Number of divisors of a given number N which are divisible by K
Given a number N and a number K. The task is to find the number of divisors of N which are divisible by K. Here K is a number always less than or equal to √(N)
Examples:
Input: N = 12, K = 3 Output: 3 Input: N = 8, K = 2 Output: 3
Simple Approach: A simple approach is to check all the numbers from 1 to N and check whether any number is a divisor of N and is divisible by K. Count such numbers less than N which satisfies both the conditions.
Below is the implementation of the above approach:
C++
// C++ program to count number of divisors// of N which are divisible by K #include <iostream>using namespace std; // Function to count number of divisors// of N which are divisible by Kint countDivisors(int n, int k){ // Variable to store // count of divisors int count = 0, i; // Traverse from 1 to n for (i = 1; i <= n; i++) { // increase the count if both // the conditions are satisfied if (n % i == 0 && i % k == 0) { count++; } } return count;} // Driver codeint main(){ int n = 12, k = 3; cout << countDivisors(n, k); return 0;} |
Java
// Java program to count number of divisors// of N which are divisible by K import java.io.*; class GFG { // Function to count number of divisors// of N which are divisible by K static int countDivisors(int n, int k){ // Variable to store // count of divisors int count = 0, i; // Traverse from 1 to n for (i = 1; i <= n; i++) { // increase the count if both // the conditions are satisfied if (n % i == 0 && i % k == 0) { count++; } } return count;} // Driver code public static void main (String[] args) { int n = 12, k = 3; System.out.println(countDivisors(n, k)); }}// This code is contributed by shashank.. |
Python3
# Python program to count number # of divisors of N which are # divisible by K # Function to count number of divisors # of N which are divisible by K def countDivisors(n, k) : # Variable to store # count of divisors count = 0 # Traverse from 1 to n for i in range(1, n + 1) : # increase the count if both # the conditions are satisfied if (n % i == 0 and i % k == 0) : count += 1 return count # Driver code if __name__ == "__main__" : n, k = 12, 3 print(countDivisors(n, k)) # This code is contributed by ANKITRAI1 |
C#
// C# program to count number // of divisors of N which are// divisible by Kusing System; class GFG{ // Function to count number // of divisors of N which// are divisible by Kstatic int countDivisors(int n, int k){ // Variable to store // count of divisors int count = 0, i; // Traverse from 1 to n for (i = 1; i <= n; i++) { // increase the count if both // the conditions are satisfied if (n % i == 0 && i % k == 0) { count++; } } return count;} // Driver codepublic static void Main () { int n = 12, k = 3; Console.WriteLine(countDivisors(n, k));}} // This code is contributed by Shashank |
PHP
<?php// PHP program to count number // of divisors of N which are // divisible by K // Function to count number of divisors// of N which are divisible by Kfunction countDivisors($n, $k){ // Variable to store // count of divisors $count = 0; // Traverse from 1 to n for ($i = 1; $i <= $n; $i++) { // increase the count if both // the conditions are satisfied if ($n % $i == 0 && $i % $k == 0) { $count++; } } return $count;} // Driver code$n = 12; $k = 3; echo countDivisors($n, $k); // This code is contributed // by Akanksha Rai(Abby_akku) |
Output
3
Time Complexity : O(N)
Efficient Approach: The idea is to run a loop from 1 to < √(N) and check whether the number is a divisor of N and is divisible by K and we will also check whether ( N/i ) is divisible by K or not. As (N/i) will also be a factor of N if i is a factor of N.
Below is the implementation of the above approach:
C++
// C++ program to count number of divisors// of N which are divisible by K#include <bits/stdc++.h>using namespace std; // Function to count number of divisors// of N which are divisible by Kint countDivisors(int n, int k){ // integer to count the divisors int count = 0, i; // Traverse from 1 to sqrt(N) for (i = 1; i <= sqrt(n); i++) { // Check if i is a factor if (n % i == 0) { // increase the count if i // is divisible by k if (i % k == 0) { count++; } // (n/i) is also a factor // check whether it is divisible by k if ((n / i) % k == 0) { count++; } } } i--; // If the number is a perfect square // and it is divisible by k if ((i * i == n) && (i % k == 0)) { count--; } return count;} // Driver codeint main(){ int n = 16, k = 4; // Function Call cout << countDivisors(n, k); return 0;} |
Java
// Java program to count number of divisors // of N which are divisible by Kimport java.io.*; class GFG { // Function to count number of divisors // of N which are divisible by K static int countDivisors(int n, int k) { // integer to count the divisors int count = 0, i; // Traverse from 1 to sqrt(N) for (i = 1; i <= Math.sqrt(n); i++) { // Check if i is a factor if (n % i == 0) { // increase the count if i // is divisible by k if (i % k == 0) { count++; } // (n/i) is also a factor // check whether it is divisible by k if ((n / i) % k == 0) { count++; } } } i--; // If the number is a perfect square // and it is divisible by k if ((i * i == n) && (i % k == 0)) { count--; } return count; } // Driver code public static void main (String[] args) { int n = 16, k = 4; System.out.println( countDivisors(n, k)); }}//This Code is Contributed by akt_mit |
Python 3
# Python 3 program to count number of # divisors of N which are divisible by Kimport math # Function to count number of divisors# of N which are divisible by Kdef countDivisors(n, k): # integer to count the divisors count = 0 # Traverse from 1 to sqrt(N) for i in range(1, int(math.sqrt(n)) + 1): # Check if i is a factor if (n % i == 0) : # increase the count if i # is divisible by k if (i % k == 0) : count += 1 # (n/i) is also a factor check # whether it is divisible by k if ((n // i) % k == 0) : count += 1 # If the number is a perfect square # and it is divisible by k # if i is sqrt reduce by 1 if ((i * i == n) and (i % k == 0)) : count -= 1 return count # Driver codeif __name__ == "__main__": n = 16 k = 4 print(countDivisors(n, k)) # This code is contributed # by ChitraNayal |
C#
// C# program to count number of divisors // of N which are divisible by Kusing System; class GFG{ // Function to count number of divisors // of N which are divisible by K static int countDivisors(int n, int k) { // integer to count the divisors int count = 0, i; // Traverse from 1 to sqrt(N) for (i = 1; i <= Math.Sqrt(n); i++) { // Check if i is a factor if (n % i == 0) { // increase the count if i // is divisible by k if (i % k == 0) { count++; } // (n/i) is also a factor check // whether it is divisible by k if ((n / i) % k == 0) { count++; } } } i--; // If the number is a perfect square // and it is divisible by k if ((i * i == n) && (i % k == 0)) { count--; } return count; } // Driver code static public void Main (){ int n = 16, k = 4; Console.WriteLine( countDivisors(n, k)); }} // This code is contributed by ajit |
PHP
<?php// PHP program to count number // of divisors of N which are // divisible by K // Function to count number // of divisors of N which // are divisible by Kfunction countDivisors($n, $k){ // integer to count the divisors $count = 0; // Traverse from 1 to sqrt(N) for ($i = 1; $i <= sqrt($n); $i++) { // Check if i is a factor if ($n % $i == 0) { // increase the count if i // is divisible by k if ($i % $k == 0) { $count++; } // (n/i) is also a factor // check whether it is // divisible by k if (($n / $i) % $k == 0) { $count++; } } } i--; // If the number is a perfect // square and it is divisible by k if (($i * $i == $n) && ($i % $k == 0)) { $count--; } return $count;} // Driver code$n = 16;$k = 4; echo (countDivisors($n, $k)); // This code is contributed// by Shivi_Aggarwal?> |
Output
3
Time Complexity :O(√(n))
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.
