feat(cp): cses
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>938 (div. 3)</a
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>—15/2/2025
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</h2>
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<div class="fold">
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<p>
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What would've been my best contest. Unfortunately, CodeForces
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decided to go down for TREE[3] centuries, which absolutely ruined
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my groove in the contest and terminated my virtual. No excuses,
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though, as I set a timer and finished up later.
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</p>
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</div>
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<p>
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What would've been my best contest. Unfortunately, CodeForces
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decided to go down for TREE[3] centuries, which absolutely ruined my
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groove in the contest and terminated my virtual. No excuses, though,
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as I set a timer and finished up later.
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</p>
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<h3>A</h3>
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<p>Brute-forced it but it still took me a few minutes.</p>
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<ul>
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@ -136,10 +134,10 @@
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</ul>
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<h3>E</h3>
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<p>
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I had mentally tapped out by this point (I submitted a TLE \(O(n^2k)\)
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solution without using my brain). I solved F first, then took a
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look at G <i>before</i> coming back to E, robbing me of 10 minutes
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that could've been the difference between another solve.
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I had mentally tapped out by this point (I submitted a TLE
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\(O(n^2k)\) solution without using my brain). I solved F first, then
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took a look at G <i>before</i> coming back to E, robbing me of 10
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minutes that could've been the difference between another solve.
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</p>
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<ul>
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<li>
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@ -188,6 +186,102 @@
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immediately disproves dp.
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</li>
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</ul>
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<h2>cses—21/2/2025</h2>
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<p>
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Everyone recommends CSES so I started with it, doing the first 8
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problems.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1068" target="_blank"
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>weird algorithm</a
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>
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</h3>
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<p>Trivial, but I forgot to print 1 at the end.</p>
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<p>
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<b>Return the exactly correct answer.</b>
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1083" target="_blank">
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missing number
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</a>
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</h3>
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<p>N/A</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1069" target="_blank">
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repetitions
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</a>
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</h3>
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<p>Use invariants.</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1094" target="_blank">
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increasing array
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</a>
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</h3>
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<p>
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Run through one iteration of the algorithm. Here, I erroneously
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added <code>x - last</code> to a quantity,
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<i>after manipulating <code>x</code></i
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>.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1070/" target="_blank"
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>permutations</a
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>
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</h3>
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<p>
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I'd seen this problem before yet struggled.
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<b>Fully understand the problem constraints</b>. In this case, While
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I understood the definition of a permissible permutation, I didn't
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fully internalize that you could place number <i>wherever</i> you
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want. Instead, I was locked in on placing some <code>x</code> at
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<code>i, i + 2, i + 4, ...</code>. Further, the fact that I didn't
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immediately recognize this solution means I need to improve at
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<b>upsolving and reviewing problems</b>.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1071" target="_blank"
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>permutations</a
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>
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</h3>
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<p>
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Absolutely disastrous. I continually just f*dged with the offsets I
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was adding to my strategy until I happened to get the answer right.
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<b>Don't guess</b>. Also,
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<b
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>don't be lazy—if an algorithm works, focus, write it out,
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and enjoy being correct</b
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>.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1072" target="_blank"
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>two knights</a
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>
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</h3>
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<p>
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Required 2 hints from Sam Altman. <b>git gud at combinatorics</b>.
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Use the paradigm "count good, remove bad." Lock in less on counting
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specifics—instead, consider what objects
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<i>mean in aggregate</i>. In this case, a \(2\times3\) grid
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represents an "area" of attack, contributing 2 bad knight pairs.
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This is much easier to digest then attempting to remove overcounting
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per-knight. Fundamentally, the problem involves placing 2 knights,
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so breaking it down 2 knights at a time is the most intuitive take.
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</p>
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<h3>
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<a href="https://cses.fi/problemset/task/1092" target="_blank"
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>two sets</a
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>
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</h3>
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<p>
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<b>Don't lock in on one approach</b>. Here, this is dp. The fact
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that I knew the idea of partitioning the first \(n\) numbers into
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two groups of size \(\frac{n(n+1)}{4}\) but failed to recognize the
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greedy approach means I didn't grasp the fundamental arithmetic of
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the problem, nor the greedy idea: every number must go into a set.
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If you add the largest number possible to set 1 to not exceed the
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target, this number can always be formed in the other set by
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choosing \(1\) and \(x-1\). <b>git gud at greedy</b>.
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</p>
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</article>
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</div>
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</main>
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