By ivan.popelyshev, 9 years ago, translation, ,
Lets start from the end.

## Problem С. Commentator problem

Let R be the distance from point А to a circle with center О and radius r. From this point the circle is observed at the angle .
So, the three stadiums are observed at the same angle if R1 / r1 = R2 / r2 = R3 / r3.
Take two different points A, B. The set of points C that AC / BC = const is either a line (perpendicular bisector of AB) or a circle with center somewhere on the line AB. This circle is easy to find. It contains two points that lie on AB and satisfy AC / BC condition.

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By MikeMirzayanov, 9 years ago, translation, ,

As some users have already noticed - contest rating has been added to Codeforces. For now it is in beta too, but it looks very adequate. Here's how it is calculated.

Each person is characterized by their rating, the number R. If person A's rating is RA, and person B's is equal to RB, then the formula

gives the probability that A will get a higher position than B in the round final standings. By the way, here everything is very close to the Elo rating.

Before updating your rating after the end of the round, for each participant his seed is calculated, that is the place that the participant is expected to take in this competition. Thus, two things are known for each participant - his seed (the expected place) and rank (the actual place). Depending on the difference between these two values, your rating increases or decreases. If the difference is higher, your rating changes more.

There are a few technical points:

• if it is the first contest for a participant, his seed is calculated as 1 + n / 2, where n is the total number of participants in the round;
• changes in the ranking of contestants are multiplied by a correction factor such that allows the sum of ratings of the participants to remain unchanged (before and after the round).

As at TopCoder all users are divided into two divisions: the first (rating over 1500 1650) and the second (rating_ not more than 1500 1650). Not rated users fall into the second division automatically.

Wish you high rating,

MikeMirzayanov

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By MikeMirzayanov, 9 years ago, translation, ,

Thank you all for participating in Codeforces Beta Round # 2. I hope you enjoyed it. You may discuss the problems and system in comments. Please express your opinion, especially if you notice any inappropriate behavior. And as always, I will read with interest the suggestions for improvement.

Congratulations to the three leaders: RAVEman, GarnetCrow and ivan.popelyshev!

See you at Codeforces Beta Round # 3.

P.S. And by the way, the round tutorial is waiting for a volunteer. It is desirable that it will be one of the leaders of today's competition. The tutorial should be in Russian and in English. It will be published on the homepage and later will be available via special link from the contest page.

Announcement of Codeforces Beta Round #2
Announcement of Codeforces Beta Round #2

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By removed1, 9 years ago, translation, ,

Problem A.

The constraint that edges of each flagstone much be parralel to edges of the square allows to analyze X and Y axes separately, that is, how many segments of length 'a' are needed to cover segment of length 'm' and 'n' -- and take product of these two quantities. Answer = ceil(m/a) * ceil(n/a), where ceil(x) is the least integer which is above or equal to x. Using integers only, it is usually written as ((m+a-1)/a)*((n+a-1)/a). Note that answer may be as large as 10^18, which does not fit in 32-bit integer.

Most difficulties, if any, contestants had with data types and operator priority, which are highly dependant on language used, so they are not covered here.

Problem B.

Let each letter representation of column number be associated with integer in radix-26, where 'A' = 0, 'B' = 1 ... 'Z'=25. Then, when converting letter representation to decimal representation, we take associated integer and add one plus quantity of valid all letter representations which are shorter than letter representation being converted. When converting from decimal representation to letter representation, we have to decide how many letters do we need. Easiest way to do this is to subtract one from number, then quantity of letter representation having length 1, then 2, then 3, and so on, until next subtraction would have produced negative result. At that point, the reduced number is the one which must be written using defined association with fixed number  of digits, with leading zeroes (i.e. 'A's) as needed.

Note that there is other ways to do the same which produce more compact code, but they are more error-prone as well.

Problem C.

The points can be vertices of regular N-polygon, if, and only if, for each pair, difference of their polar angles (as viewed from center of polygon) is a multiple of 2*pi/N. All points should lie on the circle with same center as the polygon. We can locate the center of polygon/circle [but we may avoid this, as a chord (like, say, (x1,y1)-(x2,y2)) is seen at twice greater angle from center, than it is seen from other point of a cricle (x3,y3)]. There are many ways to locate center of circle, the way I used is to build midpoint perpendiculares to segments (x1,y1)-(x2,y2) and (x2,y2)-(x3,y3) in form y = a*x + b and find their intersection. Formula y = a*x + b has drawback that it cannot be used if line is parallel to y, possible workaround is to rotate all points by random angle (using formulae x' = x*cos(a) - y*sin(a), y' = y*cos(a) + x*sin(a) ) until no segments are horizontal (and hence no perperdiculares are vertical).

