+ First contest in a while. Implementation lacked heavily but solving
+ harder problems made easier problems easier.
+
+
+
+ A: rust immediately showed with the brute force. Since \(n\cdot
+ m\leq 25\), at most \(\lceil\log_2(25)\rceil=5\) concatenations
+ need to be made. Slow down and consider constrains.
+
+
+ B: Similarly, mathematical insight and just
+ playing with parameters/quantifying the effects of
+ operations is important.
+ Test your conjectures, they may be right/helpful (ie. "I
+ must maintain the shortest length thread").
+
+
+ C: implementation weak. Simplify, step back, simplify,
+ continuously. I stopped considering altering the grid and used a
+ pair of coordinates but deriving the others inline,
+ if you trust your mathematics, is way simpler.
+
+
+ D: formalize your answer better. Understand prime factorization
+ more. Improve/memorize asymptotic bounds of factoring and prime
+ counting. Don't overcomplicate the problem—here, I
+ erroneously thought it was asking for minimum operations, not
+ possibility. In reality, all this problem asks is:
+
+ "Are the total number of each factor greater than one divisible
+ by \(n\)?"
+
+
+
+ E: dp rust. Simplify your thought process—look back into
+ SRTBOT (i.e. define the subproblem). If the subproblems depend
+ rightward, iterate right to left. That simple.
+
+
F: coming back for you!
+
I must heed the advice of Colin Galen. I rush trivial problems
because they're boring, then forget an edge case. I get overwhelmed
@@ -54,8 +95,8 @@
"this works"\(\rightarrow\) "let's simplify" \(\rightarrow\) "that
doesn't work" \(\rightarrow ...\) over and over again...
even when the right solution comes across my mind. In this
- case, choosing an invariant was just overwhelming. The simplest
- correct solution is always the right one.
+ case, choosing an invariant was just overwhelming.
+ The simplest correct solution is always the right one.