1. Define a sequence: if $a_{n}=112$, then $a_{n+1}=2$ 112, which means reading the digits of $a_{n}$ from left to right as two 1s and one 2 to form $a_{n+1}=2112$. Following this rule, we get
$$
a_{n+2}=122112, a_{n+3}=11222112 .
$$

Given $a_{1}=13255569$, and following this rule to find $a_{100}$. Then, from $a_{1}$ to $a_{100}$, the number of different digits that appear is ( ) kinds.
(A) 5
(B) 6
(C) 7
(D) 8
$-1 . B$.
According to the operation principle, if the numbers $4, 7, 8$ are to appear, they must appear in the number (let's assume this number is $a_{i}(i=3,4$, $\cdots, 100$, and $i$ is the smallest)) with the previous number $a_{i-1}$ having a continuous sequence of 4 or 7 or 8 numbers, i.e., $a_{i-1}$ has a part like $X X X X$ or $X X X X X X X$ or $X X X X X X X X X$, which contradicts the operation principle from $a_{i-2}$ to $a_{i-1}$. When $i=2$, $a_{i}=111$ 312351619, which also does not hold, so the numbers $4, 7, 8$ do not appear.

Furthermore, 0 cannot appear at all, and $a_{1}$ contains the digits $1, 2, 3, 5, 6, 9$, so the total number of different digits is 6.