Programa para contar el número de Cuadrados y Cubos distintos hasta N

Posted on by Rudeus Greyrat

Dado un número N, la tarea es encontrar el número de cuadrados perfectos y cubos perfectos desde 1 hasta un número entero dado, N (ambos inclusive).
Nota: Los números que son cuadrados perfectos y cubos perfectos deben contarse una vez.
Ejemplos: 

Entrada: N = 70 
Salida: 10 
Explicación: Números que son cuadrados perfectos o cubos perfectos 
1, 4, 8, 9, 16, 25, 27, 36, 49, 64

Entrada: N = 25 
Salida: 6 
Explicación:  Números que son cuadrados perfectos o cubos perfectos 
1, 4, 8, 9, 16, 25  

Enfoque ingenuo: 

Comprueba todos los números del 1 al N, si es un cuadrado o un cubo.

A continuación se muestra la implementación del enfoque anterior:  

C++

// C++ implementation of above approach
#include <iostream>
#include <math.h> // For sqrt() and cbrt()
using namespace std;
 
// Function to check if the
// number is a perfect square
bool isSquare(int num)
{
    int root = sqrt(num);
    return (root * root) == num;
}
 
// Function to check if the
// number is a perfect cube
bool isCube(int num)
{
    int root = cbrt(num);
    return (root * root * root) == num;
}
 
// Function to count the number
// of perfect squares and cubes
int countSC(int N)
{
    int count = 0;
    for (int i = 1; i <= N; i++) {
 
        // If a number is perfect square,
        if (isSquare(i))
            count++;
 
        // Else if the number is cube or not
        else if (isCube(i))
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int N = 20;
 
    cout << "Number of squares and cubes is " << countSC(N);
    return 0;
}

Java

// Java implementation of
// above approach
class GFG {
 
    // Function to check if the
    // number is a perfect square
    static boolean isSquare(int num)
    {
        int root = (int)Math.sqrt(num);
        return (root * root) == num;
    }
 
    // Function to check if the
    // number is a perfect cube
    static boolean isCube(int num)
    {
        int root = (int)Math.cbrt(num);
        return (root * root * root) == num;
    }
 
    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {
 
            // If a number is perfect
            // square,
            if (isSquare(i))
                count++;
 
            // Else if the number is
            // cube or not
            else if (isCube(i))
                count++;
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 20;
 
        System.out.println("Number of squares "
                           + "and cubes is " + countSC(N));
    }
}
 
// This code is contributed
// by ChitraNayal

Python3

# Python 3 implementation of
# above approach
 
# Function to check if the
# number is a perfect square
 
 
def isSquare(num):
    root = int(num ** (1 / 2))
 
    return (root * root) == num
 
# Function to check if the
# number is a perfect cube
 
 
def isCube(num):
    root = int(num ** (1 / 3))
    return (root * root * root) == num
 
# Function to count the number
# of perfect squares and cubes
 
 
def countSC(N):
    count = 0
    for i in range(1, N + 1):
 
        # If a number is perfect square,
        if isSquare(i):
            count += 1
 
        # Else if the number is cube
        elif isCube(i):
            count += 1
 
    return count
 
 
# Driver code
if __name__ == "__main__":
 
    N = 20
 
    print("Number of squares and cubes is ",
          countSC(N))
 
# This code is contributed by ANKITRAI1

C#

// C# implementation of
// above approach
using System;
 
class GFG {
 
    // Function to check if the
    // number is a perfect square
    static bool isSquare(int num)
    {
        int root = (int)Math.Sqrt(num);
        return (root * root) == num;
    }
 
    // Function to check if the
    // number is a perfect cube
    static bool isCube(int num)
    {
        int root = (int)Math.Pow(num, (1.0 / 3.0));
        return (root * root * root) == num;
    }
 
    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {
 
            // If a number is perfect
            // square,
            if (isSquare(i))
                count++;
 
            // Else if the number is
            // cube or not
            else if (isCube(i))
                count++;
        }
 
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 20;
 
        Console.Write("Number of squares and "
                      + "cubes is " + countSC(N));
    }
}
 
// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP implementation of above approach
 
// Function to check if the
// number is a perfect square
function isSquare($num)
{
    $root = (int)sqrt($num);
    return ($root * $root) == $num;
}
 
// Function to check if the
// number is a perfect cube
function isCube($num)
{
    $root = (int)pow($num, 1 / 3);
    return ($root * $root *
            $root) == $num;
}
 
// Function to count the number
// of perfect squares and cubes
function countSC($N)
{
    $count = 0;
    for ($i = 1; $i <= $N; $i++)
    {
 
