;
  
void SieveOfEratosthenes(int n, bool prime[],
                         bool primesquare[], int a[])
{
    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true. A value
    // in prime[i] will finally be false if i is
    // Not a prime, else true.
    for (int i = 2; i <= n; i++)
        prime[i] = true;
  
    // Create a boolean array "primesquare[0..n*n+1]"
    // and initialize all entries it as false. A value
    // in squareprime[i] will finally be true if i is
    // square of prime, else false.
    for (int i = 0; i <= (n * n + 1); i++)
        primesquare[i] = false;
  
    // 1 is not a prime number
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    int j = 0;
    for (int p = 2; p <= n; p++) {
        if (prime[p]) {
            // Storing primes in an array
            a[j] = p;
  
            // Update value in primesquare[p*p],
            // if p is prime.
            primesquare[p * p] = true;
            j++;
        }
    }
}
  
// Function to count divisors
int countDivisors(int n)
{
    // If number is 1, then it will have only 1
    // as a factor. So, total factors will be 1.
    if (n == 1)
        return 1;
  
    bool prime[n + 1], primesquare[n * n + 1];
  
    int a[n]; // for storing primes upto n
  
    // Calling SieveOfEratosthenes to store prime
    // factors of n and to store square of prime
    // factors of n
    SieveOfEratosthenes(n, prime, primesquare, a);
  
    // ans will contain total number of distinct
    // divisors
    int ans = 1;
  
    // Loop for counting factors of n
    for (int i = 0;; i++) {
        // a[i] is not less than cube root n
        if (a[i] * a[i] * a[i] > n)
            break;
  
        // Calculating power of a[i] in n.
        int cnt = 1; // cnt is power of prime a[i] in n.
        while (n % a[i] == 0) // if a[i] is a factor of n
        {
            n = n / a[i];
            cnt = cnt + 1; // incrementing power
        }
  
        // Calculating number of divisors
        // If n = a^p * b^q then total divisors of n
        // are (p+1)*(q+1)
        ans = ans * cnt;
    }
  
    // if a[i] is greater than cube root of n
  
    // First case
    if (prime[n])
        ans = ans * 2;
  
    // Second case
    else if (primesquare[n])
        ans = ans * 3;
  
    // Third casse
    else if (n != 1)
        ans = ans * 4;
  
    return ans; // Total divisors
}
  
// Driver Program
int main()
{
    cout << "Total distinct divisors of 100 are : "
         << countDivisors(100) << endl;
    return 0;
}

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Java

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// JAVA program to count distinct
// divisors of a given number n
import java.io.*;
  
class GFG {
  
    static void SieveOfEratosthenes(int n, boolean prime[],
                                    boolean primesquare[], int a[])
    {
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true. A value
        // in prime[i] will finally be false if i is
        // Not a prime, else true.
        for (int i = 2; i <= n; i++)
            prime[i] = true;
  
        /* Create a boolean array "primesquare[0..n*n+1]"
         and initialize all entries it as false.
         A value in squareprime[i] will finally 
         be true if i is square of prime, 
         else false.*/
        for (int i = 0; i < ((n * n) + 1); i++)
            primesquare[i] = false;
  
        // 1 is not a prime number
        prime[1] = false;
  
        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = false;
            }
        }
  
        int j = 0;
        for (int p = 2; p <= n; p++) {
            if (prime[p]) {
                // Storing primes in an array
                a[j] = p;
  
                // Update value in
                // primesquare[p*p],
                // if p is prime.
                primesquare[p * p] = true;
                j++;
            }
        }
    }
  
    // Function to count divisors
    static int countDivisors(int n)
    {
        // If number is 1, then it will
        // have only 1 as a factor. So,
        // total factors will be 1.
        if (n == 1)
            return 1;
  
        boolean prime[] = new boolean[n + 1];
        boolean primesquare[] = new boolean[(n * n) + 1];
  
        // for storing primes upto n
        int a[] = new int[n];
  
        // Calling SieveOfEratosthenes to
        // store prime factors of n and to
        // store square of prime factors of n
        SieveOfEratosthenes(n, prime, primesquare, a);
  
        // ans will contain total number
        // of distinct divisors
        int ans = 1;
  
        // Loop for counting factors of n
        for (int i = 0;; i++) {
            // a[i] is not less than cube root n
            if (a[i] * a[i] * a[i] > n)
                break;
  
            // Calculating power of a[i] in n.
            // cnt is power of prime a[i] in n.
            int cnt = 1;
  
            // if a[i] is a factor of n
            while (n % a[i] == 0) {
                n = n / a[i];
  
                // incrementing power
                cnt = cnt + 1;
            }
  
            // Calculating number of divisors
            // If n = a^p * b^q then total
            // divisors of n are (p+1)*(q+1)
            ans = ans * cnt;
        }
  
