Subtask #1:

$$x = y \text{ and } u \le v$$

The inequality can be rewritten as:

$$\left\lfloor \dfrac{x^u}{y^v} \right\rfloor = \left\lfloor \dfrac{x^u}{x^v} \right\rfloor = \left\lfloor x^{u - v} \right\rfloor$$

The result should be:

$$a + b = (2 \times \left\lfloor x^{u - v} \right\rfloor) + 1$$

Advanced assessment: The additional constraint for ~u~ and ~v~ is ~u \le v~. Therefore:

$$u - v \le 0 \Rightarrow 0 < x^{u - v} \le 1$$

Check if ~u = v~, then ~x = y \text{ and } u = v \Rightarrow \left\lfloor \dfrac{x^u}{y^v} \right\rfloor = \left\lfloor \dfrac{x^u}{x^u} \right\rfloor = 1 \Rightarrow a \le \left\lfloor \dfrac{x^u}{y^v} \right\rfloor < b \Rightarrow a \le 1 < b \Rightarrow a + b = 2 \times 1 + 1 = 3~.

Otherwise, ~u \ne v~, then ~0 < \dfrac{x^u}{y^v} < 1 \Rightarrow (a = 0) \le \left\lfloor \dfrac{x^u}{y^v} \right\rfloor < (b = 1) \Rightarrow a + b = 2 \times 0 + 1 = 1~.

Subtask #2:

$$u = v \text{ and } x \le y$$

The inequality can be rewritten as:

$$\left\lfloor \dfrac{x^u}{y^v} \right\rfloor = \left\lfloor \dfrac{x^u}{y^u} \right\rfloor = \left\lfloor \left( \dfrac{x}{y} \right)^u \right\rfloor$$

The result should be:

$$a + b = \left(2 \times \left\lfloor \left( \dfrac{x}{y} \right)^u \right\rfloor \right) + 1$$

Advanced assessment: The additional constraint for ~x~ and ~y~ is ~x \le y~. Therefore:

$$x \le y \Rightarrow \dfrac{x}{y} \le 1 \Rightarrow 0 < \left( \dfrac{x}{y} \right)^u \le 1$$

Check if ~x = y~, then ~x = y \text{ and } u = v \Rightarrow \left\lfloor \dfrac{x^u}{y^v} \right\rfloor = \left\lfloor \dfrac{x^u}{x^u} \right\rfloor = 1 \Rightarrow a \le \left\lfloor \dfrac{x^u}{y^v} \right\rfloor < b \Rightarrow a \le 1 < b \Rightarrow a + b = 2 \times 1 + 1 = 3~.

Otherwise, ~x \le y~, then ~0 < \left( \dfrac{x}{y} \right)^u < 1 \Rightarrow (a = 0) \le \left( \dfrac{x}{y} \right)^u < (b = 1) \Rightarrow a + b = 2 \times 0 + 1 = 1~.

Subtask #3: No additional constraints

$$ \dfrac{x^u}{y^v} = 10^{\log_{10}{\frac{x^u}{y^v}}} $$ $$ \log_{10}{\dfrac{x^u}{y^v}} = \log{x^u} - \log{y^v} = u \times \log{x} - v \times \log{y} $$ $$ \Rightarrow \dfrac{x^u}{y^v} = 10^{u \times \log{x} - v \times \log{y}} $$ $$ a + b = 2 \times \left\lfloor 10^{u \times \log{x} - v \times \log{y}} \right\rfloor + 1 $$