k次迭代后剩下的巧克力数量

给定一堆巧克力和一个整数“k”，即迭代次数，任务是找到 k 次迭代后剩下的巧克力数量。
注意：在每次迭代中，我们可以选择具有最大巧克力数量的一堆巧克力，然后剩下巧克力的平方根并吃掉剩下的巧克力。
例子：

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

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

注意：输出在四舍五入后打印，如第二个示例，输出将约为3.76 左右。
方法：
给出了最大数量。选择巧克力，因此请考虑总堆，因为它将是最大的。
接下来，假设在每次迭代中只剩下巧克力的正方形，以便通过考虑数学方程

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

因为这里执行了 k 次平方根，所以 (1/2) ^k是用 N 供电的。
考虑 100000000 块巧克力和没有的例子。迭代次数为 3 那么它将是

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

以下是找到剩余巧克力所需的公式：

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

C++

// C++ program to find remaining
// chocolates after k iterations
#include 
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

Javascript

// 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

输出：

Chocolates left after 3 iterations are 10