Count all perfect divisors of a number

Given a number n, count total perfect divisors of n. Perfect divisors are those divisors which are square of some integer. For example a perfect divisor of 8 is 4.
Examples:

Input  : n = 16 
Output : 3
Explanation : There are only 5 divisor of 16:
1, 2, 4, 8, 16. Only three of them are perfect 
squares: 1, 4, 16. Therefore the answer is 3

Input  : n = 7
Output : 1


Naive approach

A brute force is find all the divisors of a number. Count all divisors that are perfect squares.

C++

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// Below is C++ code to count total perfect Divisors
#include<bits/stdc++.h>
using namespace std;
  
// Utility function to check perfect square number
bool isPerfectSquare(int n)
{
    int sq = (int) sqrt(n);
    return (n == sq * sq);
}
  
// Returns count all perfect divisors of n
int countPerfectDivisors(int n)
{
    // Initialize result
    int count = 0;
  
    // Consider every number that can be a divisor
    // of n
    for (int i=1; i*i <= n; ++i)
    {
        // If i is a divisor
        if (n%i == 0)
        {
            if (isPerfectSquare(i))
                ++count;
            if (n/i != i && isPerfectSquare(n/i))
                ++count;
        }
    }
    return count;
}
  
// Driver code
int main()
{
    int n = 16;
  
    cout << "Total perfect divisors of "
         << n << " = " << countPerfectDivisors(n) << "\n";
  
    n = 12;
    cout << "Total perfect divisors of "
         << n << " = " << countPerfectDivisors(n);
  
    return 0;
}

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Java

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// Java code to count
// total perfect Divisors
import java.io.*;
  
class GFG 
{
      
// Utility function to check 
// perfect square number
static boolean isPerfectSquare(int n)
{
    int sq = (int) Math.sqrt(n);
    return (n == sq * sq);
}
  
// Returns count all 
// perfect divisors of n
static int countPerfectDivisors(int n)
{
    // Initialize result
    int count = 0;
  
    // Consider every number 
    // that can be a divisor of n
    for (int i = 1; i * i <= n; ++i)
    {
        // If i is a divisor
        if (n % i == 0)
        {
            if (isPerfectSquare(i))
                ++count;
            if (n / i != i && 
                isPerfectSquare(n / i))
                ++count;
        }
    }
    return count;
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 16;
  
    System.out.print("Total perfect "
                   "divisors of " + n);
    System.out.println(" = "
               countPerfectDivisors(n));
  
    n = 12;
    System.out.print("Total perfect "
                   "divisors of " + n);
    System.out.println(" = "
               countPerfectDivisors(n));
}
}
  
// This code is contributed by ajit

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Python3

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# Python3 implementation of Naive method 
# to count all perfect divisors
  
import math
  
def isPerfectSquare(x) :
    sq = (int)(math.sqrt(x))
    return (x == sq * sq)
  
# function to count all prefect divisors
def countPerfectDivisors(n) :
      
    # Initialize result
    cnt = 0
  
        # Consider every number that 
        # can be a divisor of n
    for i in range(1, (int)(math.sqrt(n)) + 1) :
  
            # If i is a divisor
            if ( n % i == 0 ) :
  
                if isPerfectSquare(i):
                        cnt = cnt + 1
                if n/i != i and isPerfectSquare(n/i):
                        cnt = cnt + 1
    return cnt
      
          
# Driver program to test above function 
print("Total perfect divisor of 16 = ",
    countPerfectDivisors(16))
      
print("Total perfect divisor of 12 = ",
    countPerfectDivisors(12))    
  
# This code is contributed by Saloni Gupta

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C#

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// C# code to count
// total perfect Divisors
using System;
  
class GFG
{
      
// Utility function to check 
// perfect square number
static bool isPerfectSquare(int n)
{
    int sq = (int) Math.Sqrt(n);
    return (n == sq * sq);
}
  
// Returns count all 
// perfect divisors of n
static int countPerfectDivisors(int n)
{
    // Initialize result
    int count = 0;
  