        // if a[i] is greater than cube root
        // of n
  
        // First case
        if (prime[n])
            ans = ans * 2;
  
        // Second case
        else if (primesquare[n])
            ans = ans * 3;
  
        // Third casse
        else if (n != 1)
            ans = ans * 4;
  
        return ans; // Total divisors
    }
  
    // Driver Program
    public static void main(String args[])
    {
        System.out.println("Total distinct divisors"
                           + " of 100 are : " + countDivisors(100));
    }
}
  
/*This code is contributed by Nikita Tiwari*/

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Python3

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# Python3 program to count distinct 
# divisors of a given number n
  
def SieveOfEratosthenes(n, prime,primesquare, a):
    # Create a boolean array "prime[0..n]" 
    # and initialize all entries it as 
    # true. A value in prime[i] will finally 
    # be false if i is not a prime, else true.
    for i in range(2,n+1):
        prime[i] = True
  
    # Create a boolean array "primesquare[0..n*n+1]"
    # and initialize all entries it as false. 
    # A value in squareprime[i] will finally be 
    # true if i is square of prime, else false.
    for i in range((n * n + 1)+1):
        primesquare[i] = False
  
    # 1 is not a prime number
    prime[1] = False
  
    p = 2
    while(p * p <= n):
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p] == True):
            # Update all multiples of p
            i = p * 2 
            while(i <= n):
                prime[i] = False
                i += p
        p+=1
      
  
    j = 0
    for p in range(2,n+1): 
        if (prime[p]==True): 
            # Storing primes in an array
            a[j] = p
  
            # Update value in primesquare[p*p],
            # if p is prime.
            primesquare[p * p] = True
            j+=1
  
# Function to count divisors
def countDivisors(n):
    # If number is 1, then it will
    # have only 1 as a factor. So,
    # total factors will be 1.
    if (n == 1):
        return 1
  
    prime = [False]*(n + 2)
    primesquare = [False]*(n * n + 2)
      
    # for storing primes upto n
    a = [0]*n
  
    # Calling SieveOfEratosthenes to 
    # store prime factors of n and to 
    # store square of prime factors of n
    SieveOfEratosthenes(n, prime, primesquare, a)
  
    # ans will contain total
    # number of distinct divisors
    ans = 1
  
    # Loop for counting factors of n
    i=0
    while(1): 
        # a[i] is not less than cube root n
        if(a[i] * a[i] * a[i] > n):
            break
  
        # Calculating power of a[i] in n.
        cnt = 1 # cnt is power of 
                # prime a[i] in n.
        while (n % a[i] == 0): # if a[i] is a factor of n
            n = n / a[i]
            cnt = cnt + 1 # incrementing power
  
        # Calculating number of divisors
        # If n = a^p * b^q then total 
        # divisors of n are (p+1)*(q+1)
        ans = ans * cnt
        i+=1
  
    # if a[i] is greater than
    # cube root of n
      
    n=int(n)
    # First case
    if (prime[n]==True):
        ans = ans * 2
  
    # Second case
    elif (primesquare[n]==True):
        ans = ans * 3
  
    # Third casse
    elif (n != 1):
        ans = ans * 4
  
    return ans # Total divisors
  
# Driver Code
if __name__=='__main__':
    print("Total distinct divisors of 100 are :",countDivisors(100))
  
# This code is contributed
# by mits

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C#

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// C# program to count distinct
// divisors of a given number n
using System;
  
class GFG {
  
    static void SieveOfEratosthenes(int n, bool[] prime,
                                    bool[] primesquare, int[] a)
    {
  
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true. A value
        // in prime[i] will finally be false if i is
        // Not a prime, else true.
        for (int i = 2; i <= n; i++)
            prime[i] = true;
  
        /* Create a boolean array "primesquare[0..n*n+1]"
        and initialize all entries it as false.
        A value in squareprime[i] will finally 
        be true if i is square of prime, 
        else false.*/
        for (int i = 0; i < ((n * n) + 1); i++)
            primesquare[i] = false;
  
        // 1 is not a prime number
        prime[1] = false;
  
        for (int p = 2; p * p <= n; p++) {
  
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
  
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = false;
            }
        }
  
        int j = 0;
        for (int p = 2; p <= n; p++) {
            if (prime[p]) {
  
                // Storing primes in an array
                a[j] = p;
  
                // Update value in
                // primesquare[p*p],
                // if p is prime.
                primesquare[p * p] = true;
                j++;
            }
        }
    }
  
    // Function to count divisors
    static int countDivisors(int n)
    {
  
        // If number is 1, then it will
        // have only 1 as a factor. So,
        // total factors will be 1.
        if (n == 1)
            return 1;
  
        bool[] prime = new bool[n + 1];
        bool[] primesquare = new bool[(n * n) + 1];
  
        // for storing primes upto n
        int[] a = new int[n];
  
