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 = 
= 
where ai are prime factors and pi are integral power of them.
So, for this factorization we have formula to find total number of divisor of n and that is:
C++
#include <bits/stdc++.h>
using namespace std;
int divCount(int n)
{
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;
int total = 1;
for (int p = 2; p <= n; p++) {
if (hash[p]) {
int count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
int main()
{
int n = 24;
cout << divCount(n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
static int divCount(int n)
{
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;
int total = 1;
for (int p = 2; p <= n; p++)
{
if (hash[p])
{
int count = 0;
if (n % p == 0)
{
while (n % p == 0)
{
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
public static void main(String[] args)
{
int n = 24;
System.out.print(divCount(n));
}
}
|
Python3
def divCount(n):
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;
total = 1;
for p in range(2, n + 1):
if (hh[p] == 1):
count = 0;
if (n % p == 0):
while (n % p == 0):
n = int(n / p);
count += 1;
total *= (count + 1);
return total;
n = 24;
print(divCount(n));
|
C#
using System;
class GFG
{
static int divCount(int n)
{
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;
int total = 1;
for (int p = 2; p <= n; p++)
{
if (hash[p] == false)
{
int count = 0;
if (n % p == 0)
{
while (n % p == 0)
{
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
public static void Main()
{
int n = 24;
Console.WriteLine(divCount(n));
}
}
|
PHP
<?php
function divCount($n)
{
$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;
$total = 1;
for ($p = 2; $p <= $n; $p++)
{
if ($hash[$p] == 1)
{
$count = 0;
if ($n % $p == 0)
{
while ($n % $p == 0)
{
$n = ($n / $p);
$count++;
}
$total = $total *
($count + 1);
}
}
}
return $total;
}
$n = 24;
echo divCount($n);
?>
|
Javascript
<script>
function divCount(n)
{
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;
var total = 1;
for (var p = 2; p <= n; p++) {
if (hash[p]) {
var count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = parseInt(n / p);
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
var n = 24;
document.write( divCount(n));
</script>
|
Reference : Number of divisors.
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