372. Super Pow(LeetCode)

Your task is to calculate ab 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 = 2

b = [3]

Result: 8

Example2:

a = 2

b = [1,0]

Result: 1024

思路：把数字b分拆成b’ × 10 + d，其中d为b的个位

class Solution {
private:
const int BASE = 1337;
/* a ^ d mod 1337, 0 <= d < 10 */
int digitPow(int a, int d) {
a %= BASE;
int result = 1;
for(int i=0; i<d; ++i) {
result = (result * a) % BASE;
}
return result;
}
public:
/* a ^ b mod 1337, b is an extremely large positive integer */
int superPow(int a, vector<int>& b) {
if(b.empty()) return 1;
int d = b.back();
b.pop_back();
return (digitPow(digitPow(superPow(a, b), 10) * digitPow(a, d), 1));
}
};