Skip to content
Related Articles

Related Articles

Find all divisors of a natural number | Set 2
  • Difficulty Level : Easy
  • Last Updated : 31 Mar, 2021

Given a natural number n, print all distinct divisors of it.

Examples: 

 Input : n = 10       
 Output: 1 2 5 10

 Input:  n = 100
 Output: 1 2 4 5 10 20 25 50 100

 Input:  n = 125
 Output: 1 5 25 125

We strongly recommend to refer below article as a prerequisite. 
Find all divisors of a natural number | Set 1
In the above post, we had found a way to find all the divisors in O(sqrt(n)).
However there is still a minor problem in the solution, can you guess? 
Yes! the output is not in a sorted fashion which we had got using brute-force technique.

How to print the output in sorted order? 
If we observe the output which we had got, we can analyze that the divisors are printed in a zig-zag fashion (small, large pairs). Hence if we store half of them then we can print then in a sorted order. 

Below is an implementation for the same: 



C++




// A O(sqrt(n)) program that prints all divisors
// in sorted order
#include <bits/stdc++.h>
using namespace std;
 
// function to print the divisors
void printDivisors(int n)
{
    // Vector to store half of the divisors
    vector<int> v;
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
 
            // check if divisors are equal
            if (n / i == i)
                printf("%d ", i);
            else {
                printf("%d ", i);
 
                // push the second divisor in the vector
                v.push_back(n / i);
            }
        }
    }
 
    // The vector will be printed in reverse
    for (int i = v.size() - 1; i >= 0; i--)
        printf("%d ", v[i]);
}
 
/* Driver program to test above function */
int main()
{
    printf("The divisors of 100 are: n");
    printDivisors(100);
    return 0;
}
Java



// A O(sqrt(n)) java program that prints all divisors
// in sorted order
 
import java.util.Vector;
 
class Test {
    // method to print the divisors
    static void printDivisors(int n)
    {
        // Vector to store half of the divisors
        Vector<Integer> v = new Vector<>();
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
 
                // check if divisors are equal
                if (n / i == i)
                    System.out.printf("%d ", i);
                else {
                    System.out.printf("%d ", i);
 
                    // push the second divisor in the vector
                    v.add(n / i);
                }
            }
        }
 
        // The vector will be printed in reverse
        for (int i = v.size() - 1; i >= 0; i--)
            System.out.printf("%d ", v.get(i));
    }
 
    // Driver method
    public static void main(String args[])
    {
        System.out.println("The divisors of 100 are: ");
        printDivisors(100);
    }
}
Python3



# A O(sqrt(n)) java program that prints
# all divisors in sorted order
import math
 
# Method to print the divisors
def printDivisors(n) :
    list = []
     
    # List to store half of the divisors
    for i in range(1, int(math.sqrt(n) + 1)) :
         
        if (n % i == 0) :
             
            # Check if divisors are equal
            if (n / i == i) :
                print (i, end =" ")
            else :
                # Otherwise print both
                print (i, end =" ")
                list.append(int(n / i))
                 
    # The list will be printed in reverse   
    for i in list[::-1] :
        print (i, end =" ")
         
# Driver method
print ("The divisors of 100 are: ")
printDivisors(100)
 
# This code is contributed by Gitanjali
C#



// A O(sqrt(n)) C# program that
// prints all divisors in sorted order
using System;
 
class GFG {
 
    // method to print the divisors
    static void printDivisors(int n)
    {
        // Vector to store half
        // of the divisors
        int[] v = new int[n];
        int t = 0;
        for (int i = 1;
             i <= Math.Sqrt(n); i++) {
            if (n % i == 0) {
 
                // check if divisors are equal
                if (n / i == i)
                    Console.Write(i + " ");
                else {
                    Console.Write(i + " ");
 
                    // push the second divisor
                    // in the vector
                    v[t++] = n / i;
                }
            }
        }
 
        // The vector will be
        // printed in reverse
        for (int i = t - 1; i >= 0; i--)
            Console.Write(v[i] + " ");
    }
 
    // Driver Code
    public static void Main()
    {
        Console.Write("The divisors of 100 are: \n");
        printDivisors(100);
    }
}
 
// This code is contributed
// by ChitraNayal
PHP



<?php
// A O(sqrt(n)) program that
// prints all divisors in
// sorted order
 
// function to print
// the divisors
function printDivisors($n)
{
    // Vector to store half
    // of the divisors
    $v;
    $t = 0;
    for ($i = 1;
         $i <= (int)sqrt($n); $i++)
    {
        if ($n % $i == 0)
        {
            // check if divisors are equal
            if ((int)$n / $i == $i)
                echo $i . " ";
            else
            {
                echo $i . " ";
 
                // push the second
                // divisor in the vector
                $v[$t++] = (int)$n / $i;
            }
        }
    }
 
    // The vector will be
    // printed in reverse
    for ($i = count($v) - 1;
         $i >= 0; $i--)
        echo $v[$i] . " ";
}
 
// Driver code
echo "The divisors of 100 are: \n";
printDivisors(100);
 
// This code is contributed by mits
?>
Javascript



<script>
 
// A O(sqrt(n)) program that
// prints all divisors in
// sorted order
 
// function to print
// the divisors
function printDivisors(n)
{
    // Vector to store half
    // of the divisors
    let v = [];
    let t = 0;
    for (let i = 1;
         i <= parseInt(Math.sqrt(n)); i++)
    {
        if (n % i == 0)
        {
            // check if divisors are equal
            if (parseInt(n / i) == i)
                document.write(i + " ");
            else
            {
                document.write(i + " ");
 
