RAD's blog

By RAD, 8 years ago, translation, In English,
Good evening

Yesterday evening the Saratov university delegation returned from St. Petersburg, from the ACM-ICPC NEERC 2010/11 World Programming Championship semi-finals. If you haven't seen the final standings: 4 Saratov teams received diplomas, and we (Saratov SU 2) advanced to the finals. Saratov SU 1 was also among those, who advanced, that's pretty cool for their first time, but didn't advance because of the limitation "only one team from one university".

Also we have prepared a Div. 2 round. Thanks for prompt assistance to Edvard Davtyan, Gerald Agapov and Maria Belova.

Good luck!
Artem Rakhov and Codeforces team


Unfortunately, the discrepancy between author’s solution and statement of problem E was detected. We bring our apologies to all the participants. All solutions that have not been Accepted previously were rejudged. Thanks to member xcr for detection of the issue.
 
 
 
 
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8 years ago, # |
  Vote: I like it +1 Vote: I do not like it
Someone can paste the Pretest #6 for the problem D?
I can't figure what is the problem with my solution...
  • 8 years ago, # ^ |
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    Try 1 2 and 1 4
    • 8 years ago, # ^ |
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      He means pretest?
      1 2 and 1 4 are used for hack.
    • 8 years ago, # ^ |
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      У меня бага в 1 4 оказалась. Айайай как обидно, что во время контеста не догадался его вбить.
8 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it
/ a.out 
1 2
0
1 1
1 2
1 1

./a.out 
1 4
1
1 4 1 1
1 1
1 2
1 3
1 4
1 1


  • 8 years ago, # ^ |
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    Try these tests: 4 4 , 4 5 , 5 4 , 5 5 (all the rest cases).
8 years ago, # |
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I wonder if it wouldn't be better to have a possibility to lock the problem and start hacking without passing pretests, even without submiting it.
  • 8 years ago, # ^ |
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    I think it wouldn't be better, because in your case contestant can use 2 accounts: one for stealing solutions from another, second - for submitting stolen solutions. Now for getting solutions from competitors, even cheater have to write solution which passes pretests
8 years ago, # |
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How do I prove for problem D that when both n and m are odd, we can't find a required path without using teleporting gates?
  • 8 years ago, # ^ |
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    Well, let's look at board painting. Every turn we change color of cell, in which king stays. So, for odd n and m cell at n*m turn will be black. n*m+1 turn must follow us to (1,1), which is also black. So we're used to place at least one gate.
    • 5 months ago, # ^ |
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      In the above comment, the board painting refers to visualizing the board as a chess board with the cell at (1, 1) being colored black.

8 years ago, # |
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Give me 23 test for E, please
  • 8 years ago, # ^ |
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    Big random test with 99 cars, and 100 segments for each car.
8 years ago, # |
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What's final test 25 at problem D? Thanks..
  • 8 years ago, # ^ |
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    I don't know. But I can assure you that if your code passes the following types of inputs, it should pass all the inputs,
    odd even
    even odd
    even even
    odd odd
    1 2
    2 1
    thanks to HackSon.
    • 8 years ago, # ^ |
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      I think "1 even" and "even 1" are not precise enough...
      1 2
      2 1
      1 M
      N 1
      where N, M > 2.
      • 8 years ago, # ^ |
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        it's unnecessary to check "1 M" and "N 1" when N or M odd since it same as "odd odd"
    • 8 years ago, # ^ |
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      I have it done more complicately, so it's giving me Time exceeded.. but i think it should handle it for every 1<=n, m<=100
  • 8 years ago, # ^ |
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    99 99
8 years ago, # |
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Hi to codeforces team

why you don't put test case's and problem's(in PDF) after each contest ?

it's usefull for every one ....

thanks for attention
8 years ago, # |
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Congrats on making into the finals!! 9 Problems done, that is pretty impresive.
8 years ago, # |
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Today I participated "out of competition". I have 2 questions:

-My solution got hacked by another div1 contestant. So....does this mean I'm able to hack any other div1 competitors? Cause I didn't see any one in my "room".

-Am I able to see the case that got me hacked during competition?

Thx in advance
  • 8 years ago, # ^ |
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    To see other "out of competition" contestants in your room you should check box, that doing it (in top right corner).
    Hack protocols availible at "Hacks" tab by link in "Verdict" column.
  • 8 years ago, # ^ |
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    If you participate 'out of competition', then when you enter the contest you appear in room number 1 instead of the room you were assigned to. (It's a bug.) To move into your room, select it from listbox in top right corner, it's marked with a star.
8 years ago, # |
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Can anyone explain me problem C, I couldn't program it and ended up getting -2, how about a contest analysis after every contest?
  • 8 years ago, # ^ |
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    Let C[i] = # of integers in A1...As.t. Ak = i (mod 3)
    Then answer = C[0]/2 + min(C[1], C[2])
  • 8 years ago, # ^ |
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    Previous explanation is for problem B. As regards C you just have to pair numbers a, b when:
    a % 3 = b % 3 = 0 or a % 3 = 1 and b % 3 = 2. In such cases (a + b) % 3 = 0.
8 years ago, # |
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Hi.. Can you give me the case no.3 from problem D? Thanks..
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4 years ago, # |
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can somebody tell me why using b as a char array works just fine but defining it as char* gives wrong answer??(c code)

here is the wrong answer code