By Egor, 13 years ago, translation,

This is pretty simple task - we have cycle and must direct all edges on it in one of 2 directions. We need to calculate cost of both orientation and print smallest of them.
There is a small trick - we can calculate cost of only one orientation and then cost of the other will be sum of costs of all edges minus cost of first orientation.

• +24

By Nedy88, 13 years ago,
Hey everybody,

Wellcome to Codeforces Beta Round #24. I'm author to most of the problems today. Something about me: My name is Nedyalko Prisadnikov and I'm a student from Sofia University in Bulgaria. Here are some pictures of me. Many thanks to Mike Myrzayanov and to Artem Rakhov for organizing the contest, writing alternative solutions and some of the problems.

Good luck and have fun!

UPD:
Announcement of Codeforces Beta Round 24
• +58

By RAD, 13 years ago, translation,
Hi everybody

Today the author of the majority of problems is Dmitry Zhukov, many thanks to him for this.
Also I want to thank Mike Mirzayanov for choosing problems for the contest and organizing it and Julia Satushina for the translation of the statements.

Good luck!

UPD:
Announcement of Codeforces Beta Round 23
• +25

By iakolzin, 13 years ago, translation,

Contest discussion

Problem А. Second Order Statistics

Sorting

In this problem one should find a minimal element from all elements, that are strictly greater, then the minimal one or report that it doesn't exist. Of course, there can be a lot of different solutions, but one of the simplest - to sort the given sequence and print the first element, that's not equal to the previous. If all elements are equal, then the required element doesn't exist.

• +32

By RAD, 13 years ago, translation,

Welcome all to Codeforces Beta Round #22

Note that at this time registration is possible during the round. The contest will begin at 19:00 MSK.

Today I am an author of the problems. I would like to thank Mike Mirzayanov for help in contest preparations, Edvard Davtyan and Nickolay Kuznetsov for writing the verification solutions, and Julia Satushina for translating statements into English.

Good luck on the contest!

UPD: The contest is over. Thank you all for participating!
Problems
Results
Winner Kasparyanm_Mihail gains +203 to rating after the contest!

• +23

By MikeMirzayanov, 13 years ago, translation,
I invite you to participate:) Round will be non-rated, but there will not be limited registration at this time! Please, read the rules before the contest!
• +18

By NALP, 13 years ago, translation,
Welcome to Codeforces Beta Round #19.

Authors of today's contest are Artem Rakhov and me. Thanks to Mike Mirzayanov, Edvard Davtyan and Julia Satushina for help in the organisation.

I hope, you will have fun.

Good luck!

English

UPD. The contest has finished and you can see the standings and tasks. The winner and only participant who has solved all the problems is kalinov. Congratulations!

Announcement of Codeforces Beta Round 19
• +46

By MikeMirzayanov, 13 years ago, translation,

You may read complete rules here.

The text contains only preliminary rules of Codeforces contests — something surely will change. In particular, I hope to help the community to find the illogic in the rules. However, I have already discussed the format with programming contests dinosaurs and made unrated alpha contests — it seems rules are OK.

I think than the format 5 tasks x 2 hour is quite good. For this reason, in the near future we will not change it, but of course, experiments are possible.

1. Before the competition all participants are divided into rooms, each room contains about 20 participants. For now random strategy works, but after a while it will be somehow correlated with the rating.

• +23

By Nerevar, 13 years ago, translation,
Two features were added to the interface in order to improve the view of the contest standings.
• +19

By ArtDitel, 13 years ago, translation,

Contest discussion

Problem А. Triangle

Pythagorean theorem, brute force

In this problem you should implement a function, which takes three points and checks whether they form a right-angled triangle. There are a lot of ways to do so, but the simplest one is using a Pythagorean theorem.

• +34