                // push the second
                // divisor in the vector
                v[t++] = parseInt(n / i);
            }
        }
    }
 
    // The vector will be
    // printed in reverse
    for (let i = v.length - 1;
         i >= 0; i--){
         document.write(v[i] + " ");
    }
}
 
// Driver code
document.write("The divisors of 100 are: \n");
printDivisors(100);
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output : 
The divisors of 100 are: 
1 2 4 5 10 20 25 50 100 
Time Complexity : O(sqrt(n)) 
Auxiliary Space : O(sqrt(n))
A O(sqrt(n)) Time and O(1) Space Solution : 
C++



// A O(sqrt(n)) program that prints all divisors
// in sorted order
#include <bits/stdc++.h>
using namespace std;
 
// function to print the divisors
void printDivisors(int n)
{
    for (int i = 1; i*i < n; i++) {
        if (n % i == 0)
            printf("%d ", i);
    }
    for (int i = sqrt(n); i >= 1; i--) {
        if (n % i == 0)
            printf("%d ", n / i);
    }
}
 
/* Driver program to test above function */
int main()
{
    printf("The divisors of 100 are: \n");
    printDivisors(100);
    return 0;
}
Java



// A O(sqrt(n)) program that prints all
// divisors in sorted order
import java.lang.Math;
 
class GFG{
     
// Function to print the divisors
public static void printDivisors(int n)
{
    for(int i = 1; i * i < n; i++)
    {
        if (n % i == 0)
            System.out.print(i + " ");
    }
    for(int i = (int)Math.sqrt(n);
            i >= 1; i--)
    {
        if (n % i == 0)
            System.out.print(n / i + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    System.out.println("The divisors of 100 are: ");
     
    printDivisors(100);
}
}
 
// This code is contributed by divyeshrabadiya07
Python3



# A O(sqrt(n)) program that prints all divisors
# in sorted order
from math import *
 
# Function to print the divisors
def printDivisors (n):
 
    i = 1
    while (i * i < n):
        if (n % i == 0):
            print(i, end = " ")
 
        i += 1
 
    for i in range(int(sqrt(n)), 0, -1):
        if (n % i == 0):
            print(n // i, end = " ")
 
# Driver Code
print("The divisors of 100 are: ")
 
printDivisors(100)
 
# This code is contributed by himanshu77
C#



// A O(sqrt(n)) program that prints
// all divisors in sorted order
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
// Function to print the divisors
static void printDivisors(int n)
{
    for(int i = 1; i * i < n; i++)
    {
        if (n % i == 0)
            Console.Write(i + " ");
    }
    for(int i = (int)Math.Sqrt(n); i >= 1; i--)
    {
        if (n % i == 0)
            Console.Write(n / i + " ");
    }
} 
  
// Driver code  
public static void Main(string []arg)
{
    Console.Write("The divisors of 100 are: \n");
     
    printDivisors(100);
}
}
 
// This code is contributed by rutvik_56
Javascript



<script>
 
// A O(sqrt(n)) program that prints all divisors
// in sorted order
 
// function to print the divisors
function printDivisors(n)
{
    for (var i = 1; i*i < n; i++) {
        if (n % i == 0)
            document.write(i + " ");
    }
    for (var i = Math.sqrt(n); i >= 1; i--) {
        if (n % i == 0)
            document.write(" " + n / i);
    }
}
 
// Driver program to test above function
 
    document.write("The divisors of 100 are: \n");
    printDivisors(100);
     
// This code is contributed by simranarora5sos
 
</script>
Output: 
The divisors of 100 are: 
1 2 4 5 10 20 25 50 100 
Thanks to Mysterious Mind for suggesting above solution.
This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
My Personal Notes
arrow_drop_up
Next

Find all factors of a natural number | Set 1
Recommended Articles
Page :
Find sum of divisors of all the divisors of a natural number
21, Jan 17
Find sum of inverse of the divisors when sum of divisors and the number is given
28, Jun 19
Find all divisors of first N natural numbers
07, May 20
Sum of all proper divisors of a natural number
22, Oct 16
Check if a number has an odd count of odd divisors and even count of even divisors
20, Aug 20
Sum of all proper divisors of natural numbers in an array
11, Jul 17
Divisors of n-square that are not divisors of n
11, Nov 17
Check if count of even divisors of N is equal to count of odd divisors
29, May 20
Maximum possible prime divisors that can exist in numbers having exactly N divisors
07, May 20
Minimum value exceeding X whose count of divisors has different parity with count of divisors of X
18, Nov 20
Find all factors of a natural number | Set 1
08, Mar 16
Find the number of divisors of all numbers in the range [1, n]
12, Apr 19
Check if a number is divisible by all prime divisors of another number
18, Oct 17
Program to find the Nth natural number with exactly two bits set | Set 2
02, Dec 20
Find the largest good number in the divisors of given number N
22, Oct 18
Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors
12, Apr 19
Sum of all prime divisors of all the numbers in range L-R
14, Aug 18
Sum of all divisors from 1 to N | Set 3
06, Jul 20
Sum of all divisors from 1 to N | Set 2
20, Jun 20
Find largest sum of digits in all divisors of n
21, Dec 17
Queries to print all divisors of number
02, Dec 17
Generating all divisors of a number using its prime factorization
28, Oct 19
Sum of all second largest divisors after splitting a number into one or more parts
11, Jul 19
Count all perfect divisors of a number
14, Mar 17
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Easy





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox