International Mathematical Olympiad

To solve the functional equation
$$f(\left\lfloor x\right\rfloor y) = f(x) \left\lfloor f(y) \right\rfloor$$
for all $x, y \in \mathbb{R}$, where $\left\lfloor a \right\rfloor$ denotes the greatest integer not greater than $a$, we proceed as follows:

### Step 1: Analyze the Equation for $x = 0$

Substitute $x = 0$ into the equation:
$$f(\left\lfloor 0 \right\rfloor y) = f(0) \left\lfloor f(y) \right\rfloor.$$
Since $\left\lfloor 0 \right\rfloor = 0$, we have:
$$f(0) = f(0) \left\lfloor f(y) \right\rfloor.$$
This equation implies that either $f(0) = 0$ or $\left\lfloor f(y) \right\rfloor = 1$ for all $y$.

### Step 2: Consider the Case $f(0) = 0$

If $f(0) = 0$, the equation becomes:
$$f(\left\lfloor x\right\rfloor y) = f(x) \left\lfloor f(y) \right\rfloor.$$
Substituting $y = 1$ gives:
$$f(\left\lfloor x \right\rfloor) = f(x) \left\lfloor f(1) \right\rfloor.$$
If $\left\lfloor f(1) \right\rfloor = 0$, then $f(x) = 0$ for all $x$, which is one possible solution. Thus, $f(x) = 0 \quad \forall x \in \mathbb{R}$.

### Step 3: Consider the Case $\left\lfloor f(y) \right\rfloor = 1$

If $\left\lfloor f(y) \right\rfloor = 1$ for all $y$, then:
$$1 \le f(y) < 2 \text{ for all } y.$$
In this case, the original equation simplifies to:
$$f(\left\lfloor x \right\rfloor y) = f(x).$$
For all $y \neq 0$, choosing $x = 0$ gives:
$$f(0) = f(0) \quad \text{trivial identity}.$$
For specific $y$ values like $y = n \in \mathbb{Z}$, if $1 \leq f(n) < 2$, and considering continuity or piecewise constant functions, one possible solution is that $f(x) = c \quad \forall x \in \mathbb{R}$, where $1 \leq c < 2$.

### Conclusion

Therefore, the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the given functional equation are:
$$\boxed{f(x) = 0 \quad \forall x \in \mathbb{R}, \quad f(x) = c \quad \forall x \in \mathbb{R}, \text{ where } 1 \leq c < 2}.$$