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

#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; 
} 

Java

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. 

Python

# 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 

C#

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 

PHP

<?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. 
?> 

Ouput :

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<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; 
} 

Java

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. 

Python

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 

C#

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 

PHP

<?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. 
?> 

Ouput :

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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