Number of chocolates left after k iterations

Given a pile of chocolates and an integer ‘k’ i.e. the number of iterations, the task is to find the number of chocolates left after k iterations. 
Note: In every iteration, we can choose a pile with a maximum number of chocolates, after that square root of chocolate remains and rest is eaten. 
Examples:

Input: Chocolates = 100000000, Iterations = 3
Output: 10

Input: Chocolates = 200, Iterations = 2
Output: 4

Note: Output is printed after rounding off the value as in the 2nd example, the output will be around 3.76 approx.
Approach:
It is given that maximum no. of chocolates are selected so consider total pile since it will be maximum. 
Next, it is given that in the each iteration only square of chocolates are left so that by considering the mathematics equation of 

(((number)n)n)...n for k times = (number)nk

Since here k times the square root is performed so the (1/2)^k is powered with the N.
Consider the example of 100000000 chocolates and no. of iterations is 3 then it will be as 

(((100000000)1/2)1/2)1/2 = (100000000)(1/2)3 = 10

Below is the required formula to find the remaining chocolates:

round(pow(n, (1.0/pow(2, k))))

C++

// C++ program to find remaining
// chocolates after k iterations
#include <bits/stdc++.h>
using namespace std;

// Function to find the chocolates left
int results(int n, int k)
{
    return round(pow(n, (1.0 / pow(2, k))));
}

// Driver code
int main()
{
    int k = 3, n = 100000000;

    cout << "Chocolates left after " << k
         << " iterations are " << results(n, k);

    return 0;
}

Java

// Java program to find remaining
// chocolates after k iterations
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
// Function to find the
// chocolates left
static int results(int n, int k)
{
    return (int)Math.round(Math.pow(n,
           (1.0 / Math.pow(2.0, k))));
}

// Driver code
public static void main(String args[])
{
    int k = 3, n = 100000000;

    System.out.print("Chocolates left after " + k +
               " iterations are " + results(n, k));
}
}

// This code is contributed by Subhadeep

C#

// C# program to find remaining
// chocolates after k iterations
using System;

class GFG
{
// Function to find the
// chocolates left
static int results(int n, int k)
{
    return (int)Math.Round(Math.Pow(n,
           (1.0 / Math.Pow(2.0, k))));
}

// Driver code
public static void Main()
{
    int k = 3, n = 100000000;

    Console.Write("Chocolates left after " +
                    k + " iterations are " +
                             results(n, k));
}
}

// This code is contributed
// by ChitraNayal

Python

# Python program to find
# remaining chocolates
# after k iterations

# Function to find the
# chocolates left
def results(n, k):

    return round(pow(n, (1.0 /
                 pow(2, k))))

# Driver code
k = 3
n = 100000000

print("Chocolates left after"),
print(k),
print("iterations are"),
print(int(results(n, k)))

# This code is contributed
# by Shivi_Aggarwal

PHP

<?php
// PHP program to find remaining
// chocolates after k iterations

// Function to find
// the chocolates left
function results($n, $k)
{
    return round(pow($n, (1.0 /
                 pow(2, $k))));
}

// Driver code
$k = 3;
$n = 100000000;
echo ("Chocolates left after ");
echo ($k);

echo (" iterations are ");
echo (results($n, $k));

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
// javascript program to find remaining
// chocolates after k iterations    // Function to find the
    // chocolates left
    function results(n , k) {
        return parseInt( Math.round(Math.pow(n,
        (1.0 / Math.pow(2.0, k)))));
    }

    // Driver code

        var k = 3, n = 100000000;

        document.write("Chocolates left after " +
        k + " iterations are " + results(n, k));

// This code contributed by aashish1995
</script>

Output:

Chocolates left after 3 iterations are 10

Time Complexity: O(log(n))
Auxiliary Space: O(1)