Subtask 1

Let Mike bet ~X~ dollars that the Geese will win and ~Y~ dollars that the Kangaroos will win. Then, he will bet a total of ~X+Y~ dollars, and in the worst case scenario, he will win ~\min(X \times \frac{B}{A}, Y \times \frac{D}{C})~ dollars.

We can brute force over all possible values of ~X~ and ~Y~, and check if there exists a combination such that ~X+Y < \min(X \times \frac{B}{A}, Y \times \frac{D}{C})~.

Time complexity: ~\mathcal{O}(100^2 \times T)~

Subtask 2

For Mike to guarantee a profit, we require both ~X+Y < X \times \frac{B}{A}~ and ~X+Y < Y \times \frac{D}{C}~ to be true.

Rearranging the first equation, we see that ~Y < X \times (\frac{B}{A} - 1)~.

Rearranging the second equation, we see that ~X < Y \times (\frac{D}{C} - 1)~.

Combining these equations, we obtain ~Y < (\frac{B}{A} - 1) \times (\frac{D}{C} - 1) \times Y~. Dividing both sides by ~Y~, we obtain ~1 < (\frac{B}{A} - 1) \times (\frac{D}{C} - 1)~. Therefore, if a solution exists, then ~(\frac{B}{A} - 1) \times (\frac{D}{C} - 1) > 1~ must be true. To avoid precision errors, we can instead check whether ~(B-A) \times (D-C) > A \times C~ is true.

It turns out that if this inequality is true, then a solution must exist. For example, we can set ~X = 1~ and ~Y~ to any value greater than ~\frac{1}{\frac{D}{C} - 1}~ and less than ~(\frac{B}{A} - 1)~.

Hence, we can solve each scenario in ~\mathcal{O}(1)~ time.

Time complexity: ~\mathcal{O}(T)~