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"""

* Binary Exponentiation for Powers

* This is a method to find a^b in a time complexity of O(log b)

* This is one of the most commonly used methods of finding powers.

* Also useful in cases where solution to (a^b)%c is required,

* where a,b,c can be numbers over the computers calculation limits.

* Done using iteration, can also be done using recursion

* @author chinmoy159

* @version 1.0 dated 10/08/2017

"""

def b_expo(a, b):

res = 1

while b > 0:

if b & 1:

res *= a

a *= a

b >>= 1

return res

def b_expo_mod(a, b, c):

res = 1

while b > 0:

if b & 1:

res = ((res % c) * (a % c)) % c

a *= a

b >>= 1

return res

"""

* Wondering how this method works !

* It's pretty simple.

* Let's say you need to calculate a ^ b

* RULE 1 : a ^ b = (a*a) ^ (b/2) ---- example : 4 ^ 4 = (4*4) ^ (4/2) = 16 ^ 2

* RULE 2 : IF b is ODD, then ---- a ^ b = a * (a ^ (b - 1)) :: where (b - 1) is even.

* Once b is even, repeat the process to get a ^ b

* Repeat the process till b = 1 OR b = 0, because a^1 = a AND a^0 = 1

*

* As far as the modulo is concerned,

* the fact : (a*b) % c = ((a%c) * (b%c)) % c

* Now apply RULE 1 OR 2 whichever is required.

"""