3. Given the sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ where each term is not equal to 0 or 1, and the sum $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$ satisfies $2 S_{n}=a_{n}-a_{n}^{2}(1 \leqslant n \leqslant 5)$. Then the number of sequences that satisfy the condition is ( ) .
(A) 2
(B) 6
(C) 8
(D) 16
3. B.

Substitute $n=1$ into $2 S_{n}=a_{n}-a_{n}^{2}$, we get $2 a_{1}=a_{1}-a_{1}^{2}$, which is $a_{1}^{2}+a_{1}=0$.
But $a_{1} \neq 0$, so $a_{1}=-1$.
For $n \geqslant 2$, from $2 S_{n}=a_{n}-a_{n}^{2}$, we have
$$
2 S_{n+1}=a_{n+1}-a_{n+1}^{2} \text {. }
$$

Thus, $2 a_{n+1}=2\left(S_{n+1}-S_{n}\right)$
$$
=a_{n+1}-a_{n}-a_{n+1}^{2}+a_{n}^{2} \text {, }
$$

which means $\left(a_{n+1}-a_{n}\right)\left(a_{n+1}+a_{n}\right)+\left(a_{n+1}+a_{n}\right)=0$.
Hence $a_{n+1}+a_{n}=0$

or $a_{n+1}-a_{n}+1=0$.
Substitute $n=1$ into equations (1) and (2), since no term of the sequence equals 1, we get $a_{2}=a_{1}-1=-2$ (discard $a_{2}=-a_{1}=1$ ).
Substitute $n=2$ into equations (1) and (2), we get
$$
a_{3}=-a_{2}=2 \text { or } a_{3}=a_{2}-1=-3 \text {. }
$$

Continue to substitute $n=3$ into equations (1) and (2), when $a_{3}=2$, by the fact that no term of the sequence equals 1, we have only one value $a_{4}=-a_{3}=-2$ (discard $a_{4}=a_{3}-1=1$ ).
When $a_{3}=-3$, we have two values
$$
a_{4}=-a_{3}=3, a_{4}=a_{3}-1=-4 .
$$

Finally, substitute $n=4$ into equations (1) and (2), we have three cases:
When $a_{4}=-2$, we have
$$
a_{5}=-a_{4}=2, a_{5}=a_{4}-1=-3 \text {; }
$$

When $a_{4}=3$, we have
$$
a_{5}=-a_{4}=-3, a_{5}=a_{4}-1=2 \text {; }
$$

When $a_{4}=-4$, we have
$$
a_{5}=-a_{4}=4, a_{5}=a_{4}-1=-5 \text {. }
$$

Therefore, there are 6 sequences that satisfy the conditions:
$$
\begin{array}{ll}
-1,-2,2,-2,2 ; & -1,-2,2,-2,-3 ; \\
-1,-2,-3,3,-3 ; & -1,-2,-3,3,2 ; \\
-1,-2,-3,-4,4 ; & -1,-2,-3,-4,-5
\end{array}
$$