class Solution {
public:
    double myPow(double x, int n) {
        typedef long long LL;
        double res = 1;
        LL k = abs(n);
        while (k) {
            if (k & 1) res = x * res;
            k >>= 1;
            x = x * x;
        }
        if (n < 0) return 1 / res;
        else return res;
    }
};

快速幂模板

返回a^k % p的结果
int qmi(int a, int k, int p) {
    int res = 1;
    while(k) {
        if (k & 1) res = (LL) res * a % p;
        k >>= 1;
        a = (LL) a * a % p;
    }
    return res;
}