372. Super Pow

Your task is to calculate a^b mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

Example1:a = 2b = [3]Result: 8Example2:a = 2b = [1,0]Result: 1024

One knowledge: ab % k = (a%k)(b%k)%k
Since the power here is an array, we’d better handle it digit by digit.
One observation:
a^1234567 % k = (a^1234560 % k) * (a^7 % k) % k = (a^123456 % k)^10 % k * (a^7 % k) % k
Looks complicated? Let me put it other way:
Suppose f(a, b) calculates a^b % k; Then translate above formula to using f :
f(a,1234567) = f(a, 1234560) * f(a, 7) % k = f(f(a, 123456),10) * f(a,7)%k;
Implementation of this idea:

class Solution {public:    int base = 1337;    int superPow(int a, vector<int>& b) {        if(b.empty()) return 1;        int last_digit = b.back();        b.pop_back();        return ( helper(superPow(a,b),10) * helper(a,last_digit) )% base;    }    int helper(int a ,int k){        a %= base;        int result = 1;        for (int i = 0; i < k; ++i)        {            result = (result*a)%base;        }        return result;    }};