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Codeforces Beta Round #32 (Div. 2, Codeforces format)
By RAD, 16 years ago, translation, In English

Attention, participants from the Division 1! As a test feature, you can participate in Codeforces Beta Round #32 "out of competition".


Everybody knows, that the 2nd of October - birthday of Mohandas Gandhi. We dedicate today's round to him, and many other great people who were born on October 2 :)


Round was prepared by Mike Mirzayanov, Matov Dmitry and Max Ivanov.

Special thanks to Julia Satushina for translation of statements.

Good luck!


Artem Rakhov and Codeforces team

UPD:
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Comments Comments (35)
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16 years ago, hide # |
 
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Will all the following contests be codeforces format?

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16 years ago, hide # |
 
Vote: I like it +22 Vote: I do not like it

i sacrificed my class only for codeforces  contest :(

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16 years ago, hide # |
 
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problem C's test 5 is?and the answer.thanks~

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16 years ago, hide # |
 
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C
use the long but not int in Java, longlong but not int in C++

D
three FOR loop is a good solution
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16 years ago, hide # |
 
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What is test 10 for problem C?
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16 years ago, hide # |
 
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what is the test case no-27 for problem D
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  • 16 years ago, hide # ^ |
     
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    Field 27 × 27 with 700 random stars. You need to find constellation number 1107, and it exists.
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16 years ago, hide # |
 
Vote: I like it -8 Vote: I do not like it
what is test case 10 of  problem c ... plzz post the correct answer also 
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16 years ago, hide # |
 
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Really enjoy the contest.
The first time for me to participate.
I missed several previous contests since they were in the midnight.
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16 years ago, hide # |
 
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System testing for out-of-competition participants completed. Sorry for delay.
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16 years ago, hide # |
 
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Probably the shortest solution for Problem B could be solved using java -
code.replaceAll("--","2"). replaceAll("-\\.","1").replaceAll("\\.","0");
:)
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16 years ago, hide # |
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Hello, i can't solve problem E, i can't figure it out how can i find the point(on the mirror) where those 2 see each other and then see if the "line of sight" intersects the wall(they SEE each other or they don't).
Please send me a private message with your solution or some hints. Thx:)
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  • 16 years ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it +12 Vote: I do not like it

    Let's reflect man's point A along the line M1 M2; denote this point as C (we can think of it as a "fake image" of point A).

    Then the segment BC intersects then M1 M2 segment exactly in the point necessary to you - it's the point where the ray from point A should reflect from the mirror to reach point B.

    So, in this problem you have to know how to reflect some point from a line, to check whether the point lies on a segment M1 M2, and then check that the ray haven't intersected the wall W1 W2.

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  • 16 years ago, hide # ^ |
     
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    Let's reflect man's point A along the line M1 M2; denote this point as C (we can think of it as a "fake image" of point A).

    Then the segment BC intersects then M1 M2 segment exactly in the point necessary to you - it's the point where the ray from point A should reflect from the mirror to reach point B.

    So, in this problem you have to know how to reflect some point from a line, to check whether the point lies on a segment M1 M2, and then check that the ray haven't intersected the wall W1 W2.

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16 years ago, hide # |
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Thx e-maxx!! I still have a question. If you have the kids at location (0,0) and (10,0) and the wall between (2,0) and (5,0) do they see each other? ( excepting the mirror)
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16 years ago, hide # |
 
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What is test 17 for problem C?
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16 years ago, hide # |
 
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What's test 2 in problem C? And the answer?
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16 years ago, hide # |
 
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What's the test #76 for problem E?, thanks
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16 years ago, hide # |
 
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In some contest(http://mirror.codeforces.com/contest/1/problem/C) ...when I output nothing ,they return me "Presentation error" , do anyone konw why?
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  • 16 years ago, hide # ^ |
     
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    Output
    Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point.

    "Nothing" doesn't conform to this specification, hence PE.
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16 years ago, hide # |
 
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Hi, can someone give me the test #5 for problem E? I can't pass it. Thx:)
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