I would like to summarise and extend the results of my previous answer in a new answer as I prefer to keep the original in its current form to prevent it from turning into my magnum opus.
Theorem 1:
For positive integers $a, b$ and $c$ where $ d=\text{gcd}(b,c), $ $u=c/d,$ $t= \lfloor a/b \rfloor $ and $ a \equiv \lambda \textrm{ mod } d, $ where $ 0 \le \lambda < d,$ and $ u \, | \, (t+1) $ we have
$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor = \frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-d}{2} - \lambda \right \rbrace . $
So far, the above theorem and its proof are included in my previous answer. The rest is new.
Theorem 2:
Along with the definitions in theorem 1, let $ a \equiv r \textrm{ mod } c, $ where $ 0 \le r < c $ and $ u \, | \, t $ then
$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor = \frac{t+1}{c} \left \lbrace a - \frac{tb}{2} \right \rbrace - \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor \left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace - \frac{r}{c} . $
For example, with $a=1019,b=33$ and $c=55$ we have $d=\text{gcd}(33,55)=11,$ $u=c/d=55/11=5,$ $t= \lfloor 1019/33 \rfloor = 30$ and $ 1019 \equiv 29 \textrm{ mod } 55, $ and $ 1019 \equiv 7 \textrm{ mod } 11. $ Hence $ r = 29 $ and $\lambda = 7.$
Note that the condition $ u \, | \, t $ is satisfied, and so theorem 2 gives
$ \sum_{k=0}^{30} \left \lfloor \frac{1019 - 33k}{55} \right \rfloor = \frac{31}{55} \left \lbrace 1019 - \frac{ 30 \cdot 33}{2} \right \rbrace - \frac{6}{55} \left \lbrace \frac{5(55-11)}{2} + 7 \cdot 5 \right \rbrace - \frac{29}{55} = 279.$
This is easily verified with WolframAlpha, or similar.
Both of these theorems are special cases ($ \mu=0$ and $\mu=1$) of the general result:
Theorem 3: Along with the previous definitions in theorems 1 and 2, let $ t+1 \equiv \mu \textrm{ mod } u,$ where $ 0 \le \mu < u $ then
$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor = \frac{t+1}{c} \left \lbrace a - \frac{tb}{2} \right \rbrace - \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor \left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace $ $ - \frac{1}{c} \sum_{k=0}^{\mu - 1} \left \lbrace r+k \left \lbrace \left( \left \lfloor \frac{b}{c} \right \rfloor + 1 \right) c - b \right \rbrace \textrm{ mod } c \right \rbrace .$
There are $ \mu $ terms in the last sum, so this is understood to be zero when $ \mu =0.$ The terms in the final summation of the equation are all $ \ge 0 $ and reduced modulo $c.$
A sketch of the proof runs as follows. On adding up the $t+1$ equations at the start of my previous answer, we add up $ \lfloor (t+1)/u \rfloor $ “complete” sets of residues modulo $c$ (not to be confused with a complete residue system modulo $c$) that are congruent to $ \lambda $ modulo $c.$ The term $ \sum_{k=0}^{\mu - 1} \lbrace r+k \lbrace \cdots \rbrace \textrm{ mod } c \rbrace $ (equivalent, of course, to $ \sum_{k=0}^{\mu - 1} \lbrace (r-kb) \textrm{ mod } c \rbrace $ ) is the sum of the “left over residues.”
Rearranging the equation obtained from the summation proves the theorem.
Note: In the text $ \lbrace \cdot \rbrace $ is only used for readability and does not indicate fractional part (I find doubled-up curved brackets awkward to read).
Just for fun, here is what we get when we put $ \mu = 2 $ into theorem 3.
Theorem 2(b): Along with the previous definitions, since $ \mu = 2 $ our condition here is that $ u \, | \, (t-1) .$ Now for $ r \ge b $ we obtain
$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor = \frac{t+1}{c} \left \lbrace a - \frac{tb}{2} \right \rbrace - \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor \left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace - \frac{2r-b}{c} . \quad (1) $
For example, with $a=1037,b=33$ and $c=55$ we have $d=\text{gcd}(33,55)=11,$ $u=c/d=55/11=5,$ $t= \lfloor 1037/33 \rfloor = 31$ and $ 1037 \equiv 47 \textrm{ mod } 55, $ and $ 1037 \equiv 3 \textrm{ mod } 11. $ Hence $ r = 47 $ and $\lambda = 3.$
Note that the condition $ u \, | \, (t-1) $ is satisfied, and so theorem 2(b) gives
$ \sum_{k=0}^{31} \left \lfloor \frac{1037 - 33k}{55} \right \rfloor = \frac{32}{55} \left \lbrace 1037 - \frac{ 31 \cdot 33}{2} \right \rbrace - \frac{6}{55} \left \lbrace \frac{5(55-11)}{2} + 3 \cdot 5 \right \rbrace - \frac{2 \cdot 47 - 33}{55} = 291.$
This is easily verified with WolframAlpha, or similar.
For $ r < b $ the final term in $(1)$ would be $ - \frac{2r+c-b}{c}.$
(To be continued... time and enthusiasm permitting :-) )