After the coordinates of the center are known, we use fancy function atan2, which returns angle in right quadrant: a[i] = atan2(y[i]-ycenter, x[i]-xcenter)

Area of  regular polygon increases with increasing N, so it is possible just to iterate through all possible values on N in ascending order, and exit from cycle as first satisfying N is found.

Using sin(x) is makes it easy: sin(x) = 0 when x is mutiple of pi. So, for points to belong to regular, N-polygon,

sin( N * (a[i]-a[j]) /2 )=0

because of finite precision arithmetic,

fabs( sin( N * (a[i]-a[j]) /2 ) ) < eps

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By MikeMirzayanov, 9 years ago, translation, ,
In this topic I would like to raise issues around Codeforces Beta Round # 1. What did you like? What is not liked? What seemed uncomfortable? What can be changed to make participation more comfortable?  I wonder your opinion on contest web-interface.

Please do not write about access the site with the address http://codeforces.com/ (I recommended to use http://codeforces.com:8081/). I guess the problem is around of Apache Virtual Hosts + AJP Connector. Something is configured incorrectly or doesn't work properly because , then it works so good. In short, I will correct.

I'm waiting for your comments. And, of course, welcome to Codeforces Beta Round # 2.

Also I would like to see someone who wants to write contest tutorial. This must be done in Russian and English languages. Of course you must solve the problem on either contest, or in the practice. If you have a desire to do it - write in comments. Your post will be published on the main page and later available on special link available from the contest page.

Announcement of Codeforces Beta Round #1
Announcement of Codeforces Beta Round #1

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By VladimirYakunin, 9 years ago, translation, ,

Those who has ever been to chinese market (especially if it was tourists oriented one) know that buying particular thing in China is something more than just pay-your-money-and-go-away.
Chinese sellers have a more artistic approach to their job. Who says any good has a fair price? In China fair price is a function of seller's mood, amount of leisure time the buyer has, current moon phase and Buddha's aura condition.
Here are some tips which were helpful for me and my friends in chinese markets:

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By MikeMirzayanov, 9 years ago, translation, ,

Later you'll be introduced to the rules of the Codeforces contests, which differ from those of ACM-ICPC, TopCoder, GCJ, and I hope they'll bring some difference to the world of programming competitions. Most of the official competitions will be carried out according to these rules, though there will be more traditional contests. For example, Codeforces Beta Round #1 will be carried out according to the familiar ACM-ICPC rules. For some time testing will be based on Windows, but things might change in future, fortunately, the system supports testing on different platforms, even within one contest.

At the present time the system is configured to support the following programming languages (the compilation and/or the launching line is shown for each language):

1. GNU C++ 4
g++.exe -static -DONLINE_JUDGE -lm -s -x c++ -Wl,--stack=268435456 -O2 -o {filename}.exe {file}

2. GNU C++11 4
g++.exe -static -DONLINE_JUDGE -lm -s -x c++ -Wl,--stack=268435456 -O2 -std=c++11 -D__USE_MINGW_ANSI_STDIO=0 -o {filename}.exe {file}

3. GNU C 4
gcc.exe -static -fno-optimize-sibling-calls -fno-strict-aliasing -DONLINE_JUDGE -fno-asm -lm -s -Wl,--stack=268435456 -O2 -o {filename}.exe {file}

4. MS VS C++
cl /W4 /F268435456 /EHsc /O2 /DONLINE_JUDGE {file}

5. Free Pascal 2
-n -O2 -Xs -Sgic -viwn -dONLINE_JUDGE -Cs67107839 -Mdelphi -XS {file} -o{filename}.exe

6. Delphi 7
dcc32 -Q -\$M1048576,67107839 -DONLINE_JUDGE -cc {file}

7. C# Mono 2
dmcs -define:ONLINE_JUDGE -o+ -out:{filename}.exe {file}

8. C# .NET
csc.exe /o+ /d:ONLINE_JUDGE /r:System.Numerics.dll /out:{filename}.exe {file}

9. Java 6, 7
javac -cp ".;*" {file}
и
java.exe -Xmx512M -Xss64M -DONLINE_JUDGE=true -Duser.language=en -Duser.region=US -Duser.variant=US -jar %s

