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，任务是找出 k 次迭代后剩下的巧克力数量。 注:在每次迭代中，我们可以选择巧克力数量最多的一堆，之后剩下的平方根巧克力就剩下了，剩下的吃掉。 示例:
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，如第二个例子，输出大约在 3.76 左右。 方法: 假设选择了最大数量的巧克力，那么考虑总堆数，因为它将是最大的。 接下来，假设在每次迭代中只剩下方形巧克力，因此通过考虑 的数学方程
(((number)n)n)...n for k times = (number)nk
因为这里执行 k 次平方根，所以(1/2) k 由 n 供电考虑 100000000 块巧克力的例子，迭代次数是 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配方:
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 语言(一种计算机语言，尤用于创建网站)
// 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
计算机编程语言
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<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
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