Write you own Power without using multiplication(*) and division(/) operators

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 time 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

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#include<stdio.h>
/* Works only if a >= 0 and b >= 0  */
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;
}
  
/* driver program to test above function */
int main()
{
  printf("\n %d", pow(5, 3));
  getchar();
  return 0;
}

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Java

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import java.io.*;
  
class GFG {
      
    /* Works only if a >= 0 and b >= 0 */
    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;
    }
  
    // driver program to test above function
    public static void main(String[] args)
    {
        System.out.println(pow(5, 3));
    }
}
  
// This code is contributed by vt_m.

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Python

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# Python 3 code for power
# function 
  
# Works only if a >= 0 and b >= 0 
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
  
# drive code
print(pow(5,3))
  
# this code is contributed 
# by Sam007

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C#

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using System;
  
class GFG
{
    /* Works only if a >= 0 and b >= 0 */
    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;
    }
  
    // driver program to test 
    // above function
    public static void Main()
    {
        Console.Write(pow(5, 3));
    }
}
  
// This code is contributed by Sam007

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PHP

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<?php
  
// Works only if a >= 0 
// and b >= 0 
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;
}
  
    // Driver Code
    echo( poww(5, 3));
   
// This code is contributed by nitin mittal.
?>

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Ouput :

125

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



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#include<stdio.h>
/* A recursive function to get a^b
  Works only if a >= 0 and b >= 0  */
int pow(int a, int b)
{
   if(b)
     return multiply(a, pow(a, b-1));
   else
    return 1;
}    
  
/* A recursive function to get x*y */
int multiply(int x, int y)
{
   if(y)
     return (x + multiply(x, y-1));
   else
     return 0;
}
  
/* driver program to test above functions */
int main()
{
  printf("\n %d", pow(5, 3));
  getchar();
  return 0;
}

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Java

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import java.io.*;
  
class GFG {
      
    /* A recursive function to get a^b
    Works only if a >= 0 and b >= 0 */
    static int pow(int a, int b)
    {
          
        if (b > 0)
            return multiply(a, pow(a, b - 1));
        else
            return 1;
    }
  
    /* A recursive function to get x*y */
    static int multiply(int x, int y)
    {
          
        if (y > 0)
            return (x + multiply(x, y - 1));
        else
            return 0;
    }
  
    /* driver program to test above functions */
    public static void main(String[] args)
    {
        System.out.println(pow(5, 3));
    }
}
  
// This code is contributed by vt_m.

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Python

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def pow(a,b):
      
    if(b):
        return multiply(a, pow(a, b-1));
    else:
        return 1;
       
# A recursive function to get x*y *
def multiply(x, y):
      
    if (y):
        return (x + multiply(x, y-1));
    else:
        return 0;
  
# driver program to test above functions *
print(pow(5, 3));
  
  
# This code is contributed
# by Sam007

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C#

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using System;
  
class GFG
{
    /* A recursive function to get a^b
    Works only if a >= 0 and b >= 0 */
    static int pow(int a, int b)
    {
          
        if (b > 0)
            return multiply(a, pow(a, b - 1));
        else
            return 1;
    }
  
    /* A recursive function to get x*y */
    static int multiply(int x, int y)
    {
          
        if (y > 0)
            return (x + multiply(x, y - 1));
        else
            return 0;
    }
  
    /* driver program to test above functions */
    public static void Main()
    {
        Console.Write(pow(5, 3));
    }
}
  
// This code is contributed by Sam007

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PHP

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<?php
  
/* A recursive function to get a^b
   Works only if a >= 0 and b >= 0 */
function p_ow( $a, $b)
{
    if($b)
        return multiply($a
          p_ow($a, $b - 1));
    else
        return 1;
  
/* A recursive function 
   to get x*y */
function multiply($x, $y)
{
    if($y)
        return ($x + multiply($x, $y - 1));
    else
        return 0;
}
  
// Driver Code
echo pow(5, 3);
  
// This code is contributed by anuj_67.
?>

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Ouput :

125

Please write comments if you find any bug in above code/algorithm, or find other ways to solve the same problem.



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Improved By : nitin mittal, vt_m