1. $a$ is a 1998-digit number consisting of 1998 nines, $b$ is a 1998-digit number consisting of 1998 eights, then the sum of the digits of $a \cdot b$ is ( ).
(A) 19980
(B) 19971
(C) 17982
(D) 17991
$$
\begin{array}{l}
=\underbrace{88 \cdots 87}_{1987 \uparrow 8} \underbrace{11 \cdots 12}_{11 \cdots 71 \cdot 1} . \\
\end{array}
$$

Therefore, the sum of the digits of $a b$ is
$$
8 \times 1997+7+1 \times 1997+2=17982 \text {. }
$$