10. Ruby
ruby.exe %s

11. Python 2, Python 3
python.exe %s

12. PHP 5
php.exe -n -d ONLINE_JUDGE=true -d display_errors=Off -d error_reporting=0 %s

ghc --make -O %s

14. D
dmd -L/STACK:268435456 -version=ONLINE_JUDGE -O -release -inline -noboundscheck {file}

15. OCaml
ocamlopt nums.cmxa str.cmxa -pp camlp4o -unsafe -o {filename}.exe-ocaml {file}

16. Scala
As Java

17. JavaScript V8
d8 {file}

It is not guaranteed that all the problems will have solutions in all the given languages (it's especially about the scripting ones). Probably, I'll later introduce equalizing coefficients for the working time for some languages. A "plus" next to the version name means that the testing system can use older versions. If you have suggestions about the possible ways to change the compilation or the launching line, write about them in your commentaries.

It should be mentioned that apart from standard verdicts, you can get "Denial of judgement", which usually means that your solution can't be launched, or it has unexpectedly failed. For example, is the Delphi array is too big, the compiler compiles the code, but the result will be the incorrect win32 exe-file. Solutions with the verdicts like "Compilation failed", "Denial of judgement", "Judgement failed" will be ignored while summing the results.

Moreover, pay attention, please, that the problems will be given in English as well as in Russian.

That's it, see you at Codeforces Beta Round#1.

UPD: GCC compiler has been added.

UPD 3.2: Actual compiler versions are

• Mono C# compiler version 3.2.3
• DMD32 D Compiler v2.064.2
• Delphi 7 [Borland Delphi Version 15.0]
• Free Pascal Compiler version 2.6.2
• MinGW g++.exe (GCC) 4.9.2
• Haskell Glorious Glasgow, version 7.6.1
• Java 6 javac 1.6.0_45
• Java 7 javac 1.7.0_72
• Java 8 javac 1.8.0_25
• Ocaml ocamlopt 4.00.1
• Perl v5.12.2
• PHP 5.3.8
• Python 2.7.8
• Python 3.4.1
• Ruby 2.0.0p353
• Scala compiler version 2.11.1
• MS VS C++ 2010
• JavaScript V8 3.23.0

By MikeMirzayanov, 9 years ago, translation, ,

I will briefly enumerate the changes at Codeforces:

• Codeforces beta Round #1 is announced. It will go according to the ACM-ICPC rules, but will last for 2 hours. You shouldn't expect much from the problems - in the first place it is organized to check the system function and your feedback. The problems will be in two language: Russian and English. If I finish all the preparatory work earlier, I'll bring the Round some two days forward. It's planned to have the rating system at Codeforces similar to that at TopCoder. If the beta-competition goes without system failures, the participants will get ratings. To take part you are to register first.
• The visual organization of “Recent actions” has been changed - "new comment" and "compose/edit text" can be seen in the sidebar. Blog entries are sorted by the date of events, it means the last 15 entries are displayed.
• A new “Recent actions” detailed page, from which you can clearly see who has been doing what in recent time, is introduced.
• The algorithm of your "contribution" evaluation has been changed. For more details see below. I think I might change something again later.
• I've fixed some bugs.

By MikeMirzayanov, 9 years ago, translation, ,
After a two-day stop in Beijing we're leaving for Moscow. The flight is tomorrow, a bit earlier then 2 p.m. Maybe someone else is flying with us? We land at about 6 p.m., and then we'll have to go to the Paveletsky station, it would be great to catch train 17 that leaves at 19:56. Taking into consideration the number of suitcases we have, it seems that the bus/minibus+underground variant isn't for us. Taxi is the only suitable thing, by the way, does someone know how much it costs to take taxi from the Sheremetyevo 2 Airport to the Paveletsky station?

And if everything goes according to the plan, and we catch the train, we'll arrive at Saratov on February 10 at 12:04. As Antonina Gavrilovna says "tired, but hungry". Wait for us.

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By MikeMirzayanov, 9 years ago, translation, ,

The results are announced - thanks to Ivan Romanov for being so quick. Thanks to everyone who supported us and followed the contest events. My apologies the system break-down, but if everything was stable in Codeforces, I wouldn't have called it beta. Unfortunately, I had no Internet during the day.

Congratulations to the medalists. The competition was tough.