Блог пользователя azneyes

Автор azneyes, история, 8 лет назад, По-английски

Hello everyone,

ACM-ICPC ECNA 2016 contest finished 2 months ago. 3 teams advanced to the 2017 World Finals in South Dakota, including the team from University of Waterloo:

Waterloo Royal Blue: FatalEagle,zxqfl,y0105w49

You can learn more about them from this ICPC News Facebook post. I am one of the coaches and I look forward to meeting everyone in South Dakota next May. Good luck to Royal Blue and all other teams at Finals. We will try not to place (lucky/unlucky)? 13th for 3 straight world finals!

I have added the contest to Gym: 2016-2017 ACM-ICPC East Central North America Regional Contest (ECNA 2016). Note that there are 3 clarifications in the announcements.

Enjoy.

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8 лет назад, # |
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Is there a public scoreboard available? The link on the official website does not seem to work.

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8 лет назад, # |
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13 is definitely a lucky number, it is the luckiest number of Taylor Swift!

Btw, Waterloo is colored bronze in 2016 standings as if you got bronze medal, but there was no post-mortem decision about additional bronze, right?

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    8 лет назад, # ^ |
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    Actually, that's exactly what happened!. We got a post-contest bronze medal this July. There was supposed be some news announcement from ICPC News but I can't seem to find it.

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      8 лет назад, # ^ |
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      Lols, congrats :p. So it turned out 13th place was not that unlucky :P. However it is forbidden according to rules xD (since time penalty difference was bigger than 60 mins).

      "Additional Bronze Medals may be awarded to teams solving the same number of problems as the twelfth place team and in a total elapsed time no more than 60 minutes more than the twelfth place team. Such teams will be awarded twelfth place."

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8 лет назад, # |
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Where can I find the final standings of 2015 World Finals?

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8 лет назад, # |
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I am just curious. What's the story behind the team names of Waterloo? 2014 and 2015, Waterloo Black advanced to the World Final and this year it seems Royal Blue had a lead.

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8 лет назад, # |
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Can you really coach these guys? The rating difference seems pretty big :(

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    8 лет назад, # ^ |
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    Teachers don't need to be smarter than their students.

    A football coach does not need to be a better player than the ones in the roster.

    Likewise, a ICPC coach does not need to be a better contestant than his/her pupils. Experience and guidance are the most important things to provide their teams.

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8 лет назад, # |
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Any hint on what the dp state is for problem F ?! I have tried looking through other people's submissions but couldn't get it :(

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    7 лет назад, # ^ |
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    You can try to test all possible pairs for the final two numbers after remove all numbers and for each answer get the optimal.

    Now, define the DP[i][j] as the minimum cost for remove all number between i and j where this two number are left until the end. To solve this DP the process is the following: you will remove numbers until you have tree numbers i,k,j and in that point the cost for remove k is gcd(i,j). So just iterate over all possible k in [i+1,j-1] and repeat the DP for the new two segments [i,k] and [k,j].

    How to handle the circular arrangement? Just duplicate the array of size N and, for each initial pair [i,j] with 0 <= i,j < N you will have

    ans = min (ans, gcd(i,j) + DP(i,j) + DP(j,i+N))

    Note: I needed to implement the DP in its iterative form and precalculate the GCD for all possible pairs of numbers (1000*1000) to get AC

    Hope you understand my little late explanation and english

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7 лет назад, # |
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Hi, could someone explain the solution for problem J? Is the only one that I couldn't solve yet Thanks