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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
Note that this problem is different from finding all prime factors.
A Naive Solution would be to iterate all the numbers from 1 to n, checking if that number divides n and printing it. Below is a program for the same:
C++
// C++ implementation of Naive method to print all// divisors#include <bits/stdc++.h>// function to print the divisorsvoid printDivisors(int n){ for (int i=1;i<=n;i++) if (n%i==0) printf("%d ",i);}/* Driver program to test above function */int main(){ printf("The divisors of 100 are: \n"); printDivisors(100); return 0;} |
Java
// Java implementation of Naive method to print all// divisorsclass Test{ // method to print the divisors static void printDivisors(int n) { for (int i=1;i<=n;i++) if (n%i==0) System.out.print(i+" "); } // Driver method public static void main(String args[]) { System.out.println("The divisors of 100 are: "); printDivisors(100);; }} |
Python
# Python implementation of Naive method# to print all divisors# method to print the divisorsdef printDivisors(n) : i = 1 while i <= n : if (n % i==0) : print i, i = i + 1 # Driver methodprint "The divisors of 100 are: "printDivisors(100)#This code is contributed by Nikita Tiwari. |
C#
// C# implementation of Naive method// to print all divisorsusing System;class GFG { // method to print the divisors static void printDivisors(int n) { for (int i = 1; i <= n; i++) if (n % i == 0) Console.Write( i + " "); } // Driver method public static void Main() { Console.Write("The divisors of", " 100 are: "); printDivisors(100);; }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP implementation of Naive// method to print all divisors// function to print the divisorsfunction printDivisors($n){ for ($i = 1; $i <= $n; $i++) if ($n % $i == 0) echo $i," ";}// Driver Codeecho "The divisors of 100 are:\n";printDivisors(100);// This code is contributed by ajit?> |
Javascript
<script>// Javascript implementation of Naive method to print all// divisors// function to print the divisorsfunction printDivisors(n){ for (i=1;i<=n;i++) if (n%i==0) document.write(i+ " ");}/* Driver program to test above function */ document.write("The divisors of 100 are:" + "<br>"); printDivisors(100); // This code is contributed by Mayank Tyagi </script> |
Output:
The divisors of 100 are: 1 2 4 5 10 20 25 50 100
Time Complexity : O(n)
Auxiliary Space : O(1)
Can we improve the above solution?
If we look carefully, all the divisors are present in pairs. For example if n = 100, then the various pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)
Using this fact we could speed up our program significantly.
We, however, have to be careful if there are two equal divisors as in the case of (10, 10). In such case, we’d print only one of them.
Below is an implementation for the same:
C++
// A Better (than Naive) Solution to find all divisiors#include <bits/stdc++.h>// Function to print the divisorsvoid printDivisors(int n){ // Note that this loop runs till square root for (int i=1; i<=sqrt(n); i++) { if (n%i == 0) { // If divisors are equal, print only one if (n/i == i) printf("%d ", i); else // Otherwise print both printf("%d %d ", i, n/i); } }}/* Driver program to test above function */int main(){ printf("The divisors of 100 are: \n"); printDivisors(100); return 0;} |
Java
// A Better (than Naive) Solution to find all divisorsclass Test{ // method to print the divisors static void printDivisors(int n) { // Note that this loop runs till square root for (int i=1; i<=Math.sqrt(n); i++) { if (n%i==0) { // If divisors are equal, print only one if (n/i == i) System.out.print(" "+ i); else // Otherwise print both System.out.print(i+" " + n/i + " " ); } } } // Driver method public static void main(String args[]) { System.out.println("The divisors of 100 are: "); printDivisors(100);; }} |
Python
# A Better (than Naive) Solution to find all divisiorsimport math# method to print the divisorsdef printDivisors(n) : # Note that this loop runs till square root i = 1 while i <= math.sqrt(n): if (n % i == 0) : # If divisors are equal, print only one if (n / i == i) : print i, else : # Otherwise print both print i , n/i, i = i + 1# Driver methodprint "The divisors of 100 are: "printDivisors(100)#This code is contributed by Nikita Tiwari. |
C#
// A Better (than Naive) Solution to// find all divisorsusing System;class GFG { // method to print the divisors static void printDivisors(int n) { // Note that this loop runs // till square root for (int i = 1; i <= Math.Sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // print only one if (n / i == i) Console.Write(i + " "); // Otherwise print both else Console.Write(i + " " + n / i + " "); } } } // Driver method public static void Main() { Console.Write("The divisors of " + "100 are: \n"); printDivisors(100); }}// This code is contributed by Smitha |
PHP
<?php// A Better (than Naive) Solution// to find all divisiors// Function to print the divisorsfunction printDivisors($n){ // Note that this loop // runs till square root for ($i = 1; $i <= sqrt($n); $i++) { if ($n%$i == 0) { // If divisors are equal, // print only one if ($n / $i == $i) echo $i," "; // Otherwise print both else echo $i," ", $n/$i," "; } }} // Driver Code echo "The divisors of 100 are: \n"; printDivisors(100); // This code is contributed by anuj_67.?> |
Javascript
<script>// A Better (than Naive) Solution to find all divisiors// Function to print the divisorsfunction printDivisors(n){ // Note that this loop runs till square root for(let i = 1; i <= Math.sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, print only one if (parseInt(n / i, 10) == i) document.write(i); // Otherwise print both else document.write(i + " " + parseInt(n / i, 10) + " "); } }}// Driver codedocument.write("The divisors of 100 are: </br>");printDivisors(100);// This code is contributed by divyesh072019</script> |
Output:
The divisors of 100 are: 1 100 2 50 4 25 5 20 10
Time Complexity: O(sqrt(n))
Auxiliary Space : O(1)
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 the brute-force technique. Please refer below for an O(sqrt(n)) time solution that prints divisors in sorted order.
Find all divisors of a natural number | Set 2
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
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