Cilindro circular recto más grande dentro de un cubo

Dado un cubo de longitud de lado a . La tarea es encontrar el volumen del cilindro circular recto más grande que se puede inscribir dentro de él.
Ejemplos: 
 

Input :  a = 4
Output : 50.24

Input : a = 5
Output : 98.125

Enfoque : 
Sea: 
 

• La altura del cilindro es h .
• Radio del cilindro be r .

Del diagrama es claro que: 
 

• La altura del cilindro = lado del cubo
• Radio del cilindro = lado del cubo/2

Asi que, 
 

h = a
r = a/2

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ Program to find the biggest right
// circular cylinder that can be fit within a cube
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the biggest right circular cylinder
float findVolume(float a)
{
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cylinder
    float r = a / 2;
 
    // height of right circular cylinder
    float h = a;
 
    // volume of right circular cylinder
    float V = 3.14 * pow(r, 2) * h;
 
    return V;
}
 
// Driver code
int main()
{
    float a = 5;
 
    cout << findVolume(a) << endl;
 
    return 0;
}

// Java Program to find the biggest right
// circular cylinder that can be fit within a cube
 
import java.io.*;
 
class GFG {
   
 
// Function to find the biggest right circular cylinder
 static float findVolume(float a)
{
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cylinder
    float r = a / 2;
 
    // height of right circular cylinder
    float h = a;
 
    // volume of right circular cylinder
    float V = (float)(3.14 * Math.pow(r, 2) * h);
 
    return V;
}
 
// Driver code
 
 
    public static void main (String[] args) {
            float a = 5;
 
    System.out.print(findVolume(a));
    }
}
// This code is contributed by anuj_67..

# Python3 Program to find the biggest
# right circular cylinder that can be
# fit within a cube
 
# Function to find the biggest right
# circular cylinder
def findVolume(a) :
 
    # side cannot be negative
    if (a < 0) :
        return -1
 
    # radius of right circular cylinder
    r = a / 2
 
    # height of right circular cylinder
    h = a
 
    # volume of right circular cylinder
    V = 3.14 * pow(r, 2) * h
 
    return V
 
# Driver code
if __name__ == "__main__" :
 
    a = 5
 
    print(findVolume(a))
 
# This code is contributed by Ryuga

// C# Program to find the biggest right
// circular cylinder that can be fit within a cube
 
using System;
class GFG {
 
 
// Function to find the biggest right circular cylinder
static float findVolume(float a)
{
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cylinder
    float r = a / 2;
 
    // height of right circular cylinder
    float h = a;
 
    // volume of right circular cylinder
    float V = (float)(3.14 * Math.Pow(r, 2) * h);
 
    return V;
}
 
// Driver code
 
 
    public static void Main () {
            float a = 5;
 
   Console.WriteLine(findVolume(a));
    }
}
// This code is contributed by anuj_67..

<?php
// PHP Program to find the biggest
// right circular cylinder that can
// be fit within a cube
 
// Function to find the biggest
// right circular cylinder
function findVolume($a)
{
    // side cannot be negative
    if ($a < 0)
        return -1;
 
    // radius of right circular cylinder
    $r = $a / 2;
 
    // height of right circular cylinder
    $h = $a;
 
    // volume of right circular cylinder
    $V = 3.14 * pow($r, 2) * $h;
 
    return $V;
}
 
// Driver code
$a = 5;
 
echo findVolume($a) . "\n";
 
// This code is contributed
// by Akanksha Rai

<script>
// Javascript Program to find the biggest right
// circular cylinder that can be fit within a cube
 
// Function to find the biggest right circular cylinder
 function findVolume(a)
{
 
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cylinder
    var r = a / 2;
 
    // height of right circular cylinder
    var h = a;
 
    // volume of right circular cylinder
    var V = (3.14 * Math.pow(r, 2) * h);
 
    return V;
}
 
// Driver code
var a = 5;
document.write(findVolume(a));
 
// This code is contributed by Princi Singh
</script>

98.125

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)