What is the fastest way to calculate (2^n) mod (10^9 + 7)?

Question

Is there a better solution than this :

2^n mod(m)
{  
    if n=0
        return 1;

    r=2^(n-1)mod(m);  
    if 2r < m 
        return 2r;
    if 2r > =m 
        return 2r-m;
}

That $10^9+7$ is a typical Codechef modulus :-). Anyway, have you searched the site at all. This has been explained many times over. Google for exponentiation by squaring or square-and-multiply. There is a Wikipedia page explainin how to do this fast. — Jyrki Lahtonen, Jan 01 '16 at 22:54

score 1 · Answer 1 · answered Jan 01 '16 at 22:54

This can be sped up a bit. Instead of iterating over exponents, a process called repeated squaring is used. You iteratively square and reduce mod m to obtain $a^k$ where $k$ is a power of $2$. Then you multiple together the factors you need until the exponent is correct. Remember, the base 2 expansion of the exponent tells you which factors you need.

What is the fastest way to calculate (2^n) mod (10^9 + 7)?

1 Answers1