Блог пользователя pb0207

Автор pb0207, история, 22 месяца назад, По-английски

Hello! I'm happy to announce The 1st Universal Cup. Stage 2: Hong Kong, which will be held on Feb 4th, 2023.

This contest is prepared by Zhejiang University, which has already been used as the The 47th ICPC Asia Hong Kong Regional Contest. There are 127 teams participating in the Regional. This contest also serves as Day 3 in 2023 Winter Petrozabodsk Camp on Feruary 2nd, 2023.

The contest will start soon. You can participate in the contest in the following three time windows:

  • Feb 4th 13:00 — 18:00 (UTC +8)
  • Feb 4th 19:00 — 24:00 (UTC +8)
  • Feb 5th 02:00 — 07:00 (UTC +8)

Please note that you can see two scoreboards in DOMjudge. The 'Local Scoreboard' shows the standings ONLY IN THE CURRENT TIME WINDOW. And the 'Combined Scoreboard' shows all participants, including the onsite participants, and the cup participants in the previous time windows.

Authors: Claris, Sugar_fan, Suika_predator, oipotato, qiqi20021026 ,pb0207

Testers: gisp_zjz, Roundgod, triple__a, forxen, idxcalcal, Joyemang, RDDCCD, dyxg, fr200110217102, njupt_lyy, kimoyami, yezzz, HeartFireY, gongkoufadongji, apeiya, gym580299, PaperCloud

Contest link: https://domjudge.qoj.ac/

Regional Scoreboard: https://board.xcpcio.com/icpc/47th/hongkong

Universal Cup Scoreboard: https://qoj.ac/ucup/scoreboard

About Universal Cup:

Universal Cup is a non-profit organization dedicated to providing trainings for competitive programming teams. Up to now, there are more than 200 teams from all over the world registering for Universal Cup.

A more detailed introduction: https://codeforces.net/blog/entry/111672

Register a new team: https://ucup.ac/register

UPD: The contest is in GYM now! https://codeforces.net/gym/104172

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22 месяца назад, # |
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Auto comment: topic has been updated by pb0207 (previous revision, new revision, compare).

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22 месяца назад, # |
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PB IS GOD

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21 месяц назад, # |
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How to solve problem I?

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21 месяц назад, # |
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21 месяц назад, # |
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Could someone explains the solution for E ? I don't get the part and the problem is reduced to add 1 or −1 to some interval and count the number of global minimum value

Link to the editorial https://vj.csgrandeur.cn/e276046fbf0f45b090b2e42bf0db8d55?v=1676207982