372. 超级次方

题意

题目链接

思路

二进制转化+快速幂

代码

class Solution {
public:
    int superPow(int a, vector<int>& b) {
        string tmp;
        int l = 0;
        while (l < b.size()) {
            int x = 0;
            for (int i = l; i < b.size(); i++)
            {
                int y = x * 10 + b[i];
                x = y % 2;
                b[i] = y / 2;
                if (b[i] == 0 && i == l)
                    l++;
            }
            tmp += (x ? "1" : "0");
        }
        //cout << tmp << endl;

        long ans = 1;
        long k = a % 1337;
        for (auto &index : tmp) {
            if (index == '1')
                ans = ans * k % 1337;
            k = (k * k) % 1337;
        }
        return ans;
    }
};