chokudai's blog

Tokio Marine & Nichido Fire Insurance Programming Contest 2023（ABC 304) Announcement
By chokudai, history, 17 months ago, In English

We will hold Tokio Marine & Nichido Fire Insurance Programming Contest 2023（AtCoder Beginner Contest 304).

The point values will be 100-200-300-400-475-525-625-625. We are looking forward to your participation!

  • Vote: I like it
  • +20
  • Vote: I do not like it

Comments Comments (40)
| Write comment?
»
17 months ago, # |
← Rev. 3   Vote: I like it +13 Vote: I do not like it

Is there a long queue now? 30 minutes passed and my submission is still waiting for judging.

→ Reply
»
17 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Don't DDoS AtCoder Please !!! Please !!! Please !!!

→ Reply
»
17 months ago, # |
  Vote: I like it +22 Vote: I do not like it

so much long queue :(

is there any issue with the judge? my solution to D is judging for 15mins, yet not started!

→ Reply
»
17 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What happened to the queue? Waiting for my results on F...

→ Reply
»
17 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Mine has been waiting for 20 mins.

→ Reply
»
17 months ago, # |
  Vote: I like it 0 Vote: I do not like it

It was difficult to connect AtCoder until 21:06(JST), so I could't reach rated.

→ Reply
»
17 months ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

extended by 20 minutes

→ Reply
»
17 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Extended time!

→ Reply
»
17 months ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

How does extending the contest time helps with WJ?

→ Reply
»
17 months ago, # |
  Vote: I like it +10 Vote: I do not like it

what's the point of extending time when solutions are still not judged?

→ Reply
»
17 months ago, # |
  Vote: I like it +5 Vote: I do not like it

wait if solutions are still taking a lot of time to judge how will extending contest time help? isn't the damage already done?

→ Reply
»
17 months ago, # |
  Vote: I like it +20 Vote: I do not like it

In such cases, at least do 1 optimisation that if on any test case verdict is not AC, then don't judge the further test cases

This will take way less time and avoid such a long queue.

→ Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    well to be honest this had to be done regardless of bad queue condition or not

    → Reply
    • »
      »
      »
      17 months ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      Actually, earlier this issue was not there, so in such cases it is good, we get to know how many TC got AC, WA, TLE, etc.

      As far as queue is small this strategy is good, but in these cases not.

      → Reply
      • »
        »
        »
        »
        17 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        You're correct in that point, but still we are not contesting in a point-per-testcase system, so I didn't really feel much of a need of that statistic. I think showing the testcase number where the failed verdict first happened (just like CF does) is already enough.

        → Reply
        • »
          »
          »
          »
          »
          17 months ago, # ^ |
            Vote: I like it +5 Vote: I do not like it

          in case u got 1 WA and 1 TLE, then you will try to cover the edge case as well as optimise your code in the next submission, otherwise you will be doing at least 2 submissions.

          → Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I think it is a very good idea! There should be two modes: the normal mode and the conjestion mode. The conjestion mode should not judge further when it meets a testcase without an AC verdict. Besides would you please add some system monitors like Prometheus?

    → Reply
»
17 months ago, # |
  Vote: I like it +14 Vote: I do not like it

Will today's contest become unrated?This evaluation speed has had an impact on me with a score of 114514.

→ Reply
»
17 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Does problem D require some particular Algo/data structure? I thought about a binary search sol but would have been too slow in the end. Any hints?

→ Reply
»
17 months ago, # |
  Vote: I like it +24 Vote: I do not like it

This contest is a compensation for the unrated ABC299.

→ Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    But I still think that it would be very unreasonable for this competition to be rated.

    → Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The same company who sponsored ABC299 and ABC304 is DDoS-ed again. Both contests are made unrated.

    → Reply
»
17 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it
Spoiler

What's wrong in this code? Please can someone help!

→ Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Problem D?

    → Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Your solution is correct, just use lower_bound instead of upper_bound and it should work as we also need to consider the points on the rectangle's boundary.

    → Reply
    • »
      »
      »
      17 months ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      No need to consider borders, the constraints discard this happening.

      → Reply
    • »
      »
      »
      17 months ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      i used lower bound and it still fails. Also boundary checks aren't neccessay since 0 < pi < W and 0 < qi < H.

      → Reply
      • »
        »
        »
        »
        17 months ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        My bad you guys are correct, there is no need to check boundary points.

        I think it's not working because of this:

        for(auto el : gather_points){
    mn = min(mn, el.second);
    mx = max(mn, el.second);
}

        it should be mx = max(mx, el.second);

        → Reply
        • »
          »
          »
          »
          »
          17 months ago, # ^ |
            Vote: I like it +1 Vote: I do not like it

          Bruh, i have this intense anger right now, this queue really irritated me coz half of the time i was looking for the verdict instead of actually debugging my code

          → Reply
»
17 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Problem G is interesting and the solution is reminiscent of Codeforces 1616H Keep XOR Low. Thanks!

→ Reply
»
17 months ago, # |
  Vote: I like it +3 Vote: I do not like it

I learned a lot from editorial of problem F. Thanks to the problem writers. Here is my solution, which is a little different from the editorial.

For example, if n=12, then we should consider t=1,2,3,4,6. But, in fact we don't need consider t=1,2,3, since for t=1,2, all the results must be included in t=4, while for t=3, they must be included in t=6. Therefore, it is enough to only consider t=4,6. In general, assuming that n has prime divisors, p1,p2,...,pk, we only need consider n/p1,n/p2,..,n/pk. Moreover, integers <=1e5 have at most k=6 distinct prime divisors.

Next, we could use the inclusion-exclusion principle to solve it. Assume that d1=n/p1,d2=n/p2,..,dk=n/pk, then the final answer is f(d1)+f(d2)+...+(-1)*f(gcd(d1,d2))+(-1)*f(gcd(d1,d3))+..+f(gcd(d1,d2,d3))+...

The total complexity is O(2^k*n).

→ Reply
»
17 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

It turns out that one of my AC submissions on problem D is wrong, and it fails on the following test.

test

Is it still possible to add any test now? If yes,can you add it?

→ Reply
»
17 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Are ratings updated for ABC 304 ?

→ Reply
  • »
    »
    17 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No,because the round is unrated.

    → Reply