Find the Nth Mosaic number

Given an integer N, the task is to find the Nth Mosaic number. A Mosaic number can be expressed as follows:
If N = Aa * Bb * Cc where A, B, C.. are the prime factors of N then the Nth Mosaic number will be A * a * B * b * C * c ….

Examples:

Input: N = 8
Output: 6
8 can be expressed as 23.
So, the 8th Mosaic number will be 2 * 3 = 6

Input: N = 36
Output: 24
36 can be expressed as 22 * 32.
2 * 2 * 3 * 2 = 24

Approach: We have to find all the prime factors and also the powers of the factors in the number by dividing the number by the factor until the factor divides the number. The Nth Mosaic number will then be the product of the found prime factors and their powers.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the nth mosaic number
int mosaic(int n)
{
    int i, ans = 1;
  
    // Iterate from 2 to the number
    for (i = 2; i <= n; i++) {
  
        // If i is the factor of n
        if (n % i == 0 && n > 0) {
            int count = 0;
  
            // Find the count where i^count
            // is a factor of n
            while (n % i == 0) {
  
                // Divide the number by i
                n /= i;
  
                // Increase the count
                count++;
            }
  
            // Multiply the answer with
            // count and i
            ans *= count * i;
        }
    }
  
    // Return the answer
    return ans;
}
  
// Driver code
int main()
{
    int n = 36;
    cout << mosaic(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the nth mosaic number
static int mosaic(int n)
{
    int i, ans = 1;
  
    // Iterate from 2 to the number
    for (i = 2; i <= n; i++) 
    {
  
        // If i is the factor of n
        if (n % i == 0 && n > 0)
        {
            int count = 0;
  
            // Find the count where i^count
            // is a factor of n
            while (n % i == 0)
            {
  
                // Divide the number by i
                n /= i;
  
                // Increase the count
                count++;
            }
  
            // Multiply the answer with
            // count and i
            ans *= count * i;
        }
    }
  
    // Return the answer
    return ans;
}
  
// Driver code
public static void main (String[] args)
{
      
    int n = 36;
    System.out.println (mosaic(n));
}
}
  
// This code is contributed by jit_t.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function to return the nth mosaic number
def mosaic(n):
  
    i=0
    ans = 1
  
    # Iterate from 2 to the number
    for i in range(2,n+1):
  
        # If i is the factor of n
        if (n % i == 0 and n > 0):
            count = 0
  
            # Find the count where i^count
            # is a factor of n
            while (n % i == 0):
  
                # Divide the number by i
                n //= i
  
                # Increase the count
                count+=1
              
  
            # Multiply the answer with
            # count and i
            ans *= count * i
          
  
    # Return the answer
    return ans
  
# Driver code
  
n = 36
print(mosaic(n))
  
# This code is contributed by mohit kumar 29

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the nth mosaic number
static int mosaic(int n)
{
    int i, ans = 1;
  
    // Iterate from 2 to the number
    for (i = 2; i <= n; i++) 
    {
  
        // If i is the factor of n
        if (n % i == 0 && n > 0)
        {
            int count = 0;
  
            // Find the count where i^count
            // is a factor of n
            while (n % i == 0)
            {
  
                // Divide the number by i
                n /= i;
  
                // Increase the count
                count++;
            }
  
            // Multiply the answer with
            // count and i
            ans *= count * i;
        }
    }
  
    // Return the answer
    return ans;
}
  
// Driver code
static public void Main ()
{
    int n = 36;
    Console.WriteLine(mosaic(n));
}
}
  
// This code is contributed by ajit..

chevron_right


Output:

24


My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29, jit_t