Total number of divisors for a given number
Last Updated :
15 Oct, 2024
Improve
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 the 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 the following form:
n =
// 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;
}
// 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)
# 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
// 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
<script>
// Javascript program for finding number of divisor
// program for finding no. of divisors
function divCount(n)
{
// sieve method for prime calculation
var hash = Array(n+1).fill(true);
for (var p = 2; p * p < n; p++)
if (hash[p] == true)
for (var i = p * 2; i < n; i += p)
hash[i] = false;
// Traversing through all prime numbers
var total = 1;
for (var 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
var count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = parseInt(n / p);
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
// driver program
var n = 24;
document.write( divCount(n));
</script>
<?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
?>
Output
8
