Implement pow(x, n), which calculates x raised to the power n (xn). ``` Example 1: Input: 2.00000, 10 Output: 1024.00000 ``` ``` Example 2: Input: 2.10000, 3 Output: 9.26100 ``` ``` Example 3: Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 =

test1.cpp

class Solution {
public:
    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;
    }
};