Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

double mul(double x, long n){
    
    if(n == 0){
        return 1;
    }
    
    if ((n % 2) == 1){
        x = mul(x, floor(n / 2)) * mul(x, floor(n / 2)) * x;
    }else{
        x = mul(x, floor(n / 2)) * mul(x, floor(n / 2));
    }
    
    return x;
}

double myPow(double x, int n){
        
    if(n < 0){
        return (1.0f / mul(x, (long)n*-1));
    }else{
        return mul(x, n);
    }
}