# Tag Info

I will assume you are in the honest-but-curious model. You can't represent real numbers in finite space, so I will assume all values are represented in fixed-point arithmetic, to $d$ bits of precision; thus $x$ is represented as $x = x'/2^d$ where $x'$ is an integer. Then $x \cdot y = x' \cdot y' / 2^{2d}$, so the problem is equivalent to computing $x \... 2 Using Fermat theorem,$a^p -a = 0 (\mod p) $and if a and p are co-prime, then$ a^{p−1} − 1 =1(\mod p) $So if u choose n to be a prime number(say p), then$a^{b^c} \mod p = a^{ (b^{c} \mod (p-1))} \mod p $, Then you can use Fast Exponentiation trick in two levels, once for$b^c \mod p-1 $then for$a^{b^c} \mod p \$ I also suggest you to look at cses ...