$$
\begin{aligned}
&[a \bmod b \le c]= \left\lfloor \frac{a}{b} \right\rfloor - \left \lfloor \frac{a-c-1}{b} \right \rfloor \
\end{aligned}
$$

设 $a=kb+r$

$r > c$

首先有:

$$
\begin{aligned}
&r > c \\
\Rightarrow &b-1 \ge r-c \ge 1 \\
\Rightarrow &b-2\ge r-c-1 \ge 0
\end{aligned}
$$

则:

$$
\lfloor \frac{kb+r}{b} \rfloor - \lfloor \frac{kb+r-c-1}{b} \rfloor
=\lfloor \frac{r}{b} \rfloor-\lfloor \frac{r-c-1}{b} \rfloor=0
$$

$r \le c$

首先有:

$$
\begin{aligned}
&r \le c \\
\Rightarrow &1-b\le r-c \le 0 \\
\Rightarrow &-b \le r-c-1 \le -1
\end{aligned}
$$

则有:

$$
\lfloor \frac{kb+r}{b} \rfloor - \lfloor \frac{kb+r-c-1}{b} \rfloor
=\lfloor \frac{r}{b} \rfloor-\lfloor \frac{r-c-1}{b} \rfloor=1
$$