barrett/barrettruth.com
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

feat(codeforces): 779

Browse source
This commit is contained in:
Barrett Ruth 2025-04-10 12:43:58 -04:00
parent d7c50e5739
commit dcaea06cc0
1 changed files with 100 additions and 1 deletions
Download patch file Download diff file Expand all files Collapse all files

101
posts/algorithms/competitive-programming-log.html
View file

@ -35,6 +35,104 @@
<h1 class="post-title">Competitive Programming Log</h1> <h1 class="post-title">Competitive Programming Log</h1>
</header> </header>
<article class="post-article"> <article class="post-article">
<h2>
<a href="https://codeforces.com/contest/1692" target="_blank"
>799 (div. 4)</a
>&mdash;10/4/2025
</h2>
<div>
Improvement is marginal. My desire for rating is unquenchable
(joke). Really, though, I'm improving slower than I like. 1400
performance, my best yet&mdash;I'll do one more Div. 4 then I need
to upgrade to Div. {2,3}. I think the most core realization I made
after this contest was:
<blockquote>
Separate what your code does from what you think it should do.
Conceptualize an approach, then put it aside. Implement exactly
what the approach says to do (not what you think it should say).
If the approach left something out,
<i
>step away from implementing, reconsider the approach, then
resume implementing</i
>. <b>Never reconsider and alter your strategy while coding.</b>
</blockquote>
</div>
<ol>
<li>
B: realized parity-based removal immediately. Still, I once again
didn't initially do what the problem asked (i.e. I returned the
number of operations, not array length).
</li>
<blockquote>
Never run code until you're confident it does what you want.
</blockquote>
<li>
C: Blindly plugged in incorrect deltas to detect the "X" pattern
of "#" characters before getting the right answer.
<blockquote>
I consistently tune out when I'm bored/think I've discovered the
insight/done the problem before. This results in me solving a
different problem, expressing right ideas incorrectly, or
expressing wrong ideas correctly&mdash;for example, thinking I
find a core insight, believing it sounds right, and just
<i>not checking its validity</i>. I don't know what to say
besides progressively overload your discipline and focus.
</blockquote>
</li>
<li>
D: my proof that the time must repeat was complete garbage and
complex. I've been reading AOPS and wanted to go with a math-y
approach. However, consider the following simple proof:
<blockquote>
The current time is \(s\) and you look at the clock every \(x\)
minutes. \(1440 \cdot x\equiv 0\pmod{1400}\), so the time must
repeat after at most 1440 looks.
</blockquote>
This is enough to brute force. There are alternative easier
approaches to brute-force, but for an easy problem like this speed
comes above all else (special case).
</li>
<li>
E: <b>Submitted and hoped for correctness multiple times</b>. I
was lazy and didn't want to implement the binary search approach.
Ironically, after turning my brain on, I quickly found a
\(\Theta(n)\) solution. However, I rushed to code it up thinking
about the time I had wasted and frustration from WA (more from
myself than from WA). This caused me to forget invariants and
implement what I thought was right, not the algorithm I designed
(forgetting invariants). Walking through an example (as I advised
myself yesterday) reminded me to iterate the right pointer past
any zeroes. Good, specific improvement from yesterday, but I still
should've slowed down and asked "where should the right pointer
move? what are the invariants? why? what about if if goes off the
end of the array?" <b>Only look forward</b>.
</li>
<li>
F: Came up with fancy equations because I'm taking the wrong core
lessons from AOPS. The example problems there are complex, while
these are simple&mdash;therefore, I should prioritize simplicity.
Concretely, the following phrasing massively simplifies this
problem:
<blockquote>
I'm looking for 3 numbers \(a[i],a[j],a[k], i\neq j\neq k\) such
that \(a[i]+a[j]+a[k]\equiv 3\pmod{10}\). WLOG, assume \(i\lt
j,k\). We only want a pair of remainders that satisfy the above
inequality. Since there are only 10 unique remainders, brute
force all 100 remaining modulo pairs. If such a pair exists in
\(a[i+1:]\forall\) \(i\in\{1,...n-2\}\), the answer is yes. This
can be calculated with a postfix frequency map of remainders,
with special care taken for when the remainders are equal.
</blockquote>
</li>
<li>
G: good deduction and simplicity but I once again read the
instructions too fast and skipped that the array is size \(k+1\).
Also,
<b>my sliding window implementation ability is horrendous</b>. I
spent nearly 10 minutes just debugging a "right" idea because I
couldn't code up a basic thing.
</li>
</ol>
<h2> <h2>
<a href="https://codeforces.com/contest/1791" target="_blank" <a href="https://codeforces.com/contest/1791" target="_blank"
>849 (div. 4)</a >849 (div. 4)</a
@ -75,7 +173,8 @@
>Did not read the problem statement and answered a similar (but >Did not read the problem statement and answered a similar (but
very different) problem I'd done in the past</b very different) problem I'd done in the past</b
>. By the time I saw this after impatiently submitting ~5 WA, I >. By the time I saw this after impatiently submitting ~5 WA, I
had lost mental focus. Never submit and hope for a correct answer&mdash;<i>know it</i>. had lost mental focus. Never submit and hope for a correct
answer&mdash;<i>know it</i>.
</li> </li>
<li> <li>
F: Then, I let my previous failure carry through to the next F: Then, I let my previous failure carry through to the next