BALTIC-WAY 2001 Problem 4

Let $p$ and $q$ be two different primes. Prove that

$$\left\lfloor\frac{p}{q}\right\rfloor+\left\lfloor\frac{2 p}{q}\right\rfloor+\left\lfloor\frac{3 p}{q}\right\rfloor+\ldots+\left\lfloor\frac{(q-1) p}{q}\right\rfloor=\frac{1}{2}(p-1)(q-1) .$$

(Here $\lfloor x\rfloor$ denotes the largest integer not greater than $x$.)