下面sumpower的时间复杂度是多少

double Power(double x, int n)
{
double y;
if (n > 0)
{
y = Power(x, n / 2);
y *= y;
if (n % 2)
{
y *= x;
}
}
else
{
y = 1.0;
}
return y;
}

double SumPower(double x, int n)
{
double y;
if (n > 0)
{
y = SumPower(x, n - 1) + Power(x, n);
}
else
{
y = 1.0;
}
return y;
}

nlogn吧

O(n平方)