Problem 1092. Decimation

Also, I agree with Andrew Newell's earlier comment that there should be a test case for kill_every = 1.

There are so many problems with the problem description.

1). There is no explanation as to why 10 is wrapped around to the front in the second iteration. If it is okay to have a group of one in the first iteration; why not the second?

2). On the "Third iteration" line, the 1st and 4th are chosen, instead of indexing by 3, 6, 9, ... etc.

3). Why was the array expanded to a length of 12 for the third iteration, 8 for the second iteration, and 4 for the fourth iteration?

I want to like this problem, it sounds interesting. I think any valid solution is achieved by accident rather than design. The person posting this problem should be ashamed at the poor description of the problem. This really sets the bar low, shame.

@Jason the key realization that'll address all the concerns you raise is that you shouldn't think of decimation as happening in discrete rounds to begin with.

Instead, imagine that the prisoners are standing in a circle, and Carnage Maximus is standing in the middle, pointing to each living prisoner in turn. When the number of times he has pointed is divisible by kill_every, then that prisoner is taken away to be executed, and the decimation process continues. The circle is now smaller - but Carnage Maximus is ignorant of this and simply keeps on pointing.

In other words, in the example, you start with this sequence of prisoners (this is what you'd get if you pointed to each person in the circle in term, without ever taking anyone away):

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 ...

After 3, 6 and 9 have been removed - with Carnage Maximus still pointing at the empty spot where 9 used to stand -, the circle is now

1 2 4 5 7 8 . 10 1 2 4 5 7 8 10 1 2 4 5 7 8 10 1 2 ...

and the . is where Carnage Maximus is pointing. Moving on, the 3rd and 6th prisoners he points to next are 2, 7, so the circle is now

1 4 5 . 8 10 1 4 5 8 10 1 4 5 8 10 ...

Now the 3rd and 6th prisoners he'll point to next are 1 and 8, after which the circle is

4 5 . 10 4 5 10 4 5 10 ...

Then 5 is removed next,

4 . 10 4 10 4 10 ...

and after that 10 is removed, leaving 4. I hope that this clears up what's happening and why it does, in fact, make perfect sense!