Super Pow问题及解法

问题描述：

our 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.

示例:

a = 2
b = [3]

Result: 8

a = 2
b = [1,0]

Result: 1024

问题分析：

利用欧拉准则和递归的方法，即可求得答案。

这里简单说一下欧拉准则：ab % k = (a%k)(b%k)%k

过程详见代码：

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