Sublime Sequence

We have a sequence of length $n$:

$x, -x, x, -x, x, -x, \dots$

It always starts with $x$, then alternates signs.

The important move is to stop thinking about all $n$ terms individually. Pair them up:

$$(x + (-x)) + (x + (-x)) + \dots$$

Each full pair has sum:

$$x + (-x) = 0$$

So every two terms cancel perfectly. No brute force needed, though brute force would also pass because the constraints are tiny. The math solution is just cleaner.

Case 1: $n$ is even

If $n$ is even, the whole sequence splits into pairs:

$$(x, -x), (x, -x), \dots$$

Every pair sums to $0$, so the total sum is:

$$0$$

Example with $x = 3$, $n = 6$:

$$3 - 3 + 3 - 3 + 3 - 3 = 0$$

Case 2: $n$ is odd

If $n$ is odd, all but one term can be paired away. Since the sequence starts with $x$, the unpaired leftover term is also $x$.

Example with $x = 4$, $n = 7$:

$$4 - 4 + 4 - 4 + 4 - 4 + 4 = 4$$

So the answer is:

$$\begin{cases} 0 & \text{if } n \text{ is even} \\ x & \text{if } n \text{ is odd} \end{cases}$$

That is literally the whole problem. The sequence tries to look like a sequence problem, but it is just parity wearing a fake mustache.

Complexity per test case is $O(1)$ time and $O(1)$ memory.