From 1c1346689eb51bc620210f2c8761881e48f25ca9 Mon Sep 17 00:00:00 2001 From: Barrett Ruth Date: Sun, 23 Feb 2025 18:41:56 -0500 Subject: [PATCH] feat(cses): more problems --- posts/algorithms/cp-log.html | 193 ++++++++++++++++++----------------- styles/post.css | 1 + 2 files changed, 103 insertions(+), 91 deletions(-) diff --git a/posts/algorithms/cp-log.html b/posts/algorithms/cp-log.html index fe30e69..e905a1f 100644 --- a/posts/algorithms/cp-log.html +++ b/posts/algorithms/cp-log.html @@ -191,97 +191,108 @@ Everyone recommends CSES so I started with it, doing the first 8 problems.

-

- weird algorithm -

-

Trivial, but I forgot to print 1 at the end.

-

- Return the exactly correct answer. -

-

- - missing number - -

-

N/A

-

- - repetitions - -

-

Use invariants.

-

- - increasing array - -

-

- Run through one iteration of the algorithm. Here, I erroneously - added x - last to a quantity, - after manipulating x. -

-

- permutations -

-

- I'd seen this problem before yet struggled. - Fully understand the problem constraints. In this case, While - I understood the definition of a permissible permutation, I didn't - fully internalize that you could place number wherever you - want. Instead, I was locked in on placing some x at - i, i + 2, i + 4, .... Further, the fact that I didn't - immediately recognize this solution means I need to improve at - upsolving and reviewing problems. -

-

- permutations -

-

- Absolutely disastrous. I continually just f*dged with the offsets I - was adding to my strategy until I happened to get the answer right. - Don't guess. Also, - don't be lazy—if an algorithm works, focus, write it out, - and enjoy being correct. -

-

- two knights -

-

- Required 2 hints from Sam Altman. git gud at combinatorics. - Use the paradigm "count good, remove bad." Lock in less on counting - specifics—instead, consider what objects - mean in aggregate. In this case, a \(2\times3\) grid - represents an "area" of attack, contributing 2 bad knight pairs. - This is much easier to digest then attempting to remove overcounting - per-knight. Fundamentally, the problem involves placing 2 knights, - so breaking it down 2 knights at a time is the most intuitive take. -

-

- two sets -

-

- Don't lock in on one approach. Here, this is dp. The fact - that I knew the idea of partitioning the first \(n\) numbers into - two groups of size \(\frac{n(n+1)}{4}\) but failed to recognize the - greedy approach means I didn't grasp the fundamental arithmetic of - the problem, nor the greedy idea: every number must go into a set. - If you add the largest number possible to set 1 to not exceed the - target, this number can always be formed in the other set by - choosing \(1\) and \(x-1\). git gud at greedy. -

+ +

more cses—22/2/2025

+ diff --git a/styles/post.css b/styles/post.css index fb2e73a..c883b7e 100644 --- a/styles/post.css +++ b/styles/post.css @@ -15,6 +15,7 @@ header { ul { list-style: unset; + list-style-type: "- "; } li {