Блог пользователя kondasujay2

Автор kondasujay2, история, 4 часа назад, По-английски

Thank you to everyone who participated!

We are sorry for the poor testcases on C1, along with C1 and C2's solutions being so similar to the recent Div3 E.

2263A - Min Max Game

Author: kondasujay2

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2263B - Min Matrices

Author: sukon

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2263C1 - Floor of MEX (Easy Version) / 2262A1 - Floor of MEX (Easy Version)

Author: CutSandstone

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2263C2 - Floor of MEX (Hard Version) / 2262A2 - Floor of MEX (Hard Version)

Author: kondasujay2

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2263D - Culling Game / 2262B - Culling Game

Author: kondasujay2

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2263E - Traveling the World / 2262C - Traveling the World

Author: kondasujay2

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2263F - PLUSworld / 2262D - PLUSworld

Authors: sukon & kondasujay2

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Magic
Magic Solution
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2262E - Paired Bracket Sequences

Author: kondasujay2

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2262F - Rank Removal

Authors: sukon & kondasujay2

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Разбор задач Codeforces Round 1120 (Div. 1)
Разбор задач Codeforces Round 1120 (Div. 2)
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3 часа назад, скрыть # |
 
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Auto comment: topic has been updated by kondasujay2 (previous revision, new revision, compare).

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Auto comment: topic has been updated by sukon (previous revision, new revision, compare).

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112 минут назад, скрыть # |
 
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Nice Contest

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99 минут назад, скрыть # |
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Div1A2/Div2C2 can be solved in $$$O(n\log n)$$$ via counting sort if you store the smallest interval for each right endpoint.

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92 минуты назад, скрыть # |
 
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For C2, my submission: 390474676. It works in O(nlogn) time comp. and O(n) space comp.

My DP: I have processed till i (0 <= i < n), and I want to find how many good sets exist with the largest element being i. Note till now, we are not considering intervals whose right endpoint is bigger that i.

For transition, I will be summing up the dp[j] such that j < i, j represents what is the second largest element of the set, and there is no interval completely inside [j + 1, i — 1]. j will form a contiguous range, so we can use prefix sum on dp values.

Thus, dp calc. part is O(n), but preprocessing intervals take O(nlogn) time.

Final answer: let pos represents the maximum left endpoint of any interval in which at least one number should lie. ans = dp[pos] + dp[pos + 1] + ... + dp[n — 1].

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92 минуты назад, скрыть # |
 
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"Rate the problem" appears to have all the problems sharing the same rating votes

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73 минуты назад, скрыть # |
 
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54 минуты назад, скрыть # |
 
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Not a bad contest, but it was indeed tough, thx for the rating boost, but I mean it was like:

A: 800 — 900 B: 1000 — 1100 C1: 1300, maybe 1400, because I had to use Seg Tree to quickly tell me which ranges of values needed to be excluded (but it wasn't necessary because you didn't have to do updates between queries)

C2: 1700+, I read it and it seemed kinda cooked, like I don't know how to wrap my head around all the conditions that had to be satisifed -- and how to count the number of ways to satisfy all constraints.

D: 2000+, read it and realized that your favorite pupil had never seen this kind of bullshit before.

E, F: ???

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16 минут назад, скрыть # |
 
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C1 is hard for me :<