Total number of divisors for a given number

Given a positive integer n, we have to find the total number of divisors for n.

Examples:

Input : n = 25
Output : 3
Divisors are 1, 5 and 25.

Input : n = 24
Output : 8
Divisors are 1, 2, 3, 4, 6, 8
12 and 24.

We have discussed different approaches for printing all divisors (here and here). Here task is simpler, we need to count divisors.

First of all store all primes from 2 to max_size in an array so that we should only check for the prime divisors. Now we will only wish to calculate the factorization of n in following form :
n = (a1^p1) * (a2^p2) *…….*(an^pn), where a1, a2…an all are prime factors and p1, p2, .. pn are integral power of them as pi being integral power of ai in factorization of n.
So, for this factorization we have formula to find total number of divisor of n and that is:
(p1+1) * (p2+1) *….*(pn+1).

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program for finding number of divisor
#include <bits/stdc++.h>
  
using namespace std;
  
// program for finding no. of divisors
int divCount(int n)
{
    // sieve method for prime calculation
    bool hash[n + 1];
    memset(hash, true, sizeof(hash));
    for (int p = 2; p * p < n; p++)
        if (hash[p] == true)
            for (int i = p * 2; i < n; i += p)
                hash[i] = false;
  
    // Traversing through all prime numbers
    int total = 1;
    for (int p = 2; p <= n; p++) {
        if (hash[p]) {
  
            // calculate number of divisor
            // with formula total div = 
            // (p1+1) * (p2+1) *.....* (pn+1)
            // where n = (a1^p1)*(a2^p2).... 
            // *(an^pn) ai being prime divisor
            // for n and pi are their respective 
            // power in factorization
            int count = 0;
            if (n % p == 0) {
                while (n % p == 0) {
                    n = n / p;
                    count++;
                }
                total = total * (count + 1);
            }
        }
    }
    return total;
}
  
// driver program
int main()
{
    int n = 24;
    cout << divCount(n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for finding
// number of divisor
import java.io.*;
import java.util.*;
import java.lang.*;
  
class GFG
{
// program for finding 
// no. of divisors
static int divCount(int n)
{
    // sieve method for prime calculation
    boolean hash[] = new boolean[n + 1];
    Arrays.fill(hash, true);
    for (int p = 2; p * p < n; p++)
        if (hash[p] == true)
            for (int i = p * 2; i < n; i += p)
                hash[i] = false;
  
    // Traversing through 
    // all prime numbers
    int total = 1;
    for (int p = 2; p <= n; p++) 
    {
        if (hash[p])
        {
  
            // calculate number of divisor
            // with formula total div = 
            // (p1+1) * (p2+1) *.....* (pn+1)
            // where n = (a1^p1)*(a2^p2).... 
            // *(an^pn) ai being prime divisor
            // for n and pi are their respective 
            // power in factorization
            int count = 0;
            if (n % p == 0
            {
                while (n % p == 0
                {
                    n = n / p;
                    count++;
                }
                total = total * (count + 1);
            }
        }
    }
    return total;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 24;
    System.out.print(divCount(n));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for finding 
# number of divisor
  
# program for finding 
# no. of divisors
def divCount(n):
  
    # sieve method for
    # prime calculation
    hh = [1] * (n + 1);
      
    p = 2;
    while((p * p) < n):
        if (hh[p] == 1):
            for i in range((p * 2), n, p):
                hh[i] = 0;
        p += 1;
  
    # Traversing through 
    # all prime numbers
    total = 1;
    for p in range(2, n + 1):
        if (hh[p] == 1):
  
            # calculate number of divisor
            # with formula total div = 
            # (p1+1) * (p2+1) *.....* (pn+1)
            # where n = (a1^p1)*(a2^p2).... 
            # *(an^pn) ai being prime divisor
            # for n and pi are their respective 
            # power in factorization
            count = 0;
            if (n % p == 0):
                while (n % p == 0):
                    n = int(n / p);
                    count += 1;
                total *= (count + 1);
                  
    return total;
  
# Driver Code
n = 24;
print(divCount(n));
  
# This code is contributed by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for finding
// number of divisor
using System;
  
class GFG
{
// program for finding 
// no. of divisors
static int divCount(int n)
{
    // sieve method for prime calculation
    bool[] hash = new bool[n + 1];
    for (int p = 2; p * p < n; p++)
        if (hash[p] == false)
            for (int i = p * 2;
                     i < n; i += p)
                hash[i] = true;
  
    // Traversing through 
    // all prime numbers
    int total = 1;
    for (int p = 2; p <= n; p++) 
    {
        if (hash[p] == false)
        {
  
            // calculate number of divisor
            // with formula total div = 
            // (p1+1) * (p2+1) *.....* (pn+1)
            // where n = (a1^p1)*(a2^p2).... 
            // *(an^pn) ai being prime divisor
            // for n and pi are their respective 
            // power in factorization
            int count = 0;
            if (n % p == 0) 
            {
                while (n % p == 0) 
                {
                    n = n / p;
                    count++;
                }
                total = total * (count + 1);
            }
        }
    }
    return total;
}
  
// Driver Code
public static void Main()
{
    int n = 24;
    Console.WriteLine(divCount(n));
}
}
  
// This code is contributed 
// by mits

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program for finding 
// number of divisor
  
// program for finding 
// no. of divisors
function divCount($n)
{
    // sieve method for
    // prime calculation
    $hash = array_fill(0, $n + 1, 1);
  
    for ($p = 2; 
        ($p * $p) < $n; $p++)
        if ($hash[$p] == 1)
            for ($i = ($p * 2); 
                 $i < $n; $i= ($i + $p))
                $hash[$i] = 0;
  
    // Traversing through 
    // all prime numbers
    $total = 1;
    for ($p = 2; $p <= $n; $p++)
    {
        if ($hash[$p] == 1) 
        {
  
            // calculate number of divisor
            // with formula total div = 
            // (p1+1) * (p2+1) *.....* (pn+1)
            // where n = (a1^p1)*(a2^p2).... 
            // *(an^pn) ai being prime divisor
            // for n and pi are their respective 
            // power in factorization
            $count = 0;
            if ($n % $p == 0) 
            {
                while ($n % $p == 0)
                {
                    $n = ($n / $p);
                    $count++;
                }
                $total = $total
                        ($count + 1);
            }
        }
    }
    return $total;
}
  
// Driver Code
$n = 24;
echo divCount($n);
  
// This code is contributed by mits
?>

chevron_right



Output:

8

Reference : Number of divisors.

This article is contributed by Shivam Pradhan (anuj_charm). 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Mithun Kumar, Akanksha_Rai