11. Among all the divisors of $201^{9}$, there are ( ) that are perfect squares or perfect cubes.
(A) 32
(B) 36
(C) 37
(D) 39
(E) 41
11. C.

Notice that, $201^{9}=3^{9} \times 67^{9}$, and 3, 67 are both prime numbers. First, consider the perfect square factors: choose one factor from each of the following sets:
$$
\left\{3^{0}, 3^{2}, 3^{4}, 3^{6}, 3^{8}\right\},\left\{67^{0}, 67^{2}, 67^{4}, 67^{6}, 67^{8}\right\}
$$

There are $5 \times 5=25$ ways;
Next, consider the perfect cube factors: choose one factor from each of the following sets:
$$
\left\{3^{0}, 3^{3}, 3^{6}, 3^{9}\right\},\left\{67^{0}, 67^{3}, 67^{6}, 67^{9}\right\}
$$

There are $4 \times 4=16$ ways;
Finally, consider the overlapping factors: choose one factor from each of the following sets:
$$
\left\{3^{0}, 3^{6}\right\},\left\{67^{0}, 67^{6}\right\}
$$

There are $2 \times 2=4$ ways.
In summary, the number of factors that satisfy the condition is
$$
N=25+16-4=37 .
$$