        // If a number is
        // perfect square,
        if (isSquare($i))
            $count++;
 
        // Else if the number
        // is cube or not
        else if (isCube($i))
            $count++;
    }
 
    return $count;
}
 
// Driver code
$N = 20;
 
echo "Number of squares and " .
     "cubes is " . countSC($N);
 
// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
 
    // Javascript implementation of above approach
     
    // Function to check if the
    // number is a perfect square
    function isSquare(num)
    {
        let root = parseInt(Math.sqrt(num), 10);
        return (root * root) == num;
    }
 
    // Function to check if the
    // number is a perfect cube
    function isCube(num)
    {
        let root = parseInt(Math.cbrt(num), 10);
        return (root * root * root) == num;
    }
     
    // Function to count the number
    // of perfect squares and cubes
    function countSC(N)
    {
        let count = 0;
        for (let i = 1; i <= N; i++) {
 
            // If a number is perfect square,
            if (isSquare(i))
                count++;
 
            // Else if the number is cube or not
            else if (isCube(i))
                count++;
        }
 
        return count;
    }
 
    let N = 20;
   
    document.write("Number of squares and cubes is " + countSC(N));
 
</script>
Producción

Number of squares and cubes is 5

Complejidad de tiempo: O(N)
Complejidad de espacio: O(1) 

Enfoque eficiente: 

  • El número de cuadrados de 1 a N es floor(sqrt(N)) .
  • El número de cubos de 1 a N es floor(cbrt(N)) .
  • Elimine los números que son tanto cuadrados como cúbicos (como 1, 64….) restándole piso(sqrt(cbrt(N))) .

A continuación se muestra la implementación del enfoque anterior:
 

C++

// C++ implementation of above approach
#include <iostream>
#include <math.h> // For sqrt() and cbrt()
using namespace std;
 
// Function to count the number
// of perfect squares and cubes
int countSC(int N)
{
    int res = (int)sqrt(N) + (int)cbrt(N)
              - (int)(sqrt(cbrt(N)));
 
    return res;
}
 
// Driver code
int main()
{
    int N = 20;
 
    cout << "Number of squares and cubes is " << countSC(N);
    return 0;
}

Java

// Java implementation of
// above approach
class GFG {
 
    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int res = (int)Math.sqrt(N) + (int)Math.cbrt(N)
                  - (int)(Math.sqrt(Math.cbrt(N)));
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 20;
 
        System.out.println("Number of squares "
                           + "and cubes is " + countSC(N));
    }
}
 
// This code is contributed
// by ChitraNayal

Python3

# Python implementation of
# above approach
import math  # for sqrt()
 
# Function to count the number
# of perfect squares and cubes
 
 
def cr(N):
    if pow(round(N ** (1 / 3)), 3) == N:
        return round(N ** (1 / 3))
 
    return int(N ** (1 / 3))
 
 
def countSC(N):
    res = (int(math.sqrt(N)) +
           int(cr(N)) -
           int(math.sqrt(cr(N))))
 
    return res
 
 
# Driver code
N = 20
print("Number of squares and cubes is",
      countSC(N))
 
# This code is contributed
# by vaibhav29498

C#

// C# implementation of
// above approach
using System;
public class GFG {
 
    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int res = (int)(Math.Sqrt(N)
                        + Math.Ceiling(
                            Math.Pow(N, (double)1 / 3))
                        - (Math.Sqrt(Math.Ceiling(
                            Math.Pow(N, (double)1 / 3)))));
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 20;
 
        Console.Write("Number of squares "
                      + "and cubes is " + countSC(N));
    }
}
 
/*This code is contributed by 29AjayKumar*/

PHP

<?php
// PHP implementation of above approach
 
// Function to count the number
// of perfect squares and cubes
function countSC($N)
{
    $res = sqrt($N) + pow($N, 1 / 3) -
                (sqrt(pow($N, 1 / 3)));
 
    return floor($res);
}
 
// Driver code
$N = 20;
 
echo "Number of squares and cubes is " ,
                            countSC($N);
 
// This code is contributed
// by Shashank
?>

Javascript

<script>
 
    // Javascript implementation of above approach
     
    // Function to count the number
    // of perfect squares and cubes
    function countSC(N)
    {
        let res = parseInt(Math.sqrt(N), 10) + parseInt(Math.cbrt(N), 10) - parseInt(Math.sqrt(parseInt(Math.cbrt(N), 10)), 10);
 
        return res;
    }
 
    let N = 20;
  
    document.write("Number of squares and cubes is " + countSC(N));
     
</script>
Producción

Number of squares and cubes is 5

Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por BabanGain y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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