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

  • Difficulty Level : Easy
  • Last Updated : 07 Apr, 2021

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++




// C++ code for power function
#include <bits/stdc++.h>
using namespace std;
 
/* 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 Code
int main()
{
    cout << pow(5, 3);
    return 0;
}
 
// This code is contributed
// by rathbhupendra
C



#include<stdio.h>
/* Works only if a >= 0 and b >= 0  */
int pow(int a, int b)
{
  //base case : anything raised to the power 0 is 1
  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
 
# driver 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.
?>
Javascript



<script>
    /* Works only if a >= 0 and b >= 0 */
function pow(a , b)
{
    if (b == 0)
        return 1;
         
    var answer = a;
    var increment = a;
    var 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
document.write(pow(5, 3));
 
// This code is contributed by shikhasingrajput
</script>
Output : 
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++



#include<bits/stdc++.h>
using namespace std;
 
/* A recursive function to get x*y */
int multiply(int x, int y)
{
    if(y)
        return (x + multiply(x, y - 1));
    else
        return 0;
}
 
/* 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;
}
 
// Driver Code
int main()
{
    cout << pow(5, 3);
    getchar();
    return 0;
}
 
// This code is contributed
// by Akanksha Rai
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.
Python3



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.
?>
Javascript



<script>
 
// A recursive function to get a^b
// Works only if a >= 0 and b >= 0
function pow(a, b)
{
    if (b > 0)
        return multiply(a, pow(a, b - 1));
    else
        return 1;
}
 
// A recursive function to get x*y
function multiply(x, y)
{
    if (y > 0)
        return (x + multiply(x, y - 1));
    else
        return 0;
}
 
// Driver code
document.write(pow(5, 3));
 
// This code is contributed by gauravrajput1
 
</script>
Output : 
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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