    // Consider every number 
    // that can be a divisor of n
    for (int i = 1; 
             i * i <= n; ++i)
    {
        // If i is a divisor
        if (n % i == 0)
        {
            if (isPerfectSquare(i))
                ++count;
            if (n / i != i && 
                isPerfectSquare(n / i))
                ++count;
        }
    }
    return count;
}
  
// Driver code
static public void Main ()
{
    int n = 16;
      
    Console.Write("Total perfect "
                "divisors of " + n);
    Console.WriteLine(" = "
            countPerfectDivisors(n));
      
    n = 12;
    Console.Write("Total perfect "
                "divisors of " + n);
    Console.WriteLine(" = "
            countPerfectDivisors(n));
}
}
  
// This code is contributed 
// by akt_mit

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PHP

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<?php
// PHP code to count 
// total perfect Divisors
  
// function to check 
// perfect square number
function isPerfectSquare($n)
{
    $sq = sqrt($n);
    return ($n == $sq * $sq);
}
  
// Returns count all 
// perfect divisors of n
function countPerfectDivisors($n)
{
      
    // Initialize result
    $count = 0;
  
    // Consider every number
    // that can be a divisor
    // of n
    for ($i = 1; $i * $i <= $n; ++$i)
    {
          
        // If i is a divisor
        if ($n % $i == 0)
        {
            if (isPerfectSquare($i))
                ++$count;
            if ($n / $i != $i && 
                isPerfectSquare($n / $i))
                ++$count;
        }
    }
    return $count;
}
  
    // Driver Code
    $n = 16;
    echo "Total perfect divisors of ",
         $n, " = ", countPerfectDivisors($n), "\n";
  
    $n = 12;
    echo "Total perfect divisors of ",
         $n, " = ", countPerfectDivisors($n);
  
// This code is contributed by ajit
?>

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Output:
Total Perfect divisors of 16 = 3
Total Perfect divisors of 12 = 2

Time complexity: O(sqrt(n))
Auxiliary space: O(1)

Efficient approach (For large number of queries)

The idea is based on Sieve of Eratosthenes. This approach is beneficial if there are large number of queries. Following is the algorithm to find all perfect divisors up to n numbers.

  1. Create a list of n consecutive integers from 1 to n:(1, 2, 3, …, n)
  2. Initially, let d be 2, the first divisor
  3. Starting from 4(square of 2) increment all the multiples of 4 by 1 in perfectDiv[] array, as all these multiples contain 4 as perfect divisor. These numbers will be 4d, 8d, 12d, … etc
  4. Repeat the 3rd step for all other numbers. The final array of perfectDiv[] will contain all the count of perfect divisors from 1 to n

Below is implementation of above steps.

C++

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// Below is C++ code to count total perfect
// divisors
#include<bits/stdc++.h>
using namespace std;
#define MAX 100001
  
int perfectDiv[MAX];
  
// Pre-compute counts of all perfect divisors
// of all numbers upto MAX.
void precomputeCounts()
{
    for (int i=1; i*i < MAX; ++i)
    {
        // Iterate through all the multiples of i*i
        for (int j=i*i; j < MAX; j += i*i)
  
            // Increment all such multiples by 1
            ++perfectDiv[j];
    }
}
  
// Returns count of perfect divisors of n.
int countPerfectDivisors(int n)
{
    return perfectDiv[n];
}
  
// Driver code
int main()
{
    precomputeCounts();
  
    int n = 16;
    cout << "Total perfect divisors of "
         << n << " = " << countPerfectDivisors(n) << "\n";
  
    n = 12;
    cout << "Total perfect divisors of "
         << n << " = " << countPerfectDivisors(n);
  
    return 0;
}

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Java

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// Java code to count total perfect
// divisors
class GFG
{
      
static int MAX = 100001;
  
static int[] perfectDiv = new int[MAX];
  
