3. Given a periodic sequence $\left\{x_{n}\right\}$ satisfying $x_{n}=\left|x_{n-1}-x_{n-2}\right|(n \geqslant 3)$. If $x_{1}=1, x_{2}=a \geqslant 0$, then when the period of the sequence is the smallest, the sum of the first 2002 terms of the sequence is
A. 2002 -
B. 1335
C. 1949
D. 1428
3. $\mathrm{B}$ Let the sequence $\left\{x_{n}\right\}$ have a period of $T . T \neq 1, T \geqslant 2$.

If $T=2$. Then $x_{3}=|a-1|=1$. This gives $a=0$ or 2.
By $x_{4}=\left|x_{3}-x_{2}\right|=|1-a|=a$ we get $a=\frac{1}{2}$. This contradicts $a=0$ or 2. Therefore, $T \geqslant 3$.
When $T=3$, from $x_{4}=|| a-1|-a|=1$ we get $a \geqslant 1$ or $a=0$.
From $x_{5}=|1-| a-1||=a$ we get $0 \leqslant a \leqslant 1$. Hence, $a=0$ or 1.
In this case, the sequence $\left\{x_{n}\right\}$ is either $1,0,1,1,0,1, \cdots$, or $1,1,0,1,1,0, \cdots$. Therefore, $S_{2002}=2 \times 667+1=1335$. Hence, the answer is B.