How to compute a^b^c mod p?

问题

I am trying to compute a^b^c mod p for some positive integers a,b,c,p. One possible (and obvious) way is to use fast modular exponentiation which will run in O(log(b^c))=clog(b). While I don't mind the efficiency here, the obvious downfall of this method is that you need an explicit binary representation of b^c which in itself is already exponential.

So the question for me is, if I can not represent b^c as a binary representation, is there a way I can compute a^b^c mod p from the binary representations of a,b, and c?

回答1:

(a^b^c) mod p = (((a^b) mod p)^c) mod p

So you can do

modpow(modpow(a,b,p),c,p);

Where all operands results and subresults are normal ints. As modpow you can use power by squaring in modulo p like here:

• Modular arithmetics and NTT (finite field DFT) optimizations

beware those are a bit optimized taking advantage of the properties of specific selected p so you need to change lines like

if (DWORD(d)>=DWORD(p)) d-=p;

into

d%=p;

[Example]

(2^3^5) % 6 = 
(8  ^5) % 6 =
  32768 % 6 = 2

(((2^3)%6)^5) % 6 = 
((   8 %6)^5) % 6 = 
(    2    ^5) % 6 =
    32        % 6 = 2

来源：https://stackoverflow.com/questions/46944581/how-to-compute-abc-mod-p