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Compute the value of A raise to the power B using Fast Exponentiation

Here, we are going to compute the value of A raise to the power B using Fast Exponentiation.
Submitted by Ankit Sood, on December 05, 2018

Now here we are going to learn that how to compute the value of a^b i.e. "A" raise to the power "B" using an optimized algorithm called as "fast-exponentiation"?

we could have used a brute force approach to do the required task but then it would have taken O(b) i.e a brute force approach can solve the required task in linear time whereas the same task could be done in O(log b) using FAST–EXPONENTIATION.

Fast exponentiation uses a simple recursive approach whose recurrence relation can be written as :

F(a,b) = F(a,b/2)*F(a,b/2)

From the above recurrence relation, it can be easily seen that the time complexity would be O(log b).

C++ program to find the value of A^B using fast exponentiation

#include <iostream>
using namespace std;

//function to find a^b
int fcheck(int a,int b){
	if(b == 0)
	    return 1;
	if(b == 1)
		return a;
	return fcheck(a,b/2)*fcheck(a,b/2);
}

//main code
int main(){
	int a,b;

	cout<<"Enter the first number (value of a): ";
	cin>>a;
	cout<<"Enter the second number (Value of b): ";
	cin>>b;

	int x=fcheck(a,b);
	if(b%2==0 || b==1)
		cout<<a<<" raise to the power "<<b<<" is = " <<x<<endl;
	else
		cout<<a<<" raise to the power "<<b<<" is = " <<a*x<<endl;

	return 0;
}

Output

First run:
Enter the first number (value of a): 5
Enter the second number (Value of b): 3
5 raise to the power 3 is = 125

Second run:
Enter the first number (value of a): 5
Enter the second number (Value of b): 0
5 raise to the power 0 is = 1

Third run:
Enter the first number (value of a): 5
Enter the second number (Value of b): 1
5 raise to the power 1 is = 5