How to find power of a number in python without using pow function

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Method 1 (Using Nested Loops): We can calculate power by using repeated addition. For example to calculate 5^6. 

1) First 5 times add 5, we get 25. (5^2) 
2) Then 5 times add 25, we get 125. (5^3) 
3) Then 5 times add 125, we get 625 (5^4) 
4) Then 5 times add 625, we get 3125 (5^5) 
5) Then 5 times add 3125, we get 15625 (5^6) 

C++

#include

using namespace std;

int pow(int a, int b)

{

    if (b == 0)

        return 1;

    int answer = a;

    int increment = a;

    int i, j;

    for(i = 1; i < b; i++)

    {

        for(j = 1; j < a; j++)

        {

            answer += increment;

        }

        increment = answer;

    }

    return answer;

}

int main()

{

    cout << pow(5, 3);

    return 0;

}

C

#include

int pow(int a, int b)

{

  if (b == 0)

    return 1;

  int answer = a;

  int increment = a;

  int i, j;

  for(i = 1; i < b; i++)

  {

     for(j = 1; j < a; j++)

     {

        answer += increment;

     }

     increment = answer;

  }

  return answer;

}

int main()

{

  printf("\n %d", pow(5, 3));

  getchar();

  return 0;

}

Java

import java.io.*;

class GFG {

    static int pow(int a, int b)

    {

        if (b == 0)

            return 1;

        int answer = a;

        int increment = a;

        int i, j;

        for (i = 1; i < b; i++) {

            for (j = 1; j < a; j++) {

                answer += increment;

            }

            increment = answer;

        }

        return answer;

    }

    public static void main(String[] args)

    {

        System.out.println(pow(5, 3));

    }

}

Python

def pow(a,b):

    if(b==0):

        return 1

    answer=a

    increment=a

    for i in range(1,b):

        for j in range (1,a):

            answer+=increment

        increment=answer

    return answer

print(pow(5,3))

C#

using System;

class GFG

{

    static int pow(int a, int b)

    {

        if (b == 0)

            return 1;

        int answer = a;

        int increment = a;

        int i, j;

        for (i = 1; i < b; i++) {

            for (j = 1; j < a; j++) {

                answer += increment;

            }

            increment = answer;

        }

        return answer;

    }

    public static void Main()

    {

        Console.Write(pow(5, 3));

    }

}

PHP

function poww($a, $b)

{

    if ($b == 0)

        return 1;

    $answer = $a;

    $increment = $a;

    $i;

    $j;

    for($i = 1; $i < $b; $i++)

    {

        for($j = 1; $j < $a; $j++)

        {

            $answer += $increment;

        }

        $increment = $answer;

    }

    return $answer;

}

    echo( poww(5, 3));

?>

Javascript

Output : 

125

Time Complexity: O(a * b)

Auxiliary Space:O(1)

Method 2 (Using Recursion): Recursively add a to get the multiplication of two numbers. And recursively multiply to get a raise to the power b.

C++

#include

using namespace std;

int multiply(int x, int y)

{

    if(y)

        return (x + multiply(x, y - 1));

    else

        return 0;

}

int pow(int a, int b)

{

    if(b)

        return multiply(a, pow(a, b - 1));

    else

        return 1;

}

int main()

{

    cout << pow(5, 3);

    getchar();

    return 0;

}

C

#include

int pow(int a, int b)

{

   if(b)

     return multiply(a, pow(a, b-1));

   else

    return 1;

}   

int multiply(int x, int y)

{

   if(y)

     return (x + multiply(x, y-1));

   else

     return 0;

}

int main()

{

  printf("\n %d", pow(5, 3));

  getchar();

  return 0;

}

Java

import java.io.*;

class GFG {

    static int pow(int a, int b)

    {

        if (b > 0)

            return multiply(a, pow(a, b - 1));

        else

            return 1;

    }

    static int multiply(int x, int y)

    {

        if (y > 0)

            return (x + multiply(x, y - 1));

        else

            return 0;

    }

    public static void main(String[] args)

    {

        System.out.println(pow(5, 3));

    }

}

Python3

def pow(a,b):

    if(b):

        return multiply(a, pow(a, b-1));

    else:

        return 1;

def multiply(x, y):

    if (y):

        return (x + multiply(x, y-1));

    else:

        return 0;

print(pow(5, 3));

C#

using System;

class GFG

{

    static int pow(int a, int b)

    {

        if (b > 0)

            return multiply(a, pow(a, b - 1));

        else

            return 1;

    }

    static int multiply(int x, int y)

    {

        if (y > 0)

            return (x + multiply(x, y - 1));

        else

            return 0;

    }

    public static void Main()

    {

        Console.Write(pow(5, 3));

    }

}

PHP

function p_ow( $a, $b)

{

    if($b)

        return multiply($a,

          p_ow($a, $b - 1));

    else

        return 1;

}

function multiply($x, $y)

{

    if($y)

        return ($x + multiply($x, $y - 1));

    else

        return 0;

}

echo pow(5, 3);

?>

Javascript

Output : 

125

Time Complexity: O(b)

Auxiliary Space: O(b)

Method 3 (Using bit masking) 

Approach: We can a^n (let’s say 3^5) as 3^4 * 3^0 * 3^1 = 3^5, so we can represent 5 as its binary i.e. 101

C++