Super Pow

题目来源
一道数学题…
主要用到了ab % k = (a%k)(b%k)%k
代码如下：

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