50. Pow(x, n) (中等)

    double myPow(double x, int n) {
        double ans = 1;
        unsigned long long p;
        if (n < 0) {
            p = -n;
            x = 1 / x;
        } else {
            p = n;
        }
        while (p) {
            if (p & 1)
                ans *= x;
            x 
 
*= x;
            p >>= 1;
        }
        return ans;
    }