Badminton II

The minimum amount of points that the winner can accumulate is $\text{min(winner)} = \left\lceil \dfrac{n}{2} \right\rceil \cdot k$ .

The maximum amount of points that the loser can accumulate is $\text{max(loser)} = \left(\left\lceil \dfrac{n}{2} \right\rceil - 1\right) \cdot k + \left\lceil \dfrac{n}{2} \right\rceil \left(k - 1\right)$ , which shows that the loser loses one more game than he wins with all maximum possible amounts.

For the loser to be unable to earn more points than the winner, we need $\text{min(winner)} \geq \text{max(loser)}$ , which is equivalent to solving $\left\lceil \dfrac{n}{2} \right\rceil \geq \left(2\left\lceil \dfrac{n}{2} \right\rceil - 1\right) \cdot k - \left\lceil \dfrac{n}{2} \right\rceil$ Setting $n = 2p + 1$ , where $p$ is nonnegative integer, we have $(p + 1)\cdot k \geq \left(2(p + 1) - 1\right) \cdot k - (p + 1)$ where $k \leq \dfrac{p + 1}{p}$ . But since $k > 1$ , the only choice we have is $p = 1, k = 2$ , which gives the answer for the problem.