Question

Here’s the code that does the trick. I do know the concept how to calculate m^n in O(logn) time, and I think it’s related, but how?

for(long long int i = 2; N; N /= 2, i *= i){
    if(N & 1){
        ans *= i;
    }
}

Answer 1

That’s the square and multiply algorithm.

N /= 2 has the effect of removing the lowest bit (and when all the bits remaining are zero, the algorithm is finished).

N & 1 checks if the lowest bit is odd or even