5. A.

Among single-digit numbers, there are no numbers whose digit sum is 11.
Among two-digit numbers, there are 8 numbers: $29, 38, 47, 56, 65, 74, 83, 92$.

For three-digit numbers $\overline{x y z}$, when $x=1$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers; when $x=2$, $y$ can take 10 numbers: $0, 1, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 0$, a total of 10 numbers; similarly, when $x=3, 4, \cdots, 9$, the numbers whose digit sum is 11 are $12-x$ each. Therefore, in three-digit numbers $\overline{x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+3=61$.

For four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 are equivalent to finding three-digit numbers $\overline{x y z}$ whose digit sum is 10. When $x=0$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers. Similarly, when $x=1, 2, \cdots, 9$, the three-digit numbers $\overline{x y z}$ whose digit sum is 10 have $11-x$ each. Therefore, in four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+2=63$.

In $\overline{2 x y z}$, the numbers whose digit sum is 11, from smallest to largest, are $2009, 2018$, a total of 2 numbers.
Therefore, 2018 is the
$$
8+61+63+2=134
$$

number in the sequence.
5. Arrange natural numbers whose digits sum to 11 in ascending order to form a sequence. The $m$-th number is 2018. Then $m$ is $(\quad)$.
(A) 134
(B) 143
(C) 341
(D) 413