Pow(x, n)

Input: 2.00000, 10
Output: 1024.00000

Input: 2.10000, 3
Output: 9.26100

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

#include <iostream>
#include <cmath>
using namespace std;

class Solution
{
public:
    double myPow(double x, int n)
    {
        if(abs(x-0)<0.000001)
            return 0;
        if(n==-1)
            return 1.0/x;
        else if(n==1)
            return x;
        else if(n==0)
            return 1;

        double res=myPow(x,n/2);
        res*=res;
        if(n!=n/2*2)//odd or even
        {
            if(n>0)
                return res*x;
            else
                return res/x;
        }
        else
        {
            return res;
        }
    }
};


int main()
{
    Solution s;
    cout<<s.myPow(2.0000,-10)<<endl;
    return 0;
}