leetcode 50

二分法，xn = xn/2 * x n/2 *  xn%2

class Solution {
public:
double myPow(double x, int n) {
if (n < 0) return 1.0 / power(x, -n);
else return power(x, n);
}
private:
double power(double x, int n) {
if (n == 0) return 1;
double v = power(x, n / 2);
if (n % 2 == 0) return v * v;
else return v * v * x;
}
};