// Pre-compute counts of all perfect divisors
// of all numbers upto MAX.
static void precomputeCounts()
{
    for (int i = 1; i * i < MAX; ++i)
    {
        // Iterate through all the multiples of i*i
        for (int j = i * i; j < MAX; j += i * i)
  
            // Increment all such multiples by 1
            ++perfectDiv[j];
    }
}
  
// Returns count of perfect divisors of n.
static int countPerfectDivisors(int n)
{
    return perfectDiv[n];
}
  
// Driver code
public static void main (String[] args) 
{
    precomputeCounts();
  
    int n = 16;
    System.out.println("Total perfect divisors of " +
                    n + " = " + countPerfectDivisors(n));
  
    n = 12;
    System.out.println("Total perfect divisors of " +
                        n + " = " + countPerfectDivisors(n));
}
}
  
// This code is contributed by mits

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Python3

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# Below is Python3 code to count total perfect
# divisors
MAX = 100001
   
perfectDiv= [0]*MAX
   
# Pre-compute counts of all perfect divisors
# of all numbers upto MAX.
def precomputeCounts():
  
    i=1
    while i*i < MAX:
      
        # Iterate through all the multiples of i*i
        for j in range(i*i,MAX,i*i):
   
            # Increment all such multiples by 1
            perfectDiv[j] += 1
        i += 1
   
# Returns count of perfect divisors of n.
def countPerfectDivisors( n):
  
    return perfectDiv[n]
   
# Driver code
if __name__ == "__main__":
  
    precomputeCounts()
   
    n = 16
    print ("Total perfect divisors of "
         , n , " = " ,countPerfectDivisors(n))
   
    n = 12
    print ( "Total perfect divisors of "
         ,n ," = " ,countPerfectDivisors(n))

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C#

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// C# code to count total perfect
// divisors
using System;
class GFG
{
      
static int MAX = 100001;
  
static int[] perfectDiv = new int[MAX];
  
// Pre-compute counts of all perfect 
// divisors of all numbers upto MAX.
static void precomputeCounts()
{
    for (int i = 1; i * i < MAX; ++i)
    {
        // Iterate through all the multiples of i*i
        for (int j = i * i; j < MAX; j += i * i)
  
            // Increment all such multiples by 1
            ++perfectDiv[j];
    }
}
  
// Returns count of perfect divisors of n.
static int countPerfectDivisors(int n)
{
    return perfectDiv[n];
}
  
// Driver code
public static void Main() 
{
    precomputeCounts();
  
    int n = 16;
    Console.WriteLine("Total perfect divisors of " + n + 
                       " = " + countPerfectDivisors(n));
  
    n = 12;
    Console.WriteLine("Total perfect divisors of " + n + 
                       " = " + countPerfectDivisors(n));
}
}
  
// This code is contributed by mits

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PHP

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<?php
// Below is PHP code to count total 
// perfect divisors
  
$MAX = 10001;
  
$perfectDiv = array_fill(0, $MAX, 0);
  
// Pre-compute counts of all perfect 
// divisors of all numbers upto MAX.
function precomputeCounts()
{
    global $MAX, $perfectDiv;
    for ($i = 1; $i * $i < $MAX; ++$i)
    {
        // Iterate through all the multiples
        // of i*i
        for ($j = $i * $i
             $j < $MAX; $j += $i * $i)
  
            // Increment all such multiples by 1
            ++$perfectDiv[$j];
    }
}
  
// Returns count of perfect divisors of n.
function countPerfectDivisors($n)
{
    global $perfectDiv;
    return $perfectDiv[$n];
}
  
// Driver code
precomputeCounts();
  
$n = 16;
echo "Total perfect divisors of " . $n
     " = " . countPerfectDivisors($n) . "\n";
  
$n = 12;
echo "Total perfect divisors of " . $n .
     " = " . countPerfectDivisors($n);
  
// This code is contributed by mits
?>

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Output:

Total Perfect divisors of 16 = 3
Total Perfect divisors of 12 = 2

Time complexity: O(MAX * log(log (MAX)))
Auxiliary space: O(MAX)

This article is contributed by Shubham Bansal. 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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