One way of thinking about it is that you can never lose points that you
earned on visible sets, but only your last attempt counts for invisible
sets.

This makes a lot of sense, because it means you don't have to waste time
resubmitting a solution that worked on visible sets if you get them wrong
later in the contest. So like with the old scoring system, you only get one
shot at the Large (invisible) but once the Small (visible) is solved, it's
solved.

On Mon, May 7, 2018, 05:11 Valentin Vidić, <[email protected]> wrote:

> Yes, I think you always get visible points, but hidden points only come
> from the last attempt sent.
>
> On Sat, May 5, 2018 at 4:19 PM, <[email protected]> wrote:
>
>> Okay, Round 1C is over and it's finally time to read the rules :D
>>
>> Here is the part from FAQ that confused me:
>> """
>> For example, consider a problem with four test sets: two Visible (I and
>> II) followed by two Hidden (III and IV). Suppose that your attempts, in
>> order, are as follows:
>>
>>     Passes I, but fails II.
>>     Passes I, II, and III, but fails IV.
>>     Passes I, II, III, and IV.
>>     Passes I and II, but fails III.
>>
>> In this case, we would use attempt 2 as your scoring attempt, and you
>> would earn points for the first two test sets. Your penalty time would be
>> the time of attempt 2, and you would have 1 penalty attempt to reflect the
>> 1 attempt before that. Note that it does not matter that attempt 3 would
>> have earned full points;
>> """
>> So, does it mean that, for instance, if I decided to quickly solve
>> C-visible in 10-15 mins by brute-force to gain sure points, any later
>> attempts to solve C fully would be ignored in the final scoreboard?
>>
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