me when i see arul hii arul

Surely $$$2199 - 150 \cdot 148 \cdot 146 \cdot \, ... \, \cdot 2 \lt -40$$$

Thanks a lot. I have to find it for a long time.

P/S: Firstly I wonder why you could do the animation so fast in the contest, and at last I find out that you are one of the tester, which is make sense :v.

Could you please tell me what technique are you refereeing to?

Next time, please format your code properly before commenting to ensure readability. ;)

From what I'm understanding here, in C1 you are keeping $$$dp_i$$$ as the maximum value obtainable, and $$$dx_i$$$ being the minimum one.

We can add $$$dpc_i$$$ and $$$dxc_i$$$ to supply this process with counting. At first, $$$dpc_0 = dxc_0 = 1$$$. For any $$$i \gt 0$$$:

• If $$$dx_i \ge 0$$$, $$$dxc_i = 2 \cdot dxc_{i-1}$$$, as $$$|dx_i| = dx_i$$$. Otherwise, $$$dxc_i = dxc_{i-1}$$$.
• For $$$dp_i$$$, notice that we have two different pathways: one from old $$$dx_{i-1}$$$ and one from old $$$dp_{i-1}$$$. We should combine the number of ways for both pathways if and only if $$$dp_{i-1} \ne dx_{i-1}$$$ (two pathways should have been separated before) and $$$|dx_{i-1} + v_i| = dp_{i-1} + v_i$$$ (they should yield the same $$$dp_i$$$ value). If not, pick the pathway that would result in $$$dp_i$$$, or pick any of them if $$$dx_{i-1} = dp_{i-1}$$$.
• If picking $$$dx_{i-1}$$$ for $$$dp_i$$$, $$$dpc_i = dxc_{i-1}$$$ or $$$dpc_i = 2 \cdot dxc_{i-1}$$$, depending on whether $$$dx_{i-1} + v_i \ge 0$$$.
• Similarly, if picking $$$dp_{i-1}$$$ for $$$dp_i$$$, $$$dpc_i = dpc_{i-1}$$$ or $$$dpc_i = 2 \cdot dpc_{i-1}$$$.

Sorry for the very late reply, it was midnight at my place when you asked this, and I wasn't really mentally sane to read and parse unformatted code. Now that it's morning time here and I'm a bit more conscious that I could actually read you well enough to provide an explanation catering to your own codes.

Code references (upsolved, I overcomplicated myself in-contest so I won't show you those), in case my words were overloading:

• C1: 264993147
• C2: 264993628

I had a different approach from the solution above, it seems easier to me, maybe it will be useful:

look at all not 'a' — let's say there are k of them. You will cover them with t-strings. So each t-string contains some divisor of k non-'a' letters. Iterate over all divisors d and check if strings

0-th non-'a' -> (d- 1)th non-'a',

dth -> (2*d — 1)th,

(2*d)th -> (3*d — 1)th are the same (I used hashing)

If so, we have found a good t, but it can be extended by additional a's on the left and right — this needs to be carefully calculated using amounts of a to the left from first t, to the right from last and between ts.