AtCoder Beginner Contest 272 Announcement

We will hold AtCoder Beginner Contest 272.

The point values will be 100-200-300-400-500-500-600-600.

We are looking forward to your participation!

• 99 minutes ago, # ^ | Just keep incrementing the i-th element by (i+1) and put them into a set for each m. Need to speed up the process by only caring about the cases where the number just becomes 0 and smaller than n. My solution

  • 95 minutes ago, # ^ | I'm not able to get the intuition on why it will fit in the time limit :/ Any hints regarding the same?

    • 93 minutes ago, # ^ | Because first element is used at most n/1 times, second — n/2 times and so on. So, in total maximum sum over all answers = n * (1/1 + 1/2 + ... + 1/n) ~ n * log(n).

      • 81 minute(s) ago, # ^ | but after all this process how you find mex?

        • 55 minutes ago, # ^ | take a look at this code Code

• 91 minute(s) ago, # ^ | ← Rev. 2 →   +2 in F you just implement hash + bs, G is some thinking upd. uh, just realized that TL is 2 sec (not 4 as I though) and my runs 1.4, so mostly you are right

  • 68 minutes ago, # ^ | I stalked your submission and your library for string hashing looks very cool: https://atcoder.jp/contests/abc272/submissions/35487307 Binary search to get lcp, then use lcp to lexicographically compare substrings, and then sort with those comparisons to get SA. I've never seen it done this way, I might steal this template for my personal use!

    • 50 minutes ago, # ^ | thanks and enjoy other codes of mine :) One major moment in this solution is stable_sort. It has less calls to comparator than usual sort (that gives TLE)

• 61 minute(s) ago, # ^ | It is always the f**king sqrt! Corrected version: https://atcoder.jp/contests/abc272/submissions/35512682 Please note that both sqrt/sqrtl return float. You cannot compare float with int using ==.

• 48 minutes ago, # ^ | ← Rev. 2 →   0 The euclidean distance is invariant under linear shifts. If is at square distance from , will also be at square distance from . Therefore we find all that are at distance from (which is the condition you see everywhere). Then, all neighbours of will be .

  • 38 minutes ago, # ^ | What are p and q? What i am seeing is if(i*i+j*j==m) then they are building edges of length (i,j) (-i,j) (i,-j) and (-i, -j) from each node. Can you explain more clearly please

• 50 minutes ago, # ^ | I also first enumerated the possible moves, but I did so by enumerating all s.t. , which takes to enumerate them.

• 47 minutes ago, # ^ | ← Rev. 2 →   0 Spoiler

  • 46 minutes ago, # ^ | Before reviewing it, I first pressed the upvoting button.

  • 8 minutes ago, # ^ | ← Rev. 3 →   +1 Yes, there are counterexamples. Here is my code: string s1 = s, t1 = t; s1.pop_back(); t1.pop_back(); string large = t + t1 + s + s1; In my concatenation, the large is . The first starts from No.4 (0-indexed) whose suffix is , the second starts from No.10 whose suffix is . . My algorithm would not count this pair but this pair is definitely valid. For the correct answer, please refer to the tutorial/editorial. This is the link to the original problem (ABC272F Two Strings).