11. (20 points) A complex number whose real and imaginary parts are both integers is called a "Gaussian integer". If the polynomial with integer coefficients
$$
f(z)=z^{n}+c_{n-1} z^{n-1}+\cdots+c_{2} z^{2}+3 c_{1} z+50
$$

has all distinct complex roots that are Gaussian integers, then $f(z)$ is called a "good polynomial". Find the number of such good polynomials.
11. Since complex roots occur in conjugate pairs, if $f(z)=0$ has a root $a+b\mathrm{i}$, then $a-b\mathrm{i}$ is also a root of $f(z)=0$. Therefore, $f(z)$ has a factor
$$
\begin{array}{l}
(z-(a+b \mathrm{i}))(z-(a-b \mathrm{i})) \\
=z^{2}-2 a z+\left(a^{2}+b^{2}\right) .
\end{array}
$$
$$
\text { Hence } f(z)=\prod_{k=1}^{m}\left(z^{2}-2 a_{k} z+d_{k}\right) \prod_{k=1}^{t}\left(z-h_{k}\right) \text {, }
$$

and $\prod_{k=1}^{m} d_{k} \prod_{k=1}^{t} h_{k}=50$,
where $d_{k}=a_{k}^{2}+b_{k}^{2}$, and $a_{k}, b_{k}, h_{k} \in \mathbf{Z}$.
The positive divisors of 50 are $\{1,2,5,10,25,50\}$. The factors with constant terms whose absolute values are these divisors are classified as follows:
$$
\begin{aligned}
A_{0}= & \{z-1\}, \\
A_{1}= & \left\{z+1, z^{2}+1\right\}, \\
A_{2}= & \left\{z \pm 2, z^{2} \pm 2 z+2\right\}, \\
A_{5}= & \left\{z \pm 5, z^{2} \pm 2 z+5, z^{2} \pm 4 z+5\right\}, \\
A_{10}= & \left\{z \pm 10, z^{2} \pm 6 z+10, z^{2} \pm 2 z+10\right\}, \\
A_{25}= & \left\{z \pm 25, z^{2} \pm 6 z+25, z^{2} \pm 8 z+25, z^{2}+25\right\}, \\
A_{50}= & \left\{z \pm 50, z^{2} \pm 10 z+50, z^{2} \pm 14 z+50,\right. \\
& \left.z^{2} \pm 2 z+50\right\} .
\end{aligned}
$$

In particular, $A_{0}=\{z-1\}$ represents the sign. If the constant term of the polynomial is negative, then choose $z-1$ from $A_{0}$ to multiply; otherwise, do not choose $z-1$.
(1) $50=2 \times 5^{2}$.
This requires choosing one factor from $A_{2}$ and two factors from $A_{5}$ (order does not matter), and then multiplying by $z-1$ from $A_{0}$ if necessary based on the sign of the constant term. There are $4 \mathrm{C}_{6}^{2}=60$ choices.
(2) $50=5 \times 10$.
Similarly, this requires choosing one factor from $A_{5}$ and one factor from $A_{10}$, totaling $6 \times 6=36$ choices.
(3) $50=2 \times 25$.
Similarly, this requires choosing one factor from $A_{2}$ and one factor from $A_{25}$, totaling $4 \times 7=28$ choices.
(4) $50=50$.
This requires choosing one factor from $A_{50}$, giving 8 choices.
In total, there are 132 choices.
Considering that each of these choices can be multiplied by a factor chosen from or not chosen from
$$
A_{1}=\left\{z+1, z^{2}+1\right\}
$$

there are $132 \times 4=528$ polynomials that meet the conditions.