Allanur's blog

USACO 2014-2015 First Contest
By Allanur, 10 years ago, In English

This contest is different from old contests of usaco.

For bronze division there is has 4 problem.

For silver division there is has 3 problem.

For gold division there is has 3 problem.

And each problem has different number of testcase.

Full feedback.

Maybe after contest some testcase will be add or change.

The following languages are available : en, fr, es, ru, vi (gold only), hy (silver only), ka (bronze and silver only)

GL & HF .

You can see more information here.

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10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

I registered account. How can I register the contest on Dec 12-15? Thank you

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    10 years ago, # ^ |
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    It's different from Codeforces; you don't have to register for contest. During contest, you can click a button when you are ready to start, and you will have 3 — 5 hours (time for each contest alters) after then to complete the 3 problems.

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    10 years ago, # ^ |
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    no need to register separately for contest. The contest is usually 3hrs long and you can select your own convenient time to participate(during the contest days).

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    10 years ago, # ^ |
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    Contests are available at bronze, silver and gold levels of difficulty.

    New users are in bronze level.

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10 years ago, # |
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Maybe I'm just blind, but I can't seem to find any information about starting time of a contest, any help?

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    10 years ago, # ^ |
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    You can start the contest anytime from December 12 — December 15. When the time comes, there will be a link on the website that you can click to start (it will be hard to miss). Unlike a lot of other competitions, you get to choose when to participate within a set time period. For now, just be patient for a day and wait for the USACO website to update.

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      10 years ago, # ^ |
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      Oh okay, thank you :)

      P.S. Is there penalty time or no, because of things with multiple accounts?

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        10 years ago, # ^ |
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        It has IOI rules

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          10 years ago, # ^ |
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          How does that stop me from making two accounts, solving problems on one and instasubmitting when I enter contest on real one?

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            10 years ago, # ^ |
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            It doesn't — you just don't know if it'd pass, since the results are only revealed after each contest.

            It doesn't really matter, since cheaters won't succeed in the IOI selection camp anyway, and there isn't much more to be had.

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              10 years ago, # ^ |
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              The results aren't revealed after each contest anymore. You would know if you passed all the test cases, since it gives full feedback on all the test cases now.

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                10 years ago, # ^ |
                ← Rev. 2   Vote: I like it -31 Vote: I do not like it

                In other words: full cheating.

                But considering how easy the problems are this time, I don't think it makes much of a difference. Whoever can solve them at all can also solve them in 4 hours...

                Considering the number of downvotes and the number of replies, I can only assume that many people are going to try to cheat and don't want to be caught.

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      10 years ago, # ^ |
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      How do they ensure that people do not cheat in the USACO contests ?

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10 years ago, # |
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This contest have Feedback my buddies.

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10 years ago, # |
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Is USACO really giving full feedback this time? :o

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10 years ago, # |
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The second task in Silver Division has no statement on Russian, and accidentaly I've met such problem "Grading Failure due to Network Outage; Please Resubmit in a Bit" after sending solution for second and third tasks. And yes, there was a feedback for the first problem.

UPD1: The grading now is working.

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10 years ago, # |
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How to take input via files in java ?

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10 years ago, # |
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Is it just me, or is the grading server down? It's been saying "waiting for available grading server" for around 20 minutes now, ffs...

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10 years ago, # |
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Who have any idea for solve problems ?

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    10 years ago, # ^ |
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    I don't think the contest is over yet (24 hours and 29 minutes left from time of posting).

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    10 years ago, # ^ |
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    The contest start time ended but there are some people who began late and are still taking the contest , so please don't discuss the questions now.
    UPD : Contest is extended by one extra day.

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10 years ago, # |
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How to solve usaco bronze div -4th problem ? (ok ,it's not finished now,so answer when it will be finished :) )

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    10 years ago, # ^ |
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    LOL , hey guys don't answer me now because the contest is not finished yet but it's fine to post my code now

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10 years ago, # |
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Why usaco extended by one extra day ?

Who has any idea ?

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10 years ago, # |
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Gold division

1) We can find the answer to the problem for every fixed subset of cows, then the answer to the problem is the minimum of those answers. If the total height of cows is not less than H, do binary search on the answer. To check whether we can achieve a safety factor greater than or equal to X we can try to construct a permutation of cows in the subset for which the condition of safety factor holds. For each i (i = 1, 2, ..., sizeofsubset) consider all the cows that can be in the i th position from the ground and choose the one with minimum weight.

code

2) For each checkpoint we can calculate two numbers
ai — What is our profit if we skip this checkpoint
bi — Distance between i th and (i + 1) th checkpoints
Now we can build two segment trees.
One for calculating distance between L th and R th checkpoints.
Another one for calculating the maximum profit we can have by skipping one checkpoint. max(aL + 1, ..., aR - 1)
Notice that update operations only affect at most 3 checkpoints. So we can update ai and bi for each of them.

code

3) Let ai = t * vi + xi
Then, the solution is the length of the longest decreasing subsequence of a . See task pcb of BOI 2010 and its solution.

code

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    10 years ago, # ^ |
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    In the third problem, it is needed to find longest non-increasing subsequence, isn't it?

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    10 years ago, # ^ |
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    contest hasnt ended....

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    10 years ago, # ^ |
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    If the 3rd problem is the same as silver version, then there is a linear and simpler solution.
    Let a[i] = t * v[i] + x[i]
    Answer is number of i, such that a[i] < min(a[i+1] .. a[n]). code

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