1. If you find that a pair contains the impostor, you don't need to check the two players. Checking one is enough. This means that:
2. If N is even, you can solve the problem in at most N queries by checking all the groups of two except for the last group. If you don't find the impostor in any of them, you only need to check the second to last player.
3. If N is odd and greater than 3, the previous strategy won't work since you'll end up making N + 1 queries if the impostor is the second to last player. In order to fix this, you need to check whether the impostor is one of the last three players or not, which you can do with exactly 3 checks. If all of them are in the clear, you can find the impostor in at most N — 3 queries using the previous strategy.

One is a power of two because 2 ^ 0 = 1.

Use prefix sums to efficiently calculate the answer.

You have N weapons, and each of them has an associated cost and profit. Each weapon can only be picked at most once. You want to pick the weapons in such a way that the sum of their "profits" is maximal.

Division can be expensive and imprecise, especially if you're working with floats. Use cross multiplication to sort the weapons by their costs.

You'll be given three types of numbers: the ones A can chose, the ones B can chose, the ones both can choose and the ones neither one can chose. Do you care about all of them?

Since the game only ends when neither player can choose a number and a player's score is the sum of all the numbers they picked, we can conclude the score of each player is the sum of all the numbers they can pick in that range. Can you think of an efficient way of calculating the answer?

There is one number both players can pick. Since they're playing optimally, they'll always pick it in order to maximize their score. What happens if the amount of times this number appears in the range is even? And what if it's odd?