Answer: A

Solution:

Write $x=\lfloor x\rfloor+t$, where $0\le t\lt1$. Then $$\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor,$$ so $f(x)$ depends only on the fractional part $t$.

The value of $f$ changes only when $t$ crosses a fraction $i/k$, where $2\le k\le10$ and $1\le i\lt k$. The number of distinct such fractions in $(0,1)$ is $$\begin{aligned} &\varphi(2)+\varphi(3) \\ &\quad {}+\cdots+\varphi(10) \\ &=1+2+2+4+2 \\ &\quad {}+6+4+6+4 \\ &=31. \end{aligned}$$

Including the initial value before the first breakpoint, $f$ assumes $31+1=32$ distinct values.

Thus, the correct answer is A.