joisino's blog

By joisino, history, 3 years ago, In English,

Hello Codeforces!

The 16th Japanese Olympiad in Informatics Final Round Online Contest will be held on 12 Feb. (00:30 — 04:30 GMT).

The contest duration is 4 hours and there will be 5 problems, which is the same as the 16th JOI Final Round. Problem statements will be provided both in Japanese and English this year.

Details are available in contest information .

Good luck and have fun!

UPD: Registration is available now.

UPD2: Judge server is open now. You can submit your solutions. You need remake your account to use judge.

UPD3: Editorial will be provided only in Japanese. Judge data and sample source codes will also be provided.

 
 
 
 
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3 years ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

Wow! I'll participate it (on-site).
By the way, it is not "final" round because there are JOI spring camp :-)

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    3 years ago, # ^ |
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    Somehow it's always called "final" round. Perhaps it's because the spring camp is not Japanese Olympiad in Informatics itself, but the selection camp for IOI.

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3 years ago, # |
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Contest starts in < 1 hour.

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3 years ago, # |
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Auto comment: topic has been updated by joisino (previous revision, new revision, compare).

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3 years ago, # |
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Will there be an editorial?

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3 years ago, # |
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How to solve Problem 4 and 5?

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

    5: the papers will be grouped into 2 stack of papers, for every paper (by paper i mean the length 1 ropes) lets write which stack it will end up at, i'll claim that the way it'll be will be like A or B ->
    A: size Odd 1's size even 0's size even 1's size even 0's ... size even "something" size ((n — 1) % 2) "something" ^ 1, basically the first one is odd and all others are even, except the last group which its sizes parity will be determined by n's parity
    B: size even 1's size even 0's size even 0's ... size even "something" size (n % 2) "something" ^ 1, this begins with even and ends with something either odd or even that is determined by n's parity
    i don't have a proof for this, but i got accepted using this assumption
    now i could see the problem like this, if the first group is odd, then the 2nd and 3rd guy must end up in the same stack of paper, 4th and 5th also must end up in the same, 6th and 7th same...
    if its even 1st and 2nd must end up in the same 3rd and 4th must end up in the same ....
    so now, i'll just do a for on the color we want the 1st stack to be, i'll chose every "matching" block (the 2 guys that have to be in the same stack of paper) that has at least one of the chosen color greedily (we don't need the useless guys we will give them to the other stack of paper), and i'll color the second paper, with the color that has appeared the most in the other "matching" blocks
    i am aware that my english isn't really good, so you will probably have some questions and need clarifications, don't be afraid to ask :D

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3 years ago, # |
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How to solve problem 2?

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3 years ago, # |
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Auto comment: topic has been updated by joisino (previous revision, new revision, compare).

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3 years ago, # |
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Where I can find problems?

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3 years ago, # |
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How can I find problems since I didn't register before the contest ? thanks in advance

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3 years ago, # |
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For question 4 (Soccer):

Step 1: We will start off by finding for each 1<=i<=500,1<=j<=500 in the grid how far the closest player is who can come here just by moving.

Step 2: Now, assume that nth player is at bottom right corner -- it is an assumption that makes the explanation a bit easier. Then we shall do a dp[a][b] which denotes min. fatigue for ball to go from a,b -> bottom right corner. So, we can either move one with our current player (dp[a+1][b] or dp[a][b+1]) or kick the ball to someone. We can loop through where we kick the ball (hint: it will always be in the rectangle formed by a..n and b..m and either a or b must be fixed), and then use the function we calculated in Step 1 to transition the dp.