WF23 Luxor _(Long blog ahead. Reader discretion advised.)_ I feel we have a rather interesting WF story (note: we did the 47th finals, as the University of Maryland, <b><i>as 2 people</i></b>... explained below), so I figured I'd tell it. Exposition ---------- It actually starts way before WF, too. Our teams were decided on (some form of) merit based on an internal qualifier. My team for the regional and NAC (North America superregional) ended up as {me, [user:Navick,2024-04-24], [user:mr.banana,2024-04-24]}. We did [fairly](https://nasouth22d1.kattis.com/contests/nasouth22d1/standings) [well](https://nac.icpc.global/wp-content/uploads/2023/06/NAC23-scoreboard.pdf) in those. However, for WF, my two teammates only had single-entry visas for the US, meaning that if they left, they couldn't come back into our country. So they couldn't really go with me... Luckily, in such situations, ICPC lets you just replace your teammates. So we held another internal qualifier, and ended up with two ...

Полный текст и комментарии »

Problems that I authored so far Since competitive programming is dying, and I'm almost retired, so it's time to review the problems I authored. Hi everyone! I wanted to write such a blog for a long time, motivated by similar blogs, [by](https://codeforces.me/blog/entry/113093) [user:antontrygubO_o,2023-02-25], [by](https://codeforces.me/blog/entry/108940) [user:adamant,2023-02-25] and [by](https://codeforces.me/blog/entry/108595) [user:tibinyte,2023-02-25]. This is not a super-comprehensive list. I set many shit problems that I don't want to share for some local contests. It can be long, and I'm not sure if I have finished half of them yet. The number of asterisks after the label indicates the recommendation levels. One asterisk means this problem is worth reading. Two asterisks mean this problem is one of my favorite problems, Three asterisks mean this problem is one of my best problems. | # | Date | Problem | Contest | Comment | | ---- | ----------- | --------------------------------------------...

Полный текст и комментарии »

Problem Topics Good Day to you! I've been asked to make some topic-wise list of problems I've solved. Even though I couldn't involve all problems, I've tried to involve at least "few" problems at each topic I thought up (I'm sorry if I forgot about something "easy"). I've alredy made such list once anyway I've tried to include more problems now &mdash; so here it is: <spoiler summary="aho"> http://www.spoj.com/problems/ADAJOBS/ URI 2226 (5) //[NICE][NUMBERS][DP] http://www.spoj.com/problems/SUB_PROB/en/ http://codeforces.me/contest/696/problem/D 8 http://www.spoj.com/problems/AHOCUR/ 5 //Aho-Corassic + DP https://www.codechef.com/problems/LYRC (5) //Sample aho-brute-force http://codeforces.me/problemset/problem/346/B //Proposed by [user:bradyawn,2019-08-03] </spoiler> <spoiler summary="automat"> 6861 [LA] //CYK UVA 10679 //Suffix Automat http://www.spoj.com/problems/STRMATCH/ //Suffix Automat &mdash; trie might do too http://www.spoj.com/problems/NSUBST...

Полный текст и комментарии »

Codeforces Raif Round 1 Editorial [problem:1428A] ------------------ Setter: [user:bensonlzl,2020-10-17] Prepared by: [user:errorgorn,2020-10-17] <spoiler summary="Hint 1"> Consider when $x_1=x_2$ </spoiler> <spoiler summary="Hint 2"> Consider when $x_1 \ne x_2$ </spoiler> <spoiler summary="Solution"> We consider 2 cases. The first is that the starting and ending point lie on an axis-aligned line. In this case, we simply pull the box in 1 direction, and the time needed is the distance between the 2 points as we need 1 second to decrease the distance by 1. The second is that they do not lie on any axis-aligned line. Wabbit can pull the box horizontally (left or right depends on the relative values of $x_1$ and $x_2$) for $|x_1-x_2|$ seconds, take 2 seconds to move either above or below the box, then take another $|y_1-y_2|$ seconds to move the box to $(x_2,y_2)$. </spoiler> <spoiler summary="Code (C++)"> ```c++ #include <bits/stdc++.h> using namespace std; int main(){ ios_base::...

Полный текст и комментарии »

Codeforces Round 1085 (Div. 1 + Div. 2) Editorial <center> <table class="tex-tabular bordertable" style="border-left: none; border-right: none; border-top: 1px solid; border-bottom: 1px solid; border-collapse: collapse;"> <tbody> <tr> <td class="tex-tabular-text-align-center" style="border-left: none; border-right: none; border-top: none; border-bottom: none;"> <span class="tex-font-size-small"> <span class="tex-font-style-it"> <a href="https://www.youtube.com/watch?v=HCYKLnT0UNU">Trophy Presentations — Asuka Ota, Ryo Nagamatsu, Mario Kart Wii</a> </span> </span> </td> </tr> </tbody> </table> </center> **UPD 1:** added hints, problem credits, more specific acknowledgements and some remarks. Implementations are on the way, sorry for making y'all wait! **UPD 2:** Implementations are finally here. [problem:2207A] Author: [user:awang11,2026-03-09] Preparers: [user:awang11,2026-03-09], [user:IceSerpent,2026-03-09] Anal...

Полный текст и комментарии »

Editorial Round 1 of Yandex.Algorithm 2018 <spoiler summary="Police Patrol"> First, let's try to find the largest block that two patrols can cover (hint, it's not k). For instance, we can have one patrol cover $1,2,\ldots,k$. and the other cover $2,3,\ldots,k+1$. So, we can split $n$ houses into blocks of length $k+1$. Now, we have to look at what to do with the leftover houses. If there are more than two empty blocks, we need two patrols to cover both blocks. If there is only one empty block, we only need one patrol. Thus, the answer can be computed by the formula in the code below <spoiler summary="code"> ~~~~~ n,k = map(int, raw_input().split()) print (n/(k+1))*2 + min(2, n%(k+1)) ~~~~~ </spoiler> </spoiler> <spoiler summary="Alphabetic Subsequence"> We can check that this is equivalent to counting the number of permutations of $0,1,2,\ldots,9$ that appear in $s$. There are two solutions: <spoiler summary="solution 1"> Just do it. This is O(10! * |s|) ~ 3*10^8. This is fast enough in some languages. ...

Полный текст и комментарии »