        // Calling SieveOfEratosthenes to
        // store prime factors of n and to
        // store square of prime factors of n
        SieveOfEratosthenes(n, prime, primesquare, a);
  
        // ans will contain total number
        // of distinct divisors
        int ans = 1;
  
        // Loop for counting factors of n
        for (int i = 0;; i++) {
  
            // a[i] is not less than cube root n
            if (a[i] * a[i] * a[i] > n)
                break;
  
            // Calculating power of a[i] in n.
            // cnt is power of prime a[i] in n.
            int cnt = 1;
  
            // if a[i] is a factor of n
            while (n % a[i] == 0) {
                n = n / a[i];
  
                // incrementing power
                cnt = cnt + 1;
            }
  
            // Calculating number of divisors
            // If n = a^p * b^q then total
            // divisors of n are (p+1)*(q+1)
            ans = ans * cnt;
        }
  
        // if a[i] is greater than cube root
        // of n
  
        // First case
        if (prime[n])
            ans = ans * 2;
  
        // Second case
        else if (primesquare[n])
            ans = ans * 3;
  
        // Third casse
        else if (n != 1)
            ans = ans * 4;
  
        return ans; // Total divisors
    }
  
    // Driver Program
    public static void Main()
    {
        Console.Write("Total distinct divisors"
                      + " of 100 are : " + countDivisors(100));
    }
}
  
// This code is contributed by parashar.

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PHP

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<?php 
// PHP program to count distinct 
// divisors of a given number n
  
function SieveOfEratosthenes($n, &$prime,
                             &$primesquare, &$a)
{
    // Create a boolean array "prime[0..n]" 
    // and initialize all entries it as 
    // true. A value in prime[i] will finally 
    // be false if i is not a prime, else true.
    for ($i = 2; $i <= $n; $i++)
        $prime[$i] = true;
  
    // Create a boolean array "primesquare[0..n*n+1]"
    // and initialize all entries it as false. 
    // A value in squareprime[i] will finally be 
    // true if i is square of prime, else false.
    for ($i = 0; $i <= ($n * $n + 1); $i++)
        $primesquare[$i] = false;
  
    // 1 is not a prime number
    $prime[1] = false;
  
    for ($p = 2; $p * $p <= $n; $p++)
    {
        // If prime[p] is not changed, 
        // then it is a prime
        if ($prime[$p] == true)
        {
            // Update all multiples of p
            for ($i = $p * 2; 
                 $i <= $n; $i += $p)
                $prime[$i] = false;
        }
    }
  
    $j = 0;
    for ($p = 2; $p <= $n; $p++) 
    {
        if ($prime[$p]) 
        {
            // Storing primes in an array
            $a[$j] = $p;
  
            // Update value in primesquare[p*p],
            // if p is prime.
            $primesquare[$p * $p] = true;
            $j++;
        }
    }
}
  
// Function to count divisors
function countDivisors($n)
{
    // If number is 1, then it will
    // have only 1 as a factor. So,
    // total factors will be 1.
    if ($n == 1)
        return 1;
  
    $prime = array_fill(false, $n + 1, NULL);
    $primesquare = array_fill(false, 
                          $n * $n + 1, NULL);
      
    // for storing primes upto n
    $a = array_fill(0, $n, NULL);
   
    // Calling SieveOfEratosthenes to 
    // store prime factors of n and to 
    // store square of prime factors of n
    SieveOfEratosthenes($n, $prime
                        $primesquare, $a);
  
    // ans will contain total
    // number of distinct divisors
    $ans = 1;
  
    // Loop for counting factors of n
    for ($i = 0;; $i++) 
    {
        // a[i] is not less than cube root n
        if ($a[$i] * $a[$i] * $a[$i] > $n)
            break;
  
        // Calculating power of a[i] in n.
        $cnt = 1; // cnt is power of 
                  // prime a[i] in n.
        while ($n % $a[$i] == 0) // if a[i] is a
                                 // factor of n
        {
            $n = $n / $a[$i];
            $cnt = $cnt + 1; // incrementing power
        }
  
        // Calculating number of divisors
        // If n = a^p * b^q then total 
        // divisors of n are (p+1)*(q+1)
        $ans = $ans * $cnt;
    }
  
    // if a[i] is greater than
    // cube root of n
  
    // First case
    if ($prime[$n])
        $ans = $ans * 2;
  
    // Second case
    else if ($primesquare[$n])
        $ans = $ans * 3;
  
    // Third casse
    else if ($n != 1)
        $ans = $ans * 4;
  
    return $ans; // Total divisors
}
  
// Driver Code
echo "Total distinct divisors of 100 are : ".
                    countDivisors(100). "\n";
  
// This code is contributed
// by ChitraNayal
?>

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Output :

Total distinct divisors of 100 are : 9

Time Complexity: O(n1/3)

This article is contributed by Rahul Agrawal. 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.

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