查找第N个镶嵌号

给定整数N ，任务是找到第N^个镶嵌数。镶嵌数可以表示为：
如果N = A ^a * B ^b * C ^c … ，其中A ， B ， C ..是N的素数，则第N个镶嵌数将为A * a * B * b * C * c… 。

例子：

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

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

方法：通过将数字除以因子，直到因子除以数字，我们必须找到数字中的所有素因子以及因子的幂。然后，第N^个镶嵌数将是找到的素数及其功效的乘积。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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;
}

Java

// 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.

Python3

# 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

C#

// 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..

Javascript

 
// Javascript implementation of the approach
 
// Function to return the nth mosaic number
function mosaic(n)
{
    let 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)
        {
            let count = 0;
 
            // Find the count where i^count
            // is a factor of n
            while (n % i == 0)
            {
                 
                // Divide the number by i
                n = parseInt(n / i, 10);
 
                // Increase the count
                count++;
            }
 
            // Multiply the answer with
            // count and i
            ans *= count * i;
        }
    }
 
    // Return the answer
    return ans;
}
 
// Driver code
let n = 36;
document.write(mosaic(n));
 
// This code is contributed by mukesh07
 

输出：

24

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。