1. Given that $a$, $b$, and $c$ are integers, and satisfy $(a-b)^{10}+(a-c)^{10}=1$.
Then $|a-b|+|b-c|+|c-a|=(\quad)$.
(A)1
(B) 2
(C) 3
(D) 4
$-1 . B$
From the problem, we know that $a-b$ and $a-c$ are both integers.
According to the given equation, $|a-b|$ and $|a-c|$ are one 0 and one 1.
Therefore, $|a-b|+|b-c|+|c-a|$ $=1+|(a-b)-(a-c)|=2$.