Cara menggunakan cubes python

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Given a number N, the task is to find No. of perfect Squares and perfect Cubes from 1 to a given integer, N ( Both Inclusive).
Note: Numbers which are both perfect Square and perfect Cube should be counted once.
Examples: 

Input: N = 70 
Output: 10 
Explanation: Numbers that are perfect Square or perfect Cube 
1, 4, 8, 9, 16, 25, 27, 36, 49, 64

Input: N = 25 
Output: 6 
Explanation: Numbers that are perfect Square or perfect Cube 
1, 4, 8, 9, 16, 25  

Naive Approach: 

Check all numbers from 1 to N, if it is a square or cube.

Below is the implementation of above approach:  

C++

#include <iostream>

#include <math.h> // For sqrt() and cbrt()

using namespace std;

bool isSquare(int num)

{

    int root = sqrt(num);

    return (root * root) == num;

}

bool isCube(int num)

{

    int root = cbrt(num);

    return (root * root * root) == num;

}

int countSC(int N)

{

    int count = 0;

    for (int i = 1; i <= N; i++) {

        if (isSquare(i))

            count++;

        else if (isCube(i))

            count++;

    }

    return count;

}

int main()

{

    int N = 20;

    cout << "Number of squares and cubes is " << countSC(N);

    return 0;

}

Java

class GFG {

    static boolean isSquare(int num)

    {

        int root = (int)Math.sqrt(num);

        return (root * root) == num;

    }

    static boolean isCube(int num)

    {

        int root = (int)Math.cbrt(num);

        return (root * root * root) == num;

    }

    static int countSC(int N)

    {

        int count = 0;

        for (int i = 1; i <= N; i++) {

            if (isSquare(i))

                count++;

            else if (isCube(i))

                count++;

        }

        return count;

    }

    public static void main(String[] args)

    {

        int N = 20;

        System.out.println("Number of squares "

                           + "and cubes is " + countSC(N));

    }

}

Python3

def isSquare(num):

    root = int(num ** (1 / 2))

    return (root * root) == num

def isCube(num):

    root = int(num ** (1 / 3))

    return (root * root * root) == num

def countSC(N):

    count = 0

    for i in range(1, N + 1):

        if isSquare(i):

            count += 1

        elif isCube(i):

            count += 1

    return count

if __name__ == "__main__":

    N = 20

    print("Number of squares and cubes is ",

          countSC(N))

C#

using System;

class GFG {

    static bool isSquare(int num)

    {

        int root = (int)Math.Sqrt(num);

        return (root * root) == num;

    }

    static bool isCube(int num)

    {

        int root = (int)Math.Pow(num, (1.0 / 3.0));

        return (root * root * root) == num;

    }

    static int countSC(int N)

    {

        int count = 0;

        for (int i = 1; i <= N; i++) {

            if (isSquare(i))

                count++;

            else if (isCube(i))

                count++;

        }

        return count;

    }

    public static void Main()

    {

        int N = 20;

        Console.Write("Number of squares and "

                      + "cubes is " + countSC(N));

    }

}

PHP

<?php

function isSquare($num)

{

    $root = (int)sqrt($num);

    return ($root * $root) == $num;

}

function isCube($num)

{

    $root = (int)pow($num, 1 / 3);

    return ($root * $root *

            $root) == $num;

}

function countSC($N)

{

    $count = 0;

    for ($i = 1; $i <= $N; $i++)

    {

        if (isSquare($i))

            $count++;

        else if (isCube($i))

            $count++;

    }

    return $count;

}

$N = 20;

echo "Number of squares and " .

     "cubes is " . countSC($N);

?>

Javascript

<script>

    function isSquare(num)

    {

        let root = parseInt(Math.sqrt(num), 10);

        return (root * root) == num;

    }

    function isCube(num)

    {

        let root = parseInt(Math.cbrt(num), 10);

        return (root * root * root) == num;

    }

    function countSC(N)

    {

        let count = 0;

        for (let i = 1; i <= N; i++) {

            if (isSquare(i))

                count++;

            else if (isCube(i))

                count++;

        }

        return count;

    }

    let N = 20;

    document.write("Number of squares and cubes is " + countSC(N));

</script>

Output

Number of squares and cubes is 5

Time Complexity: O(N)
Space Complexity: O(1) 

Efficient Approach: 

• Number of squares from 1 to N is floor(sqrt(N)).
• Number of cubes from 1 to N is floor(cbrt(N)).
• Eliminate the numbers which are both square and cube ( like 1, 64…. ) by subtracting floor(sqrt(cbrt(N))) from it.

Below is the implementation of the above approach:
 

C++

#include <iostream>

#include <math.h> // For sqrt() and cbrt()

using namespace std;

int countSC(int N)

{

    int res = (int)sqrt(N) + (int)cbrt(N)

              - (int)(sqrt(cbrt(N)));

    return res;

}

int main()

{

    int N = 20;

    cout << "Number of squares and cubes is " << countSC(N);

    return 0;

}

Java

class GFG {

    static int countSC(int N)

    {

        int res = (int)Math.sqrt(N) + (int)Math.cbrt(N)

                  - (int)(Math.sqrt(Math.cbrt(N)));

        return res;

    }

    public static void main(String[] args)

    {

        int N = 20;

        System.out.println("Number of squares "

                           + "and cubes is " + countSC(N));

    }

}

Python3

import math 

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

N = 20

print("Number of squares and cubes is",

      countSC(N))

C#

using System;

public class GFG {

    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;

    }

    public static void Main()

    {

        int N = 20;

        Console.Write("Number of squares "

                      + "and cubes is " + countSC(N));

    }

}

PHP

<?php

function countSC($N)

{

    $res = sqrt($N) + pow($N, 1 / 3) -

                (sqrt(pow($N, 1 / 3)));

    return floor($res);

}

$N = 20;

echo "Number of squares and cubes is " ,

                            countSC($N);

?>

Javascript

<script>

    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>

Output

Number of squares and cubes is 5

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