Flooded Floors

The problem is actually based on a single conjecture which is
$\bigg\lfloor \frac{x}{2} \bigg\rfloor + \bigg\lfloor \frac{x+1}{2} \bigg\rfloor = \lfloor x \rfloor \text{ } \forall x \in \mathbb{R}$ Firstly, I will prove the conjecture.

Let $\displaystyle \lfloor x \rfloor = I$ and $\displaystyle \{x\} = f$ , where $I$ is an integer and $f$ is a fraction $\displaystyle ( 0 \leq f < 1)$ .
Case $1$ : $I$ is even, i.e., $I = 2p$ for some integer $p$ .
$\displaystyle \Rightarrow \big\lfloor \frac{x}{2} \big\rfloor = \big\lfloor \frac{I+f}{2} \big\rfloor = \big\lfloor \frac{2p+f}{2} \big\rfloor = \big\lfloor p +\frac{f}{2} \big\rfloor = p$

$\displaystyle \Rightarrow \big\lfloor \frac{x+1}{2} \big\rfloor = \big\lfloor \frac{I+f+1}{2} \big\rfloor = \big\lfloor \frac{2p+f+1}{2} \big\rfloor = \big\lfloor p + \frac{f+1}{2} \big\rfloor = p$

$\displaystyle \Rightarrow \big\lfloor \frac{x}{2} \big\rfloor + \big\lfloor \frac{x+1}{2} \big\rfloor = p+p =2p = I = \lfloor x \rfloor$

Case $2$ : $I$ is odd, i.e., $I = 2p+1$ for some integer $p$ .
$\displaystyle \Rightarrow \big\lfloor \frac{x}{2} \big\rfloor = \big\lfloor \frac{I+f}{2} \big\rfloor = \big\lfloor \frac{2p+1+f}{2} \big\rfloor = \big\lfloor p +\frac{f+1}{2} \big\rfloor = p$

$\displaystyle \Rightarrow \big\lfloor \frac{x+1}{2} \big\rfloor = \big\lfloor \frac{I+f+1}{2} \big\rfloor = \big\lfloor \frac{2p+f+2}{2} \big\rfloor = \big\lfloor p+1 + \frac{f}{2} \big\rfloor = p+1$

$\displaystyle \Rightarrow \big\lfloor \frac{x}{2} \big\rfloor + \big\lfloor \frac{x+1}{2} \big\rfloor = p+p+1 =2p+1 = I = \lfloor x \rfloor$ .

Hence, Proved. Now we would apply this repeatedly by replacing with different values.

$\begin{array}{c}& & \\ x \to z & \Rightarrow \bigg\lfloor \frac{z}{2} \bigg\rfloor + \bigg\lfloor \frac{z+1}{2} \bigg\rfloor = & \lfloor z \rfloor \\ x \to \frac{z}{2} & \Rightarrow \bigg\lfloor \frac{z}{4} \bigg\rfloor + \bigg\lfloor \frac{z+2}{4} \bigg\rfloor = & \bigg\lfloor \frac{z}{2} \bigg\rfloor \\ x \to \frac{z}{4} & \Rightarrow \bigg\lfloor \frac{z}{8} \bigg\rfloor + \bigg\lfloor \frac{z+4}{8} \bigg\rfloor = & \bigg\lfloor \frac{z}{4} \bigg\rfloor \\ \vdots & \vdots & \vdots \\ \end{array}$

Adding all of these, we obtain, $\bigg\lfloor \frac{z+1}{2} \bigg\rfloor + \bigg\lfloor \frac{z+2}{4} \bigg\rfloor + \bigg\lfloor \frac{z+4}{8} \bigg\rfloor + \dotsb + = \lfloor z \rfloor$ $\displaystyle \Rightarrow \sum_{r=1}^\infty \bigg\lfloor \frac{z+2^{r-1} }{2^r} \bigg\rfloor = \lfloor z \rfloor$

Therefore, $\displaystyle f(x) = \lfloor x \rfloor \Rightarrow f(20.14) = \boxed{20}$