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<h1 class="post-title">Competitive Programming Log</h1>
</header>
<article class="post-article">
<h2>
<a href="https://codeforces.com/contest/2093" target="_blank"
>1016 (div. 3)</a
>&mdash;8/4/2025
</h2>
<div>
Horrendous competition but I refrain from cringing for the sake of
improvement.
</div>
<ul>
<li>A: trivial</li>
<li>
B: took me a while and I still don't get the proof. Good
resilience in trying to formally prove it then detecting a pattern
(remove all non-zero/zero digits after/before the last non-zero
digit).
</li>
<li>
C:
<b>math skills remain weak but pattern recognition is improving</b
>. I "guess and checked" that any \(x+x\) is not prime when
\(x\neq1\). However,
<b
>after catching 2 edge cases \(x=k=1\), \(k=1\), I gave up,
ignoring \(x=1,k=2\),</b
>
causing a WA. This is the downside of lacking a formal
proof/understanding the math&mdash;while in retrospect I can say
"consider a few more edge cases," I can't tell when to stop
investigating. Still, I should've separated out cases \(k=1\),
\(k\neq1\) and exhaustively proved \(k=1\), which is remarkably
facile.
<b
>I still don't know how to factorize numbers in
\(O(\sqrt{n})\)</b
>
and copy from AI (allegedly).
</li>
<li>
D: got the idea but it was very abstract. Rushed to
implementation, wasted time&mdash;same old story. Got it after
contest.
<b
>After failing to implement for a long time, abandon the
approach and start from scratch&mdash;99% of the time you're not
going to get it.</b
>
</li>
<li>
E: failed implementation. At least I saw binary search after
stepping back. I tried to find an upper bound on \(x\) by creating
<i>exactly</i> \(k\) groups but for ease of implementation I
should've went for \(\geq k\) since any sequence with MEX \(x\)
can be extended to have MEX \(\geq x\) with the addition of any
number.
<blockquote>
Specifically define <i>everything</i>. What am I binary
searching over (in this case, forming some \(x\) with
<i>at least</i> \(k\) groups)? What are the bounds? Is the
search space monotonic? Why?
</blockquote>
</li>
<li>
F: I had seen a bitwise trie before and didn't review the problem.
I didn't upsolve then, so I couldn't upsolve now, and getting this
problem right would've made a massive difference in my
performance. These are the consequences&mdash;<b>upsolving is goated</b>.
</li>
</ul>
<h2>
<a href="https://codeforces.com/contest/1873/" target="_blank"
>898 (div. 4)</a